XtraEdge Feb 2011

XtraEdge for IIT-JEE 1 FEBRUARY 2011

Dear Students, It's the question you dreamed about when you were ten years old. It's the question our parents nagged you about during high school. It's the question that stresses most of us out more and more the older we get. "What do you want to be when you grow up?" There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. if you work for 10 hours each day, you're going to end up spending over 50% of your awake life at work. Personally, I think it's important that we spend that 50% of your awake life at work. Personally, I think it's important that we spend that 50% wisely. But how can you make sure that you do? Here are some cool tips for how to decide that you really want to be when you grow up. • Relax and Keep an Open Mind: Contrary to popular belief, you don't have

to "choose a career" and stick with it for the rest of your life. You never have to sign a contract that says, "I agree to force myself to do this for the rest of my life" You're free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an open mind.

• Notice Your Passions: Every one of us is born with an innate desire to do something purposeful with our lives. We long to do something that we're passionate about; something that will make a meaningful impact on the world.

• Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal trainer. Making a positive impact on the world will not only ensure that you are successful financially, it will also make you feel wonderful. It's proven principle: The more you give to the world, the more the world will give you in return.

• Figure Our How You Can Benefit: Once you've figured out what your passions are and how you can use those passions to add value to the world & to yourself, it's time to take the last step: figure out how you can make great success doing it. my most important piece of advice about this last step is to remember just that: It's the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for that field. No amount of money can compensate for a life wasted at a job that makes you miserable. However, that's not to say that the money isn't important. Money is important, and I'm a firm believer in the concept that no matter what it is that you love doing, there's at least one way to make extraordinary money doing it. So be creative!

No matter how successful you become, how great your life is, or how beautiful you happen to be... there will still be times when you simply feel like you're an ugly mess. But when those times come, remember that all you need to get yourself back on track is a positive outlook, a dash of self confidence, and the willingness to make yourself feel better as soon as you know how. Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself. Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Volume-6 Issue-8 February, 2011 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics,, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

IIT-JEE Mock Test Paper with Solution

AIEEE & BIT-SAT Mock Test Paper with Solution

S

Success Tips for the Months

• "The way to succeed is to double your error rate."

• "Success is the ability to go from failure to failure without losing your enthusiasm."

• "Success is the maximum utilization of the ability that you have."

• We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.

• Along with success comes a reputation for wisdom.

• They can, because they think they can.

• Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.

• Keep steadily before you the fact that all true success depends at last upon yourself.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 Dr. Abdul Kalam's Message to Every Indian Two Mumbai CAT toppers are from IIT-Bombay

IITian ON THE PATH OF SUCCESS 6 Mr. Vineet Buch

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 50

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper Mock Test-3 (CBSE Board Pattern) [Class # XII] 72 Solution of Mock Test-2 (CBSE Pattern) Solution of Mock Test-3 (CBSE Pattern)

Regulars ..........

DYNAMIC PHYSICS 15

8-Challenging Problems [Set# 9] Students’ Forum Physics Fundamentals Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics CATALYSE CHEMISTRY 31

Key Concept Carbonyl Compounds Co-ordination Compound & Metallurgy Understanding : Physical Chemistry

DICEY MATHS 38

Mathematical Challenges Students’ Forum Key Concept

Integration Trigonometrical Equation

Study Time........

Test Time ..........


Dr. Abdul Kalam’s Message to Every Indian What does a system consist of? Very conveniently for us it consists of our neighbours, other households, other cities, other communities and the government. But definitely not me and YOU. When it comes to us actually making a positive contribution to the system we lock ourselves along with our families into a safe cocoon and look into the distance at countries far away and wait for a Mr.Clean to come along & work miracles for us with a majestic sweep of his hand or we leave the country and run away. ‘ASK WHAT WE CAN DO FOR INDIA AND DO WHAT HAS TO BE DONE TO MAKE INDIA WHAT AMERICA AND OTHER WESTERN COUNTRIES ARE TODAY’

Two Mumbai CAT toppers are from IIT-Bombay Three people, who appeared for CAT from the city, scored 100 percentile. Two of them are from the computer science department of the IIT-B. The other is a faculty member of a city-based coaching institute. Shashank Samant, 22, who scored 99.98 percentile on his last attempt in 2008, gave up the IIM seat to take up a job. “After a year and a half job in an investment firm, I was finally prepared to get in to an IIM. Though the 100 percentile was unexpected.” “After clearing the CAT, I spoke to my peers and seniors at IIT-B and decided that work experience would be important before getting in to the IIMs,” said Samant. He also added that IITs help in developing the aptitude to clear any competitive exams. Samant had graduated in computer science from IIT-B in 2009. About his choice of IIMs, Samant said, “I am ready to get in to any of the IIMs though I prefer IIM-

Ahmedabad and Bangalore over others.” Gaurav Malpani, a fourth-year student of computer science at IIT-B has appeared for the entrance exam from Mumbai, he is originally from Kolkata. He managed to score100 percentile, without any coaching. “I have never focused on developing my technical knowledge. I was only polishing my problem solving skills. I also focused on my vocabulary,” said 21-year-old Malpani. He insisted that he had never studied exclusively for CAT during the year. “I always knew that I had the aptitude to score well, but scoring 100 percentile was not expected,” he added. “I would love to join IIM-Ahmedabad or Bangalore. Since I am from Kolkata, I will also consider seeking admission there. I am interested in pursuing an MBA in either finance or human resources,” he added. The faculty member of a city-based coaching institute Jose D’Abreu also got a perfect score.

IIT Techfest 2011: The Robots Raged It was the perfect way to end a very well-planned event. On the first two days of Techfest, IIT Bombay was buzzing with exhibits of some cool robots and a few other inventions as well. On the final day, the robots became restless and just wanted to have a go at each other. What followed was a lengthy battle fought hard and long. Mars Manoeuvre This tournament had two robots moving around a grid collecting blocks. The team that collected the most blocks won. The last battle was between C2R and Black Beast, from Thailand and Australia respectively. Black Beast: In all its glory Black Beast was the creation of second year students from the Department of

Electrical & Computer Science at Swinburne University of Technology in Melbourne. Plagued with exams and other academic diseases, they still managed to build this in about two weeks.

Kids at Swinburne Uni, came second with a smile Their robots communicated through radio frequency waves and everything, right from the wheels to the circuit boards was custom made. One robot would go and measure the dimensions of the grid. The second, after getting the information would start its mission of picking up blocks.

Can it say, "Thai, Robot"? C2R was made by third year students of the computer engineering department of the Kasetsart University in Kapmphaeng Saen, Thailand. These kids spent two months and about Rs. 1,47,000 (100,000 Thai Bath) to make these robots.

Kasetsart University students with the C2R


Unlike the kids of Oz, they chose to use only one robot which would find its way and collect blocks. These robots also used radio frequency technology to communicate and had sensors, so that the robot never drifted away from the grid lines. The battle ensued and it was clear that spending more time with your robot makes them strong and obedient. In fact, they can also win you competitions! C2R won the Mars Manoeuvre competition and prize money of Rs. 1,50,000. The kids demonstrated how they won the battle. Check out the video below.

IIT-Bombay gets $3 million gift More than 40 years ago a quiet student named Victor Menezes graduated from the Indian Institute of Technology Bombay (IIT-B). He went on to become, among other things, the senior vice-chairperson of Citigroup Inc. His “small way to say thank you” to the institute has translated into a $3 million

(about Rs 13.5 crore) towards a state-of-the-art convention centre on the institute’s Powai campus

“I received priceless education from IIT Bombay and this is a small way to say thank you”, said Menezes. “I hope the centre will help support the exchange of ideas at IIT Bombay.”

IIT-JEE candidates to get performance cards now Students appearing for the next Joint Entrance Examination (JEE) for admission to IITs will get performance cards specifying marks and the ranks secured by them in the test. However, as per the new provision, they cannot seek regrading or re-totalling.

For the first time, the JEE Board would issue performance cards which can be considered as certificates by many other institutions wanting to give admission to JEE candidates. The board will also put out the answers of the questions on its website to help students make assessment of their performance.

IIT Guwahati Director Prof Gautam Baruah said the board had urged for issuing such performance cards which would serve as certificates for the students. “Many other institutes, which want to take JEE candidates, can give admission to students on the basis of these performance cards,” Baruah said.

Indian institute of Science will start management course Indian Institute of Science (IISc.) is planning to start a two-year Master programme in management from this academic session. The new courses will be very advance as it will concentrate more on technology management and business analytics.

IISc registrar R Mohan Das said the course would concentrate on synergies between managing science and technology. Das said, “India, in recent times, has emerged as one of the global hubs of technology and research and development (R&D) units. Such technology-based and R&D-intensive industries need executives with exposure and training in technology management and business analytics. The program has been specially designed to train students in technology management and business analytics.”

The course will be conducted by the department of management studies, which was established in the year 1848, and is one of the oldest schools in the country. An official at the dept. said that application forms for the course will be available from the month of February. Candidates who have passed the Joint Entrance Test (JMET) with first class BE/B.Tech degree/equivalent is eligible for the course. The department will conduct group discussion and personal interview before selecting students for the course.

IIT Mandi to formulate plan for solving technical problems pertaining to agriculture in Himachal Pradesh Shimla: Shri Ram Subhag Singh, Secretary, Agriculture and Information and Public Relations said that H.P. Agriculture Department and IIT Mandi

would formulate a long term scheme for solving technical problems pertaining to agriculture and a Joint Working Group at State level would be formed for solving the problems relating to farm technology.

Star Donor of the Month - Mr. Rajesh Achanta [1987/BT/ME] I have been donating off and on as a way of keeping the connection with IITM going & also to express gratitude for the many ways in which the institute shaped me in my formative years. I'll be transiting through Chennai in early January - I would like to stop by at IITM & relive old memories for a little while!

Orissa CM confers award to IIT-Kanpur Prof. Dr Devi Prasad Mishra In recognition of his research work, Dr. Mishra received Sir Rajendranath Mookerjee Memorial and Aerospace Engineering Division Prize from The Institution of Engineers (India), Kolkata, India. Dr. Mishra has more than 15 years of teaching and research experience. He has served as Visiting Professor in 2002 at the Tokyo-Denki University, Japan. Presently, he is working as an Associate Professor in the Department of Aerospace Engineering at Indian Institute of Technology (IIT) Kanpur, Kanpur, India where he was instrumental in establishing a combustion laboratory.

His areas of research interest include combustion, computational fluid dynamics, atomization, nanomaterial synthesis etc. He is an Associate Editor, Journal of Natural Gas Science and Engineering, Elsevier, USA and Assistant Editor, International Journal of Hydrogen Energy, Elsevier, USA. Currently he is serving as Editor, Asia Pacific Conference on Combustion, 2010. Dr. Mishra has four Indian patents and more than 154 publications in referred Journals and in conference proceedings to his credit.


Vineet Buch still remembers 10 June 1987. Bhopal. The Indian Institute of Technology All India Joint Entrance Exam (IIT-JEE) results were announced. Buch, then a 15-year-old dabbling with career choices, scanned through the rank-holders list. Then he scanned it again. Soon he made up his mind. He would try and finish No. 1 in the entrance exam. “It seemed like a cool thing to do.”

Every year thousands of Indian students aspire to get into an IIT. Close to 400,000 candidates lined up this year. One in 65 made the cut. Twenty years ago, the number of applicants wasn’t as staggering but there were fewer seats. Golfers will tell you that the odds of an amateur pulling off a hole-in-one are 1 in 12,750. Still, that’s a doddle compared to what Buch was up against.

“Hardly anyone in Bhopal even wrote the JEE, let alone got in,” says Buch, 37, a venture capitalist based in San Francisco. “I found it tough to get the right books, like a Russian physics book by IE Irodov. My parents [who were IAS officers] requested the Indian embassy in Moscow to photocopy the book and send it across.”

In June 1989, Buch was declared No. 1 in the IIT JEE exam, arguably the most challenging and competitive exam in the world. Only around 50 Indians have experienced the feeling—the numbness, the ecstasy, the dizziness.

Once every year, JEE toppers appear on television and newspapers carry congratulatory messages. You see mug shots of students, interviews with parents, and advertisements for coaching centres. We spend a lot of time celebrating their success, but rarely do we look further.

What becomes of these brilliant 17-year-olds? What are the challenges they encounter? Do any of them pursue unconventional careers? These were some of the questions Open set out with while tracking down the very elite group of JEE toppers.

IT HELPS TO BE NO. 1 During his days in IIT Kanpur, Buch was a long-distance athlete, weightlifter and footballer. He competed in both the 5,000 and 10,000 metres. But in August 1993, a doctor at Delhi’s All India Institute of Medical Sciences diagnosed the 20-year-old with ankylosing spondylitis, a progressively crippling disease without a cure.

Buch suffered inflammation of the eyes and internal organs. “Sometimes it was so hard for me to even sit, stand or sleep,” he recalls. Things got progressively worse over his two-year graduate program at Cornell University. “When I finished in 1995, I was immobilised throughout much of my body. A doctor advised me to stop working and apply for disability payments.”

Buch refused. He moved to San Francisco and started a self-directed rehabilitation programme. He began with long sessions of swimming and gradually started to walk, bike and hike. In 2001, he successfully undertook the Death Ride over five alpine passes on the Sierra Nevada mountain range in California, US. But biking hurt his knees. Searching for a sport that didn’t tax his legs, he discovered surf skiing, one that uses a long, narrow, lightweight kayak with an open cockpit and a foot-pedal controlled rudder. On 17 May, Buch took part in the 2009 Molokai Challenge in Hawaii, a 32-mile surf ski race between Molokai and Oahu, in rough waters swarming with tiger sharks. He finished the race.

“I thought being No 1 in JEE was tough,” says Buch. “But overcoming this disease has been something else. The JEE effort definitely helped with this—I knew the levels of determination I was capable of and refused to give up.”

Success Story Success Story This article contains storie/interviews of persons who succeed after graduation from different IITs

Mr. Vineet Buch B-Tech from IIT-Kanpur (A venture capitalist based in san francisco)


PHYSICS

1. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled with a monoatomic gas of molar mass MA. Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB. Both gases are at the same temperature. [IIT- 2002]

(a) If the frequency of the second harmonic of the fundamental mode in pipe A is equal to the frequecy of the third harmonic of the fundamental mode in pipe B, determine the value of MA/MB.

(b) Now the open end of pipe B is also closed (so that the pipe is closed at bout ends). FInd the ratio of the fundamental frequency in pipe A to that in pipe B.

Sol. (a) Second harmonic in pipe A = 2 [(v0)A] Third harmonic of pipe B = 3 [(v0)B]

= 2

l2v = 3

l4v

= l

1 A

A

MRTγ =

l43

B

B

MRTγ

A Gas (Monoatomic) MA

l

BGas (Diatomic)MB

l Given that second harmonic in pipe A = Third

harmonic of pipe B

⇒ l

1 A

A

MRTγ =

l43

B

B

MRTγ

⇒ B

A

MM =

189400 [γA = 1.67 and γβ = 1.4]

(b) (v0)A = A

A

MRTγ (v0)B =

B

B

MRTγ

∴ B

A

vv

)()(

0

0 = B

B

A

A MM γ

×γ =

43

2. A non-conducting disc of radius a and uniform positive surcface charge density σ is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q/m = ε0g/σ.

(a) Find the value of H if the particle just reaches the disc.

(b) Sketch the potential energy of the particle as a function of its height and find its equilibirum position. [IIT- 1999]

Sol. (a) Given that : a = radius of disc, σ = surface charge density, q/m = 4ε0g/σ

The K.E. of the particle, when it reaches the disc can be taken as zero.

Potential due to a charged disc at any axial point situated at a distance x from 0.

V(x) = 02ε

σ ]–[ 22 xxa + ]

Hence, V(H) 02ε

σ ]–[ 22 HHa +

and V(O) = 02ε

σa

According to law of conservation of energy, Loss of gravitation potential energy = gain in electric potential energy

H (m,q)

a O

H

mgH = qDV = q[V(0) – V(H)]

mgH = g[a – )( 22 Ha + – H] 02ε

σ …(1)

From the given relatuion : 02ε

σq = 2 mg (given)

Putting this is equation (1), we get,

MgH = 2mg[a – HHa –)( 22 + ]

or H = 2[a + H – )( 22 Ha + ]

or H = 2a + 2H – 2 )( 22 Ha +

or 2 )( 22 Ha + = H + 2a or 4a2 + 4H2 = H2 + 4a2 + 4aH

or 3H2 + 4aH or H = 3

4a

[Q H = O is not valid]

KNOW IIT-JEE By Previous Exam Questions


(b) Total potential energy of the particle at height h

U(x) = mgx + qV(x) = mgx + )–(2

22

0xxaq

+εσ

= mgx + 2mg ]–[ 22 xxa +

= mg ]–)2[ 22 xxa + …(2)

For equilibrium : dxdU = 0

This gives : x = 3

a

From equation (2), graph between U(x) and x is as shown above.

O a / 3 H = 4a/3 X

U

2 mga

3 mga

3. A wheel of radius R having charge Q, uniformly

distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by ligh inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T0. If the breaking tension of the stringas

are 2

3 0T, find the maximum angular velocity ω0

with which the wheel can be rotated. [IIT-2003]

T0 T0

d

B

ω0

Sol. From above figure, when the ring is not rotating wt. of ring = Tension in string mg = 2T0

∴ T0 = 2

mg …(1)

When the ring is rotating, we can treat it as a current carrying loop. The magnetic mement of this loop

M = iA = TQ × πr2 =

π2Q ω × πr2

This current carrying loop will create its own magnetic field which will interact with the given vertical magnetic field in such a way that the tensions in the strings will become unequal. Let the tension in the string be T1 and T2.

For translational equilibrium T1 + T2 = mg …(2) For rotational equilibrium Torque acting on the ring about the centre of ring

→τ =

→M ×

→B

t = M × B × sin 90º

= π2

Qω × πr2 × B =

2

2BrQω

For rotational equilibrium, the torque about the centre of ring should be zero.

∴ T1 × 2D – T2 ×

2D =

2

2BrQω

⇒ T1 – T2 = DBrQ 2ω …(3)

On solving (2) and (3) we get

T1 = 2

mg + DBrQ

2

2ω

But the maximam tension is 2

3 0T

∴ 2

3 0T = T0 +

DBrQ

2

2maxω

=

20mgTQ

∴ ωmax = 20

BQrDT

4. An object is moving with velocity 0.01 m/s towards a convex lens of focal length 0.3 m. Find the magnitude of rate of separation of image from the lens when the object is at a distance of 0.4m From the lens. Also calculated the magnitude of the rate of change of the lateral magnification. [IIT-2004]

Sol. f = 0.3 m, u = – 0.4 m Using lens formula

v1 –

4.0–1 =

3.01

⇒ v = 1.2 m


Now we have

v1 –

u1 =

f1 , differentiating w.r.t. t

we have – 2v1

dtdv + 2

1u dt

du = 0

given dtdu = 0.01 m/s

⇒

dtdv = 2

2

)4.0()120( × 0.01 = 0.09 m/s

So, rate of seperation of the image (w.r.t. the lens) = 0.09 m/s

Now, m = uv ⇒

dtdm =

2udt

vdu–dt

udv

2)4.0()01.0)(2.1(–)09.0)(4.0( = – 0.35

So magnitude of the rate of change of lateral magnification = 0.35.

5. A particle of charge equal to that of an electron, –e, and

mass 208 times the mass of the electron (called a nu-meson) moves in a circular orbit around a nucleus of charge + 3e. (Take the mass of the nucleus to be infinite). Assuming that the bohr model of the atom is applicable to this system.

(i) Derive an expression for the radius of the nth Bohr orbit.

(ii) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.

(iii) Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit of the first orbit. [IIT-1988]

Sol. (i) Let m be the mass of electron. Then the mass of mu-meson is 208 m. According to Bohr's postualte, the angular momentum of mu-meson should be an integral multiple of h/2π.

+3e

e

r

∴ (208 m) vr = π2

nh

∴ v = mr

nh2082 ×π

= mr

nhπ416

…(1)

Since mu-meson is moving in a circular path therefore it needs centripetal force which is provided by the electrostatic force between the nucleus and mu-meson.

∴ r

v)m208( 2 =

041πε

× 23

ree×

∴ r = 2

0

2

mv2084e3

×πε

Substituting the value of v from (1) we get

r = 220

2

20844164163

hnmrmre

×πεπ×π×

⇒ r = 20

22

624 mehnπ

ε …(2)

(ii) The radius of the first orbit of the hydrogen atom

= 2

20

meh

π

ε …(3)

To find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for hydrogen atom, we equate equation (2) and (3)

20

22

624 mehnπ

ε = 2

20

meh

π

ε ⇒ n = 624 ≈ 25

(iii) λ1 = 208 R × z2

22

21

1–1nn

⇒ λ1 = 208 × 1.097 × 107 × 32

22 31–

11

⇒ λ = 5.478 × 10–11 m

CHEMISTRY

6. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ..... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ? [IIT-1996]

Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in fig. Three such cells form one hcp unit.

For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside, hence

Number of atoms per unit cell = 88 + 1 = 2

N b

a

O

60º

Area of the base = b × ON = b × a sin 60º = 23 a2

( Q b = a) Volume of the hexagonal cell

= Area of the base × height = 23 a2. c


But c = 322 a

c b α β

a γ

∴ Volume of the hexagonal cell

= 23 a2 .

322 a = a3 2

and radius of the atom, r = a/2 Hence, fraction of total volume of atomic packing

factor = cellhexagonaltheofVolume

atoms2ofVolume

= 2

342

3

3

a

rπ× =

223

42

3

3

a

a

π×

= 23

π

= 0.74 = 74% ∴ The percentage of void space = 100 – 74 = 26% 7. (The standard reduction potential of Ag+/Ag

electrode at 298 K is 0.799V. Given that for AgI, Ksp = 8.7 × 10–17, evaluate the potential of Ag+/Ag

electrode in a saturated solution of AgI. Also calculate the standard reduction potential of I– electrode. [IIT-1994]

Sol. In the saturated solution of AgI, the half cell reactions are

At anode : Ag → Ag+ + e–

At cathode : AgI + e– → Ag + I– Cell reaction AgI → Ag+ + I–

On applying Nernst equation

Ecell = Eºcell – n0591.0 log [Ag+] [I–]

For electrode

Ag+ + e– → Ag

∴ Ag/AgE + = Ag/AgºE + – n

0591.0 log]Ag[

1+

Ksp of AgI = [Ag+] [I–]

Q [Ag+] = [I–]

∴ Ksp of AgI = [Ag+]2

∴ [Ag+] of AgI = spK of AgI

[Ag+] = 17107.8 −×

= 9.3 × 10–9 M

So Ag/AgE + = 0.799 – 1

0591.0 log 9103.91

−×

= + 0.799 + 0.0591 log 9.3 – 0.0591 × 9 log 10

= + 0.799 + 0.0591 × 0.9785 – 0.0591 × 9

= 0.325 V

For above cell reaction

Ecell = Eºcell – n

0591.0 log [Ag+] [I–]

= Eºcell – n0591.0 log (Ksp of AgI)

At equilibrium Ecell = 0

∴ Eºcell = 1

0591.0 log(8.7 × 10–17) = –0.95 volt

Eºcell = Eºcathode + Eºanode

–0.95 = –0.799 + EºAg/AgI/I–

(In form of cell reaction)

EºAg/AgI/I– = – 0.95 + 0.799 = –0.151 V

or EºI–/AgI/Ag = + 0.151 V

8. An organic compound A, C6H10O, on reaction with

CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C.

[IIT-2000] Sol. The given reactions are as follows.

O

CH3MgBr

OMgBr CH3

H+

–H2O

CH3

HBr

CH3Br

(A) (B) (E)

CH3

O O

COCH3

OBase

COCH3

(D) (C)


The conversion of C into D may involve the following mechanism.

COCH3

(C)

CH2 O

–BH+ B+

COCH3

HC O

COCH3

HC O–

–BBH+

COCH3

OH

–BH+ +B

COCH3

OH – –OH–

COCH3

(D)

9. A colourless solid (A) on heating gives a white solid (B) and a colourless gas (C). (B) gives off reddish-brown fumes on treating with H2SO4. On treating with NH4Cl, (B) gives a colourless gas (D) and a residue (E). The compound (A) on heating with (NH4)2SO4 gives a colourless gas (F) and white residue (G). Both (E) and (G) impart bright yellow colour to Bunsen flame. The gas (C) forms white powder with strongly heated Mg metal which on hydrolysis produces Mg(OH)2. The gas (D) on heating with Ca gives a compound which on hydrolysis produces NH3. Identify compounds (A) to (G) giving chemical equations involved.

Sol. The given information is as follows : (i) A →Heat B + C Colourless Solid Colourless Solid gas (ii) B + H2SO4 →∆ Reddish brown gas (iii) B + NH4Cl →∆ D + E Colourless gas (iv) A + (NH4)2SO4 →∆ F + G olourless gas White Residue (v) E and G imparts yellow colour to the flame. (vi) C + Mg →Heat White powder → OH2 Mg(OH)2 (vii) D + Ca →Heat Compound → OH2 NH3 Information of (v) indicates that (E) and (G) and also

(A) are the salts of sodium because Na+ ions give yellow coloured flame. Observations of (ii) indicate that the anion associated with Na+ in (A) may be NO3

–. Thus, the compound (A) is NaNO3. The reactions involved are as follows : (i) 2NaNO3 →∆ 2NaNO2 + O2 ↑ (A) (B) (C) (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 (B) Dil. 3HNO2 → HNO3 + H2O + 2NO↑ 2NO + O2 → 2NO2 ↑ Reddish brown Fumes

(iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (B) (E) (D)

(iv) 2NaNO3 + (NH4)2SO4 →∆ Na2SO4 + 2NH3 (A) (G) (F) 2HNO3

(v) O2 + 2Mg →∆ 2MgO → OH2 Mg(OH)2 (C)

(vi) N2 + 3Ca →∆ Ca3N2 (D) Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ Hence, (A) is NaNO3, (B) is NaNO2, (C) is O2, (D) is N2, (E) is NaCl, (F) is NH3 and (G) is Na2SO4. 10. An alkyl halide X, of formula C6H13Cl on treatment

with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y and Z. [IIT-1996]

Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes.

X136 ClHC

HCl–;

butoxidetK

∆

−− → 126HCZY +

Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.

Y and Z (C6H12) Ni

H2→ CH3 – CH – CH – CH3

CH3 CH3

2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3

CH3 CH32,3-dimethyl

butene-2 (Y)

and CH3 – CH – C = CH2

CH3 CH32,3-dimethyl butene-1

(Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3

CH3 CH3(Y)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

CH3 – CH – C = CH2

CH3 CH3(Z)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

Both, Y and Z can be obtained from following alkyl

halide :


CH3 – C – CH – CH3

CH3 CH3 2-chloro-2,3-dimethyl butane

(X)

K-t-butoxide

∆; –HCl

CH2 = C — CH – CH3

CH3 CH3

Cl

+ CH3 – C = C – CH3

CH3 CH3

(Z) 20% (Y) 80%

Hence, X,

CH3 – C – CH – CH3

CH3 CH3

Cl

Y, CH3 – C = C – CH3

CH3 CH3

Z, CH3 – CH – C = CH2

CH3 CH3

MATHEMATICS

11. The curve y = ax3 + bx2 + cx + 5, touches the x-axis at P(–2, 0) and cuts the y axis at a point Q, where its gradient is 3. Find a, b, c. [IIT-1994]

Sol. It is given that y = ax3 + bx2 + cx + 5 touches x-axis at P(–2, 0) which implies that x-axis is tangent at (–2, 0) and the curve is also passes through (–2, 0).

The curve cuts y-axis at (0, 5) and gradient at this point is given 3 therefore at (0, 5) slope of the tangent is 3.

Now, dxdy = 3ax2 + 2bx + c

since x-axis is tangent at (–2, 0) therefore

2−=xdx

dy = 0

⇒ 0 = 3a(–2)2 + 2b(–2) + c ⇒ 0 = 12a – 4b + c ...(1) again slope of tangent at (0, 5) is 3

⇒ )5,0(dx

dy = 3

⇒ 3 = 3a(0)2 + 2b(0) + c ⇒ 3 = c ...(2) Since, the curve passes through (–2, 0), we get 0 = a(–2)3 + b(–2)2 + c(–2) + 5 0 = – 8a + 4b – 2c + 5 ...(3) from (1) and (2), we get 12a – 4b = –3 ...(4)

from (3) and (2), we get – 8a + 4b = 1 ...(5) adding (4) and (5), we get 4a = –2 ⇒ a = –1/2 Putting a = –1/2 in (4), we get 12(–1/2) – 4b = –3 ⇒ – 6 – 4b = –3 ⇒ – 3 = 4b ⇒ b = –3/4 Hence, a = –1/2, b = –3/4 and c = 3 12. In a triangle ABC, the median to the side BC is of

length 3611

1

− and it divides the angle A into

angles 30º and 45º. Find the length of the side BC. [IIT-1985]

Sol. Let AD be the median to the base BC = a of ∆ABC let ∠ADC = θ then

+

22aa cot θ =

2a cot 30º –

2a cot 45º

⇒ cot θ = 2

13 −

Applying sine rule in ∆ADC, we get

CD a/2 a/2B

A

45º30º

θ

)º45sin(

AD−θ−π

= º45sin

DC

⇒ )º45sin(

AD+θ

= 2/12/a

⇒ AD = 2

a sin 45º cosθ + cos45ºsinθ

⇒ AD =

θ+θ

2sincos

2a =

2a (cos θ + sin θ)

⇒ 3611

1

−=

−+

−

−

328

2

328

132a

⇒ a = 3611)13(

3282

−+

−

⇒ a = 3611)13(

32822 −+

−


⇒ a = )3611)(324(

3282

−+

−

⇒ a = 23632232444

328

−+−

− = 2328

328

−

− = 2

13. Without using tables, prove that

(sin 12º) (sin 48º) (sin 54º) = 81 [IIT-1982]

Sol. (sin 12º) (sin 48º) (sin 54º)

= 21 (2 sin 12º sin 48º) sin 54º

= 21 cos (36º) – cos (60º)sin 54º

= 21

−

21º36cos sin 54º

= 41 2 cos 36º sin 54º – sin 54º

= 41 (sin 90º + sin 18º – sin 54º)

= 41

+

−−

+4

154

151

= 41

−−−

+4

15151

= 41

−

211 =

81

14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001]

Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws.

Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements

in the range. Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0 ⇒ M = 3n – 3.2n + 3 Hence the total number of favourable cases = (3n – 3.2n + 3). 6C3

⇒ Required probability = n

nn C6

)32.33( 36×+−

15. Evaluate ∫π

π−

π

+−

+π3/

3/

3

3||cos2

4

x

x dx [IIT-2004]

Sol. Let,

I = ∫π

π−

π

+−

π3/

3/

3||cos2 x

dx + 4 ∫π

π−

π

+−

3/

3/

3

3||cos2 x

dxx

Using ∫−

a

axf )( dx =

=−

−=−

∫ )()(,)(2

)()(,0

0xfxfdxxf

xfxfa

∴ I = 2 ∫π

π

+−

π3/

0

3||cos2 x

dx + 0

π

+−∫

π

π−

3/

3/

3

3||cos2

oddisx

dxxas

I = 2π ∫π

π+−

3/

0 )3/cos(2 xdx

= 2π ∫π

π −

3/2

3/ cos2 tdt , where x +

3π = t

= 2π ∫π

π +

3/2

3/ 2

2

2tan31

2sec

t

dtt

= 2π ∫ +

3

3/1 2312

udu =

34π . 3

3/11 3tan3 u−

= 3

4π (tan–1 3 – tan–11) = 3

4π tan–1

21

∴ ∫π

π−

π

+−

+π3/

3/

3

3||cos2

4

x

x dx = 3

4π tan–1

21 .



1. Two capacitors C1 and C2, can be charged to a

potential V/2 each by having

V S1 S2

R R

C2 C1

O

(A) S1 closed and S2 open (B) S1 open and S2 closed (C) S1 and S2 both closed (D) cannot be charged at V/2

2. Energy liberated in the de-excitation of hydrogen atom from 3rd level to 1st level falls on a photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen like gas, excited to 2nd energy level, it is found that the de-Broglie wavelength of the fastest photoelectrons, now ejected has decreased by a factor of 3. For this new gas, difference of energies of 2nd Lyman line and 1st Balmer line if found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s)

(A) The gas is lithium (B) The gas is helium (C) The work function of photo-cathode is 8.5eV (D) The work function of photo-cathode is 5.5eV

3. In the figure shown there exists a uniform time varying magnetic field B = [(4T/s) t + 0.3T] in a cylindrical region of radius 4m. An equilateral triangular conducting loop is placed in the magnetic field with its centroide on the axis of the field and its plane perpendicular to the field.

+ + + +

+ + + +

+ + + +

++++

+ + + +

+ + + + +

+ ++

++

++

A

C B (A) e.m.f. induced in any one rod is 16V (B) e.m.f. induced in the complete ABC∆ is V348 (C) e.m.f. induced in the complete ABC∆ is 48V (D) e.m.f. induced in any one rod is V316

4. 6 parallel plates are arranged as shown. Each plate has an area A and distance between them is as shown. Plate 1-4 and plates 3-6 are connected equivalent capacitance across 2 and 5 can be writted

as d

nA 0∈. Find mininum value of n. (n, d are

natural numbers) 1

23

4 2d

5 6 d

d

dd

5. Match the following

Column – I Column – II (A) A light conducting (P) Magnetic field B circular flexible is doubled. loop of wire of radius r carrying current I is placed in uniform magnetic field B, the tension in the loop is doubled if (B) Magnetic field at a (Q) Inductance is point due to a long increased by four straight current times. carrying wire at a point near the wire is doubled if (C) The energy stored (R) Current I is in the inductor will doubled become four times (D)The force acting on a (S) Radius r is moving charge, doubled moving in a constant magnetic field will be doubled if (T) Velocity v is Doubled

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 10


Passage # (Q. No. 6 to Q. No. 8 ) A solid, insulating ball of radius ‘a’ is surrounded by

a conducting spherical shell of inner radius ‘b’ and outer radius ‘c’ as shown in the figure. The inner ball has a charge Q which is uniformly distribute throughout is volume. The conducting spherical shell has a charge –Q.

Answer the following questions. –Q

b c

a Q

6. Assuming the potential at infinity to be zero, the

potential at a point located at a distance a/2 from the centre of the sphere will be :

(A)

−

πε b1

a2

4Q

0

(B)

−

πε b1

a811

4Q

0

(C)

−

πε b1

a1

4Q

0

(D) None of these

7. Work done by external agent in taking a charge q slowly from inner surface of the shell to surface of the sphericalball will be :

(A)

−

c1

a1kQq

(B)

−

a1

b1kQq

(C)

−

b1

a1kQq

(D)

−

a1

c1kQq

8. Now the outer shell is grounded, i.e., the outer surface is fixed to be zero. Now the charge on the inner ball will be :

(A) zero (B) Q

(C)

−+

b1

c1

a1

CQ (D)

−+

b1

c1

a1

bQ

Regents Physics You Should Know

Nuclear Physics : • Alpha particles are the same as helium nuclei and

have the symbol .

• The atomic number is equal to the number of protons (2 for alpha)

• Deuterium ( ) is an isotope of hydrogen ( )

• The number of nucleons is equal to protons + neutrons (4 for alpha)

• Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator.

• Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays)

• A loss of a beta particle results in an increase in atomic number.

• All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2)

• Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers).

• Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.

• Rutherford discovered the positive nucleus using his famous gold-foil experiment.

• Fusion requires that hydrogen be combined to make helium.

• Fission requires that a neutron causes uranium to be split into middle size atoms and produce extra neutrons.

• Radioactive half-lives can not be changed by heat or pressure.

• One AMU of mass is equal to 931 meV of energy (E = mc2).

• Nuclear forces are strong and short ranged.


1. Option [C] is correct Magnetic field due to infinite current carrying sheet

is given by ,2

JB 0µ

= where J is linear current

density.

2J0µ

I IV

(b)(a)

2J0µ

2J0µ

2J0µ

Fig. (a) and (b) represent the direction of magnetic field due to current carrying sheets. For x < a,

2)J4(

2)J3(

2)J2(J

2J

B 0000ttanresul

µ+

µ−

µ−

µ=

For a < x < 2a,

J2

)J4(2

)J3(2

)J2(2

JB 0

0000ttanresul µ−=

µ+

µ−

µ−

µ=

For 2a < x < 3a,

02

)J4(2

)J3(2

)J2(2

JB 0000

ttanresul =µ

−µ

−µ

+µ

=

So, the required curve is

2. A → P,Q,S ; B → P,Q,R,S C → P,Q,R,S ; D → Q i. Velocity of the particle may be constant, if forces

of electric and magnetic fields balance each other. Then, path of particle will be straight line. Also, path of particle may be helical if magnetic and electric fields are in same direction. But path of particle cannot be circular. Path can be circular if only magnetic field is present, or if some other forces is present which can cancel the effect of electric field.

ii. Here, all the possibilities are possible depending upon the combinations of the three fields.

iii. This situation is similar to part (i) iv. In a uniform electric field, path can be only

straight line or parabolic.

3. A → Q B → R C → P D → Q

i. At t = 1s, flux is increasing in the inward direction, hence induced e.m.f. will be in anticlockwise direction.

ii. At t = 5s, there is no change in flux, so induced e.m.f. is zero

iii. At t = 9s, flux is increasing in upward direction hence induced e.m.f. will be in clockwise direction.

iv. At t = 15s, flux is decreasing in upward direction, so induced e.m.f. will be in anticlockwise direction.

4. Option [A,B,D] is correct Rate of work done by external agent is de/dt = BIL.dx/dt = BILv and thermal power dissipated in resistor = eI = (BvL) I clearly both are equal, hence (A). If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increase, hence (B). Since, I = e/R On doubling R, current and hence required power become half. Since, P = BILv Hence (D)

5. Option [A] is correct

)j(2)j(5.1 21

∧→∧→×µ=×µ

]j)jdic[(2j)jbia(5.1∧∧∧∧∧∧

×+=×+

∧∧

= kc2ka5.1

34

5.120

ca

==

Solution Physics Challenging Problems

Set # 9

8 Questions were Published in January Issue


6. Option [A] is correct

f sin60º

f f cos60º

I2 f –

+

I2

x u = -f cos60º f = +f

º60cosf

1v1

f1

−−=

f2

v1

f1

+=

v1

f2

f1

=−

v = -f

º60cosxf

=

xº60cos

f=

x = 2f ∴ final image will formed at optical centre of first

lens.

7. Option [C] is correct Cv = (3 + 2T)R dQ = dU + PdV adiabatic process dQ = 0 0 = Rn (3 + 2T)dT + PdV

dVV

nRTdT)T23(Rn0 ++=

∫ ∫

+

=− dTT

T23V

dV

-log V = 3 logT + 2T + C -logV – logT3 = 2T + C log VT3 = 2T + C VT3 = e2T VT3e-2T = C

8. Option [A] is correct 2

0 VPP α−= PV = RT

20 VP

VRT

α−=

RV

RVP

T3

0 α−=

0dVdT

=

0RV3

RP 2

0 =α

−

α

=3P

V 0 Now put V in T.

WHAT ARE EARTHQUAKES?

Earthquakes like hurricanes are not only super destructive forces but continue to remain a mystery in terms of how to predict and anticipate them. To understand the level of destruction associated with earthquakes you really need to look at some examples of the past.

If we go back to the 27th July 1976 in Tangshan, China, a huge earthquake racked up an official death toll of 255,000 people. In addition to this an estimated 690,000 were also injured, whole families, industries and areas were wiped out in the blink of a second. The scale of destruction is hard to imagine but earthquakes of all scales continue to happen all the time.

So what exactly are they ? Well the earths outer layer is made up of a thin crust divided into a number of plates. The edges of these plates are referred to as boundaries and it’s at these boundaries that the plates collide, slide and rub against each other. Over time when the pressure at the plate edges gets too much, something has to give which results in the sudden and often violent tremblings we know as earthquakes.

The strength of an earthquake is measured using a machine called a seismograph. It records the trembling of the ground and scientists are able to measure the exact power of the quake via a scale known as the richter scale. The numbers range from 1-10 with 1 being a minor earthquake (happen multiple times per day and in most case we don’t even feel them) and 7-10 being the stronger quakes (happen around once every 10-20 years). There’s a lot to learn about earthquakes so hopefully we’ll release some more cool facts in the coming months.


1. A trolley initially at rest with a solid cylinder placed on its bed such that cylinder axis makes angle θ with direction of motion of trolley as shown in Figure starts to move forward with constant acceleration a. If initial distance of mid point of cylinder axis from rear edge of trolley bed is d, calculate the distance s which the trolley goes before the cylinder rolls off the edge of its horizontal bed. Assume dimensions of cylinder to be very small in comparison to other dimensions. Neglect slipping.

d

θ

Calculate also, frictional force acting on the cylinder. Sol. Since, axis of cylinder is inclined at angle θ with the

direction of motion of trolley, therefore components of acceleration a of trolley are acosθ along axis of cylinder and asinθ normal to axis of the cylinder. Cylinder rolls backward due to this normal component asinθ.

Let mass and radius of cylinder be m and r respectively and let angular acceleration of cylinder be α.

Due to angular acceleration, cylinder axis has acceleration relative to trolley bed, which will be equal to rα normal to cylinder axis. But component of acceleration of trolley normal to cylinder axis is asinθ. Therefore, net acceleration of cylinder axis is (asinθ – rα) normal to axis.

Consider free body diagram of the cylinder as shown figure

Note : There are two components of friction (i) F1 (normal to cylinder axis) and

(ii) F2 (along cylinder axis). F2 prevents cylinder from sliding along axis or acosθ component of acceleration of cylinder along axis is due to F2.

mg

l.α

F1

N m(a.sin θ –rα)

∴ F2 = ma cos θ F2 is not shown in the free body diagram because in

this diagram forces action normal to cylinder axis are shown.

For horizontal forces, F1 = m (a sin θ – rα) …(1)

F1r = I α where I = 2

2mr

∴ F1 = 21 mrα …(2)

Form equation (1) and (2),

rα = 32 a sin θ

The cylinder will roll off the edge of trolley bed when its centre of mass reaches the edge. Since. cylinder axis is inclined at an angle 'θ' with direction of motion of trolley, therefore, its centre of mass follows a straight line path relative to the trolley bed, and that straight line is normal to cylinder axis. Hence, displacement of centre of mass of the cylinde, relative to trolley is equal to (d. cosec θ).

considereing motion of cylinder relative to the trolley,

u = 0, acceleration = rα = 32 a sin θ, s = d cosec θ,

t = ?

Using, s = ut + 21 at2, or t =

θ2sin3

ad

Now considering motion of trolley during this interval ofd time,

u = 0, acceleration a , t = θ2sin

3a

d , s = ?

Using, s = ut + 21 at2, s =

23 d cosec2 θ Ans.

F1 = 21 m. rα =

31 ma sin θ

Total frictional force acting on the cylinder is

F = 22

21 FF +

= 31 ma θ+θ 22 cos9sin Ans.

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICS


2. A particle of mass m is placed on centre of curvature of a fixed, uniform semi-circular ring of radius R and M as shown in Figure. Claculate

m

M

R

(i) interaction force between the ring and the particle

and (ii) work required to displace the particle from centre

of curvature to infinity. Sol. To calculate, interaction force, consider two equal are

lengths R dθ each of the semi-circular ring as shown in figure

Rdθ

x

dθ

dθ θ θ

Rdθ

Mass of each arc, dM = R

Mπ

Rdθ = π

θmd

Gravitational force exerted by each arc on the particle,

dF = 2RGmdM = 2R

GMmπ

dθ

Since, force exerted by each arc is directly towards the arc, therefore, resultant of these two forces is along negative x-axis and the resultant force = dF1 cos θ

= 22

RGMmπ

cos θ dθ

Total force on the particle is

F = 22

RGMmπ

∫π=θ

=θ

θθ2/

0

cos d

or F = 22

RGMmπ

Ans. (i)

Work done during displacement of particle from centre of the curvature to infinity is used to increase gravitational potential energy of the system.

Initial gravitational potential energy of particle with each arc is

dU = –RdMGm. = –

RGMm

πdθ

∴ Total initial potential energy,

U1 = – R

GMmπ ∫

π

π=θ

θ2/

2/–

d

or U1 = – R

GMm

When separation between particle and semicircular ring becomes large, potential energy becomes U2 = 0

∴ Work done = U2 – U1 = R

GMm Ans.(ii)

3. A long round conductor of radius a is made of a

material whose thermal conductivity depends on distance r from axis of the conductor as K = cr2 , where c is a constant. Calculate

(i) thermal resistance per unit length of such a conductor and

(ii) temperature gradient if rate of heat flow through the rod is H.

Sol. Since, thermal conductivity of material of the conductor depends upon distance from its axis, therefore, conductivity at every point of a co-axial cylindrical surface will be the same. To calculate thermal resistance of the given conductor, it may be assumed to be composed of thin co-axial cylindrical shells which are in parallel combination with each other.

Consider a thin co-axial cylindrical shell of radius x, radial thickness dx and of unit length as shown in figure

Its cross sectional area, A = 2πx.dx Thermal conductivity K = cx2 and length l = 1 m

∴ Its thermal resistance, dR = KAl =

dx.x2)cx( 2 π

l

or dR = dxcx .2

13π

Since, such cylindrical shells are in parallel with each other, therefore, equivalent resistance R per unit length is given by

R1 = ∫ dR

1 = ∫=

=

πax

x

dxcx0

3.2

or R = 42caπ

Ans.(i)

Since, temperature gradient is temperature difference per unit length, therefore, temperature gradient = rate of heat flow × resistance per unit length

or dtdθ = H × R = 4

2caH

π Ans. (ii)

4. Switch S of circuit shown in Figure is in position 1

for a long time. At instant t = 0, it is thrown from position 1 to 2. Calculate thermal power P1(t) and P2(t) generated across resistance R1 and R2 respectively.


C

R2

R1

2 S

1

+ –

E Sol. Since, initially the switch was in position 1 for a long

time, therefore, initially the capacitor was fully charged or potential difference across capacitor at t = 0 was equal to emf E fo the battery.

∴ Initial charge on capacitor, q0 = CE When switch is thrown to position 2, capacitor starts

to discharge through resistance R1 and R2. To calculate thermal power P1(t) and P2(t) generated across R1 and R2 respectively, current I at time t through the circuit must be known.

Let at instant t, charge remaining on the capacitor be q and let current through the circuit be I.

Applying Kirchhoff's voltage law on the mesh in the circuit of figure

C

R2

R1

I

q + –

I

Cq – IR2 – IR1 = 0

or I = CRR

q)( 21 +

...(1)

Since, the capacitor is discharging, therfore,

I = – dtdq

∴ From equation (1),

q

dq = – CRR

dt)( 21 +

...(2)

Knowing that at t = 0, q = q0 = CE, integrating equation (2),

∫=

=

?q

CEqq

dq = – ∫=

+

t

tCRR

dt

0 21 )(

∴ logCEq = –

CRRt

)( 21 +

or CRRtCEeq )/(– 21+=

But I = – dtdq ,

therefore, I = )( 21 RR

E+

CRRte )/(– 21+ ...(3)

Hence, thermal power across R1 is

P1 = I2R1

or P1 = CRR

RE)( 21

12

+CRRte )/(2– 21+ Ans.

Similarly, thermal power across R2, P2 = I2R2

or P2 = CRRteRRRE )/(2–

221

22

21

)(+

+ Ans.

5. Two plane mirrors, a source S of light, emitting

mono-chromatic rays of wavelength λ and a screen are arranged as shown in figure. If angle θ is very small, calculate fringe width of interference pattern formed on screen by reflected rays.

ba

S

Screen

θ

θ

Sol. Since, interference is due to reflected rays, therefore,

images S1 and S2 of the source S behave like two coherent sources as shown in figure

ba

SOθ

θ

M

NRd

D Distance of source S from each mirror = a cos θ ∴ SS1 = SS2 = 2 × a cos θ Distance between S1 and S2, d = SS1 sin θ + SS2 sin θ = 4a cos θ sin θ But θ is very small, therefore cos θ ≈ 1 and sin θ ≈ θ ∴ d = 4aθ Distance RS = SS1 cos θ = 2a.cos2θ ≈ 2a ∴ Distance of screen from two coherent sources S1

and S2 is D = RO = RS + SO or D = (2a + b) Now the arrangement is similar to Young's double

slit arrangement.

∴ Fringe width, ω = d

Dλ = θ

λ+a

ba4

)2( Ans.


Matter Waves : Planck's quantum theory : Wave-particle duality - Planck gave quantum theory while explaining the

radiation spectrum of a black body. According to Planck's theory, energy is always exchanged in integral multiples of a quanta of light or photon.

Each photon has an energy E that depends only on the frequency ν of electromagnetic radiation and is given by :

E = hν .....(1) where h = 6.6 × 10–34 joule-sec, is Planck's

constant. In any interaction, the photon either gives up all of its energy or none of it.

From Einstein's mass-energy equivalence principle, we have

E = mc2 .....(2) Using equations (1) and (2), we get ;

mc2 = hν or m = 2chν .....(3)

where m represents the mass of a photon in motion. The velocity v of a photon is equal to that of light, i.e., v = c.

According to theory of relativity, the rest mass m0 of a photon is given by :

m0 = 2

2

cv1m −

Here, m = 2chν and v = c

Hence, m0 = 0 ....(4) i.e., rest mass of photon is zero, i.e., energy of

photon is totally kinetic.

The momentum p of each photon is given by :

p = mc = 2chν × c =

chν =

ν/ch =

λh ......(5)

The left hand side of the above equation involves the particle aspect of photons (momentum) while the right hand side involves the wave aspect (wavelength) and the Planck's constant is the bridge between the two sides. This shows that electromagnetic radiation exhibits a wave-

particle duality. In certain circumstances, it behaves like a wave, while in other circumstances it behaves like a particle.

The wave-particle is not the sole monopoly of e.m. waves. Even a material particle in motion according to de Broglie will have a wavelength. The de Broglie wavelength λ of the matter waves is also given by :

λ = mvh =

ph =

mKh

2

where K is the kinetic energy of the particle. If a particle of mass m kg and charge q coulomb

is accelerated from rest through a potential difference of V volt. Then

21 mv2 = qV or mv = mqV2

Hence, λ = mqVh

2 =

V34.12 Å

Photoelectric effect : When light of suitable frequency (electromagnetic

radiation) is allowed to fall on a metal surface, electrons are emitted from the surface. These electrons are known as photoelectrons and the effect is known as photoelectric effect. Photoelectric effect, light energy is converted into electrical energy.

Laws of photolectric effect : The kinetic energy of the emitted electron is

independent of intensity of incident radiation. But the photoelectric current increases with the increase of intensity of incident radiation.

The kinetic energy of the emitted electron depends on the frequency of the incident radiation. It increases with the increase of frequency of incident radiation.

If the frequency of the incident radiation is less than a certain value, then photoelectric emission is not possible. This frequency is known as threshold frequency. This threshold frequency varies from emitter to emitter, i.e., depends on the material.

There is no time lag between the arrival of light and the emission of photoelectrons, i.e., it is an instantaneous phenomenon.

Matter Waves, Photo-electric Effect

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Failure of wave theory : Wave theory of light could not explain the laws of

photoelectric effect. According to wave theory, the kinetic energy of

the emitted electrons should increase with the increase of intensity of incident radiation.

Kinetic energy of the emitted electron does not depend on the frequency of incident radiation according to wave theory.

Wave theory failed to explain the existence of threshold frequency.

According to wave theory there must be a time lag between the arrival of light and emission of photoelectrons.

Einstein's theory of photoelectric effect : Einstein explained the laws of photoelectric effect

on the basis of Planck's quantum theory of radiation.

Einstein treated photoelectric effect as a collision between a photon and an atom in which photon is absorbed by the atom and an electron is emitted.

According to law of conservation of energy,

hν = hν0 + 21 mv2

where hν is the energy of the incident photon; hv0 is the minimum energy required to detach the electron from the atom (work function or ionisation energy) and (1/2) mv2 is the kinetic energy of the emitted electron.

The above equation is known as Einstein's photoelectric equation. Kinetic energy of the emitted electron,

= 21 mv2 = h(ν – ν0) = hν – W

Explanation of laws of photoelectric effect : (a) The KE of the emitted electron increases with the

increase of frequency of incident radiation since W (work function) is constant for a given emitter. KE is directly proportional to (ν – ν0)

(b) Keeping the frequency of incident radiation constant if the intensity of incident light is increased, more photons collide with more atoms and more photoelectrons are emitted. The KE of the emitted electron remains constant since the same photon collides with the same atom (i.e., the nature of the collision does not change). With the increase in the intensity of incident light photoelectric current increases.

(c) According to Einstein's equation, if the frequency of incident radiation is less than certain minimum value, the photoelectric emission is not possible. This frequency is known as threshold frequency. Hence, the frequency of incident radiation below which photoelectric emission is not possible is known as threshold frequency or cut-off frequency. It is given by :

ν0 = h

mv)2/1(h 2−ν

On the other hand, if the wavelength of the incident radiation is more than certain critical value, then photoelectric emission is not possible. This wavelength is known as threshold wavelength of cut-off wavelength. It is given by :

λ0 = ]mv)2/1(h[

hc2−ν

(d) Since Einstein treated photoelectric effect as a collision between a photon and an atom, he explained the instantaneous nature of photoelectric effect.

Some other important points : Stopping potential : The negative potential

applied to the collector in order to prevent the electron from reaching the collector (i.e., to reduce the photoelectric current to zero) is known as stopping potential.

eV0 = 2.maxmv

21 = hν – W = h(ν – ν0)

Millikan measured K.E. of emitted electrons or stopping potentials for different frequencies of incident radiation for a given emitter. He plotted a graph with the frequency on x-axis and stopping potential on y-axis. The graph so obtained was a straight line as shown in figure.

ν0

Frequency of incident light

V0(s

topp

ing

pote

ntia

l)

Millikan measured the slope of the straight line

(=h/e) and calculated the value of Planck's constant.

I

Full intensity

75% intensity 50% intensity 25% intensity

V0– +

Potential difference


The intercept of V0 versus ν graph on frequency axis is equal to threshold frequency (ν0). From this, the work function (hν0) can be calculated.

Graphs in photoelectric effect : (a) Photoelectric current versus potential difference

graphs for varying intensity (keeping same metal plate and same frequency of incident light) : These graphs indicate that stopping potential is independent of the intensity and saturation current is directly proportional to the intensity of light.

I

(V0)1 – +

Potential difference (V0)2

ν2 ν1

ν2>ν1

(b) Photoelectric current versus potential difference

graphs for varying frequency (keeping same metal plate and same intensity of incident light) : These graphs indicate that the stopping potential is constant for a given frequency. The stopping potential increases with increase of frequency. The KE of the emitted electrons is proportional to the frequency of incident light.

A1 A2 A3

ν0

B1

B2

B3

Frequency

Stop

ping

pot

entia

l

(c) Stopping potential versus frequency graphs for

different metals : These graphs indicate that the stops is same for all metal, since they are parallel straight lines. The slope is a universal constant (=h/e). Further, the threshold frequency varies with emitter since the intercepts on frequency axis are different for different metals.

1. (i) A stopping potential of 0.82 V is required to stop

the emission of photoelectrons from the surface of a metal by light of wavelength 4000 Å. For light of wavelength 3000 Å, the stopping potential is 1.85 V. Find the value of Planck's constant.

(ii) At stopping potential, if the wavelength of the incident light is kept at 4000 Å but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer.

Sol. (i) We have 1λ

hc = eV1 + W

and 2λ

hc = eV2 + W

⇒

λ

−λ 12

11hc = e(V2 – V1)

or h =

λ

−λ

−

12

12

11)(

e

VVe =

×−

××

−×

−−

−

778

19

1041

1031103

)82.085.1(106.1

= 6.592 × 10–34 Js (ii) No, because the stopping potential depends only

on the wavelength of light and not on its intensity. 2. A small plate of a metal (work function = 1.17 eV) is

plated at a distance of 2m from a monochromatic light source of wavelength 4.8 × 10–7 m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. If a constant magnetic field of strength 10–4 tesla is parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectrons.

Sol. Energy of one photon = λhc = 7

834

108.4103106.6

−

−

×

×××

= 4.125 × 10–19 J Number of photons emitted per second

= 1910125.40.1

−× = 2.424 × 1018

Number of photons striking the plate per square metre per second

= 2

18

)2(14.3410424.2××

× = 4.82 × 1016

Maximum kinetic energy of photoelectrons emitted from the plate

Emax = λhc – W

= 4.125 × 10–19 – 1.17 × 1.6 × 10–19 = 2.253 × 10–19 J

Solved Examples


3. A monochromatic light source of frequency ν illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atom in ground state. When the whole experiment is repeated with an incident radiation of frequency (5/6) ν, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. Find the work function of the metal and the frequency ν.

Sol. In the first case, Emax = Ionization energy = 13.6 eV = 21.76 × 10–19 J So, hν = 21.76 × 10–19 J ....(1) In the second case,

E'max = λhc

= 10

834

101215103106.6

−

−

×

×××

=16.3×10–19 J

So, 6

5 hν = 16.3 × 10–19 + W ...(2)

Dividing Eq.(1) by Eq.(2)

56 =

W103.16W1076.21

19

19

+×

+×−

−

Solving, we get W = 11.0 × 10 – 19 J = 6.875 eV

From Eq.(1) ν = 34

1919

106.6100.111076.21

−

−−

×

×+×

= 5 × 1015 Hz 4. The radiation, emitted when an electron jumps from

n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1/320 T in a radius of 10–3 m. Find (i) the kinetic energy of electrons, (ii) wavelength of radiation and (iii) the work function of metal.

Sol. (i) Speed of an electron in the magnetic field,

v = m

Ber

Kinetic energy of electrons

Emax = 21 mv2 =

mreB

2

222

= 2

3201

× 31

23219

101.92)10()106.1(

−

−−

××

××

= 1.374 × 10–19 J = 0.8588 eV

(ii) Energy of the photon emitted from a hydrogen atom

hν = λhc =

− 22 3

121

= 1.888 eV Wavelength of radiation,

λ = 19

834

106.1888.11031062.6

−

−

××

×××

= 6.572 × 10–7m = 6572 Å (iii) Work function of metal W = hν – Emax = 1.8888 – 0.8588 = 1.03 eV 5. X-rays are produced in an X-ray tube by electrons

accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half of its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

Sol. Initial kinetic energy of the electron = 50.0 keV Kinetic energy after first collision = 25.0 keV Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

λ1 = 1E

hc = 319

834

100.25106.1103106.6×××

×××−

−

= 0.495 × 10–10 m = 0.495 Å Kinetic energy of the electron after second collision = 12.5 eV Energy of the photon produced in the second

collision, E2 = 25.0 – 12.5 = 12.5 keV Wavelength of this photon

λ2 = 2E

hc = 319

834

105.12106.1103106.6×××

×××−

−

= 0.99 × 10–10 m = 0.99 Å Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision,

E3 = 12.5 – 0 = 12.5 keV This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å.



Thermal Expansion :

.(a) When the temperature of a substance is increased, it expands. The heat energy which is supplied to the substance is gained by the constituent particles of the substance as its kinetic energy. Because of this the collisions between the constituents particles are accompanied with greater force which increase the distance between the constituent particles.

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; V' = V(1 + γ∆T) (b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at

higher temperature clearly ρ' < ρ for substances which have positive value of γ

* β = 2α and γ = 3α Water has negative value of γ for certain temperature

range (0º to 4ºC). This means that for that temperature range the volume decreases with increase in temperature. In other words the density increases with increase in temperature.

0 ml 5 ml 10 ml 15 ml 20 ml 25 ml 30 ml

If a liquid is kept in a container and the temperature

of the system is increased then the volume of the liquid as well as the container increases. The apparent change in volume of the liquid as shown by the scale is

∆Vapp = V(γ – 3α) ∆T Where V is the volume of liquid at lower temperature ∆Vapp is the apparent change in volume γ is the coefficient of cubical expansion of liquid α is the coefficients of linear expansion of the

container. Loss or gain in time by a pendulum clock with

change in temperature is ∆t = 21

α(∆T) × t

Where ∆t is the loss or gain in time in a time interval t ∆T is change in temperature and d is coefficient of

linear expansion. If a rod is heated or cooled but not allowed to expand

or contract then the thermal stresses developed

AF = γα∆T.

If a scale is calibrated at a temperature T1 but used at a temperature T2, then the observed reading will be wrong. In this case the actual reading is given by

R = R0(1 + α∆T) Where R0 is the observed reading, R is the actual

reading. For difference between two rods to the same at all

temperatures l 1α1 = l2α2. Thermodynamics

According to first law of thermodynamics q = ∆U + W

For an isothermal process (for a gaseous system) (a) The pressure volume relationship is ρV = constt. (b) ∆U = 0 (c) q = W (d) W = 2.303 nRT log10

i

f

VV

= 2.303 nRT log10 f

ipp

(e) Graphs T2 > T1

T2T1

P

V

P

T

V

T These lines are called isotherms (parameters at

constant temperature) For an adiabatic process (for a gaseous system)

(a) The pressure-volume relationship is PVγ = constt. (b) The pressure-volume-temperature relationship is

TPV = constt.

(c) From (a) and (b) TVγ–I = constt. (d) q = 0 (e) W = –∆U

Thermal Expansion, Thermodynamics

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


(f) ∆U = ncv∆T where cv = 1−γ

R

(g) W = 1−γ

− ffii VpVp =

1)(

−γ

− fi TTnR

(h) Graphs

V

P

P

T

V

T Please note that P-V graph line (isotherm) is

steeper. For isochoric process

(a) P ∝ T (b) W = 0 (c) q = ∆U

(d) ∆U = nCv∆T where Cv = 1−γ

R

(e) Graphs

P

V

P

T

V

T For isobaric process

(a) V ∝ T (b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti) (c) ∆U = nCv∆T (d) q = nCp∆T (e) Graphs

P

V

P

T

V

T For a cyclic process

(a) ∆U = 0 ⇒ q = W (b) Work done is the area enclosed in p-V graph.

For any process depicted by P-V diagram, area under the graph represents the word done.

Kirchoff's law states that good absorbers are good emitters also.

Problem solving Strategy : Thermal Expansion Step 1: Identify the relevant concepts: Decide

whether the problem involves changes in length (linear thermal expansion) or in volume (volume thermal expansion)

Step 2: Set up the problem using the following steps: Eq. ∆L = αL0∆T for linear expansion and Eq. ∆V = βV0∆T for volume expansion. Identify which quantities in Eq. ∆L = αL0∆T or

∆V = βV0∆T are known and which are the unknown target variables.

Step 3: Execute the solution as follows: Solve for the target variables. Often you will be

given two temperatures and asked to compute ∆T. Or you may be given an initial temperature T0 and asked to find a final temperature corresponding to a given length or volume change. In this case, plan to find ∆T first; then the final temperature is T0 + ∆T.

Unit consistency is crucial, as always. L0 and ∆L (or V0 ∆V) must have the same units, and if you use a value or α or β in K–1 or (Cº)–1, then ∆T must be in kelvins or Celsius degrees (Cº). But you can use K and Cº interchangeably.

Step 4: Evaluate your answer: Check whether your results make sense. Remember that the sizes of holes in a material expand with temperature just as the same way as any other linear dimension, and the volume of a hole (such as the volume of a container) expands the same way as the corresponding solid shape.

Problem solving strategy : Thermodynamics Ist Law Step 1: Identify the relevant concepts : The first law

of thermodynamics is the statement of the law of conservation of energy in its most general form. You can apply it to any situation in which you are concerned with changes in the internal energy of a system, with heat flow into or out of a system, and/or with work done by or on a system.

Step 2: Set up the problem using the following steps Carefully define what the thermodynamics system is. The first law of thermodynamics focuses on

systems that go through thermodynamic processes. Some problems involve processes with more than one step. so make sure that you identify the initial and final state for each step.

Identify the known quantities and the target variables.

Check whether you have enough equations. The first law, ∆U = Q – W, can be applied just once to each step in a thermodynamic process, so you will often need additional equations. These often

include Eq. ∫=2

1

V

V

dVpW for the work done in a

volume change and the equation of state of the material that makes up the thermodynamic system (for an ideal gas, pV = nRT).

Step 3: Execute the solution as follows : You shouldn't be surprised to be told that

consistent units are essential. If p is a Pa and V in m3, then W is in joules. Otherwise, you may want to convert the pressure and volume units into units of Pa and m3. If a heat capacity is given in terms of calories, usually the simplest procedure is to convert it to joules. Be especially careful with moles. When you use n = mtot/M to convert


between total mass and number of moles, remember that if mtot is in kilograms, M must be in kilograms per mole. The usual units for M are grams per mole; be careful !

The internal energy change ∆U in any thermodynamic process or series of processes in independent of the path, whether the substance is an ideal gas or not. This point is of the utmost importance in the problems in this topic. Sometimes you will be given enough information about one path between the given initial and final states to calculate ∆U for that path. Since ∆U is the same for every possible path between the same two states, you can then relate the various energy quantities for other paths.

When a process consists of several distinct steps, it often helps to make a chart showing Q, W, and ∆U for each step. Put these quantities for each step on a different line, and arrange them so the Q's, W's, and ∆U's form columns. Then you can apply the first law to each line ; in addition, you can add each column and apply the first law to the sums. Do you see why ?

Using above steps, solve for the target variables. Step 4: Evaluate your answer : Check your results for

reasonableness. In particular, make sure that each of your answers has the correct algebraic sign. Remember that a positive Q means that heat flows into the system, and that a negative Q means that heat flows into the system, and that a negative Q means that heat flows out of the system. A positive W means that work is done by the system on its environment, while a negative W means that work is done on the system by its environment.

1. A metallic bob weighs 50 g in air. It it is immersed

in a liquid at a temperature of 25ºC, it weighs 45 g. When the temperature of the liquid is raised to 100ºC, it weighs 45.1 g. Calculate the coefficient of cubical expansion of the liquid given that the coefficient of linear expansion of the metal is 2 × 10–6(ºC)–1.

Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm Weight of liquid displaced at 25ºC = V25ρ25g ∴ 5 = V25ρ25g ...(1) Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2) From eq.(1) & (2) we get,

9.4

5 = 100

25

100

25 .ρρ

VV

Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75) = V25(1 + 3 × 12 × 10–6 × 75) or V100 = V25(1 + 0.0027) = V25 × 1.0027 Also, ρ25 = ρ100(1 + γ × 75) where, γ = Required coefficient of expansion of the liquid

9.4

5 = 100

100

25

25 )751(0027.1 ρ

γ+ρ×

×VV =

0027.1751 γ+

or γ = 3.1 × 10–4 (ºC)–1

2. A one litre flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given that the coefficient of linear expansion of glass = 9 × 10–6(ºC)–1 and coefficient of volume expansion of mercury = 1.8 × 10–4 (ºC–1).

Sol. Let V = Volume of the vessel V' = Volume of mercury For unoccupied volume to remain constant increase

in volume of mercury should be equal to increase in volume of vessel.

∴ V' γm∆T = Vγg∆T or V' = m

gVγ

γ×

∴ V' = 4

6

108.110271000

−

−

×

×× = 150 cm3

3. A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20ºC and loses 12 seconds each day when the temperature is 40ºC. Find the coefficient of linear expansion of the metal.

Sol. Time taken for one oscillation of the pendulum is

T = gL

π2 or T2 = 4π2 × gL .....(1)

Partially differentiating, we get

2T∆t = 4π2 × gL∆ .....(2)

Dividing (2) by (1), we get

TT∆ =

LL

2∆ =

LtL

2∆α = t∆α

21

where ∆t is the change in temperature. Now, One day = 24 hours = 86400 sec Let t be the temperature at which the clock keeps

correct time. At 20ºC, the gain in time is

6 = 21

α × (t – 20) × 86400 ....(3)

At 40ºC, the loss in time is

12 = 21

α× (40 – t) × 86400 ...(4)

Dividing (4) by (3), we have

6

12 = 20

40−

−t

t

which gives t = 3

80 ºC.

Using the value in equation(3), we have

6 = 21 × α ×

− 20

380 × 86400

which gives α = 2.1 × 10–5 perºC

Solved Examples


4. A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ?

Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber.

∴ V0 + ∆V = η(V0 – ∆V)

or ∆V = 011 V

+η−η

P0,v0,T0 P0,v0,T0

As piston is moved slowly therefore, change in

kinetic energy is zero. By work-energy theorem, we can write

Wgas in right chamber + Wgas in left chamber + extAgentW = ∆KE

extAgentW = (Wgas(R) + Wgas(L))

We know that in isothermal process, work done is given by

W = nRT ln

i

f

VV

∴ Work done by gas in left chamber (WL)

= P0V0 ln

∆+

0

0

VVV = P0V0 ln

+ηη

12

Similarly, work done by gas in right chamber (WR)

= P0V0 ln

∆−

0

0

VVV = P0V0 ln

+ηη

12

extAgentW = –P0V0 ln

+ηη

12 – P0V0 ln

+ηη

12

= P0V0 ln 2

41

η+η

5. A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one. The combined mass of the two pistons is equal to m. The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l.

P0

P0

Sol. Let A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string P = Gas pressure m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston

PA1 m1g

P0 A1

From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1 Similarly, consider FBD of lower piston

PA2

m2g P0 A2

T

∴ P0A2 + T = m2g + PA2 Eliminating T, we get

P = P0 + 21

21 )(AA

gmm−

+

According to problem m = m1 + m2 and ∆S = A1 – A2

∴ P = P0 + S

mg∆

Now, PV = RT

or P∆V = R∆T or ∆T = R

VP∆

But ∆V = (A1 – A2)l = ∆S. l

∴ ∆T =

∆+

SmgP0 ∆S.l

ll

l


Reduction of Aldehydes and Ketones by Hydride Transfer :

H3B – H + C = OR

R´ H – C – O

R

R´

– H – C – O – H

R

R´

H – OHδ+ δ–

Hydride transfer Alkoxide ion Alcohol

These steps are repeated until all hydrogen atoms attached to boron have been transferred.

Sodium borohydride is a less powerful reducing agent than lithium aluminum hydride. Lithium aluminum hydride reduces acids, aldehydes, and ketones but sodium borohydride reduces only aldehydes and ketones :

O

C O– R

O

C OR´ R

O

C R´ R

O

C HR

< < <

Reduced by LiAlH4

Reduced by NaBH4

Ease of reduction

Lithium aluminum hydride reacts violently with water, and therefore reductions with lithium aluminum hydride must be carried out in anhydrous solutions, usually in anhydrous ether. (Ethyl acetate is added cautiously after the reaction is over to decompose excess LiAlH4; then water is added to decompose the aluminum complex.) Sodium borohydride reductions, by contrast, can be carried out in water or alcohol solutions.

The Addition of Ylides : The Wittig reaction : Aldehydes and ketones react with phosphorus ylides

to yield alkenes and triphenylphosphine oxide. (An ylide is a neutral molecule having a negative carbon adjacent to a positive heteroatom.) Phosphorus ylides are also called phosphoranes :

C = O + (C6H5)3P – C

R

R

..+ R´´

R´´´

C = C R´´

R´´´

R

R´+ O =P(C6H5)3

Aldehyde orketone

Phosphorus ylide or phosphorane

Alkene [(E) and(Z) isomers]

Triphenyl phosphineoxide

This reaction, known as the Wittig reaction, has proved to be a valuable method for synthesizing alkenes. The Wittig reaction is applicable to a wide variety of compounds, and although a mixture of (E) and (Z) isomers may result, the Wittig reaction offers a great advantage over most other alkene syntheses in that no ambiguity exists as to the location of the double bond in the product. (This is in contrast to E1 eliminations, which may yield multiple alkene products by rearrangement to more stable carbocation intermediates, and both E1 and E2 elimination reactions, which may produce multiple products when different β hydrogens are available for removal.)

Phosphorus ylides are easily prepared from triphenylphosphine and primary or secondary alkyl halides. Their preparation involves two reactions :

General Reaction Reaction 1

(C6H5)3P : + CH – X → (C6H5)3P – CH X–

R´´

R´´´

+

Triphenylphosphine

R´´

R´´´

An alkyltriphenylphosphoniumhalide

Reaction 2

(C6H5)3P – C – H : B– → (C6H5)3P – C :– + H:B

R´´

R´´´

+R´´

R´´´

A phosphorus ylide

+

Specific Example Reaction 1

(C6H5)3P : + CH3Br → (C6H5)3P – CH3Br –C6H6+

Methyltriphenylphosphoniumbromide (89%)

Organic Chemistry

Fundamentals

CARBONYL COMPOUNDS

KEY CONCEPT


Reaction 2

(C6H5)3P – CH3 + C6H5Li →

(C6H5)3P – CH2 :– + C6H6 + LiBr+

+

Br–

The first reaction is a nucleophilic substitution

reaction. Triphenylphosphine is an excellent nucleophile and a weak base. It reacts readily with 1º and 2º alkyl halide by an SN2 mechanism to displace a halide ion from the alkyl halide to give an alkyltriphenylphosphonium salt. The second reaction is an acid-base reaction. A strong base (usually an alkyllithium or phenyllithium) removes a proton from the carbon that is attached to phosphorus to give the ylide.

Phosphorus ylides can be represented as a hybrid of the two resonance structures shown here. Quantum mechanical calculations indicate that the contribution made by the first structure is relatively unimportant.

(C6H5)3P = C

R´´

R´´´ (C6H5)3P – C :–

R´´

R´´´

+

The mechanism of the Wittig reaction has been the

subject of considerable study. An early mechanistic proposal suggested that the ylide, acting as a carbanion, attacks the carbonyl carbon of the aldehyde or ketone to form an unstable intermediate with separated charges called a betaine. In the next step, the betaine is envisioned as becoming an unstable four-membered cyclic system called an oxaphosphetane, which then spontaneously loses triphenylphosphine oxide to become an alkene. However, studies by E. Vedejs and others suggest that the betaine is not an intermediate and that the oxaphosphetane is formed directly by a cycloaddition reaction. The driving force for the Wittig reaction is the formation of the very strong (∆Hº = 540 kJ mol–1) phosphorus –oxygen bond in triphenylphosphine oxide.

R–C + –:C–R´ R – C – C – R´´´

R ´

–:O:

R ´´

P(C6H5)3..+

R – C – C – R´´´

R ´

:O –

R ´´

P(C6H5)3..

Aldehyde or ketone

R ´ R ´´

P(C6H5)3 +

:O:

Ylide Betaine (may not be formed)

Oxaphosphetane

C = C + O = P(C6H5)3R´´

R´´´

R´

R

Alkene (+diastereomer)

Triphenylphosphineoxide

Specific Example :

O + :CH2 – P(C6H5)3 CH2

O P(C6H5)3+–

CH2

O –P(C6H5)3

CH2 + O=P(C6H5)3

Methylenecyclohexane(86%)

– +

Michael Additions : Conjugate additions of enolate anions to

α-β-unsaturated carbonyl compound are known generally as Michael additions. An example is the addition of cyclohexanone to C6H5CH=CHCOC6H5 :

O O

O–

C6H5CH=CH–CC6H5

O –

O CH

C6H5

CHδ–

C—Oδ–

C6H5

O CH

C6H5

C

C = O

C6H5

H

H

+H3O+

OH–

The sequence that follows illustrates how a conjugate aldol addition (Michael addition) followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the Robinson anulation (ring-forming) reaction (after the English chemist, Sir Robert Robinon, who won the Nobel Prize in chemistry in 1947 for his research on naturally occurring compounds) :

O

+ CH2 = CHCCH3

O CH2

CH3

CH2

CH3C

O

2-Methylcyclo- hexane-1, 3-dione

CH3

O

O

O O

CH3

O (65%)

OH–

CH3OH (conjugateaddition) Methyl vinyl

ketone

aldol condensation

base (–H2O)


Inorganic Chemistry

Fundamentals

Tetragonal distortion of octahedral complexes (Jahn-Teller distortion) : The shape of transition metal complexes are affected

by whether the d orbitals are symmetrically or asymmetrically filled.

Repulsion by six ligands in an octahedral complex splits the d orbitals on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a completely regular octahedron. The symmetrical arrangements of d electrons are shown in Table.

Symmetrical electronic arrangements :

Electronic configuration

t2g eg

d5

d6

d8

d10 All other arrangements have an asymmetrical

arrangement of d electrons. If the d electrons are asymmetrically arranged, they will repel some ligands in the complex more than others. Thus the structure is distorted because some ligands are prevented from approaching the metal.

as closely as others. The eg orbitals point directly at the ligands. Thus asymmetric filling of the eg orbitals in some ligands being repelled more than others. This causes a significant distortion of the octahedral shape. In contrast the t2g orbitals do not point directly at the ligands, but point in between the ligand directions. Thus asymmetric filling of the t2g orbitals has only a very small effect on the stereochemistry. Distortion caused by asymmetric filling of the t2g orbitals is usually too small to measure. The electronic arrangements which will produce a large distortion are shown in Table.

The two eg orbitals 22 yxd

− and 2z

d are normally

degenerate. However, if they are asymmetrically filled then this degeneracy is destroyed, and the two

orbitals are no longer equal in energy. If the 2zd

orbital contains one. Asymmetrical electronic arrangements :

Electronic configuration

t2g eg

d4

d7

d9 more electron than the 22 yx

d−

orbital then the ligands

approaching along +z and –z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octahedron along the z axis. This is called tetragonal distortion. Strictly it should be called tetragonal elongation. This form of distortion is commonly obsered.

If the 22 yxd − orbital contains the extra electron, then

elongation will occur along the x and y axes. This means that the ligands approach more closely along the z-axis. Thus there will be four long bonds and two short bonds. This is equivalent to compressing the octahedron along the z axis, and is called tetragonal compression, and it is not possible to predict which will occur.

For example, the crystal structure of CrF2 is a distorted rutile (TiO2) structure. Cr2+ is octahedrally surrounded by six F–, and there are four Cr–F bonds of length 1.98 – 2.01 Å, and two longer bonds of length 2.43 Å. The octahedron is said to be tetragonally distorted. The electronic arrangement in Cr2+ is d4. F– is a weak field ligand, and so the t2g level contains three electrons and the eg level contains one electron. The 22 yx

d−

orbital has four lobes whilst

the 2zd orbital has only two lobes pointing at the

ligands. To minimize repulsion with the ligands, the single eg electron will occupy the 2z

d orbital. This is equivalent to splitting the degeneracy of the eg level so that 2z

d is of lower energy, i.e. more stable, and

22 yxd

− is of higher energy, i.e. less stable. Thus the

CO-ORDINATION COMPOUND &

METALLURGY

KEY CONCEPT


two ligands approaching along the +z and –z directions are subjected to greater repulsion than the four ligands along +x, –x, +y and –y. This causes tetragonal distortion with four short bonds and two long bonds. In the same way MnF3 contains Mn3+ with a d4 configuration, and forms a tetragonally distorted octahedral structure.

Many Cu(+II) salts and complexes also show tetragonally distorted octahedral structures. Cu2+ has a d9 configuration :

t2g

eg

To minimize repulsion with the ligands, two

electrons occupy the 2zd orbital and one electron

occupies the 22 yxd

− orbital. Thus the two ligands

along –z and –z are repelled more strongly than are the other four ligands.

The examples above show that whenever the 2zd and

22 yxd

− orbitals are unequally occupied, distortion

occurs. This is know as Jahn–Teller distortion. Leaching : It involves the treatment of the ore with a suitable

reagents as to make it soluble while impurities remain insoluble. The ore is recovered from the solution by suitable chemical method. For example, bauxite ore contains ferric oxide, titanium oxide and silica as impurities. When the powdered ore is digested with an aqueous solution of sodium hydroxide at about 150ºC under pressure, the alumina (Al2O3) dissolves forming soluble sodium meta-aluminate while ferric oxide (Fe2O3), TiO2 and silica remain as insoluble part.

Al2O3 + 2NaOH → 2NaAlO2 + H2O Pure alumina is recovered from the filtrate NaAlO2 + 2H2O → Al(OH)3 + NaOH

2Al(OH)3 )autoclave(Ignited → Al2O3 + 3H2O

Gold and silver are also extracted from their native ores by Leaching (Mac-Arthur Forrest cyanide process). Both silver and gold particles dissolve in dilute solution of sodium cyanide in presence of oxygen of the air forming complex cyanides.

4Ag + 8NaCN + 2H2O + O2 → 4NaAg(CN)2 + 4NaOH Sod. argentocyanide 4Au + 8NaCN + 2H2O + O2 → 4NaAu(CN)2 + 4NaOH Sod. aurocyanide Ag or Au is recovered from the solution by the

addition of electropositive metal like zinc.

2NaAg(CN)2 + Zn → Na2Zn(CN)4 + 2Ag ↓ 2NaAu(CN)2 + Zn → Na2Zn(CN)4 + 2Au ↓ Soluble complex Special Methods : Mond's process : Nickel is purified by this method.

Impure nickel is treated with carbon monoxide at 60–80º C when volatile compound, nickel carbonyl, is formed. Nickel carbonyl decomposes at 180ºC to form pure nickel and carbon monoxide which can again be used.

Impure nickel + CO 60–80ºCNI(CO)4

Ni + 4CO

180ºC

Gaseous compound

Zone refining or Fractional crystallisation : Elements such as Si, Ge, Ga, etc., which are used as

semiconductors are refined by this method. Highly pure metals are obtained. The method is based on the difference in solubility of impurities in molten and solid state of the metal. A movable heater is fitted around a rod of the impure metal. The heater is slowly moved across the rod. The metal melts at the point of heating and as the heater moves on from one end of the rod to the other end, the pure metal crystallises while the impurities pass on the adjacent melted zone.

Molten zonecontainingimpurity

Pure metalMoving circular

heater

Impure zone

Different metallurgical processes can be broadly

divided into three main types. Pyrometallurgy : Extraction is done using heat

energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr, Hg, etc., which are found in nature in the form of oxides, carbonates, sulphides are extracted by this process.

Hydrometallurgy : Extraction of metals involving aqueous solution is known as hydrometallurgy. Silver, gold, etc., are extracted by this process.

Electrometallurgy : Extraction of highly reactive metals such as Na, K, Ca, Mg, Al, etc., by carrying electrolysis of one of the suitable compound in fused or molten state.


1. The critical temperature and pressure for NO are 177

K and 6.485 MPa, respectively, and for CCl4 these are 550 K and 4.56 MPa, respectively. Which gas (i) has smaller value for the van der Walls constant b; (ii) has smaller value of constant a; (iii) has larger critical volume; and (iv) is most nearly ideal in behaviour at 300 K and 1.013 MPa.

Sol. We have Tc(NO) = 177 K Tc(CCl4) = 550 K pc(NO) = 6.485 MPa pc(CCl4) = 4.56 MPa

(i) Since c

c

Tp

= Rb27/a8b27/a 2

= b8

R therefore, b = c

c

p8RT

Thus,

b(NO) = )MPa485.6)(8(

)molKcmMPa314.8)(K177( 11–3 −

= 28.36 cm3 mol–1 and

b(CCl4) = )MPa56.4)(8(

)molKcmMPa314.8)(K550 113 −−

= 125.35 cm3 mol–1 Hence b(NO) < b(CCl4) (ii) Since a = 27pcb2 therefore a(NO) = (27) (6.485 MPa) (28.36 cm3 mol–1)2 = 140827 MPa cm6mol–2 ≡ 140.827 kPa dm6 mol–2 a(CCl4) = (27) (4.56 MPa) (125.35 cm3 mol–1)2 = 1934538 MPa cm6 mol–2 ≡ 1934.538 KPa dm6mol–2 Hence a(NO) < a(CCl4) (iii) Since Vc = 3b therefore, Vc(NO) = 3 × (28.36 cm3 mol–1) = 85.08 cm3 mol–1 Vc(CCl4) = 3 × (125.35 cm3 mol–1) = 376.05 cm3 mol–1 Hence Vc(NO) < Vc(CCl4) (iv) NO is more ideal in behaviour at 300 K and

1.013 MPa, because its critical temperature is less than 300 K, whereas for CCl4 the corresponding critical temperature is greater than 300 K.

2. Potassium alum is KA1(SO4)2.12H2O. As a strong

electrolyte, it is considered to be 100% dissociated into K+, Al3+, and SO4

2–. The solution is acidic because of the hydrolysis of Al3+, but not so acidic as might be expected, because the SO4

2– can sponge up some of the H3O+ by forming HSO4

–. Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm3 of solution, calculate its [H3O+] :

(a) Considering the hydrolysis

Al3+ + 2H2O Al(OH)2+ + H3O+ with Kh = 1.4 × 10–5 M (b) Allowing also for the equilibrium HSO4

– + H2O H3O+ + SO42–

with K2 = 1.26 × 10–2 M

Sol. (a) Amount of alum = 1molg38.474g4.11

− = 0.024 mol

Molarity of the prepared solution = 3dm1.0mol024.0

= 0.24 M Hydrolysis of Al3+ is Al3+ + 2H2O Al(OH)2+ + H3O+

Kh = ]Al[

]OH][)OH(Al[3

32

+

++

If x is the concentration of Al3+ that has hydrolyzed, we have

Kh = xM24.0

)x)(x(−

= 1.4 × 10–5 M

Solving for x, we get [H3O+] = x = 1.82 × 10–3 M (b) We will have to consider the following equilibria. Al3+ + 2H2O Al(OH)2+ + H3O+ H3O+ + SO4

2– HSO4– + H2O

Let z be the concentration of SO42– that combines

with H3O+ and y be the net concentration of H3O+ that is present in the solution. Since the concentration z of SO4

2– combines with the concentration z of H3O+, it is obvious that the net concentration of H3O+ produced in the hydrolysis reaction of Al3+ is (y + z). Thus, the concentration (y + z) of Al3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are

zyM24.0

3Al−−

+ + 2H2O zy

2)OH(Al+

+ + y

3OH +

y

3OH + + zM48.0

24SO

−

− z

4HSO− + H2O

Thus, Kh = )zyM24.0(

)y)(zy(−−

+ = 1.4 × 10–5 M ...(i)

K2 = )zM48.0(y

z−

= M1026.1

12−×

...(ii)

From Eq. (ii), we get

z = y)M1026.1(

y)M48.0(2 +× −

Substituting this in Eq. (i), we get

UNDERSTANDINGPhysical Chemistry


+×−−

+×+

−

−

y)M1026.1(y)M48.0(y24.0

yy)M1026.1(

y)M48.0(y

2

2

= 1.4 × 10–5

Making an assumption that y <<1.26 × 10–2 M, and then solving for y, we get

[H3O+] = y = 2.932 × 10–4 M 3. A solution contains Na2CO3 and NaHCO3.10 ml of

this requires 2.0 ml of 0.1 M H2SO4 for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M H2SO4 was required. Calculate the strength of Na2CO3 and NaHCO3 in solution. [IIT-1978]

Sol. Step 1.

Equivalent mass of Na2CO3 = 2

massMolecular

= 2

106 = 53

Meq. of Na2CO3 in solution = 53m1 × 1000

Step 2.

Equivalent mass of NaHCO3 = 1

massMolecular

= 84

Meq. of NaHCO3 in solution = 84m2 × 1000

Step 3. Meq. of H2SO4 used with phenolphthalein = Valency factor × Molarity × Volume (ml) = 2 × 0.1 × 2.0 = 0.4 2Na2CO3 + H2SO4 → 2NaHCO3 + Na2SO4 Meq. of H2SO4 used with phenolphthalein

= 21 Meq. of Na2CO3 ∴

21 Meq. of Na2CO3 = 0.4

Step 4. Meq. of H2SO4 used with methyl orange = Valency factor × molarity × volume(ml) = 2 × 0.2 × 2.5 = 1 Meq. of H2SO4 used with methyl orange

= Meq. of NaHCO3 + 21 Meq. of Na2CO3

∴ Meq. of NaHCO3 + 21 Meq. of Na2CO3 = 1

∴ Meq. of NaHCO3 = 1 – 0.4 = 0.6 and Meq. of Na2CO3 = 2 × 0.4 = 0.8 Step 5.

53m1 × 1000 = 0.8 or m1 =

1000538.0 × = 0.0424

∴ Strength of Na2CO3 solution = 10

10000424.0 ×

= 4.24 g L–1

Step 6.

84m2 × 1000 = 0.6 or m2 =

1000846.0 × = 0.0504

∴ Strength of NaHCO3 solution = 10

10000504.0 ×

= 5.04 g L–1 4. From the standard potentials shown in the following

diagram, calculate the potentials º1E and º

2E .

BrO3– 0.54 V BrO– 0.45 V

21 Br2

1.07 V Br–

0.17 V

E2º

E1º

Sol. The reaction corresponding to the potential Eº1 is

BrO3– + 3H2O + 5e– =

21 Br2 + 6OH– ...(1)

This reaction can be obtained by adding the following two reduction reactions:

BrO3– + 2H2O + 4e– = BrO– + 4OH– ...(2)

BrO– + H2O + e– = 21 Br2 + 2OH– ...(3)

Hence the free energy change of reaction (1) will be º

)1(reactionG∆ = º)2(reactionG∆ + º

)3(reactionG∆

Replacing ∆Gºs in terms of potentials, we get – 5FE1º = – 4F(0.54 V) – 1F (0.45 V) = (–2.61 V) F

Hence E1º = 5

V61.2 = 0.52 V

Now the reaction corresponding to the potential E2º is BrO3

– + 2H2O + 6e– = Br– + 6OH– ...(4) This reaction can be obtained by adding the

following three reactions. BrO3

– + 2H2O + 4e– = BrO– + 4OH– (Eq.2)

BrO– + H2O + e– = 21 Br2 + 2OH– (Eq.3)

21 Br2 + e– = Br– ...(5)

Hence º

)4(reactionG∆ = º)2(reactionG∆ + º

)3(reactionG∆

+ º)5(reactionG∆

or – 6F(E2º) = – 4F(0.54 V) – 1F(0.45 V) – 1F (1.07 V) = (– 3.68 V) F

or E2º = 668.3 = 0.61 V.


5. The freezing point of an aqueous solution of KCN containing 0.189 mol kg–1 was – 0.704 ºC. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution became –0.530ºC. Assuming that the complex is formed according to the equation

Hg(CN)2 + x CN– → –x2x)CN(Hg +

Find the formula of the complex. Sol. Molality of the solution containing only KCN is

m = f

f

K)T(–∆

= )molkgK86.1(

)K704.0(1– = 0.379 mol kg–1

This is just double of the given molality ( = 0.189 mol kg–1) of KCN, indicating complete dissociation of KCN. Molality of the solution after the formation of the complex

m = f

f

K)T(–∆

= )molkgK86.1(

)K530.0(1– = 0.285 mol kg–1

If it be assumed that the whole of Hg(CN)2 is converted into complex, the amounts of various species in 1 kg of solvent after the formation of the complex will be

n(K+) = 0.189 mol, n(CN–) = (0.189 – x) mol ))CN(Hg(n –x

2x+ = 0.095 mol Total amount of species in 1 kg solvent becomes ntotal = [0.189 + (0.189 – x) + 0.095] mol = (0.473 – x) mol Equating this to 0.285 mol, we get (0.473 – x) mol = 0.285 mol i.e. x = (0.473 – 0.285) = 0.188

Number of CN– units combined = mol095.0mol188.0 = 2

Thus, the formula of the complex is –24)CN(Hg .

TRUE OR FALSE

1. The magnitude of charge on one gram of electrons is 1.60 × 10–19 coulomb.

2. Chromyl chloride test of Cl– radical is not given by HgCl2.

3. The energy levels in a hydrogen atom can be compared with the steps of a ladder placed at equal distance.

4. In SN1 mechanism, the leaving group in the molecule, leaves the molecule, well before joining of an attacking group.

5. Metamerism is special type of isomerism where isomers exist simultaneously in dynamic equilibrium.

6. Addition of HCN with formaldehyde is an example of electrophilic addition reaction.

7. Ligroin is essentially petroleum ether containing aliphatic hydrocarbons and is generally used in dry cleaning clothes.

Sol. 1. [False] Thomson through his experiment

determined the charge to mass ratio of an electron and the value of e/m is equal to 1.76 × 108 coulomb/gm. Hence one gm of electrons have charge 1.76 × 108 C.

2. [True] 3. [False] 4. [True] SN1 reaction mechanism takes place in

two steps as :

R—X →Slow R+ + X–

R+ + OH– →Fast ROH

5. [False] In metamerism isomers differ in structure due to difference in distribution of carbon atoms about the functional group.

For example : CH3CH2–O–CH2CH3 and CH3–OCH2CH2CH3 Conditions mentioned in the statement are

associated with phenomenon of trautomerism. 6. [False]

H – C = O + H+ CN–

H

H – C = OH

H

CN 7. [True]

MEMORABLE POINTS

• Parsec is the unit of Distance

• Estimated radius of universe is 1025 m

• Estimated age of Sun is 1018 s

• 18/5 km h–1 equal to 1 ms–1

• 1 femtometre (1 fm) is equal to 10–15 m

• Dot product of force and velocity is Power

• Moment of momentum is equal to

Angular momentum

• Rocket propulision is based on the principle of

Conservation of linear momentum

• The largest of astronomical unit, light year and parsec is Parsec



WHAT MAKES A STAR?

So you're out one night and you look up into the sky. Assuming you aren't in a city with tons of smog or clouds, you will probably see a sky filled with little dots of light. Those dots (this should not be a surprise) are stars. Some are only a few hundred light years away and some are thousands of light years away. They all have some things in common. You see, stars are huge balls of fire. They aren't just any fire. That fire is from a constant number of nuclear reactions.

1. Let f(x) = sinx and

g(x) =

π>π≤≤≤≤

xxxforxttf

;2/sin0;0);(max

2

Discuss the continuity and differentiability of g(x) in (0, ∞)

2. Is the inequality sin2 x < x sin(sin x) true for 0 < x < π/2 ? Justify your answer.

3. A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-cream cones if

(i) they are all of different flavours;

(ii) they are not necessarily of different flavours;

(iii) they contain only 3 different flavoures;

(iv) they contain only 2 or 3 different flavoures ?

4. Using vector method, show that the internal (external) bisector of any angle of a triangle divides the opposite side internally (externally) in the ratio of the other two sides containing the triangle.

5. Prove that

(a) cos x + nC1 cos 2x + nC2 cos 3x + ............

...... + nCn cos(n + 1)x = 2n. cosnx/2. cos

+ xn

22

(b) sin x + nC1 sin 2x + nC2 sin 3x + ............... ....... + nCn sin(n + 1)x = 2n . cosn x/2 . sin

+ xn

22

6. In a town with a population of n, a person sands two letters to two sperate people, each of whom is asked to repeat the procedure. Thus, for each letter received, two letters are sent to separate persons chosen at random (irrespective of what happened in the past). What is the probability that in the first k stages, the person who started the chain will not receive a letter ?

7. Prove the identity :

∫ −x zzxe0

2dz = ∫ −x zx ee

0

44 22 dz, deriving for the

function f(x) = ∫ −x zzxe0

2dz a differential equation

and solving it.

8. Prove that ∫ θθsecsin n dθ

= –1

)1cos(2−

θ−n

n – ∫ θθθ dn sec)2–sin( dθ.

Hence or otherwise evaluate

∫π

θθθ2/

0 cos3sin5cos dθ.

9. Find the latus rectum of parabola

9x2 – 24 xy + 16y2 – 18x – 101y + 19 = 0.

10. A circle of radius 1 unit touches positive x-axis and positive y-axis at A and B respectively. A variable line passing through origin intersects the circle in two points in two points D and E. Find the equation of the lines for which area of ∆ DEB is maximum.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

10Set


1. as φ (a) = φ (b) = φ (c) so by Rolle’s theorem there must exist at least a point

x = α & x = β each of intervals (a, c) & (c, b) such that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem, there must exist at least a point x = µ such that α < µ < β where φ′(µ) = 0

so )()(

)(2caba

af−−

+ )()(

)(2abcb

bf−−

+ )()(

)(2bcac

cf−−

– f ′′ (µ) = 0

so )()(

)(caba

af−−

+ )()(

)(abcb

bf−−

+ )()(

)(bcac

cf−−

= 21 f ′′ (µ)

where a < µ < b. 2. Required probability

1 . 65 .

65 .

65 ........

65 .

61 =

2r

65 −

.

61 (r – 2) times

Note : any number in 1st loss same no. does not in 2nd (any other comes). Now 3rd is also diff. (and in same r − 2 times) Now (r − 1)th & r th must be same. 3. 2s = a + b + c ON = − BN + BO Let BN = x 2BN + 2CN + 2AR = 2s x + (a − x) + (b − a + x) = s x = s − b

R

O C

N B

M

A

r

I (h,k)

so h = ON =

2a − (s − b)

= 2

22 bas ++− = 2

cb − & r = k.

so r = k = s∆ =

scsbsass ))()(( −−−

r = k = s

csbsass ))()(( −−−

2sk = ))()(( cbacbaass −++−−

= )2)(2)(( xaxaass +−−

2sk = )4)(( 22 haass −− required locus is 4s2y2 = A(a2 – 4x2)

⇒ s2y2 + Ax2 = 4

2Aa

where A is = s (s – a) here h2 < as so it is an ellipse

4. f (0) = c f (1) = a + b + c & f (−1) = a − b + c solving these,

a = 21 [f (1) + f (−1) − 2 f (0)] ,

b = 21 [f (1) − f (−1)] & c = f (0)

so f (x) = 2

)1( +xx f (1) + (1− x2) f (0) +2

)1( −xx f(−1)

2 | f (x) | < | x | | x + 1 | + 2| 1 − x2 | + | x | | x − 1| ; as | f (1) | , | f (0) |, | f (−1) | ≤ 1. 2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x2) + | x | (1 − x) as

x ∈ [−1, 1]

so 2 | f (x) | ≤ 2 (|x| + 1 − x2) ≤ 2 . 45 so | f (x) | ≤

45

Now as g (x) = x2 f (1/x) = 21 (1 + x) f (1)

+ (x2 − 1) f (0) + 21 (1 − x) f (−1)

so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x2 | + | 1 − x| ⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x2) | + 1 − x ; as x ∈ [−1, 1] ⇒ 2 | g (x) | < − 2x2 + 4 ≤ 4. ⇒ |g (x) | ≤ 2.

5. Oil bed is being shown by the plane A′ PQ. θ be the angle between the planes A′ PQ & A′ B′ C′. Let A′ B′ C′ be the x − y plane with x-axis along A′ C′ and origin at A′. The P.V.s of the various points are defined as follows

MATHEMATICAL CHALLENGES SOLUTION FOR JANUARY ISSUE (SET # 9)


B

A C

B´

A´ P

Q

C´

point C′ : b i , point B′ : cos A i + c sin A j ,

point Q : b i – z k , point P : cos A i + c sin A j – y k normal vector to the plane A′ B′ C′ = 1n

r = bc sin A k

normal vector to the plane A'PQ = 2nr

= cz sin A i + (by – cz cos A) j + bc sin A k

so cos θ = ||||

.

11

21nn

nnrr

rr

= 2/12222222 ]sin)cos(sin[sin

AcbAczbyAzcAbc

+−+

cos θ = 2/12222222 )]cos2(sin[sin

AbyczybzcAcbAcb

−++

so tan θ = Abc

Abyczybzcsin

]cos2[ 2/12222 −+

so tan θ . sin A = Abcyz

cy

bz cos2

2

2

2

2−+

6. ∫ +−

xxx

5cos217cos8cos .

xx

5sin25sin2 dx

= ∫ ++−−)10sin5(sin2

2sin12sin3sin13sinxx

xxxx dx

= ∫ +−+

)10sin5(sin212sin–3sin2sin13sin

xxxxxx dx

= ∫−

25cos

215sin.2.2

29cos

215sin2

211cos

215sin2

xx

xxxx

dx

= ∫−

25cos2

29cos

211cos

x

xx

dx

= ∫−

25cos2

2sin5sin2

x

xxdx

= − 2 ∫

2sin

25sin xx dx

= ∫

−

24cos

26cos xx dx

= ∫ − dxxx )2cos3(cos

= 33sin x −

22sin x + C

7. 2

2

dxyd = 2 ∫

x

dttf0

)(

integrate using by parts method

dxdy = 2

− ∫∫

xx

dxxfxdttfx00

)(.)(

= 2

−∫

x

dttftx0

)()(

again integrating,

y = 2

−−− ∫∫∫ dxdttfxdttftxx

xxx

000

0)()()(

=2

+−− ∫∫∫ dxxfxdttfxdttftxx

xxx

0

2

0

2

0

)(2

)(2

)()(

= ∫ −x

dttfxtx0

2 )()(2 − ∫x

dttfx0

2 )( + ∫x

dttft0

2 )(

y = ∫ +−x

dttftxtx0

22 )()2( = ∫ −x

dttftx0

2 )()(

8. To prove that αα

+

/1

1ba <

ββ

+

/1

1ba

Let ba = c > 0

so (cα + 1)1/α < (cβ + 1)1/β. Let f (x) = (cx + 1)1/x ; x > 0

f ′(x) = (cx + 1)1/x ln (cx + 1)

− 2

1x

+x1 (cx + 1) x

1 –1. cx ln c

= 2

11

)1(x

c xx −+ ])1()1([ xxxx cnccnc ll +++− < 0

so f (x) is decreasing function so f (α) < f (β). Hence proved. 9. Point P (x, 1/2) under the given condition are length

PB = OB


B (t, 1)

A (t – 1)

C O

P θ

rθ = t ; so θ = t

from ∆PAB : 2

PB = PA sin 2θ

⇒ PB = 2 sin 2t ........(1)

Now ∠ PBC = 2θ =

2t ;

so from ∠ PCB ; 2θ =

2t

so from ∆ PCB ; PB

2/1 = sin 2t ........(2)

from (1) & (2) PB = 1 ; so θ = t = π/3

thus | PB |2 = (t − x)2 + 41 = 1.

| t − x | = 23 ; t − x =

23 ; as t > x

so x = 3π −

23

10. Let xn = 1−n + 1+n be rational, then

nx

1 = 11

1++− nn

is also rational

nx

1 = 2

11 −−+ nn is also rational

1+n − 1−n is also rational

as 1+n + 1−n & 1+n − 1−n are rational

so 1+n + 1−n must be rational i.e. (n + 1) & (n – 1) are perfect squares. This is not possible as any two perfect squares differe

at least by 3. Hence there is not positive integer n for

which 1−n + 1+n is a rational.

1. Emeralds have been produced synthetically in

labs since 1848 and can be virtually indistinguishable from the genuine article.

2. In the last 200 years the use of metals has

increased as scientists have discovered new ones: until the 17th Century only 12 metals were known - there are now 86.

3. The only person to have an element named

after him while still alive was Glenn Seaborg, the most prolific of all the element hunters.

4. Traffic lights with red and green gas lights were

first introduced in London in 1868. Unfortunately, they exploded and killed a policeman. The first successful system was installed in Cleveland, Ohio in 1914.

5. In 1998, design student Damini Kumar at South

Bank University patented a teapot with a special grooved spout, which she claims virtually rules out dribbling.

6. Even though most items in the home today are

technologically up to date, most of us are still using the standard light bulb designed in 1928!

7. A chest x-ray is comprised of 90,000 to

130,000 electron volts. 8. The strength of early lasers was measured in

Gillettes, the number of blue razor blades a given beam could puncture.

9. The first commercial radio station in the United

States, KDKA Pittsburgh, began broadcasting in November 1920.

10. A British rocket attack on US soldiers is

celebrated in the lyrics of the US National Anthem.

11. Until the late 1800s, it was forbidden for

women in the United States to obtain a patent, so if a woman had invented something she would file for a patent under her husband or father's name. For this reason, the number of early female inventors remains a mystery.

12. Milt Garland, a 102 year old engineer, invented

a technology that forms ice on the exterior of a casing instead of inside it, which is used to create indoor ice rinks.


1. Evaluate ∫ −+1

0)1( nxtx dx where n ∈ I+ and t is a

parameter independent of x. Hence show that

∫ −−1

0)1( knk xx = [nCk(n + 1)]–1

Sol. ∫ +−1

0]1)1[( nxt =

)1)(1(11

+−−+

ntt n

= 1

1+n

[1 + t + t2 + t3 + ..... + tk + .... + tn] ...(1)

Also ∫ −+1

0)]1([ nxtx dx = ∫ +−

1

0])1[( ntxx dx

= ∫ −1

00 )1( dxxC nn + t ∫ −−

1

0

11 )1( dxxxC nn

+ .... + tk nCk ∫ −−1

0)1( knx xk dx ...... ...(2)

As R.H.S of (1) = R.H.S. of (2) compare equ. of tk on both sides

⇒ 1

1+n

= nCk ∫ −−1

0)1( knx xkdx

∫ −−1

0)1( knx xkdx =

)1(1

+nCkn Hence proved.

2. Let point A describes a curve C such that the difference between its distances from the points (0, 0) and (3, 4) is 5. Then find the no. of points at which the circle x2 + y2 = 4 and c intersect.

Sol. Locus of the point A is curve C which is satisfying |AB – AP| = PB

where P : (0, 0), B : (3, 4) curve C represents two rays BA or PA and it is clear from diagram that curve C and given circle are intersecting at only one point.

A

P(0, 0)

B(3, 4)

A

3. Let [x] stands for the greatest integer function find

the derivative of f(x) = xxxx sin3 2])1[( +++ , where it

exists in (1, 1.5). Indicate the point(s) where it does not exist. Give reason(s) for your conclusion.

Sol. The greatest integer [x3 + 1] takes jump from 2 to 3 at 3 2 and again from 3 to 4 at 3 3 in [1, 1.5] and therefore it is discontinuous at these two points. As a result the given function is discontinuous at 3 2 and hence not differentiable.

To find the derivative at other points we write :

in (1, 3 2 ), f(x) = xxx sin2)2( ++

⇒ f ´(x) = 1sin2)2( −++ xxx

x2 + sin x + (x + 2) (2x + cos x) log (x + 2)

in ( 3 2 , 3 3 ), f(x) = xxx sin2)3( ++ ,

f ´(x) = 1sin2)3( −++ xxx x2 + sin x

+ (2x + cos x) (x + 3) × loge (x + 3)

in ( 3 5 , 1.5), f(x) = xxx sin2)4( ++ ,

f ´(x) = 1sin2)4( −++ xxx , x2 + sin x + (2x + cos x)

(x + 4) × loge(x + 4) 4. For three unit vectors a , b and c not all collinear

given that ca ˆˆ× = bc ˆˆ× and ab ˆˆ × = ca ˆˆ× . Show that cosα + cos β + cos γ = –3/2, where α, β and γ are the angles between a and b , b and c and c and a respectively.

Sol. a × c = bc ˆˆ× ⇒ ( a + b ) × c = →0

⇒ c is collinear with a + b ⇒ a + b = λ c for same λ ∈ R

Similarly b + c = µ a for some scalar u

Now a + b = λ c ⇒ a + b + c = (λ + 1) →c Similarly ⇒ a + b + c = (µ + 1) a Hence (λ + 1) c = (µ + 1) a , either λ + 1 = µ + 1 = 0 or c is collinear with a . But c can not be collinear to a other wise ac ˆˆ × = 0 ⇒ bc ˆˆ × = 0 ⇒ b is collinear to with c

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumMATHS


⇒ a b and c are collinear. Hence c is not collinear to a ⇒ λ + 1 = µ + 1 = 0 ⇒ λ ± µ = –1

Hence b + c = µ a

⇒ a + b + c = →0

⇒ ( a + b + c ) . ( a + b + c ) = 0

⇒ 1 + 1 + 1 + 2 ( a . b + b . c + c . a) = 0

⇒ a . b + b . c + c . a = –23

⇒ cos α + cos β + cos γ = – 23

5. Let S be the coefficients of x49 in given expression

f(x) and if P be product of roots of the equation

f(x) = 0, then find the value of PS , given that :

f(x) = (x – 1)2

− 2

2x

−

21x

− 3

3x

−

31x ,

.........

− 25

25x

−

251x

Sol. Here we can write f(x) as :

f(x) =

−

−

−− 25

25...3

32

2)1( xxxx

×

−

−

−−

251...

31

21)1( xxxx

Now roots of f(x) = 0 are;

12, 22, 32, ..... , 252 and 1, 21 ,

31 , .....,

251

Now f(x) is the polynomial of degree 50, So coefficient of x49 will be : S = – (sum of roots)

= – (12 + 22 + ... + 252) –

++++

251....

31

211

= –

+

×× K6

512625 where, K = ∑=

25

1

1

n n

⇒ S = –(K + 5525). Product of roots :

12 . 22 . 32 .... 252 . 1 . 21 .

31 ....

251 = 1 . 2 . 3 ...25

∴ P = 25 !

Hence PS =

!25)5525K( +− , where K = ∑

=

25

1

1

n n

6. A man standing at a distance 5m in front of the base

of a building 10m high on which a flagstaff is mounted observes that the top of the building and the top of a mountain behind the building are along the same straight line. When he recedes by a distance of 48 m he observes that now the top of the flagstaff and the top of the mountain are along the same straight line. If at both the locations, the flagstaff subtends the same angle at the man’s eye, find the height of mountain.

Sol. CD : Flagstaf DE : Building KF : Mountain (height = h say) The figure illustrates the situation. Since, ∠CBD = ∠CAD = α say, points A, B, C and

D are concyclic. ⇒ ∠ABD = ∠ACD = 90º – (α + β) ⇒ ∠ABC = 90º – (α + β) + α = 90º – β = ∠KCH

K

H

G

FE A 548B

90º – β

α β

α 10

D

C 90º – β

β

Now, h = KH + HF = (CH) tan (90º – β) + (BE) tan(90º – β) (Q HF = CE) = [DG + (BA + AE) cot β = [KG cot β + (48 + 5)] cot β ⇒ h = [(h – 10)cotβ + 53] cot β (Q KG = KF – GF)

Putting cot β = 105 =

21 , we get

h =

+

− 53210h

21

⇒ 4h = h – 10 + 106 ⇒ 3h = 96 ⇒ h = 32 m


Integration :

If dxd f(x) = F(x), then ∫ )(xF dx = f(x) + c, where c

is an arbitrary constant called constant of integration.

1. ∫ dxxn = 1

1

+

+

nxn

(n ≠ –1)

2. ∫ dxx1 = log x

3. ∫ dxex = ex

4. ∫ dxa x = a

a

e

x

log

5. ∫ dxxsin = – cos x

6. ∫ dxxcos = sin x

7. ∫ dxx2sec = tan x

8. ∫ dxxec2cos = – cot x

9. ∫ sec x tan x dx = sec x

10. ∫ cosec x cot x dx = – cosec x

11. ∫ sec x dx = log(sec x + tan x) = log tan

π

+42

x

12. ∫ cosec x dx = – log (cosec x + cot x) = log tan

2x

13. ∫ tan x dx = – log cos x

14. ∫ cot x dx = log sin x

15. ∫− 22 xa

dx = sin–1

ax = – cos–1

ax

16. ∫ + 22 xadx =

a1 tan–1

ax = –

a1 cot–1

ax

17. ∫− 22 axx

dx = a1 sec–1

ax = –

a1 cosec–1

ax

18. ∫ − 221

ax =

a21 log

axax

+− , when x > a

19. ∫ − 221

xadx =

a21 log

xaxa

−+ , when x < a

20. ∫− 22 ax

dx = log

−+ 22 axx = cos h–1

ax

21. ∫+ 22 ax

dx = log

++ 22 axx = sin h–1

ax

22. ∫ − 22 xa dx = 21 x 22 xa − +

21 a2 sin–1

ax

23. ∫ − 22 ax dx = 21 x 22 ax −

– 21 a2log

−+ 22 axx

24. ∫ + 22 ax dx = 21 x 22 ax +

+ 21 a2 log

++ 22 axx

25. ∫ )()´(

xfxf dx = log f(x)

26. ∫ )()´(

xfxf dx = 2 )(xf

Integration by Decomposition into Sum : 1. Trigonometrical transformations : For the

integrations of the trigonometrical products such as sin2x, cos2x, sin3x, cos3x, sin ax cos bx, etc., they are expressed as the sum or difference of the sines and cosines of multiples of angles.

2. Partial fractions : If the given function is in the form of fractions of two polynomials, then for its integration, decompose it into partial fractions (if possible).

Integration of some special integrals :

(i) ∫ ++ cbxaxdx

2

This may be reduced to one of the forms of the above formulae (16), (18) or (19).

INTEGRATION Mathematics Fundamentals M

ATHS


(ii) ∫++ cbxax

dx2

This can be reduced to one of the forms of the above formulae (15), (20) or (21).

(iii) ∫ ++ cbxax2 dx

This can be reduced to one of the forms of the above formulae (22), (23) or (24).

(iv) ∫ +++

cbxaxdxqpx

2)( , ∫

++

+

cbxax

dxqpx2

)(

For the evaluation of any of these integrals, put px + q = A differentiation of (ax2 + bx + c) + B

Find A and B by comparing the coefficients of like powers of x on the two sides.

1. If k is a constant, then

∫ dxk = kx and ∫ dxxfk )( = k ∫ dxxf )(

2. ∫ ± dxxfxf )()( 21 = ∫ dxxf )(1 ± ∫ dxxf )(2

Some Proper Substitutions :

1. ∫ f(ax + b) dx, ax + b = t

2. ∫ f(axn + b)xn–1dx, axn + b = t

3. ∫ fφ(x) φ´(x) dx, φ(x) = t

4. ∫ dxxfxf)()´( , f(x) = t

5. ∫ − 22 xa dx, x = a sin θ or a cos θ

6. ∫ + 22 xa dx, x = a tan θ

7. ∫ +−

22

22

xaxa dx, x2 = a2 cos 2θ

8. ∫ ± xa dx, a ± x = t2

9. ∫ +−

xaxa dx, x = a cos 2θ

10. ∫ − 22 xax dx, x = a(1 – cos θ)

11. ∫ − 22 ax dx, x = a sec θ

Substitution for Some irrational Functions :

1. ∫ ++ baxqpxdx)(

, ax + b = t2

2. ∫+++ cbxaxqpx

dx2)(

, px + q = t1

3. ∫ +++ baxrqxpxdx

)( 2 , ax + b = t2

4. ∫++ caxrpx

dx22 )(

, at first x =t1 and then a + ct2 = z2

Some Important Integrals :

1. To evaluate ∫ β−α− ))(( xxdx , ∫

−β

α−x

x dx,

∫ −βα− ))(( xx dx. Put x = α cos2θ + β sin2θ

2. To evaluate ∫ + xbadx

cos, ∫ + xba

dxsin

,

∫ ++ xcxbadx

sincos

Replace sin x =

+

2tan1

2tan2

2 x

x

and cos x =

+

−

2tan1

2tan1

2

2

x

x

Then put tan 2x = t.

3. To evaluate ∫ +++

xcxbaxqxp

sincossincos dx

Put p cos x + q sin x = A(a + b cos x + c sin x) + B. diff. of (a + b cos x + c sin x) + C A, B and C can be calculated by equating the

coefficients of cos x, sin x and the constant terms.

4. To evaluate ∫ ++ xcxxbxadx

22 sincossin2cos,

∫ + bxadx2cos

, ∫ + xbadx

2sin

In the above type of questions divide Nr and Dr by cos2x. The numerator will become sec2x and in the denominator we will have a quadratic equation in tan x (change sec2x into 1 + tan2x).

Putting tan x = t the question will reduce to the form

∫ ++ cbtatdt

2

5. Integration of rational function of the given form

(i) ∫ +++

424

22

akxxax dx, (ii) ∫ ++

−424

22

akxxax dx, where

k is a constant, positive, negative or zero. These integrals can be obtained by dividing

numerator and denominator by x2, then putting

x – x

a2 = t and x +

xa2

= t respectively.

Integration of Product of Two Functions :

1. ∫ f1(x) f2(x) dx = f1(x) ∫ f2(x) dx – [ ]∫ ∫ dxxfxf )()(( 2'

1 dx

Proper choice of the first and second functions : Integration with the help of the above rule is called


integration by parts, In the above rule, there are two terms on R.H.S. and in both the terms integral of the second function is involve. Therefore in the product of two functions if one of the two functions is not directly integrable (e.g. log x, sin–1x, cos–1x, tan–1x etc.) we take it as the first function and the remaining function is taken as the second function. If there is no other function, then unity is taken as the second function. If in the integral both the functions are easily integrable, then the first function is chosen in such a way that the derivative of the function is a simple functions and the function thus obtained under the integral sign is easily integrable than the original function.

2. ∫ + )sin( cbxeax dx

= 22 baeax

+[a sin (bx + c) – b cos (bx + c)]

= 22 ba

eax

+sin

−+ −

abcbx 1tan

3. ∫ + )cos( cbxeax dx

= 22 baeax

+[a cos (bx + c) + b sin(bx + c)]

= 22 ba

eax

+cos

−+ −

abcbx 1tan

4. ∫ ekxkf(x) + f '(x) dx = ekxf(x)

5. ∫ xelog = x(logex – 1) = x loge

ex

Integration of Trigonometric Functions : 1. To evaluate the integrals of the form

I = ∫ sinmx cosnx dx, where m and n are rational

numbers. (i) Substitute sin x = t, if n is odd; (ii) Substitute cos x = t, if m is odd; (iii) Substitute tan x = t, if m + n is a negative even

integer; and

(iv) Substitute cot x = t, if 21 (m + 1) +

21 (n – 1) is an

integer.

2. Integrals of the form ∫ R (sin x, cos x) dx, where R is

a rational function of sin x and cos x, are transformed into integrals of a rational function by the substitution

tan 2x = t, where –π < x < π. This is the so called

universal substitution. Sometimes it is more

convenient to make the substitution cot2x = t for

0 < x < 2π. The above substitution enables us to integrate any

function of the form R (sin x, cos x). However, in practice, it sometimes leads to extremely complex rational functions. In some cases, the integral can be simplified by –

(i) Substituting sin x = t, if the integral is of the form

∫R (sin x) cos x dx.

(ii) Substituting cos x = t, if the integral is of the form

∫R (cos x) sin x dx.

(iii) Substituting tan x = t, i.e. dx = 21 tdt+

, if the

integral is dependent only on tan x. Some Useful Integrals :

1. (When a > b) ∫ + xbadx

cos

= 22

2

ba −tan–1

+−

2tan x

baba + c

2. (When a < b) ∫ + xbadx

cos

= –22

1

ab −log

baaxab

baaxab

++−

+−−

tan

tan

3. (when a = b) ∫ + xbadx

cos=

a1 tan

2x + c

4. (When a > b) ∫ + xbadx

sin

= 22

2

ba − tan–1

−

+

22

2tan

ba

bxa + c

5. (When a < b) ∫ + xbadx

sin

= 22

1

ab − log

22

22

2tan

2tan

abbxa

abbxa

−++

−−+

+ c

6. (When a = b) ∫ + xbadx

sin=

a1 [tan x – sec x] + c


Functions with their Periods :

Function Period

sin (ax + b), cos (ax + b), sec (ax + b), cosec (ax + b)

2π/a

tan(ax + b), cot (ax + b) π/a

|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, |cosec (ax + b)|

π/a

|tan (ax + b)|, |cot (ax + b)| π/2a

Trigonometrical Equations with their General Solution:

Trgonometrical equation General Solution

sin θ = 0 θ = nπ

cos θ = 0 θ = nπ + π/2

tan θ = 0 θ = nπ

sin θ = 1 θ = 2nπ + π/2

cos θ = 1 θ = 2nπ

sin θ = sin α θ = nπ + (–1)n α

cos θ = cos α θ = 2nπ ± α

tan θ = tan α θ = nπ + α

sin2θ = sin2α θ = nπ ± α

tan2θ = tan2α θ = nπ ± α

cos2θ = cos2α θ = nπ ± α

*coscossinsin

α=θα=θ

θ = 2nπ + α

*tantansinsin

α=θα=θ

θ = 2nπ + α

*coscostantan

α=θα=θ

θ = 2nπ + α

* If α be the least positive value of θ which satisfy two given trigonometrical equations, then the general value of θ will be 2nπ + α.

Note : 1. If while solving an equation we have to square it,

then the roots found after squaring must be checked whether they satisfy the original equation or not. e.g. Let x = 3. Squaring, we get x2 = 9, ∴ x = 3 and – 3 but x = – 3 does not satisfy the original equation x = 3.

2. Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of x for which ∞ = ∞ will not be a solution as it is an indeterminate form.

3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or

y = z or both. But xy =

xz ⇒ y = z only and not

x = 0, as it will make ∞ = ∞. Similarly, if ay = az, then it will also imply y = z only as a ≠ 0 being a constant.

Similarly, x + y = x + z ⇒ y = z and x – y = x – z ⇒ y = z. Here we do not take x = 0 as in the above because x is an additive factor and not multiplicative factor.

4. When cos θ = 0, then sin θ = 1 or –1. We have to verify which value of sin θ is to be chosen which

satisfies the equation cos θ = 0 ⇒ θ =

+

21n π

If sin θ = 1, then obviously n = even. But if sin θ = –1, then n = odd.

Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 or –1.

If cos θ = 1, then n is even and if cos θ = –1, then n is odd.

5. The equations a cos θ ± b sin θ = c are solved as follows :

Put a = r cos α, b = r sin α so that r = 22 ba + and α = tan–1 b/a.

The given equation becomes

r[cos θ cos α ± sin θ sin α] = c ;

cos (θ ± α) = rc provided

rc ≤ 1.

TRIGONOMETRICAL EQUATION

Mathematics Fundamentals MATHS


Relation between the sides and the angle of a triangle: 1. Sine formula :

a

Asin = b

Bsin = c

Csin = R21

Where R is the radius of circumcircle of triangle ABC.

2. Cosine formulae :

cos A = bc

acb2

222 −+ , cos B = ac

bca2

222 −+ ,

cos C = ab

cba2

222 −+

It should be remembered that, in a triangle ABC If ∠A = 60º, then b2 + c2 – a2 = bc If ∠B = 60º, then a2 + c2 – b2 = ac If ∠C = 60º, then a2 + b2 – c2 = ab 3. Projection formulae : a = b cos C + c cos B, b = c cos A + a cos C c = a cos B + b cos A Trigonometrical Ratios of the Half Angles of a Triangle:

If s = 2

cba ++ in triangle ABC, where a, b and c are

the lengths of sides of ∆ABC, then

(a) cos2A =

bcass )( − , cos

2B =

acbss )( − ,

cos2C =

abcss )( −

(b) sin2A =

bccsbs ))(( −− , sin

2B =

accsas ))(( −− ,

sin2C =

abbsas ))(( −−

(c) tan2A =

)())((

asscsbs

−−− ,

tan2B =

)())((

bsscsas

−−− , tan

2C

)())((

cssbsas

−−−

Napier's Analogy :

tan2

CB − = cbcb

+− cot

2A , tan

2AC − =

acac

+− cot

2B

tan2

BA − = baba

+− cot

2C

Area of Triangle :

∆ = 21 bc sin A=

21 ca sin B =

21 ab sin C

∆ =)sin(

sinsin21 2

CBCBa

+=

)sin(sinsin

21 2

ACACb

+=

)sin(sinsin

21 2

BABAc

+

sin A = bc2 ))()(( csbsass −−− =

bc∆2

Similarly sin B = ca∆2 & sin C =

ab∆2

Some Important Results :

1. tan2A tan

2B =

scs − ∴ cot

2A cot

2B =

css−

2. tan2A + tan

2B =

sc cot

2C =

∆c (s – c)

3. tan2A – tan

2B =

∆− ba (s – c)

4. cot2A + cot

2B =

2tan

2tan

2tan

2tan

BA

BA+

= cs

c−

cot2C

5. Also note the following identities : Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 Solution of Triangles : 1. Introduction : In a triangle, there are six

elements viz. three sides and three angles. In plane geometry we have done that if three of the elements are given, at least one of which must be a side, then the other three elements can be uniquely determined. The procedure of determining unknown elements from the known elements is called solving a triangle.

2. Solution of a right angled triangle : Case I. When two sides are given : Let the

triangle be right angled at C. Then we can determine the remaining elements as given in the following table.

Given Required

(i) a, b tanA =

ba , B = 90º – A, c =

Aa

sin

(ii) a, c sinA =

ca , b = c cos A, B = 90º – A

Case II. When a side and an acute angle are given – In this case, we can determine

Given Required

(i) a, A B = 90º – A, b = a cot A, c =

Aa

sin

(ii) c, A B = 90º – A, a = c sin A, b = c cos A


a

PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. Two men B and C are watching man A. B watches A to be stationary and C watches A moving. Then -

(A) Man A may be at absolute rest (B) Man B may be at absolute rest (C) Man C may be at absolute rest (D) None of these 2. A particle of mass m is placed on the centre of a fixed

uniform semi-circular ring of radius R and mass M as shown. Then work required to displace the particle slowly from centre of ring to infinity is : (Assume only gravitational interaction of ring and particle)

R

M

m

(A) R

GMm (B) – R

GMm

(C) R

GMmπ

(D) – R

GMmπ

3. A ideal diatomic gas occupies a volume V1 at a pressure P1. The gas undergoes process in which the pressure is proportional to the volume. At the end of process the rms sped of the gas molecules has double from its initial value then the heat supplied to the gas in the given process is -

(A) 7 P1V1 (B) 8 P1V1 (C) 9 P1V1 (D) 10 P1V1 4. An electron gum T emits electron accelerated by a

potential difference U in a vacuum in the direction of the line a as shown in figure. Target M is placed at a distance d as shown in figure. Find the magnetic field perpendicular to the plane determine by line a and the point M in order that electron hit the target M –

aElectron gun

d

M

Target

α

(A) de

Ume αsin22 (B)

deUme

2sin2 α

(C) de

Ume αsin2 (D)

deUme αsin2

8

5. When 24.8 KeV x-rays strike a material, the photoelectrons emitted from K shell are abserved to move in a circle of radius 23 mm in a magnetic field of 2 × 10–2 T. The binding energy of K-shell electrons is -

(A) 6.2 KeV (B) 5.4 KeV (C) 7.4 KeV (D) 8.6 KeV

IIT-JEE 2011

XtraEdge Test Series # 10

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with one or more than one correct asnwer. +4 marks will be

awarded for correct answer and –1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly marked answer in any row.


6. In the circuit shown the cell is ideal. The coil has an inductance of 4H ans zero resistance. F is a fuse of zero resistance and will blow when the current through it reaches 5A. The switch is closed at t = 0. The fuse will blow -

F

L = 4H2V

S

+–

(A) after 5 sec (B) after 2 sec (C) after 10 sec (D) almost at once 7. In an insulating medium (K = 1) volumetric charge

density varies with y-coordinates according the law ρ = a. y. A particle of mass m having positive charge q is at point A(0, y0) and projected with velocity

vr = ^

0 iv as shown in figure. At y = 0 electric field is zero. Neglect the gravity and fractional resistance, the slope of trajectory of the particle as a function of y(E is only along y-axis) is –

y

A v0

x (0,y0)

(A) )y–y(vm

qa 30

3200ε

(B) )y–y(vm3

qa 30

3200ε

(C) 200

30

3

vm5)y–y(qa

ε (D) 2

00

30

3

vm2)y–y(qa

ε

8. If E denotes electric field in a uniform conductor, I

corresponding current through it, vd-drift velocity of electrons and P denotes thermal power produced in the conductor, then which of the following graph is incorrect -

(A)

vd

E

(B)

P

E

(C)

P

vd

(D)

P

I

9. Find the de Broglie wavelength of Earth. Mass of

Earth is 6 × 1024 kg. Mean orbital radius of Earth around Sun is 150 × 106 km -

(A) 3.7 m (B) 3.7 × 10–63 (C) 3.7 × 1063 m (D) 3.7 × 10–63 cm

Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

10. A pendulum of length l is suspended on a flat car that is moving with a velocity u on the horizontal road. If the car is suddenly stopped, then : (Assume bob of pendulum does not collide anywhere)

u

l θ

(A) the maximum angle θ with the initial vertical line

through which the pendulum swing is

lg2usin 1–

(B) the maximum angle θ with the initial vertical line through which the pendulum swing is

lg2usin2 1–

(C) If maximum angle is 60º, l = 5 m and g = 9.8 m/s2 then the initial speed of car u is 7 m/s

(D) If maximum angle 60º, l = 5 m and g = 9.8 m/s2, then the initial speed of car u is 6 m/s

11. A parallel plate air capacitor is connected to a battery. If plates of the capacitor are pulled further apart, then which of the following statements are correct?

(A) Strength of electric field inside the capacitor remain unchanged, if battery is disconnected before pulling the plate

(B) During the process, work is done by an external force applied to pull the plates whether battery is disconnected or it remains connected

(C) Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart

(D) None of the above

12. 12Ω 2Ω

2Ω6Ω

2H

6V

(A) Its time constant is 41 sec

(B) Its time constant is 4 sec (C) In steady state current through battery will be

equal to 0.75 A (D) In steady state current through inductance will be

equal to 0.75 A


13. In passing through a boundary refraction will not take place if -

(A) light is incident normally on the boundary (B) the indices of refraction of the two media are

same (C) the boundary is not visible (D) angle of incidence is lesser than angle of

refraction but greater then

µµ

D

R1–sin

14. A body moves in a circular path of radius R with

deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment t = 0, the velocity of body is v0 then the velocity of body will be -

(A) v =

+R

tv1

v

0

0 at time.t

(B) v = R/S–0ev after it has moved S meter

(C) v = SR–0ev after it has moved S meter

(D) None of these This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) A narrow tube is bent in the form of circle of radius R

as shown. Two small holes S and D are made in the tube at the positions right angles to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts. One part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at point D where a detector is placed.

D

RS

15. Maximum intensity produced at D is given by - (A) 4I0 (B) 2I0 (C) 3I0 (D) I0 16. The maximum value of wavelength λ to produce a

maximum at D its given by - (A) πR (B) 2πR

(C) 2Rπ (D)

23 Rπ

17. The maximum value of wavelength λ to produce a minimum at D is given by -

(A) πR (B) 2πR

(C) 2Rπ (D)

23 Rπ

Passage # 2 (Ques. 18 to 20) A metal sphere of radius R, carrying charge q1 is

surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge.

R

a

b 18. Find the surface charge density σ at r = R, r = a, r = b -

(A) σR = 24 Rq

π, σa = 24 a

qπ

, σb = 24 bq

π

(B) σR = 24 Rq

π, σa = 24

–aq

π, σb = 24 b

qπ

(C) σR = 24–Rq

π, σa = 24 a

qπ

, σb = 24 bq

π

(D) σR = 24 Rq

π, σa = 24 a

qπ

, σb = 24–

bq

π

19. The potential at the centre, using infinity at the

reference point : (potential is zero at infinity)

(A)

πε aq

Rq –

41

0 (B)

++

πε bq

aq

Rq

041

(C)

πε Rq

041 (D)

+

πε Rq

aq

bq –

41

0

20. Now the outer surface is touched to a grounding wire,

which lowers its potential to zero. Now the potential at the centre : (Assume at infinity also potential is zero)

(A)

πε aq

Rq –

41

0 (B)

++

πε bq

aq

Rq

041

(C)

πε Rq

041 (D)

+

πε Rq

aq

bq –

41

0

This section contains 2 questions (Questions 21, 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

21. Match Column-I with Column-II in the light of possibility of occurrence of phenomena listed in Column-I using the systems in Column-II

Column-I Column-II (A) Interference (P) Non-mechanical waves (B) Diffraction (Q) Electromagnetic waves (C) Polarisation (R) Visible light waves (D) Reflection (S) Sound waves

22. A satellite is revolving around earth in a circular orbit of m radius r0 with velocity v0. a particle of mass is projected from satellite in forward direction with

relative velocity v =

1–

45 v0. During subsequent

motion of particle match the following (assume M = mass of earth)

Column-I Column-II

(A) Magnitude of total energy of (P) 08

3r

GMm

Patrticle (B) Minimum distance of particle (Q) r0

from earth

(C) Maximum distance of particle (R) 53 r0

from eath

(D) Minimum kinetic energy of (S) 0r8

GMm5

particle

CHEMISTRY


1. The equilibrium constant for the reaction in aqueous solution –

H3BO3 + glycerin (H3BO3 – glycerin) is 0.90. How many moles of glycerin should be added per litre of 0.10 M H3BO3 so that 80% of the H3BO3 is converted to the boric acid glycerin complex ?

(A) 0.08 (B) 4.44 (C) 4.52 (D) 3.6

2. If optical rotation produced by the compound (i) is –30°, then rotation produced by compound (ii) is

(i)

CH3

H OHHO H

CH3

(ii)

CH3

H OHH OH

CH3

(A) + 30° (B) –30° (C) zero (D) unpredictable

3. A mixture of CO and CO2 having a volume of 20 ml is mixed with x ml of oxygen and electrically sparked. The volume after explosion is (16 + x) ml under the same conditions. What would be the residual volume if 30 ml of the original mixture is treated with aqueous NaOH ?

• (A) 12 ml (B) 10 ml • (C) 9 ml (D) 8 ml

4. Rutherford’s experiment, which estabilished the nuclear model of the atom, used a beam of -

(A) β–particles, which impinged on a metal foil and got absorbed

(B) γ–rays, which impinged on a metal foil and ejected electrons

(C) helium atoms, which impinged on a metal foil and got scattered

(D) helium nuclei, which impinged on a metal foil and got scattered

5. The correct order of acidic strength is – (A) Cl2O7

>SO2>P4O10

(B) CO2 >N2O5 <SO3

(C) Na2O >MgO >Al2O3

(D) K2O >CaO >MgO

6. A reaction follows the given concentration (C) vs time graph. The rate for this reaction at 20 seconds will be –

0.50.4

0.30.20.1

0 20 40 60 80 100Time/second

(A) 4 × 10–3 Ms–1 (B) 8 × 10–2 Ms–1 (C) 2 × 10–2 Ms–1 (D) 7 × 10–3 Ms–1

7. The potential of the Daniell cell,

Zn)M1(

ZnSO4

)M1(CuSO4 Cu was reported by Buckbee,

Surdzial, and Metz as Eº = 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T2, where

T is the celcius temperature. Calculate ∆Sº for the cell reaction at 25º C –


(A) – 45.32 (B) – 34.52 (C) – 25.43 (D) – 54.23 8. In a hypothetical solid C atoms form CCP lattice with

A atoms occupying all the Tetrahedral voids and B atoms occupying all the octahedral voids. A and B atoms are of the appropriate size such that there is no distortion in the CCP lattice. Now if a plane is cut (as shown) then the cross section would like –

Plane

CCP unit cell

(A)

C C

C C

B

B B B A

A

B

(B)

C C

C

BB B

C C

CC

(C)

C

C

B B B

C C

C C

A

A A

A

(D)

C C

C

BB B

C C

CC

9. The favourable conditions for a spontaneous

reactions are- (A) T ∆S > ∆H, ∆H = ⊕ , ∆S = ⊕ (B) T ∆S > ∆H, ∆H = ⊕ , ∆S = Θ (C) T ∆S = ∆H, ∆H = Θ , ∆S = Θ (D) T ∆S = ∆H, ∆H = ⊕ , ∆S = ⊕ Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

10. A sample of water has a hardness expressed as 77.5 ppm Ca2+. This sample is passed through an ion exchange column and the Ca2+ is replaced by H+. Select correct statement(s)

(A) pH of the water after it has been so treated is 2.4 (B) Every Ca2+ ion is replaced by one H+ ion (C) Every Ca2+ ion is replaced by two H+ ions (D) pH of the solution remains unchanged 11. Consider a sample of He gas and Ne gas both at 300

K and 1 atmosphere. Assuming ideal behaviour which of the following quantities are equal for two samples ?

(A) Root mean square speed of molecules (B) Mean translational kinetic energy of molecules (C) Number density of molecules (D) Kinetic energy of molecules 12. Which of the following samples of reducing agents

is/are chemically equivalent to 25 mL of 0.2 N KMnO4, to be reduced to Mn2+ + H2O ?

(A) 25 mL of 0.2 M FeSO4 to be oxidized to Fe3+ (B) 50 mL of 0.1 MH3AsO3 to be oxidized to H3AsO4 (C) 25 mL of 0.2 M H2O2 to be oxidized to H+ and O2 (D) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+ 13. Which of the following statement is/are correct ? (A) [Ni(CO)4] is tetrahedral, paramagnetic, sp3

hybridised (B) [Ni(CN)4]2– is square planar, diamagnetic, dsp2

hybridised (C) [Ni(CO)4] is tetrahedral, diamagnetic, sp3

hybridised (D) [NiCl4]2– is tetrahedral, paramagnetic, sp3

hybridised 14. Consider the reaction

C – OH

O

EtOH

)(NH,Na 3 → l A 22

23

ClCH

SMe,O → B + C

D∆

Identify the correct representation of structure of the products -

(A) A is

COOH

(B) The intermediate formed in the conversion of B to D is enol

(C) The structure of C is

O O

(D) A can also be formed from the reaction

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.






Passage # 1 (Ques. 15 to 17)

218

85

15

56473 K

573 K

647 K

673 K

J

G

F

C

H A

1

Pressure in atm

Vm (in cm3/mol)

EB

In the given figure P-Vm isotherm of H2O is shown.

The line (……) represent, the vanderwaals plot for H2O at 473 K. The vanderwaals constant of H2O is represented by a and b.

15. What is the equation of the dotted line (- - - )

A I C F B ?

(A) 0=

TmdVdP and 0

dVPd

T2

m

2=

(B) P =

+

mm Vb

Va 21

2

(C) P =

mm Vb

Va 2–1

2

(D) TmdV

Pd

2

2 = 0

16 As per the vanderwaals line I H G F (- - - -) which of

the following section against the behaviour of gas- (A) I H (B) H G (C) G F (D) All of the given 17. For H2O which of the following is / are correct- (A) For H2O, compressibility factor (Zc) is equal to

0.23. (B) For H2O, compressibility factor (Zc) is lesser than

0.375 because of stronger intermolecular attraction among H2O molecules.

(C) For H2O if reduced pressure, reduced volume and reduced temperature are 20, 0.6 and 2 respectively then intermolecular force of repulsion predominate over intermolecular H-bonding among H2O molecules.

(D) All of the above are correct.

Passage # 2 (Ques. 18 to 20) A useful method to convert oxime to substituted

amide is Beckmann rearrangement which occurs through following steps,

Ph

C = NCH3 OH

)I(

H→+

Ph C = N

Me OH2+

)II(

OH2 →− PhNCCH3 −=−⊕

)III(

OH2 →

CH3 –C = N–Ph

OH(IV)

CH3 –C–NH–Ph

O

18. Rate determining step in Beckmann rearrangement is (A) I (B) II (C) III (D) IV

19. The compound Me

C = N Ph OH

when treated

with H2SO4 and hydrolysed the products formed are (A) CH3COOH and PhNH2 (B) CH3NH2 and PhCOOH (C) PhCH2NH2 and CH3COOH (D) PhCH2COOH and CH3NH2 20. In the following sequence of reaction

Ph – C – –CH3

O

64pH

OHNH2

−= → I

∆ → 5PCl P

the product P may be (A) PhCOOH

(B) CH3– –C–NH2

O

(C) Ph – C – NH – – CH3

O

(D) Ph – C – NH2

O



A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


21. Column –I Column II (A) Decomposition (P) 10 t1/2 of H2O2

(B) K298

K308

kk

(Q) 1st order

(C) Arrhenius eq. (R) Temperature coefficient

(D) t99.9% (S) log 1

2

kk =

R303.2Ea

21

12

TTT–T

22. Column-I

(A)

Me

Me Me

Me

Me

Me

Me

Me

Me

Me

Cl

(B)

Me

Me

Me

Me

Me

Me

Me

Me

Me Cl

Cl

(C)

Me

Me

Me

Br

Me

Me

Br

Me

Me

Cl

Me

Cl

(D)

Me

Me

Me

Br

Me

Me

Br

Me

Me Cl

Cl

Me

Column-II (P) Optically active (Q) Cis compound (R) Trans compound (S) Optically inactive

MATHEMATICS


1. If α, β are the roots of the equation; 6x2 + 11x + 3 =0 then : (A) both cos–-1 α and cos–1 β are real (B) both cosec–1 α and cosec–1 β are real (C) both cot–1α and cot–1β are real (D) None of these

2. If for the differential equation y′ =xy + φ

yx the

general solution is y =||log Cx

x then f (x / y) is given

by - (A) – x2 / y2 (B) y2 / x2 (C) x2 / y2 (D) – y2 / x2

3. Two flagstaffs stand on a horizontal plane. A and B are two points on the line joining their feet and between them. the angles of elevation of the tops of the flagstaffs as seen from A are 30º and 60º and as seen from B are 60º and 45º. If AB is 30m, the distance between the flagstaffs in meters is

(A) 30 + 15 3 (B) 45 + 15 3

(C) 60 – 15 3 (D) 60 + 15 3

4. If the probability of choosing an integer k out of 2m integers 1, 2, 3, ...., 2m is inversely proportional to k4(1 ≤ k ≤ 2m), then the probability that chosen number is odd, is

(A) equal to 1/2 (B) less than 1/2 (C) greater than 1/2 (D) less than 1/3

5. The line x + y = 1 meets x-axis at A and y-axis at B.P is the mid-point of AB (fig.) P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3 is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn–1, then OPn =

A x

P M1

M2

O P3 P2 P1

B

y


(A) 1/2 (B) 1/2n (C) 1/2n/2 (D) 1/ 2

6. The sum ∑ ∑≤ ≤<0 10

10 )(ji

jC (jCi) is equal to

(A) 210 – 1 (B) 210 (C) 310 – 1 (D) 310

7. Reflection of the line za + za = 0 in the real axis is (A) za + az = 0

(B) aa =

zz

(C) (a + a ) (z + z ) = 0 (D) None of these 8. If g(x) is a polynomial satisfying g(x) g(y) = g(x) +

g(y) + g(xy) – 2 for all real x and y and g(2) = 5 then

3lim

→xg(x) is -

(A) 9 (B) 25 (C) 10 (D) none of these 9. The domain of definition of

f(x) =

+ 51–log 4.0 x

x × 36–

12x

is

(A) (– ∞, 0) ~ – 6 (B) (0, ∞) ~ 1, 6 (C) (1, ∞) ~ 6 (D) [1, ∞)~ 6 Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

10. The

→ 38

0

1limx

xx

(where [x] is greatest integer

function) is (A) a nonzero real number (B) a rational number (C) an integer (D) zero

11. If l = dxxx

xxxxxe xxx

+−∫ +

22

24cossin

coscossincos

then l equals -

(A) ex sin x + cos x

−

xxx sec + C

(B) ex sin x + cos x

−

xxxx cossin

(C) ex sin x + cos x

−

xx

xx sec

tan + C

(D) xex sin x+cos x –

−

−∫ +

xxxxxe xxx

22cossin

cossincos1 dx

12. Let f(x) = xxxxx x

π+++

sin)31(–)7()3–4.3)(2log–)1(log(

2/13/1

1–, x ≠1

The value of f(1) so that f is continuous at x = 1 is (A) an algebraic number (B) a rational number (C) a trance dental number

(D) – π9 log 4e

13. The solution of y1(x2y3 + xy) = 1 is

(A) 1/x = 2 – y2 + C 2/2ye− (B) the solution of an equation which is reducible to

linear equation (C) 2/x = 1 – y2 + e–y/2

(D)

+

− 22/ 212y

xxe y = C

14. Suppose a, b, c are positive integers and

f(x) = ax2 – bx + c = 0 has two distinct roots in (0, 1), then -

(A) a ≥ 5 (B) b ≥ 5 (C) abc ≥ 25 (D) abc ≥ 250 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) At times the methods of co-ordinates becomes

effective in solving problems of properties of triangles. We may choose one vertex of the triangle as origin and one side passing through this vertex as x-axis. Thus, without loss of generality, we can assume that every triangle ABC has a vertex B situated at B(0, 0), vertex C at (a, 0) and A as (h, k).

15. If in ∆ABC, AC = 3, BC = 4, medians AD and BE

are perpendicular, then area of triangle ABC must be equal to

(A) 7 (B) 11

(C) 22 (D) None of these


16. Suppose the bisector AD of the interior angle A of ∆ABC divides side BC into segments BD = 4, DC = 2. Then we must have

(A) b > 6 and c < 4 (B) 2 < b < 6 and c < 1 (C) 2 < b < 6 and 4 < c < 12 (D) None of these

17. If altitudes CD = 7, AE = 6 and E divides BC such

that ECBE =

43 , then c must be

(A) 32 (B) 35 (C) 3 (D) 34 Passage # 2 (Ques. 18 to 20) Among several applications of maxima and minima

is finding the largest term of a sequence. Let <an> be a sequence. Consider f(x) obtained by replacing x by

n e.g. let an = 1+n

n consider f(x) = 1+x

x on [1, ∞]

f ´(x) = 2)1( +xx > 0 for all x.

Hence max f(x) = )(lim xfx ∞→

= 1.

18. The largest term of an = n2/(n3 + 200) is (A) 29/453 (B) 49/543 (C) 43/543 (D) 41/451 19. The largest term of the sequence an = n/(n2 + 10) is (A) 3/19 (B) 2/13 (C) 1 (D) 1/7

20. If f(x) is the function required to find largest term in Q. 14 then

(A) f is increase for all x (B) f decreases for all x (C) f has a maximum at x = 3 400 (D) f increases on [0, 9] This section contains 2 questions (Questions 21, 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row. 21. Column –I Column II (A) The period of sin πx + (P) 22n – 1

tan 2xπ + sin 22

xπ + .... +

sin 12 −πnx + tan n

x2π

(B) g(x) = 3 + 4x, the value of (Q) 22n

gn(0) = gog .... o g(0) is

(C) f(x) = x3 + 2nx2 + bx + c is (R) 2n bijection if and only if 3b ≥ d where d is equal to

(D) (22n – 1)∑=

−n

r

r

0

)1( nCr (S) 2n + 1

+++ infinityupto.....

27

23

21

r3

r

r2

r

r

22. Centre of circle Column-I Column-II (A) |z – 2|2 + |z – 4i|2 = 20 (P) 1 – i

(B) 11

+−

zz = 2 (Q) 5/3 + 0i

(C) z z – (1 + i)z (R) – 4 – i

– (1 – i) z + 7 = 0

(D) arg

−+++

iziz

2543 (S) 1 + 2i



PHYSICS


1. A particle is projected towards north with speed 20 m/s at angle 45º with horizontal. Ball get horizontal acceleration of 7.5 m/s2 towards east due to wind. Range of ball is -

(A) 40 m (B) 70 m (C) 50 m (D) 60 m 2. A cylinder of mass m1 is kept over a block of mass m2

kept over smooth inclined plane shown in figure. Surface between cylinder and block is rough. Friction on cylinder-

m1

m2

θ

(A) is in upward direction (B) is in downward direction (C) is zero (D) will depend on angle of inclination and

coefficient of friction between cylinder and block

3. A over head tank of capacity 10 K liter (10000 liter) is kept at top of building 15 m high. Water level is at depth 5m from ground. Water falls in tank with

velocity 25 m/s. The tank has to filled in 21 hr. If

efficiency of pump is 67.5%, electric power used is approximately -

(A) 4 kW (B) 5 kW (C) 2 kW (D) 2.5 kW

4. There ia a sphere of radius 'R'. Let E1 and E2 be gravitational field at distance r1 and r2 from centre-

(A) If r1 < R < r2 then E1 must be less than E2 (B) If r1 < r2 < R then E1 must be greater than E2 (C) If R < r1 < r2 then E1 must be less than E2 (D) If r1 = R – k and r2 = R + k (where k < R) E1 must

be greater than E2

5. A block of mass m is placed at top of frictionless wedge of mass 'M' placed on frictionless surface as shown in figure. Velocity of block on wedge at the time it slips off the wedge is u. Velocity of wedge at that instant is -

θ

(A) M

um (B) M

um θcos

(C) Mm

um+

θcos (D) Mm

um+

θ2

cos

IIT-JEE 2012

XtraEdge Test Series # 10

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with one or more than one correct asnwer. +4 marks will be

awarded for correct answer and –1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly marked answer in any row.


6. A rectangular plate is kept in y-z plane. Which of the following is correct for this plate?

(A) Iz = Ix + Iy (B) Iy = Ix + Iz (C) Ix = Iy + Iz (D) All of these 7. A glass of water is to be cooled using an ice-cube.

For which of following position water will be cooled fastest -

(A) Ice is left floating (B) Ice is kept just submerged in water (C) Ice is kept bottom of glass (D) Water will be cooled at same rate no matter

where ice is kept 8. Shape of string carrying transverse wave at t = 0 and

t = 1 sec is given by y = 1

12 +x

and y = 342

12 ++ xx

respectively, where 'x' is distance in meter. Wave velocity is -

(A) 1 m/s in positive x-direction (B) 2 m/s in negative x-direction (C) 1 m/s in negative x-direction (D) 50 cm/ sec in negative x-direction 9. A body of mass 200 gm is heated up. Graph shows

change in temperature as heat is supplied to body. Specific heat capacity of body is (in J/kg/ºC) –

∆H(in kJ)

∆H

(in ºC

)

30º

(A) 3

103 (B) 35 × 103

(C) 3 × 103 (D) 3 Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

10. A particle is moving along straight line with velocity v = t2 – 3t + 2 m/s. Particle will retard for time t -

(A) t < 1 (B) 1 < t < 1.5 (C) 1.5 < t < 2 (D) t > 2

11. Two identical rods P and Q are placed on frictionless horizontal surface. Two identical mass hit two rods separately and comes at rest after hitting. Mass hits rod 'P' at its centre while rod 'Q' is hit by mass a little distance away from centre -

(A) Rod P and Q will have same speed (B) Q will have greater kinetic energy

(C) The two rods have same kinetic energy but linear kinetic energy of 'B' will be less than that of 'A'

(D) The kinetic energy of 'B' will depend on the distance from centre where the mass hit

12. Which of the following is true, for a sample of gas

according to kinetic theory of gases - (A) Net velocity of the gas molecules is zero (B) Net momentum of the gas molecules is zero (C) Net speed of the gas molecules is zero (D) Net kinetic energy of gas molecules is zero 13. A cylinder is floating in a liquid kept in container.

Coefficient of cubical expansion of cylinder is 'γ'. Expansion of liquid and container are negligible. Upon increasing temperature -

(A) Level of liquid in container will increase (B) Level of liquid in container will remain same (C) Volume of cylinder inside water will increase (D) Volume of cylinder inside water will remain

same 14. Length of kundt's tube is 1m. When tuning fork is

vibrated and brought near rod of the kundt's tube, the powder keeps on moving. If velocity of sound is 320 m, frequency of tuning fork cannot be -

(A) 880 Hz (B) 900 Hz (C) 960 Hz (D) 1040 Hz This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.


h

2h

A cylindrical container of cross-sectional area 'A' and

height '5h' is kept at height '2h' above ground. It contains a liquid of density '2ρ' till height 'h'. The cylinder is filled with light piston as show in figure.

15. Where should a hole is made in the container so that liquid, strikes ground farthest ?

(A) At bottom of container (B) At height h/3 above bottom of container (C) At height h/2 above bottom of container (D) Liquid will strike ground at same distance

irrespective of position of hole


16. A block of mass M is kept over piston and hole is made at a distance 'h/2' from piston. Velocity of efflux is -

(A) gh (B) ghA

M

+

ρ

(C) ghA

M

+

ρ2 (D) gh

AM

+

ρ22

17. A liquid of density 'ρ' is poured in container till

height h above container. Velocity of efflux from hole at distance 'h/2' below piston is -

(A) gh3 (B) gh23

(C) gh2 (D) gh


A B

C

D

l

l

A T- shape iron frame of mass m free to rotate in

vertical plane about one of its end as shown in figure. The two rods AB and CD making T-shape are identical. Initially the frame is in the position shown in figure. The frame is left to rotate freely in vertical plane.

18. Moment of inertia of frame about axis of rotation -

(A) 3

m2 2l (B) 12m17 2l

(C) 24m17 2l (D)

12m5 2l

19. Angular acceleration of frame when rod AB is

making angle 'θ' with vertical is -

(A) l17

sin18 θ . g (B) l17

sin24 θ . g

(C) l5

sin12 θ . g (D) l2

sin9 θ . g

20. Force due to axis on frame when frame becomes

vertical -

(A) 1736 mg (B)

532 mg

(C) 1739 mg (D)

1744 mg


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row. 21. Column-I contains molar heat capacity for certain

process for an ideal gas and column II contains corresponding processes. α,β and a are constant and γ is adiabatic exponent. Match the correct one :

Column-I Column-II

(A) C = Tα (P) V exp

α

RT– = const.

(B) C = CV + αT (Q) V – aT = const.

(C) C = CV + βV (R)PVγexp

γα

PV)1–(– = const.

(D) C = CV + aP (S) T exp

βVR = const.

22. A horizontal plane support a vertical cylinder of radius 20 cm and a disk of mass 2 kg is attached to the cylinder by a horizontal thread of length π/5 m. The disk can move frictionlessly on the table. An initial velocity 1 m/s is imparted to the disk. Consider

a situation when 20π m length of string is wrapped

on cylinder. Column-I Column-II

(A) Angular velocity of disk (P) 10

2π

(in rad/sec)

(B) Time taken (in sec) (Q) π3

40

(in wrapping 20π meter)


(C) Tension in string (in N) (R) π3

20

(D) Time taken (in sec) after (S) 1607 2π

which disk will hit cylinder

CHEMISTRY


1. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is -

(A) 0.7 (B) 0.5 (C) 0.30 (D) 0.10 2. If the threshold frequency of a metal for photoelectric

effect is v0 then which of the following will not happen ?

(A) If frequency of the incident radiation is v0, the kinetic energy of the electrons ejected is zero.

(B) If frequency of the incident radiation is v, the kinetic energy of the electrons ejected will be hv – hv0

(C) If frequency is kept same at v but intensity is increased, the number of electrons ejected will increase.

(D) If frequency of the incident radiation is further increased, the number of photo-electrons ejected will be increase.

3. Which of the following is violation of Pauli's

exclusion principle ?

(A) (B)

(C) (D)

4. The IUPAC name of compound HO – C = O

NH2 – C ==== C ––– C – H

NH2

CH3

Cl

is –

(A) 2, 3 diamino-4-chloro-2-pentenoic acid (B) 4-chloro-3, 3-diamino-2-pentenoic acid (C) 3, 3–diamino-3-chloro-pentenoic acid (D) All of the above

5. If optical rotation produced by the compound (i) is –30°, then rotation produced by compound (ii) is

(i)

CH3

H OHHO H

CH3

(ii)

CH3 H OHH OH

CH3

(A) + 30° (B) –30° (C) zero (D) unpredictable

6. 16 mL of a gaseous aliphatic CnH3nOm was mixed with 60 mL O2 and sparked, the gas mixture on cooling occupied 44 mL. After treatment with KOH solution the volume of gas remaining was 12 mL. Formula of compound is -

(A) C2H6O (B) C3H8O (C) CH4O (D) None of the above

7. Most stable free radical is

(A)

CH3

(B)

(C)

(D)

8. At constant pressure P, A dissociates on heating according to the equation

A(g) B(g) + C(g) The equilibrium partial pressure of A at T K is 1/9 P,

the equilibrium Kp at TK is

(A) 98 P (B)

964 P (C)

916 P (D) 9 P

9. Calculate the pH of 6.66 × 10–3 M solution of Al(OH)3. Its first dissociation is 100% where as second dissociation is 50% and third dissociation is negligible.

(A) 2 (B) 12 (C) 11 (D) 13

Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

10. The IUPAC name of the following compound is - OH

CNBr (A) 3-Bromo-3-cyano phenol (B) 3-Bromo-5-hydroxy benzonitrile (C) 3-Cyano-3-hydroxybromo benzene (D) 5-Bromo-3-hydroxy benzonitrile


11. Which of the following is/are correct regarding the periodic classification of elements ?

(A) The properties of elements are the periodic function of their atomic number

(B) Non metals are lesser in number than metals (C) The first ionization energies of elements in a

period do not increase with the increase in atomic numbers

(D) For transition elements the d-subshells are filled with electrons monotonically with the increase in atomic number

12. Identify the correct statements -

(A) The compound

H3C CH3

COOH O

fails to undergo

decarboxylation (B) A Grignard reagent can be successfully made

from the following dibromide Br

Br

(C) Cyclopentan –1, 2– dione exists almost 100% in the enol form whereas diacetyl (CH3COCOCH3) can exist in the keto form as well as the enol form

(D) Among the following resonance structure given below, (ii) will be the major contributor to the resonance hybrid.

13. Which of the following are possible products from

aldol condensation of 6-oxoheptanal ?

(A)

C O CH3

(B)

C

O H

CH3

(C)

O

(D)

OCH3

14. Which of the metal is/are used in flash bulbs? (A) Be (B) Mg (C) Ca (D) Ba

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) According to molecular orbital theory all atomic

orbital combine to form molecular orbital by LCAO (Linear combination of atomic orbital) method. When two atomic orbitals have additive (constructive) overlapping, they form bonding molecular orbital (BMO) which have lower energy than atomic orbitals whereas when atomic orbitals overlap subtractively, higher energy antibonding molecular orbitals (AMO) are formed. Each M.O. occupies two electrons with opposite spin. Distribution of electrons in M.O. follows Aufbau principle as well as Hund's rule. M.O. theory can successfully explain magnetic behaviour of molecules.

15. Which of the following is/are not paramagnetic ? (A) NO (B) B2 (C) CO (D) O2 16. Bond strength increases when (A) bond order increases (B) bond length increases (C) antibonding electrons increases (D) bond angle increases

17. –22O will have

(A) bond order equal to H2 and diamagnetic (B) bond order equal to H2 but paramagnetic (C) bond order equal to N2 and diamagnetic (D) bond order higher than O2


C (Resolvable)

HBr,Peroxide

A(C6H11Br) Decolourise Br2 water and connot be resolved

alc. KOH

a single possibleproduct E

HBr,R2O2(Excess)

GResolvable

F non-resolvable

(Non-resolvable)

HBr

B

Zn,Heat

D(C6H12)O3;Zn,H2O

O ||

CH3–C–CH3


18. Organic compound 'A' is –

(A)

CH2Br

(B)

Br

(C)

CH2Br

(D)

Br

19. The resolvable orgainc compound 'C' is –

(A) CH2Br

Br

(B)

Br

Br *

(C)

Br Br (D)

Br

Br

• • 20. The resolvable organic compound, G is – •

(A) CH3

CH3 CH3 Br Br

CH3

(B) Br

CH2CH3 Br H H

CH2CH3

(C)

H

CH2Br H CH3

CH3 CH2Br

(D) Br

CH3 Br H H

CH

CH3 CH3


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


21. Match the following : Column -I Column-II

(A) Compound show Geometrical isomerism

(P)

Me Me

(B) Compound is chiral (Q) = C

Me H

MeH

(C) Compound having plane of symmetry

(R) C = C

Me

H MeH

(D) Compound having centre of symmetry

(S)

MeH Me

H

22. Column-I Column-II (Ionic species) (Shapes) (A) XeF5

+ (P) Tetrahedral (B) SiF5

– (Q) Square planar (C) AsF4

+ (R) Trigonal bipyramidal (D) ICl4

– (S) Square pyramidal

MATHEMATICS


1. If sinx + sin2x + sin3x = 1, then cos6x– 4cos4x + 8 cos2x is equal to - (A) 0 (B) 2 (C) 4 (D) 8 2. A line meets the coordinate axes in A and B. A circle

is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is

(A) m(m + n) (B) m + n (C) n(m + n) (D) (1/2) (m + n) 3. The line joining A(b cos α, b sin α) and B (a cos β,

a sin β) is produced to the point M(x, y) so that

AM : MB = b : a, then x cos2

β+α + y sin2

β+α =

(A) –1 (B) 0 (C) 1 (D) a2 + b2 4. The equation

143 −−+ xx + 168 −−+ xx = 1 has (A) no solution (B) only one solution (C) only two solution (D) more than two solutions


5. Equation of the line of shortest distance between the

lines 132zyx

=−

= and 2

251

32 +

=−−

=− zyx is -

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) 3/1

)3/31(3/131

3/1)3/62( +

=−

=− zyx

(C) 3/1

)3/32(3/1

)3/92(3/121 +

=−

=− zyx

(D) 3/11

3/13

3/12 −

=+

=− zyx

6. The set of all x satisfying the equation

10)(loglog 23

23 −+ xxx = 1/x2 is

(A) 1, 9 (B) 1, 9, 1/81 (C) 1, 4, 1/81 (D) 9, 1/81 7. In a certain test there are n questions. In this test 2k

students gave wrong answers to at least (n – k) questions, where k = 0, 1, 2, ...... , n. If the total number of wrong answers is 4095, then value of n is

(A) 11 (B) 12 (C) 13 (D) 15 8. Equation of the locus of the pole with respect to the

ellipse 2

2

ax + 2

2

by = 1, of any tangent line to the

auxiliary circle is the curve 4

2

ax + 4

2

by = λ2 where

(A) λ2 = a2 (B) λ2 = 1/a2 (C) λ2 = b2 (D) λ2 = 1/b2 9. The number of values of x ∈[0, nπ], n ∈ I that satisfy

log|sinx|(1 + cos x) = 2 is (A) 0 (B) n (C) 2n (D) none Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE THAN ONE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 10. If x2 + 2hxy + y2 = 0 represents the equations of the

straight lines through the origin which make an angle α with the straight line y + x = 0, then

(A) sec 2α = h (B) cos α = hh

21+

(C) 2 sin α = h

h+1 (D) cot α = 1

1−+

hh

11. If the numerical value of tan (cos–1(4/5) + tan–1(2/3) is a/b then

(A) a + b = 23 (B) a – b = 11 (C) 3b = a + 1 (D) 2a = 3b

12. Let E =

++

+

502

31

501

31 + ... upto 50 terms, then -

(A) E is divisible by exactly 2 primes (B) E is prime (C) E ≥ 30 (D) E ≤ 35 13. If PQ is a double ordinate of the hyperbola

2

2

ax – 2

2

by = 1 such that OPQ is an equilateral

triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola, satisfies

(A) 1 < e < 2/ 3 (B) e = 2/ 3

(C) e = 3 /2 (D) e > 2/ 3 14. If z1, z2, z3, z4 are the vertices of a square in that order,

then (A) z1 + z3 = z2 + z4 (B) |z1 – z2| = |z2 – z3| = |z3 – z4| = |z4 – z1| (C) |z1 – z3| = |z2 – z4| (D) (z1 – z3)/(z2 – z4) is purely imaginary This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) In ∆ABC, a = 14, b = 15, c = 13, P be a point with in

the triangle such that ∠PAB = ∠PBC = ∠PCA = α

and tan α = nm , where m and n are relatively prime

positive integers. Let PA = x, PB = y, PC = z 15. The area of triangle ABC is

(A) 21 sin α (cx + ay + bz) (B)

21 (x2 + y2 + z2) tan α

(C) 21 (xy + yz + zx) (D) None of these

16. tan α must be equal to

(A) 222 cba ++

∆ (B) 2222

cba ++

∆

(C) 2224

cba ++∆ (D) None of these


17. m + n must be equal to (A) 461 (B) 463 (C) 465 (D) 365 Passage # 2 (Ques. 18 to 20) A(3, 7) and B(6, 5) are two points. C : x2 + y2 – 4x – 6y – 3 = 0 is a circle. 18. The chords in which the circle C cuts the members of

the family S of circles through A and B are concurrent at

(A) (2, 3) (B) (2, 23/3) (C) (3, 23/2) (D) (3, 2)

19. Equation of the member of the family S which bisects the circumference of C is

(A) x2 + y2 – 5x – 1 = 0 (B) x2 + y2 – 5x + 6y – 1 = 0 (C) x2 + y2 – 5x – 6y – 1 = 0 (D) x2 + y2 + 5x – 6y – 1= 0

20. If O is the origin and P is the centre of C, then difference of the squares of the lengths of the tangents from A and B to the circle C is equal to

(A) (AB)2 (B) (OP)2 (C) |(AP)2 – (BP)2| (D) None of these


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


21. Value of x when Column-I Column-II (A) 52 54 56 ... 52x = (0.04)–28 (P) 3 log3 5

(B) x2 =

+++ ...

161

81

41log 5

)2.0( (Q) 4

(C) x =

+++ ...

31

31

31log 325.2

)16.0( (R) 2 (D) 3x–1 + 3x–2 + 3x–3 + ... (S) 7

= 2

+++++ ...

51

51155 2

2

22. If a and b are two units vectors inclined at angle α to each other then

Column –I Column-II

(A) |a + b| < 1 if (P) 3

2π < α < π

(B) |a – b| = |a + b| if (Q) π/2 < θ ≤ π

(C) |a + b| < 2 (R) α = π/2

(D) |a – b| < 2 (S) 0 ≤ θ < π/2

Interesting Science Facts

• The dinosaurs became extinct before the Rockies or the Alps were formed.

• Female black widow spiders eat their males after mating.

• When a flea jumps, the rate of acceleration is 20 times that of the space shuttle during launch.

• The earliest wine makers lived in Egypt around 2300 BC.

• If our Sun were just inch in diameter, the nearest star would be 445 miles away.

• The Australian billy goat plum contains 100 times more vitamin C than an orange.

• Astronauts cannot belch - there is no gravity to separate liquid from gas in their stomachs.

• The air at the summit of Mount Everest, 29,029 feet is only a third as thick as the air at sea level.

• One million, million, million, million, millionth of a second after the Big Bang the Universe was the size of a …pea.

• DNA was first discovered in 1869 by Swiss Friedrich Mieschler.

• The molecular structure of DNA was first determined by Watson and Crick in 1953.

• The thermometer was invented in 1607 by Galileo.

• Englishman Roger Bacon invented the magnifying glass in 1250.


PHYSICS 1. Distinguish between ‘point to point’ and ‘broadcast’

communication modes. Give one example of each.

2. Two identical prisms made of the same material placed with their bases on opposite sides (of the incident white light) and faces touching (or parallel) neither deviate nor disperse. Can this arrangement produce a parallel displacement of the beam ?

3. What is the formula for the magnifying power of a compound microscope ?

4. Sketch two equipotential surfaces for a point charge.

5. What are superconductors ?

6. Draw a labelled diagram of Hertz' experiment for producing E.M. waves.

7. What is the phase difference between voltage and current in a series LCR circuit at resonance connected with an AC source ?

8. Explain the difference between ‘Hard’ and ‘Soft’ X-rays.

9. The given graph show the variation of photo electric current (I) with the applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiations.

I

V

1 3

2 4

10. Four nuclei of an element fuse together to form a heavier nucleus. If the process is accompanied by release of energy, which of the two-the parent or the daughter nucleus would have a higher binding energy/nucleon ?

11. Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How does it effects the

(i) Width of the depletion layer ? (ii) Junction field ?

General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted.

General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

MOCK TEST-3

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solut ions published in same issue


12. Oil floating on water looks coloured due to interference of light. What should be the approximate thickness of the film for such effects to be visible ?

13. Draw a graph showing the variation of intensity with angle in a single slit diffraction experiment.

14. A parallel plate air filled capacitor has capacitance 5 µF. If plate separation made twice and whole space is filled with medium, capacitance becomes 20 µF. Find dielectric constant of the medium.

15. Suppose you have two bars of identical dimensions, one made of paramagnetic substance and the other of diamagnetic substance. If you place these bars along a uniform magnetic field, show diagramatically, what modifications in the field pattern would take place in each case.

16. In a plane e.m. wave, the electric field oscillates with a frequency of 2 × 1010s–1 and an amplitude of 40 Vm–1. (i) What is the wavelength of the wave and (ii) What is the energy density due to the electric field ?

17. A solenoid has self-inductance 2 mH and current in it is 5 amp. Find magnetic energy stored in it.

18. Find power factor of the adjacent circuit. L = 30 mH

~

R = 4 Ω

V = 2200 sin 100 t

19. Experimental observations have shown that X-rays (i) travel in vacuum with a speed of 3 × 108 ms–1, (ii) exhibit the phenomenon of diffraction and can

the polarized. What conclusion can be drawn about the nature of

X-rays from each of these observations ?

20. A radioactive material is reduced to 161 of its

original amount in 4 days. How much material should one begin with so that 4 × 10–3 kg of the material is left after 6 days.

21. Why are apertures of camera lenses so small while the apertures of telescopes are as large as feasible ?

22. In adjacent circuit, if current in 5 Ω resistance is zero, find resistance R.

R 6Ω

5Ω

10Ω 20Ω

1Ω

12V

23. The self inductance of a solenoid is 5 mH and current flowing in it depends on time t as i = t2.

(where i → In Amp., t → In second). Find induced emf in it at t = 4 s.

24. Derive an expression for magnetic field inside a long solenoid.

25. What is Wheatstone bridge ? Deduce the condition

for which Wheatstone bridge is balanced.

26. Explain the differences between diamagnetic, paramagnetic and ferromagnetic substances.

27. Describe the method to obtain Reverse Bias characterstics of a P-N junction diode. Define reverse resistance. Draw necessary circuit diagram and also the reverse characterstic curve.

28. A student has to study the input and output characteristics of a n-p-n silicon transistor in the Common Emitter configuration. What kind of a circuit arrangement should she use for this purpose ? Draw the typical shape of input characteristics likely to be obtained by her. What do we understand by the cut off, active and saturation states of the transistor? In which of these states does the transistor not remain when being used as a switch ?

OR Input signals A and B are applied to the input

terminals of the ‘dotted box’ set-up shown here. Let Y be the final output signal from the box.

Draw the wave forms of the signals labelled as C1 and C2 within the box, giving (in brief) the reasons for getting these wave forms, Hence draw the wave form of the final output signal Y. Give reasons for your choice.

What can we state (in words) as the relation between the final output signal Y and the input signals A and B ?

Au1

0 1 2 3 4

0 1 2 3 4

B

B

C2

C1 A

B


29. Draw a labelled ray diagram of an astronomical telescope. Write mathematical expression for its magnifying power. How does the magnifying power get affected on increasing the aperture of the objective lens and why ?

30. Derive an expression for the energy density of a capacitor.

OR An electric flux of –6 × 103 Nm2/C passes normally

through a spherical Gaussian surface of radius 10 cm, due to a point charge placed at the centre.

(i) What is the charge enclosed by the Gaussian surface ?

(ii) If the radius of the Gaussian surface is doubled, how much flux would pass through the surface ?

CHEMISTRY

1. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body centered cubic and (iii) hexagonal close packed lattice ?

2. What is meant by 'specific surface area' of a solid ?

3. Give the IUPAC name of Li [AlH4]

4 What is formula of siderite ore ?

5. Name the monomer units of Bakelite

6. What are antiseptics. Give two example :

7. What are basic amino acids. Give a example ?

8. Arrange the following in the order of their increasing reactivity in nucleophilic substitution reactions :

CH3F, CH3I, CH3Br, CH3Cl

9. The half life for radioactive decay of 14C is 5730 y. An archaeological artefact contained wood had only 80% of the 14C in a living tree. Estimate the age of the sample.

10. What is the effect of temperature on the rate constant of reaction ? How can this temperature effect on rate constant ? Explain using collision theory ?

11. Explain the following :- (i) S.H.E.; (ii) Kohlrausch's law

12. Why Actinides show much higher oxidation states as compared to Lanthanides ?

13. How group I radicals like Ag+ and Hg22+ are

seperated by complex formation with NH4OH ?

14. What is Roasting ?

15. Give the structure and monomer units of biodegradable polymer PHBV ?

16. Name the purine and pyrimidine bases in DNA and RNA .

17. What are detergents. Give a example of Cationic & Anionic detergents

18. A sweet smelling organic compound 'A' is slowly oxidised by air in the presence of light to a highly poisonous gas. On warming with silver powder, it forms gaseous substance 'B' which is also formed by the action of calcium carbide on water. Identify 'A' and 'B' and write the equations of the reactions involved .

19. (a) Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of a side of the cell ?

(b) Classify each of the following as being either a p-type or an n-type semiconductor

(i) Ge doped with In (ii) Si doped with As.

20. Two liquids A and B form ideal solution at 323 K. A liquid mixture containing one mole of A and two moles of B has a vapour pressure of 250 bar. If one more mole of A is added to the solution, the vapour pressure increases to 300 bar. Calculate the vapour pressures of liquids A and B at 323 K.

21. (a) In which of the following does adsorption take place and why ?

(i) Silica gel placed in the atmosphere saturated with water.

(ii) Anhydrous CaCl2 placed in the atmosphere saturated with water.

(c) Give an example of shape-selective catalysis

22. Explain the order of basic character in hydrides of nitrogen family ?

23. Give structure of Cr2O72– ?

24. How will you convert ? (i) Phenol to p-hydroxyazobenzene (ii) Ethyl alcohol to methyl alcohol.

25. Write the IUPAC name of the following : CH3 – O – C (CH3)3

26. Give a suitable colour reaction test to distinguish between

(i) 2-Pentanone and 3-Pentanone (ii) Acetone and acetaldehyde ?

27. An organic compound A(C3H6O) is resistant to oxidation but forms compound B(C3H8O) on reduction which reacts with HBr to form the bromide (C). C forms a Grignard reagent which reacts with A to give D (C6H14O). Give the structures of A, B, C and D and explain the reactions involved.


28. (a) What are ideal and non-ideal solutions ? What type of non-idealities are exhibited by cyclohexane-ethanol and acetone-chloroform mixtures ? Give reasons for your answers.

(b) A solution containing 30 g of a nonvolatile solute exactly in 90g water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to solution, the new vapour pressure becomes 2.9 kPa at 298 K. Calculate,

(i0 Molecular mass of the solute; (ii) Vapour pressure of water at 298 K.

29. Give structure of :- (a) Hypophosphorus acid (b) Pyrophosphoric acid (c) Dithionic acid (d) Marshall acid (e) Hypophosphoric acid

30. (a) An optically inactive compound (A) having molecular formula C4H11N on treatment with HNO2 gave an alcohol (B). (B) on heating at 440 K gave an alkene (C). (C) on treatment with HBr gave an optically active compound (D) having the molecular formula C4H9Br. IdentifyA, B, C and D and write down their structural formulae. Also write equations involved.

(b) Explain why Alkyl amines are stronger bases than arylamines.

MATHEMATICS

Section A

1. Show that relation R on the set A = 1, 2, 3 given by R = (1, 2), (2, 1) is symmetric but neither reflexive nor transitive.

2. If x2/3 + y2/3 = a2/3 then find dxdy .

3. Evaluate ∫ dxex xsinlog3cos .

4. Solve : (x + y)2

dxdy = a2

5. The projection of a vector on the coordinate axes are 6, –3, 2. Find its length and direction cosines.

6. Find the values of x for which the angle between the vectors a

r= 2x2 i + 4x j + k and b

r= 7 i –2 j + x k is

obtuse.

7. A plane meets the coordinate axes in A, B, C such that the centroid of triangle ABC is the point (p, q, r). Show that the equation of the plane is

px +

qy +

rz = 3.

8. If

+ab

ba5

2=

8526

find a, b.

9. Find matrix X and Y if

X + Y =

9025

, X –Y=

−1063

10. Using determinant, find k so that points (k, 2 –2k), (–k + 1, 2k) and (– 4 – k, 6 –2k) are collinear.

Section B 11. In two successive throws of a pair of dice, determine

the probability of getting a total of 8, each time. 12. If f : R → R is given by f(x) = sin2 x + sin2(π/3 + x) + cos x . cos (π/3 + x) ∀

x ∈ R. g : R → R be such that g(5/4) = 1 then prove that gof is constant function.

13. f(x) =

≤<+≤≤+<≤++

84,5242,2320,2

xbaxxxxbaxx

f(x) is continuous on [0, 8] then find a, b.

14. y = tan–1 (sec x + tan x) then find dxdy ;

where –2π < x <

2π .

15. Differentiate tan–1

−++

−−+22

22

11

11

xx

xx w.r.t,cos–1 x2

16. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm/sec. How fast is the area decreasing when the two sides are equal to the base.

OR Use lagrange's Mean Value theorem to determine a

point P on the curve y = 2x − where the tangent is parallel to the chord joining (2, 0) and (3, 1).

17. Evaluate ∫−−

dxxx

x421

2

OR

Evaluate : ∫ ++ 2/xe.

xcos1xsin2 .dx

18. Evaluate ∫ ++ dx

xx

164

4

2


19. Solve : dxdy –2y = cos 3x.

20. For any two vectors ar

and br

, show that (1 + | a

r|2) (1 + | b

r|2) = (1 – a

r. b

r)2

+ | ar

+ br

+ ( ar

× br

)|2

21. Find the foot of the perpendicular from the point

(0, 2, 3) on the line 5

3+x = 2

1−y = 3

4+z . Also,

find the length of the perpendicular. OR

Find the particular solution of the differential

equation dydx + y cot x = 2x + x2cot x, x ≠ 0 given

that y = 0, when x = 2π

22. Show that xyzzxyyzx

111

= 2

2

2

111

zzyyxx

and hence factorize. OR

If a, b and c are real numbers and

accbbacbbaacbaaccb

+++++++++

= 0

Show that either a + b + c = 0 or a = b = c.

Section C

23. Two persons A and B throw a die alternately till one

of them gets a 'six' and wins the game. Find their respectively probabilities of winning.

24. Find the area bounded by the curves y = x and y = x3.

25. Find the shortest distance between the lines

2

1−x = 3

2−y = 4

3−z and 3

2−x = 4

4−y = 5

5−z .

26. If A =

−

−

120312021

, find A–1. Using A–1, solve the

system of linear equations x – 2y = 10 2x + y + 3z = 8 –2y + z = 7

27. Show that sinp θ cosq θ attains a maximum value when θ = tan–1 qp / .(where p, q > 0)

28. Evaluate ∫ +3

1

2 )5( dxxx , by first principle method

OR

Evaluate ∫−

1

1

dxex by first principle method

29. There is a factory located at each of two places P and Q. From these locations, a certain commodity is delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are 8 and 6 units respectively. The cost of transportation per unit is give below.

From To

COST (In Rs.)

A B C P 16 10 15 Q 10 12 10

How many units should be transported from each factory to each depot in order that the transportation cost is minimum. Formulate the above as a linear programming problem.

OR A brick manufacturer has two depots, A and B, with

stocks of 30,000 and 20,000 bricks respectively. He receives orders from three builders P, Q and R for 15, 000, 20,000 and 15000 bricks respectively. The cost in Rs. transporting 1000 bricks to the builders from the depots are given below.

From To

P Q R

A 40 20 30 B 20 60 40

How should the manufacturer fulfill the orders so as to keep the costs of transportation minimum?

• Pluto lies at the outer edge of the planetary system of our sun, and at the inner edge of the Kuiper Belt, a belt of icy comets that are the remnants of the formation of the solar system.

• Gamma ray bursts - mysterious explosions at the edge of the Universe - were first detected in 1969 by military satellites monitoring the Test Ban Treaty.

• Titan is the largest moon of Saturn and the second largest moon in the entire solar system.


PHYSICS

1. p = q × 2l It's a vector quantity

2. Sensitivity ∝ gradientPotential

1

3. Gamma rays, X-rays, ultraviolet, Infrared

4. λ = ph =

mkh

2

∴ The proton will have a higher K.E. (mass of proton is slightly less than that of the neutron)

5. The ionization energy of silicon gets (considerably) reduced compared to that of carbon. Silicon (a semi-conductor), therefore, becomes a (much) better conductor of electricity than carbon (an insulator)

6. (0 to t1), (t3 to t4)

7. No, when the refractive index of prism material is same as that of the surrounding, then there is no dispersion.

8. As, P ∝ f1 , so lens of smaller focal length is more

powerful and more magnifying power.

9. F = qvB sinθ (i) θ = 90°, Fmax = q v B (ii) θ = 0°, 180°, F = 0

10. S = g

g

iiGi

–×

11. V1 = 2 V V2

6 µF 12 µF

V

V1 = 2 = 126

12+

× V

V = 3 volt

OR

d

A0∈ = 8

C' =

2

)5(0d

A∈ = 10 × d

A0∈

= 10 × 8 = 80

12. The region containing the uncompensated acceptor and donor ions is called depletion region there is a barrier at the junction which opposes the movement of majority charge carriers.

– ° – ° – ° – – + + + + +

– ° – ° – °

– ° – ° – °

– ° – ° – °

– –

– –

– –

+ +

+ +

+ +

+ + +

+ + +

+ + +

P N

Depletion region

Formation of depletion region in PN junction diode The physical distance from one side of the barrier to the other is called the width of the barrier. The width of the depletion region or barrier depends upon the nature of the material. Its typical value is nearly 10–6 m. The difference of potential from one side of the barrier to the other side is called potential barrier or height of the barrier. Its value is nearly 0.7 V for a silicon PN junction and 0.3 V for a germanium diode.

13. Reasons : (i) Size of antenna (ii) Effective power radiated by the antenna

14. The activity of a radioactive element at any instant, equals its rate of decay at that instant. Its SI unit is Becquerel (Bq) (= 1 decay per second)

Activity R = –dtdN = λN =

Te 2log N

∴ 2

1

RR =

1

1

TN +

2

2

TN =

12

21

TNTN

MOCK TEST-2 (SOLUTION) MOCK TEST– 2 (PAPER) PUBLISHED IN JANUARY ISSUE


15.

Information Source

Transmitter Transmission channel

Receiver

Message Signal

Transmitted Signal

Received Signal

16. The permitted stationary orbits for the electron in a

hydrogen atom are those for which the angular momentum of the electron is an integral multiple of h/2π

m vn rn = n π2h

∴ 2πrn = n mvh

But nmv

h = λn the associated de Broglie

wavelength for electron in its nth orbit Hence 2πrn = n λn or circumference of nth permitted orbit = n × de Broglie wavelength associated with the electron in the nth orbit.

17. For telescope, focal length & aperture of objective has to be maximum i.e., lens A and eyepiece has to smaller focal length and smaller aperture i.e., lens D. Then magnifying power will be maximum, i.e.

M.P. = ef

f0 = 5

100 = 20

and l = f0 + fe = 105 cm.

18. The interference pattern due to different component colours of white light makes interference pattern of different colours and overlap on each other, so the central fringe is white. As violet wavelength is minimum and red has maximum so violet fringes are closer & red are farther. And there are different lines of different colours.

19.

r θ

P

90°–θ r '

i

y

dy

At point P

dB = π

µ4

0 2)'()–90(sin

ryid θ°

= π

µ4

0 2)'r(cosidy θ

∴ y = r tanθ ∴ dy = r sec2 θdθ

and cosθ = 'r

r ∴ r ' = θcos

r = r secθ

∴ dB = π

µ4

0 2

2

)sec(cos)sec(

θθθθ

rdri =

ri

πµ4

0 cosθ dθ

∴ B = r4i0

πµ

θθ∫π=θ

π=θ

d2/

2/–

cos = ri

πµ2

0

20. (a) τ = NiAB sin θ here θ = 0 ∴ τ = 0

(b) F = evdB , vd = neA

i

21. (a) K = L

V0 = xteRrR

R++

× Lε

= 48020

20+

× 105 (r = 0)

= 2 × 10–2 V/m (b) ∴ V = Kl = 2 × 10–2 × 6 = 12 × 10–2 volt

22. ∴ R = Alρ

∴ If l = 1m, A = 1 m2

∴ ρ=R

S.I. unit, ρ = R l

A = ohm-metre

m

ne τ=ρ

2'

23. (i) A → Capacitive circuit B → Inductive only

(ii) For device A ; i = CX

V = V ω C

∴ I ∝ ω For device B :

i = LX

V = L

Vω

∴ I ∝ ω1


24. Output not symmetric for A,B = (0,1) and (1,0) Not gate in one input (i) has three zeroes NOR gate Thus

A B

Y

A B

Y

(ii) has three one's ⇒ OR gate Thus

25. λ = ph =

mvh

∴ λe = 631–

34–

103109106.6

×××× = 2.44 × 10–10 m

λball = 100103

106.62–

34–

××× = 2.2 × 10–34 m

λe = size of atom, λball << size of ball

26. ν = λ1 = R

21

22

1–1nn

∴ λ =

21

22

1–1

1

nn

R =

21

22

1–1970

nn

Å

Let us take n2 = 1 (Lyman series of hydrogen spectrum)

Here can take values )4/3(

Å970 , )9/8(

Å970 , )16/15(

Å970 …..

.........,1

Å970

(corresponding to n1 = 2, 3, 4, …….. ∞) ∴ Permitted values of λ are 1293.3 Å, 1091Å,

1034.6 Å, ………970 Å Let us next take n2 = 2 (Balmer series of hydrogen

spectrum) Here λ can take values

36/5Å970 ,

16/3Å970 ,

100/21Å970 , ………

4/1Å970

(Corresponding to n1 = 3, 4, 5 ……….∞ ) Possible values of λ are 6984 Å, 5173. 3Å, 4619 Å, …… 3880 Å Hence λ = 824 Å, 1120Å, 2504 Å, 6100 Å, of the given lines, cannot belong to the hydrogen atom spectrum.

27. Directions of different minima in the diffraction pattern are given by

θn = Dxn =

dnλ

or, xn = dDnλ

Width of secondary maximum,

β = xn – xn–1 = dnn )1–(– λD =

dDλ

width of central maximum,

β0 = 2x1 = dDλ2 = 2β

Thus the central maximum is twice as wide as any secondary maximum

Again, β ∝ d1

So, as slit width is increased, the secondary maxima get narrower.

28. (a) fr = LCπ2

1

(b) Z = 22 )–( CL XXR +

At resonance, XL = Xc

∴ RZ =

∴ i = ZV =

RV

(c) VR = irms R =

RV × R = V

VL = irms XL = RV × XL

VC = irms XC = RXV C

(d) The voltage across L,C,R are not in phase. OR

(i) p

s

NN

= 100, NP = 100 ⇒ Ns = 100 Np = 10,000

(ii) Vp = 220 Vp ip = 1100 220 × ip = 1100

ip = 22

110 = 5 Amp.

(iii) p

s

VV =

p

s

NN

= 100

Vs = 100 Vp = 100 × 220 = 22000 volt (iv) Vsis = Vpip 22000 is = 1100


∴ is = 220001100 =

201 Amp.

(v) Power in secondary Vsip = Vp is = 1100 W

29.

Ei K

E0 + – – +

Due to dielectric, electric field between plates decreased, so p.d. decreased, consequently capacitance increased

Net Electric field between plates E = E+ + E–

= K02ε

σ + K02ε

σ

= K0ε

σ

E = AKq

0ε

∴ dU∆ =

AKq

0∈

∴ Uq

∆ =

dKA0∈ ⇒ C =

dKA0∈

30. A

i δ1 δ2

e r1 r2

When light passes through a prism, it gets refracted twice from its two non-parallel refracting surfaces such that net deviation is given by sum of deviations produced by each surface, δ = δ1 + δ2 = (i – r1) + (e – r2) = i + e – (r1 + r2)

Aei –+=δ

At minimum deviation, r1 = r2 = r (say) & i = e so, A = r1 + r2 = 2r ⇒ r = A/2

δmin = 2i – A ⇒ i = 2

min A+δ

As, µ = rsinisin =

2/sin2

sin

A

Am

+δ

L

O v0

E

β

α

Fe

A''

B''F0

u0

A

B

D

B'

A'

f0

P

Magnifying power,

M = αβ

tantan =

DPBDBA

/''

''''

= AB

BA '''' = ''''''

BABA .

ABBA ''

= me m0 =

+

efD1 .

0

0

uv = –

0fL

+

efD1

CHEMISTRY

1. NCl5 is not found because nitrogen do not have vacant d-orbitals

2. Ores of aluminium are : (i) Bauxite ore : Al2O3. 2H2O. (ii) Diaspore : Al2O3. H2O

3. Enzymes are protein natured specific biocatalyst which increases the rates of reaction by lowering the energy of activation.

4. Coordination No. of B.C.C. = 8 H.C.P. = 12 C.C.P = 12 S.C. = 6

5. K0 = 0.25 hrM

x = K0t t = 30 min (a0 – at) = K0t a0 = 0.25 × 0.5 + 0.075 = 0.20 M

6. Ca+2 + 2e– → Ca 2mol 1 mol 1F → ½ mol = 20 g

7. In absorption association over the surface takes place

∴ It is an exothermic process.


8. CH3Cl, Na, Dry ether; Wurtz Fittig reaction

9. Carbon monoxide is poisonous as it combines with haemoglobin of blood forming carboxyhaemoglobin due to which deficiency of oxygen occurs in blood.

10. The principle oxidation state of lanthanides is +3. However, some lanthanides also show oxidation state of +2 and +4.

For example, Eu shows oxidation state of +2 and Cerium shows oxidation state of + 4 .

11. Bayer's process is used when bauxite ore contains ferric oxide as chief impurity.

The powdered ore is first roasted at low temperature to convert ferrous oxide into ferric oxide. It is then digested with a concentrated solution of sodium hydroxide.

The aluminium oxide dissolves in caustic soda (NaOH) forming soluble sodium metal aluminate (NaAlO2) while ferric oxide and silica remains insoluble and settle down. These are removed by filtration.

Al2O3 . 2H2O + 2NaOH → 2NaAlO2 + 3H2O (soluble)

The sodium metaaluminate solution is agitated and it undergoes hydrolysis with formation of Al(OH)3 as precipitate.

NaAlO2 + 2H2O → NaOH + Al(OH)3 (Precipitate)

The precipitate is washed and dried

12. (i)Colligative Properties : The properties which depends upon no. of particles but do not depends upon nature of particles are called colligative properties.

(ii) Reverse Osmosis : When we apply pressure greater than osmotic pressure on the conc. side of the two solution which are separated by semipermeable membrane. This results in movement of solvent molecules from high conc. to low conc.

use : purification of water

13. The minimum additional energy which is required by the reactant molecule to participate in the chemical rxn is called activation energy. Catalyst reduces the activation energy as it offer an additional path to the chemical rxn with rise in temperature the activation energy is not affected but more and more molecules will have that minimum energy which is required to participate in chemical rxn

14. log 1

2

KK =

REa

303.2

−

21

11TT

ln 2 = 30.8303.2 ×

aE

−

3101

3001

0.3010 = 314.8303.2 ×

aE

×30031010

Ea = 2.303 × 8.314 × 0.3010 × 31 × 300 = 53.598 kJ

15. 0Ag/AgE + = 0.80 V 0

Zn/Zn 2E + = 0.76 V

E0cell = 0

Ag/AgE + – 0Zn/Zn 2E +

= 0.8 – (–0.76) = 1.56 V

∆Gºr = –nF Eºcell = – 2 × 96500 × 1.56; = – 301.08 kJ

16.

H2 (5atm)

O2 (5atm)

NaOH + KOHPorous C-electrode

⊕ Cathode

Anode

Impregnatedwith Pt or Ni

Chemical rxn At anode

H2(g) + 2OH¯ → 2H2O(g) + 2e¯ At cathode

21 O2(g) + H2O(l) + 2e– → 2OH¯ (aq)

overall rxn H2(g) + 21 O2(g) → H2O(l)

17. )Z(56

)Y(

–256

)X(256 IHCClNHCNHHC →→ +

18. (i)

C2H5Br Mg

etherC2H5MgBr

O

C2H5 – CH2 – CH2MgBr

H2O / H+

CH3CH2CH2CH2OHButan-1-ol


(ii) CH3MgBr + CH3 – C – CH3 CH3 – C – CH3

O

CH3

OMgBr

H2O / H+

CH3 – C – CH3 + Mg

CH3

OH OH

Br

Methyl magnesium bromide Propanone

2-Methylpropan-2-ol

19. The preparation of K2Cr2O7 from chromite ore is given in following steps :

Step – I : Preparation of sodium chromite : 4FeCr2O4 + 16NaOH + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8H2O

Step – II : Conversion of sodium chromate into sodium dichromate : 2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

Step – III : Conversion of sodium dichromate into potassium dichromate.

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl on increasing pH value, dichromate ions (Cr2O7

2–) get converted into chromate ions (CrO4

2–).

20. (i) [CoCl2(en)2] Cl Dichloridobis (ethane –1, 2–diamine) cobalt (III)

chloride. (ii) Potassium tetrahydroxozincate (II) (iii) Tetraammine aqua chloridocobalt (III) chloride

21. Polymers are macro molecules with number of repeating units called monomers

Terylene

– O – CH2 – CH2 – O – C – – C –

O O

( )n

Nylon 6, 6 ( HN – (CH2)6 – NH – CO – (CH2)4 –

CO )n

22. Dissaccharide are sugar containing two monosaccharide unit (i) maltose = αD Glucose + αD Glucose

(ii) Sucrose = αD Glucose + βD Fructose (iii) Lactose = βD Galactose + βD glucose

23. Antibiotics are naturally produced chemical substances which kill or arrest the growth of bacteria

(i) Bactericidal – These kill bacteria Ex – Penicillin, Ofloxacin

(ii) Bacteriostatic – These stop the growth of bacteria

Ex – Chloremphenicol, Erythromycin

24. Edge of unit cell a = 288 × 10–10 cm. V = a3 = 23.9 × 10–24 cm3

d = AN

MaZ

×3

7.2 = AN××

×−24109.23522

NA = 2.7109.23

52224– ××

× = 6.04 × 1023

25. (a) 0OH2

p = 12.3 kPa

In 1 molal solution

m = Kg

B

wn

nB = 1 OH2n =

181000 = 55.5

OH2x =

15.555.55+

= 0.982

p = 0.982 × 12.3 = 12.08 kPa

v.p. of solution pA = 0.8 0Ap

(b) let mass of solute be W g

moles of solute = 40W

moles of octane n0 = 114114 = 1

xB = 140

W40

W

+

0App∆

= xB 0A

0A

0A

pp8.0–p

= 140

W40

W

+

⇒ W = 8.0402.0 × = 10 g

26. (a) The associative colloide is that colloide which is form due association of large no. of particles. These particles form true solution at lower conc. but as conc. became greater than C.M.C. (critical micelles conc.) their association results in the formation of colloidal solution Ex. Soap, detergents

(b) Hardy-Schulz rule states that for the coagulation of colloidal solution active ions are required and active ions are those ions which are having opposite charge more is the charge greater will be the coagulation tendency.


(c) Cellophane bag

H2O H2O

⊕

Dialysis is a phenomenon in which removal of dissolved impurities from the colloidal solution by means of diffussion through a suitable semipermeable membrane.

In this process colloidal solution is placed in cellophane bag and impurities get removed through small pores of the bag. To enhance the removel of electrolyte it is placed in electric field which is called electro – dialysis.

27. (a) The reactivity of aldehydes and ketones toward nucleophilic addition depends upon (i) + I effect (ii) steric hinderance. Hence the order is Di-tert butyl ketone < methyl-tert-butyl ketone < Acetone < Acetaldehyde

(b) The acidic strength depends upon (i) nature of + I effect (ii) nature of atom / group attached (iii) position of substituent on the chain. Hence, (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

(c) 4-Methoxy benzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid

28. Hydrolysis of trichlorosilanes gives cross – linked silicones.

Cl – Si – Cl + 3H2O –3HCl

HO – Si – OH

R

Cl

R

OH

n HO – Si – OH

R

OH

–(n–1) H2O

-- O – Si – O – Si – O --

R R

O O

-- O – Si – O – Si – O --

R R

Cross linked silicon 29. A is C6H5CONH2 ; B is C6H5CN; C is C6H5CH2OH

the sequence of reactions is C6H5CONH2

P2O5

–H2O C6H5CN

Red C6H5CH2NH2

Benzamide Benzonitrile Benzylamine

–N2 – H2O HNO2

C6H5CH2OHOxide [O]

C6H5COOH Benzoic acid Benzyl alcohol

30. Formic acid H – C – OH

O

contains both an

aldehyde – C = O

H

as well as carboxyl

group – C – OH

O

but acetic acid contain

only a carboxyl group. Formic acid behaves as reducing agent whereas acetic acid does not.

(a) Formic acid reduces Tollen's reagent to metallic silver but acetic acid does not.

HCOOH + 2[Ag(NH3)2]+ + 2(OH)¯ Tollen's reagent

2Ag ↓ + CO2 ↑ + 2H2O + 4NH3

Silver mirror

No silver mirror is formed with acetic acid. (b) Formic acid reduced Fehling solution to red ppt.

of Cu2O but acetic acid does not. HCOOH+2Cu2+ + 4(OH)¯ → Cu2O ↓ + CO2 ↑ + 3H2O Fehling solution Red ppt (cuprous oxide)

MATHEMATICS

Section A

1. tr (A) = a11+a22+a33 = 14 + (–5) + (–2) ⇒ = 14 – 7 = 7 2. x + 10 = 3x + 4 and y2 + 2y = 3 and y2 –5y = –4 ⇒ 2x = 6 ⇒ (y +3) (y – 1) = 0 y2 – 5y + 4 = 0

⇒ (y – 4) (y – 1) = 0

⇒ x = 3 ⇒ y = –3, 1 y = 1, 4

⇒ x = 3, y = 1 3. sin 10° cos 80° – sin 80° (–cos 10°) = sin 10° cos 80° + sin 80° cos 10° = sin (10 + 80)° = sin 90° = 1 4. Diff. w.r. to x

ex + ey

dxdy = ex+y

+

dxdy1

⇒ (ey – ex+y) dxdy = ex+y – ex

dxdy =

)1()1(

xy

yx

eeee

−−


5. f(f(x)) = 3)(22)(3

−−

xfxf =

332232

232233

−

−−

−

−−

xxxx

= 96466469

+−−+−−

xxxx = x

6. = ∫

−

2sin2

2cos

2sin21

2 x

xx

ex dx

= ∫

−

2xcot

2xcosec

21e 2x dx

= – III 2

xcosece21dx

2xcote 2xx∫ ∫+ dx

= –

−− ∫ dxe

21.

2xcosece.

2xcot x2x

+ 21

∫ dx2xcosece 2x

= –ex cot 2x –

21

∫ dx2xcosece 2x +

21

∫ dx2xcosece 2x + C

= – ex cot 2x + C

7. We are given that cos2 x dxdy + y = tan x

⇒ dxdy + (sec2x)y = tan x . sec2 x ….(i)

This is a linear differential equation of the form

dxdy + Py = Q, where P = sec2 x and Q = tan x sec2 x

∴ I = I.F. = ∫ dxxsec2

e = etanx

Multiplying both sides of (i) by I.F. = xe tan , we get

etan x

dxdy + sec2 x etanx. y = etanx. tan x sec2 x

Integrating both sides w.r.t. x, we get

yetan x = ∫ xtane . tan x sec2 x dx + C

[Using : y (I.F.)= dx.)F.I(Q∫ + C]

⇒ yetan x = III

,Cdttet +∫ where t = tan x

⇒ yetan x = tet – dtet∫ + C [Integrating by parts]

⇒ yetan x = tet – et + C

⇒ yetan x = etanx (tan x –1) + C, which is the required solution.

8. We have,

( ar

× br

)2 = | ar

× br

|2

⇒ ( ar

× br

)2 = | ar

| | br

| sin θ2

⇒ ( ar

× br

)2 = | ar

|2 | br

|2 sin2 θ

⇒ ( ar

× br

)2 = | ar

|2 | br

|2 (1 – cos2 θ)

⇒ ( ar

× br

)2 = | ar

|2 | br

|2 – | ar

|2 | br

|2 cos2θ

⇒ ( ar

× br

)2 = ( ar

. ar

) ( br

. br

) – ( ar

. br

) ( ar

. br

)

[Q ar

. br

= | ar

| | br

|cos θ]

⇒ ( ar

× br

)2 =bbbabaaarrrr

rrrr

..

..

9. Let ar

= i +2 j + 3 k and br

= 3 i –2 j + k . The vector

area of the parallelogram whose adjacent sides are

represented by the vectors ar

and br

is ar

× br

.

Now, ar

× br

= 123321

ˆˆˆ

−

kji

= (2 + 6) i – (1 – 9) j + (–2 – 6) k = 8 i +8 j –8 k

So, area of the parallelogram = | ar

× br

|

= 222 )8(88 −++

= 38 square units

10. We know that the angle θ between the line

rr

= ar

+ λ br

and the plane rr

. nr

= d is given by

sin θ = ||||

.nb

nbrr

rr

Here, br

= i – j + k and nr

= 2 i – j + k

∴ sin θ = 222222 1)1(21)1(1

)ˆˆˆ2().ˆˆˆ(

+−++−+

+−+− kjikji

= 63112 ++ =

234 =

322

⇒ θ = sin–1

322


Section B

11. R3 → R3 + R2, R1 → R1 + R2

∆ = cbcbcb

acbacbababa

+++−−++−

+−++

)(

)(

= (a + b) (b + c) 111

111

−−++−−

acbac

R1 → R1 + R3

= (a + b) (b + c) . 111

020

−−++− acbac

= (a + b) (b + c) (–2) 11ac

−−−

= (a + b) (b + c). 2 (c + a)

12. A → Integer chosen is divisible by 6 B → integer chosen is divisible by 8 n (A) = 33, n (B) = 25, n (A ∩ B) = 8, n(S) = 200 P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 20033 +

20025 –

2008 =

20050 =

41

OR Let E : Candidate Reaches late A1 = Candidate travels by bus A2 : Candidate travels by scooter A3 : Candidate travels by other modes of transport

P(A1) = 103 , P(A2) =

101 , P(A3) =

53

P(E/A1) = 41 , P(E/A2) =

31 , P(E/A3) = 0

∴ By Baye's Theorem P(A1/E) =

)A/E(P)A(P)A/E(P)A(P)A/E(P)A(P)A/E(P)A(P

332211

11

++

= 0

301

403

41

103

++

× =

139

13. Let y = tan–1

−+xx 11 2

z = tan–1 x

put x = tan θ dxdz = 21

1x+

y = tan–1

θθ−

sincos1

y = tan–1

θθ

θ

2cos

2sin2

2sin2 2

y = 21 tan–1 x

dxdy =

21 . 21

1x+

∴ dzdy = 2

2

1/11/12/1x

x+

+× = 21

14. dtdx = a

− 2

11t

dtdy = a

+ 2

11t

dxdy =

−

+

2

2

11

11

ta

ta

= )1()1(

2

2

−

+

tata =

−

+

ttat

ttat

1

1

dxdy =

yx

OR xpyp = (x + y)p + q Take log on both sides p log x + q log y = (p + q) log (x + y)

xp +

yq .

dxdy =

+

++

dxdy

yxqp 1

or xp –

yxqp

++ =

dxdy

−

++

yq

yxqp

or )( yxx

qxpxpypx+

−−+ =

+

−−+)( yxy

qyqxqypydxdy

or x

qxpy − =

−y

qxpydxdy

or xy =

dxdy

15. Let f(x) = logex, x ∈ [a, b] ↓ continuous & differentiable

∴ f ′(c) = ab

afbf−− )()(

c1 =

abab

−− loglog

Q a < c < b

⇒ b1 <

c1 <

a1

⇒ b1 <

abab

−− loglog <

a1

⇒ b

ab − < log b – log a < a

ab −


16. Q differentiable at x = c ⇒ continuous at x = c ⇒ f(c) = f(c+) ⇒ c2 =

0lim→h

a(c + h) + b

⇒ c2 = ac + b …(1) Now f ′(c+) = f ′ (c–)

⇒ 0

lim→h h

cbhca 2])([ −++

= 0

lim→h h

chc−

−− 22 ])[(

⇒ 0

lim→h h

cahbac 2)( −++

= 0

lim→h h

chh−− 22

⇒ 0

lim→h h

cahc 22 −+ = 0

lim→h

(2c – h) [From (1)]

a = 2c …(2) from (1), (2) a = 2c, b = – c2

17. f(1) = 2

11+ = 1

f(2) = 22 = 1

many-one function If n → odd natural number then 2n –1 is also odd

number

f(2n –1) = 2

112 +−n = n

If n → even natural number then 2n is also an even natural number

f(2n) = 2

2n = n

⇒ f is onto function.

18. We have,

I = ∫ xx

4sinsin dx = ∫ xx

x2cos2sin2

sin dx

= ∫ xxxx

2coscossin4sin dx

⇒ I = 41 ∫ xx 2coscos

1 dx = 41

∫ xxx

2coscoscos2 dx

⇒ I = 41

∫ −− )sin21()sin1(cos

22 xxx dx

Putting sin x = t and cos x dx = dt, we get

I =41

∫ −− )21()1( 22 ttdt

Let t2 = y. Then,

)21()1(

122 tt −−

= )21()1(

1yy −−

Let )21()1(

1yy −−

= y

A−1

+ y

B21−

. Then,

1 = A (1 –2y) + B (1 – y) ….(i)

Putting y = 1 and y = 21 respectively in (i), we get A

= –1 and B = 2

∴ )21()1(

1yy −−

= y−

−1

1 + y21

2−

⇒ )21()1(

122 tt −−

= – 211t−

+ 2212

t−

⇒ I = 41

∫ −− )21()1( 22 ttdt

= 41

∫

−− 21

1t

+

− 2212

tdt

⇒ I = –41

∫ − 211t

dt + 42

∫ − 2)2(11

tdt

⇒ I = –41 .

21 log

tt

−+

11

+21 .

221 log

tt

2121

−

+ + C

⇒ I = – 81 log

tt

−+

11 +

241 log

tt

2121

−

+ + C

⇒ I = –81 log

xx

sin1sin1

−+ +

241 log

xx

sin21sin21

−

+ + C

19. We have, I = ∫ θ+θ cottan dθ

⇒ I = ∫

θ+θ

tan1tan dθ

⇒ I = θθ

+θ∫ d

tan1tan

Let tan θ = x2. Then, d (tan θ) = d(x2) ⇒ sec2θ dθ = 2x dx

⇒ dθ = θ2sec

2 dxx = θ+ 2tan1

2 dxx = 412

xdxx

+

∴ I = ∫+2

2 1

x

x . 412

xdxx

+= 2 ∫ +

+11

4

2

xx dx

= 2 ∫ ++

22

2

/1/11

xxx dx

⇒ I = 2 dxxx

x∫ +−

+2)/1(

/112

2

2

1 1


= 2 dxxx

x∫ +−

+22

2

)2()/1(/11

⇒ I = 2 ∫ + 22 )2(udu =

22 tan–1

2u + C,

where x – x1 = u

⇒ I = 2 tan–1

−

2/1 xx + C

⇒ I = 2 tan–1

−

xx

212

+ C

⇒ I = 2 tan–1

θ

−θ

tan21tan + C

OR

∫ −+++

)5()3()4)(1(

22

22

xxxx dx

Consider

)5()3()4)(1(

22

22

−+++

xxxx =

)5)(3()4)(1(

−+++

tttt where t = x2

= 1 + )5)(3(

197−+

+tt

t

Consider

)5)(3(

197−+

+tt

t = 3+t

A + 5−t

B

A = 41 , B =

427

∴ ∫ −+++

)5()3()4)(1(

22

22

xxxx dx

= ∫dx + ∫ ∫ −+

+ 5427

341

22 xdx

xdx

= x +34

1 tan–1

3x +

5827 log

55

+

−

xx + c

20. Let P(x, y) be any point on the curve. The equation of

the normal at P (x, y) to the given curve is

Y – y = –dx/dy

1 (X –x) … (i)

It is given that the normal at each point passes through (2, 0). Therefore, (i) also passes through (2, 0). Putting Y = 0 and x = 2 in (i), we get

0 – y = – dxdy /

1 (2 –x)

⇒ ydxdy = 2 – x

⇒ ydy = (2 – x) dx [On integrating both sides]

⇒ 2

2y = – 2

)2( 2x− + C

⇒ y2 = – (2 – x)2 + 2C … (ii) This passes through (2, 3). Therefore,

9 = 0 + 2C ⇒ C = 29

Putting C = 29 in (ii), we get

y2 = – (2 – x)2 + 9 This is the equation of required curve. 21. We have, (2 i + 6 j + 27 k ) × ( i + λ j + µ k ) = 0

r

⇒ µλ12762kji

= 0r

⇒ (6µ –27λ) i – (2µ –27) j + (2λ – 6) k = 0r

⇒ 6µ –27λ = 0, 2µ –27 = 0 and 2λ – 6 = 0

⇒ λ = 3 and µ = 2

27

22. The equation of a plane passing through the

intersection of the given planes is (4x – y + z –10) + λ(x + y – z –4) = 0 ⇒ x(4 + λ) + y (λ –1) + z (1 –λ) –10 – 4λ = 0 This plane is parallel to the line with direction ratios

proportional to 2, 1,1 ∴ 2(4 + λ) + 1(λ –1) + 1(1 –λ) = 0 ⇒ λ = – 4 Putting λ = – 4 in (i), we obtain 5y – 5z – 6 = 0 This is the equation of the required plane. Now, length of the perpendicular from (1, 1, 1) on (ii)

is given by

d = 22 )5(5

61515

−+

−×−× = 5

23

OR Given line

5

1−x = 2

3 y− = 4

1+z

or, 5

1−x = 23

−−y =

4)1(−−z ....(i)

is passing through (1, 3, –1) and has D.R. 5, –2, 4. Equations of line passing through (3, 0, –4) and

parallel to given line is

5

3−x = 20

−−y =

44+z ...(ii)

Vector equations of line (i) & (ii)

→r = i + 3 j – k + λ(5 i – 2 j + 4 k )


→r = 3 i – 4 j + µ (5 i – 2 j + 4 k )

∴ →

2a – →

1a = 2 i – 3 j – 3 k

→b = 222 )4()2()5( +−+

= 45 = 53

Also →b ×

−

→→

12 aa = 332

425

ˆˆˆ

−−−

kji

= 18 i + 23 j – 11 k

∴

−×

→→→

12 aab = 222 )11()23()18( ++ = 974

∴ Distance between two parallel lines.

= →

→→→

−×

b

aab 12

= 45

974 units

Section C

23. We have,

∫b

a

dxxf )(

= 0

lim→h

h[f(a) + f(a + h) + f(a+2h) +….+

f (a + (n –1)h)]

=0

lim→h

h

−+

2sin

2sin

2)1(sin

h

nhhna

=0

lim→h

h

−+

2sin

2sin

22sin

h

nhhnha

=0

lim→h

h

−

−

−+

2sin

2sin

22sin

h

abhaba

[Q nh = b –a]

=0

lim→h

−

−

+×

2sin

22sin2

2sin

2 abhbah

h

=0

lim→h

2sin

2h

h

.0

lim→h

2 sin

−

+22hba sin

−

2ab

= 2 sin

+

2ba sin

−

2ab

= cos a –cos b [Q 2 sin A sin B = cos (A –B) – cos (A + B)] 24. A → getting a white ball from 1st bag. B → getting a black ball from 1st bag

C → getting a white ball from 2nd bag D → getting a black ball from 2nd bag

P(A) = 64 , P(B) =

62 , P(C) =

83 , P(D) =

85

(A) P (both are white) = P(A). P(C) = 41

(B) P (one is white and one is black)

= P(A). P(D) + P(B). P(D) = 2413

25. Given curves are x2 + y2 = 16 and x2 = 6y Solving these two equations

y

xO(0, 0)

D(0, 2) B(0, 4)

A(2 3 , 2) (–2 3 , 2)C

x2 +36

2x = 16

x4 + 36x2 –576 = 0 (x2 + 48) (x2 –12) = 0 x2 –12 = 0

x = ± 32 ∴ y = 2 ∴ Required area = 2(area of shaded portion) Reqd. area = 2[area of OADO + area of DABD]

= 2

−+∫ ∫

2

0

4

2

2166 dyydyy

= 2

+−+ −

4

2

1220

2/3

4sin

21616

21)(6

32 yyyy

=

π+

316

334 sq. units

.


26. The given lines are r

r= ( i + 2 j + 3 k ) + λ (2 i + 3 j + 4 k ) …(i)

and, rr

=(2 i + 4 j + 5 k ) + 2µ (2 i + 3 j + 4 k ) …(ii) Equation (ii) can re-written as

rr

= (2 i + 4 j + 5 k )+ µ′(2 i + 3 j + 4 k ) …(iii)

where µ′ = 2µ These two lines passes through the points having

position vectors 1ar

= i + 2 j + 3 k & 2ar

= 2 i +4 j +5 k respectively and both are parallel to the vector br

=2 i + 3 j + 4 k .

∴ shortest distance = ||

|)(| 12

bbaa

r

rrr×−

…(iv)

We have,

( 2ar

– 1ar

) × br

= ( i + 2 j + 2 k ) × (2 i + 3 j + 4 k )

⇒ ( 2ar

– 1ar

) × br

= 432221

ˆˆˆ kji

= (8 – 6) i – (4 – 4) j + (3 – 4) k = 2 i – 0 j – k

⇒ |( 2ar

– 1ar

)× br

| = 104 ++ = 5 and

| br

| = 1694 ++ = 29

Substituting the values of |( 2ar

– 1ar

) × br

| and | br

| in

(iv), we get shortest distance = 295

27. The given data may be put in the following tabular form:

Refinery High grade Medium grade Low grade Cost per day

A 100 300 200 Rs.400

B 200 400 100 Rs.300

Minimumrequirement 12,000 20,000 15,000

Suppose refineries A and B should run for x and y

days respectively to minimize the total cost. The mathematical form of the above LPP is Minimize Z = 400x + 300y Subject to 100x + 200y ≥ 12,000 300x + 400y ≥ 20,000 200x + 100y ≥ 15,000 and, x, y ≥ 0 The feasible region of the above LPP is represented by

the shaded region in fig. The corner points of the feasible region are

A2 (120, 0), P(60, 30) and B3(0, 150). The value of the objective function at these points are given in the following table:

Points (x, y) Value of the objective function Z = 400x + 300y A2 (120, 0) Z = 400 × 120 + 300 × 0 = 48,000 P (60, 30) Z = 400 × 60 + 300 × 30 = 33, 000 B3(0, 150) Z = 400 × 0 + 300 × 150 = 45,000

Clearly, Z is minimum when x = 60, y = 30. Hence the machine A should run for 60 days and the machine B should run for 30 days to minimize the cost while satisfying the constraints.

• O • •

•

• A3(75,0)

A1

0,

3200

100x + 200y = 12000 X

B1(0,50)

300x + 400y = 20000

P(60,30)

B2(0,60)

A2(120,0)

22x + 100y = 15000

B3(0,150)

y

28. A11 = 3211

−−

= –3 + 2 = –1;

A12 = –3112

= – (6 –1) = –5;

A13 = 2112

−−

= –4 + 1 = –3;

A21 = – 3211

−= – (3 + 2) = –5

A22 = 3111

= 3 – 1 = 2;

A23 = – 21

11−

= – (–2 –1) = 3

A31 = 1111

−= 1 + 1 = 2;

A32 = – 1211

= – (1 –2) = 1

A33 = 12

11−

= –1 –2 = –3

∴ adj A = ′

−−

−−−

312325351

=

−−−

−−

333125251

Also |A| = 1 (–1) + 1 (–5) + 1(–3) = –1 –5 – 3 = – 9 ≠ 0 ∴ A–1 exists.

A–1 = |A|Aadj = –

91

−−−

−−

333125251

The given system of equations can be written as


AX = B where X =

zyx

, B =

223

∴ X = A–1 B

i.e.

zyx

= –91

−−−

−−

333125251

223

= –91

−+−++−+−−

66924154103

=

111

Thus x = 1, y = 1, z = 1. OR

A11 = 73

13−

−= 21 – 3 = 18

A12 = – 72

11−

= – (–7 –2) = 9

A13 = 3231

−−

= 3 + 6 = 9

A21 = – 73115

−−−

= – (35 + 33) = – 68

A22 = 72114

−−

= –28 + 22 = – 6,

A23 = – 3254 −

= –(12 + 10) = –22

A31 = 13115

−−−

= – 5 – 33 = – 38

A32 = – 11114 −

= – (4 + 11) = –15

A33 = 3154

−−

= –12 + 5 = – 7

∴ adj A = ′

−−−−−−

71538226689918

=

−−−−−−

72291569386818

Also |A| = 4(18) + (–5) 9 + (–11)9 = 72 – 45 – 99 = 72 – 144 = –72 ≠ 0 ∴ A–1 exists.

∴ A–1 = |A|Aadj

= – 721

−−−−−−

72291569386818

Further the given system of equations can be written as

AX = B where X =

zyx

, B =

21

12

∴ X = A–1 B

i.e.

zyx

= –721

−−−−−−

72291569386818

21

12

= –721

−−−−−−

14221083061087668216

= –721

727272

=

−−−

111

∴ x = –1, y = –1, z = –1 29. 1st part OC = X

AC = 22 XR − DC = R + X Volume

V=31

π(R2–X2) (R + X)

dXdV =

3π [R2 –2RX –3X2] = 0

⇒ (R – 3X) (R +X) = 0 ⇒ X = R/3 (Q R + X ≠ 0)

2

2

dXVd = π/3 (–2R –6X)

3/RX

2

2

dXVd

=

= – 34

πR < 0 ⇒ maximum volume

Put X = R/3 in V, we have V = 8/27 (volume of sphere) OR f ′(x) = cos x – sin 2x = 0 = cos x [1 –2 sin x] = 0 ⇒ x = π/2, π/6 Now f ′′(x) = – sin x –2 cos 2x f ′′(x)|x= π/2 = –1 + 2 = 1 > 0 minimum

f ′′(x)|x = π/4 = – 21 – 2

21 = –

23 < 0

maximum

∴ f(0) =21

f(π/2) = 1 –21 =

21 ⇒ minimum value =

21

f(π/6) =21 +

21 ×

21 maximum value =

43

= 43

X

O R

R R

C A B

D

•


ADAPTIVE TESTING (An initiative by a2zExam.com)

What is Adaptive Testing? Adaptive testing in the tertiary environment also has a long history as it has existed since the time of the oral examination, and responding adaptively to students. This was totally lost in the “paper based fixed test for everyone”, because it was not feasible to conduct oral test of each and every student in such a short time period. But due to the development in technology, it is again possible to have adaptive testing for everyone. Computer Based Adaptive Testing is the way going forward, as it gives the benefits of Adaptive Testing and also do not require too much resource.

Adaptive Testing Process: • Computer Based-adaptive test (CAT) is a form of assessment where the level of the questions administered to

individual test-takers is dynamically tailored to their skill and knowledge levels.

• First set of Papers are fixed for each part of syllabus, student take this test and system judges the level of the student based on their performance in this papers

• Next papers onward system creates the paper especially for the student based on the student’s performance in previous papers.

• Thus, each paper adapted the students level and thus is known as adaptive testing

It is important to differentiate between online assessment tools, those readily delivered through learning management systems, although having the advantages of collation and immediate results are fixed rather than dynamic, as they are not adaptive.


Performance Analysis by Feedback Report:

• Analyze the bigger picture then go deep to find exact cause

Overall Paper’s Danger Zone Analysis: Danger zone analysis tells where the student lies out of three zone viz. Danger, Normal and Safe Zone represented by a white circle. Overall Paper’s Danger Zone analysis tells where the student lies in the distribution graph as per total marks obtained.

Student or circle lies in “Red” area means student is in Danger Zone and needs to work very hard to achieve the goal of selection

Student or circle lies in “Blue” area means student is in Normal Zone and should regularize and do systematic study to achieve the goal

Student or circle lies in “Green” area means student is in Safe Zone but needs to keep improving to achieve the goal

Subject Wise Comparative Analysis: This analysis compares student’s subject wise marks with highest and

average marks in the subject. It also tells percentage of question attempted by student correctly, incorrectly & percentage of question student did not attempted.

Graph represents Student’s, Highest & Average Marks in each subject

Graph implies that student is doing much better in Physics than Maths & Chemistry

Table shows the Attempt Status & Marks of each subject in the test paper.

Table shows that in Chemistry, student has attempted very few questions, which implies either he was short of time or he doesn’t knew how to answer


Similarly, for Maths student has very high (%) of Incorrect Question, which implies either student, has misconception or has made lot of silly mistakes.

• Help you know your learning gaps & how to improve or fill them

Skill-Wise Personal Analysis: Personal analysis is done to find out the skill wise performance. Whether student is able to Direct Theory or Formula based Questions, whether he has good concepts or problem solving skills, etc.

Inferences and Suggestions for Improvements: Based on the observation made above we deduct inferences and suggest methods for improvement

Observation Inference Suggestion For Improvement

You have got only 50 % of Easy and Direct Questions Correct.

You are expected to grab all the Easy and Direct problems

Your real problem lies in your unsystematic self study, Revise lectures on day to day basis and plan mega revision (1hrs) of theory on weekends.

• Help you focus on specific weak areas

Topic Wise Danger Zone Analysis: For each topic the analysis is done and suggestion based on student’s zone is given in next column. Student should figure out the weak topics and should work according to the suggestion.

The circle o in the graph represents student’s position

Question wise Detailed Analysis: Each question is analyzed based on the student’s attempt status and compared with overall attempt status.

Question wise analysis table shows Knowledge Area & Skills which the question belongs, along with

Students attempt status

(%) of student attempted the question

(%) of students among attempted which does it correctly Each paper has some tricky questions, which most people attempt, but does it wrong. Understanding how to solve those questions, improve student’s concept & learning on those topic. But the important thing is how to figure out those questions. Each paper has hard questions, which most people has left and if you have attempted you might have taken a lot of time; this type of question should be tried in the end.


The above table helps you in recognizing those questions and if you keep this in mind you will be able to save time and marks by not attempting questions which are meant to be left.

Benefits of Adaptive Testing

• Adaptive test encourages student to bring out their maximum output by providing them the questions with levels close to their skills and knowledge level, instead of very easy or hard which most of the time de-motivates student from attempting.

• Adaptive testing helps evaluator to measure the accurate skill level of students, even with small number of questions.

Better Analysis of Students Performance can be judged and provided to students using adaptive testing

• Adaptive Testing targets the student weak areas and motivates them to improve on those areas by going to their level and upgrading as students improve.

This also brings students focus on those areas where they need improvement

• The experience of taking an adaptive test is like participating in a high-jump event. The high-jumper, regardless of ability, quickly reaches a challenging level where there is about an equal chance of clearing the bar or knocking it down. The ‘score’ is the last height that was successfully cleared and is earned without having to jump all possible lower heights or trying the higher levels.

Research shows that adaptive testing has improved the students learning by more than 22% compare to student taking fixed test during their preparation.

How you can get Adaptive Testing for your preparation of Entrance Exams like IIT-JEE, AIEEE, BITSAT etc.

• Visit a2zExam.com and Register.

• Select the Course according to exam you are taking and get the Adaptive Set of papers for the preparation of the exam.

• Everything will be done by them

a2zExam - Adaptive Testing with added advantage

First time in India, a2zExam.com brings “Adaptive Testing” for the preparation of Engineering Entrance Exams like IIT-JEE, AIEEE, BIT-SAT etc. a2zExam not just provide adaptive testing but also provide Feedback Report, which is used by the system to adapt to your level for testing, so that you can work on your weak areas before taking the next test.

Adaptive testing is like High Jump but if every jump of high jumper (participant) is recorded and shown before the next jump pointing out the mistakes or good things done by them. Think how much beneficial it will be for them and the improvement in their records.

Similarly, after every test you will be provided here with the feedback report pointing your strength and weakness along with the suggestions for improvement, so that you can work on those areas before taking the next test and improve your performance.


• Adaptive Testing

Get Adaptive Testing for preparation of IIT-JEE, AIEEE, BITSAT etc.

Give edge to your preparation and improve your ranking or chance of selection

Don’t leave any stone unturned for such an important event of your or your child’s life.

• Detailed Performance Analysis

Subject Wise, Topic Wise Personal and Comparative Analysis

Skill Wise Personal Analysis

Inferences and Suggestions for Improvement

Question Selection Strategy

• Anytime Anywhere Access

You can take the test anytime (Day or Night), anywhere (from Home or Cyber Café or any place with Computer and Internet)

As soon as the syllabus for a test is completed take the test, no need to wait for the specific Sunday for test to be conducted.


PHYSICS

1. Point to point : Communication over a link between a single transmitter and received Example : Telephone

3. (i) When final image is formed at infinity

M = – 0

0

uv ×

euD

(ii) When final image is formed at distance D

M = –0

0

uv

+

fD1

4. S1 and S2 are the two desired surfaces.

q

S1 S2

5. Superconductors are those material which resistivity is zero below a certain temperature.

6.

~

Induction coil

Conducting Plate I Sphere I Sphere II Conducting Plate II

Detector

7. Zero

8. High energy X-rays are known as hard X-rays and low energy X-rays are known as soft X-rays. These terms are relative.

9. (1, 3) – (2, 4)

10. The daughter element (release of energy is accompanied by an increase of

B.E)

11. (i) ‘Depletion layer’ width decreases, (ii) Junction field becomes very high

12. The approximate thickness of the film should be of the order of wavelength of the light.

13.

IntensityIO

θ '2θ θ2

'1θ

θ1

yn

'1P

P1

P2 '2P

14. d

A0∈ = 5 µF ………. (i)

dKA

20∈ = 20 µF ……. (ii)

(ii) ÷ (i) 2K =

520

∴ 8=K

15.

Uniform magnetic field

Paramagnetic substance

Diamagnetic substance

16. (i) λ = νc

(ii) (Uav)E = 41 ∈0 2

0E

MOCK TEST-3 (SOLUTION) MOCK TEST– 3 PUBLISHED IN SAME ISSUE


17. U = 21 Li2 =

21 × 2 × 10–3 (5)2 J

= 2.5 × 10–2 J

18. XL = ωL = 100 × 30 × 10–3 = 3 Ω R = 4 Ω

V = 200 2 sin 100 t

Z = 2L

2 XR +

= 22 34 +

= 5 Ω

∴ cosφ = ZR =

54 = 0.8

19. (i) X-rays are e.m. waves (ii) X-rays are transverse in nature

20. Reduction factor = 161 = 42

1 in 4 days.

Hence life = 1 day

∴ For 6 days reduction factor would be 621 =

641

∴ original amount = 4 × 10–3 × 64kg = 0.256 kg

21. A telescope views large objects at large distances; a microscope views small objects at small distances. Both need a small field of view. A camera views objects of ordinary sizes at fairly close distances. Here the field of view is required much more (compare 45° for a camera with about 1° for a microscope objective and something similar for a telescope, a moon subtends about 0.5° at the earth)

Thus rays entering a camera lens are far from being paraxial and aberrations will be large and images will be blurred if the apertures are not very small. For a telescope, on the other hand, the important thing is its ability to resolve distant abjects (i.e., see them as distinct). We have seen that the resolving power increases with increase in aperture. Therefore, telescopes have as large an aperture as feasible.

22. Q Current in 5 Ω is zero ∴ bridge is balance

∴

RR

+661 =

2010

or RR

66+ =

21

∴ R = 3 Ω

23. |ε| = L dtdi =

dtdL (t2) = 2 L t

∴ At t = 4, ε = 2 × 5 × 10–3 × 4 = 40 × 10–3 V

24.

P Q B

S R a

b

Let i → current N → No. of turns

l → length

n = l

N

Applying Ampere's Law along PQRS,

→→

∫ ldB . = µ0 inet

∫→→

Q

P

dB l. + ∫→→

R

Q

dB l. + ∫→→

S

R

dB l. + ∫→→

P

S

dB l. = µ0 inet

B.a + 0 + 0 + 0 = µ0 n (a) i

∴ niB 0µ=

25.

Ig

G

I2 – Ig Q

C AP

SI1

R

DI1 + Ig

–

+

B

I

I2

+

–

+

–

+

– –

+

It is an arrangement of four resistances used for measuring unknown resistance.

Applying KVL in loop ADBA –I1 R + Ig G + I2 P = 0 …. (i) and in loop DCBD, – (I1 + Ig) S + (I2 – Ig) Q – Ig G = 0 for balance bridge, Ig = 0 ∴ I1 R = I2P …….. (iii) I1 S = I2 Q …….. (iv)

(iii) ÷ (iv) we get SR

QP

=


26. Diamagnetic substances are feebly magnetised in opposite direction to that of magnetising field

Paramagnetic substances are feebly magnetised in the direction of magnetic field. Ferromagnetic substances are strongly magnetised in the direction of magnetic field.

27. V + –

P N + –

Reverse biasing

In the experimental set, the P and N terminals of a P-N diode are connected to the negative and positive potential point's of a potential divider respectively. As the reverse bias current a feeble current measurable in micro amperes so a micro-ammeter is used to measure it. To plot reverse bias characteristic, we note down reverse currents corresponding to various different reverse voltages on the diode with help of the potential divider. After obtaining it, the applied voltages are plotted along X-axis and corresponding reverse currents along Y-axis of a graph as shown in the fig.

I (mA)

+ V

–

Characteristic of a P-N diode Reverse bias resistance- The ratio of small change

in reverse voltage (before break down voltage) to the corresponding change in reverse current for a P-N diode is known as its reverse bias resistance, i.e, Reverse bias resistance

=

currentreverseinchangeingCorrespond

voltagereverseinchangeSmall

28. Circuit diagram for drawing the input and output characteristics.

R1

Vcc VCE + –

Ic

– + mA C

E IB

IB

+ – µA VBE VBB

R2 B

Typical shape of the input characteristics.

29.

α βα Q'

P' B

fe

E Q''

P''

O

Astronomical telescope (i) When the final image is formed on the nearest

distance of clear vision D

M = – ef

f0

+Dfe1

(ii) When the final image is formed at infinity

M = – ef

f0

On increasing the aperture of the objective lens the magnifying power of telescope will increase.

30. U = ∫qVdq

0

= ∫

q

Cq

0 dq

= C

q2

2

= 21 CV2

= 21

∈d

A0 (Ed)2

= 21 ∈0E2 (Ad)

∴ Energy density, AdU =

21 ∈0E2

OR (i) From Gauss' theorem

φ = 0∈inq

∴ qin = φ × ∈0 = – 6 × 103 × 8.85 × 10–12 = –5.31 × 10–8C (ii) Flux remains the same

+ –

d

E


CHEMISTRY

1. H.C.P has highest 74% efficiency

2. When in Fe(OH)3 ppt FeCl3 is added Fe+3 ions are adsorb over the surface of Fe(OH)3 it results in the formation of Fe(OH)3 solution.

3. Lithium tetrahydrido aluminate (III)

4. FeCO3 is siderite ore.

5. Phenol & Formaldehyde

6. Antiseptics are germicides which can be applied on wounds

Ex. Soframycin, Tincture iodine

7. Amino acid in which amino group are more then – COOH group are called basic amino acid.

Lysine ( R is (CH2)3 – NH2 )

8. CH3F < CH3Cl < CH3Br < CH3I

9. t1/2 = 5730

λ = 2/1

693.0t

= 1.21 × 10–4 yr–1

λ = t303.2 ln

t

0

aa

t = 41021.1303.2

−× ln

810

= 1845 yr

10. With the increase in temperature rate constant increases. It is found that with 10 K rise in temperature the rate of reaction become 2 – 3 times.

With the increase in temperature (1) More no. of collisions occur between the

molecules. (2)Only those molecules which are having

minimum sufficient energy to participate in the chemical rxn, reacts with each other and form product

(3) For effective collision activated molecule must collide in the proper orientation

∴ Rate of rxn = P × Z . RTEae /− P = Orientation factor Z = No. of collisions

RTEae /− = No. of activated molecules

11. (a) Standard Hydrogen electrode - When H2 gas at 1 atm pr is supplied on Pt sheet dipped in the aqueous solution of an acid having molarity 1M

H2 l atm

Pt

1 M aq acidsolution

Following Chemical rxn takes place H2(g, l atm) 2H+ (aq, 1M) + 2e–

the potential of this half electrode = 0.0 V (b) Kohlrausch's law states conductivity of a

solution at infinite dilution is equal to sum of molar conductivity of all the ions present in the solution.

0mΛ = γ⊕ λ⊕º + γ –º λ–º

12. Actinides show much higher oxidation states than Lanthanides because energy difference between 5f, 6d and 7s orbitals is less and hence electrons also participate from 5f orbital also.

13. The seperation of Ag+ and Hg22+ in group – I is

carried out by dissolving the precipitate of AgCl in NH4OH, AgCl forms a soluble complex with NH4OH. Whereas Hg2Cl2 forms a black water insoluble complex.

⇒ AgCl + NH4OH → OH2Cl])NH(Ag[ 223 +

(water soluble) ⇒ Hg2Cl2 + NH4OH → Hg(NH2)Cl + Hg + HCl black percipitate + H2O

14. The concentrated ore is heated with excess of air to remove water and carbon dioxide to remove sulphur and arsenic impurities and to oxidise ferrous to ferric oxide for eg;

2Fe2O3.3H2O → 2Fe2O3 + 3H2O FeCO3 → FeO + CO2 ↑ S + O2 → SO2 ↑ 4As + 3O2 → 2As2O3 4Fe + O2 → 2Fe2O3

15. PHBV has 3hydroxybutanoic acid & 3 hydroxypentanoic acid.

O – CH – CH2 – CO – O – CH – CH2 – CO

CH3 CH2 – CH3

n


16. Purine bases in DNA and RNA are Adenine & Guanine. Pyrimidine bases in DNA are Cytosine & Thymine while in RNA are Cytosine & Uracil

17. Detergents are sodium or potassium salts of sulphonic acid.

Cationic detergent – Cetyl trimethyl ammonium bromide

Anionic – Alkyl benzene sulphonate

18. (A) CHCl3 (B) HC ≡ CH

19. (a) Radius of gold r = 0.144 nm

F.C.C. 4 r = 2 a

a = 2r4 = 2 2 r

= 2 × 1.414 × 0.144 = 0.407 nm Edge length a = 0.407 nm. (b) (i) When

)p4s4( 22Ge is doped with

)p5s5( 12In 13th gr

element all the 3e– get bonded and fourth bond of Ge contain only one e– and hence an e– deficient bond or a hole is formed and p type semiconductor is formed.

(ii) Similarly when Si is doped with As 4s24p3 4e–'s get bonded and fifth e– remain unbonded ∴ n-type semiconductor is formed.

20. p = A0A xp + B

0Bxp

nA = 1 nB = 2 mol.

1Tp = 250 bar

250 = 0Ap ×

31 + 0

Bp ×32 ..... (1)

nA = 2 nB = 2 mol

2Tp = 300 = 0

Ap . 21 + 0

Bp × 21 ......(2)

solving (1) & (2)

0Ap = 450 bar 0

Bp = 150 bar

21. (a) (i) When silica gel is placed is atmosphere saturated with water adsorption of moisture takes place

(ii) CaCl2 adsorbs H2O. (b) Zeolite is a shape selective catalyst which are

metal alumino silicates Mx/n(AlO2)x(SiO2)y.mH2O. When zeolite is heated pores are generated, these pores are having size 260 – 740 pm which can absorb molecule of definite size. Therefore it is called shape selective catalyst

22. In case of nitrogen family, basic character of hydrides decreases from NH3 to BiH3 because with the increase in size of central element lone pair density about it decreases and tendency of proton (H+) to coordinate with it decreases and hence basic character of hydrides decreases.

23.

O

Cr

O

O¯

OCr

O

O¯O

131º

Cr2O72–

24.

25. 2-Methoxy-2-methylpropane

26.

27. The compound A can be either an aldehyde or a ketone. Since it resists oxidation it must be a ketone. i.e., acetone (CH3COCH3)

The reactions involved are :

CH3 – C – CH3Reduction

O

2[H]CH3 – CH – CH3

OH

(B) 2-Propanol

HBr–H2O

CH3 – CH – CH3

Br

(C)2-Bromopropane

MgCH3 – C – CH3 + CH3 – CH – CH3

MgBr O

(B)

(A)

CH3 – C – CH3

OMgBr

CH(CH3)2

H2OCH3 – C – CH3

OH

CH(CH3)2

2, 3 – Dimethyl-2-butanol

28. (a) The ideal solution is that solution which follows the Raoult's law i.e for ideal solution :-

(i) ∆Hmixing = 0 (ii) ∆Vmixing = 0 Non ideal solution is that solution which does not

follow Raoult's law and for which : (i) ∆Hmixing ≠ 0 (ii) ∆Vmixing = 0 In case of cyclohexane – ethanol a solution with

+ve deviation is obtained In case of a acetone – Chloroform


–ve deviation is obtained

Cl

Cl

Cl

C H O C

CH3

CH3

+δ –δ +δ

(b) Moles of solute nB = M30

Moles of H2O ( OH2n ) =

1890 = 5

xA = M/305

5+

= M6

M+

pA = A0Axp

pA = 0Ap .

M6M+

...... (1)

When 18g of H2O is added to solution

OH2x =

M/3066

+ =

M5M+

p'A = 0Ap

M5M+

...... (2)

from (1) & (2)

M6M5

++ =

9.28.2

⇒ M = 23 g/mol

29. (a) Hypophosphorus acid

H OH

P

O

H (b) Pyrophosphoric acid

HO O

P

O

OH OH

P

O

OH (c) Dithionic acid

HO — S — S — OH

O O

O O (d) Marshall acid

HO — S — O — O — S — OH

O O

O O

(e) Hypophosphoric acid

HO

P

O

OHOH

P

O

OH

30. (a) Since compound (A) is optically inactive and contains nitrogen which gives alcohol with HNO2, it is primary amine. The reactions may be given as

CH3CH2CH2CH2NH2 → 2HNO CH3CH2CH2CH2OH+N2+H2

O (A) (B) 1-Aminobutane 1-Butanol

CH3CH2CH2CH2OH OH– 2

K440,SOH 42 → CH3CH2CH = CH2

(B) (C) 1-Butene

CH3CH2CH=CH2 →HBr CH3CH2CH–CH3 (C) | Br (D) 2-Bromobutane(optically active)

MATHEMATICS

Section A

1. Q (1, 1), (2, 2), (3, 3) ∉ R ⇒ R is not reflexive. again (1, 2) ∈ R ⇒ (2, 1) ∈ R ⇒ R is symmetric again (1, 2) ∈ R, (2, 1) ∈ R but (1, 1) ∉ R ⇒ R is not transitive.

2. On differentiating

32 x–1/3 +

32 y–1/3

dxdy = 0

dxdy = –

3/1

3/1

xy

3. We have,

I = ∫ dxex xsinlog3cos = ∫ dxxxsincos3

Putting cos x = t and – sin x dx = dt or, sin x dx = – dt, we get

I = – ∫ dtt3 = 4t 4− + C = –

4cos4 x + C


4. Let x + y = v. Then

1 + dxdy =

dxdv

⇒ dxdy =

dxdv –1

Putting x + y = v and dxdy =

dxdv –1 the given

differential equation, we get v2

−1

dxdv = a2s

⇒ v2

dxdv = a2 + v2

⇒ v2 dv = (a2 + v2) dx

⇒ 22

2

avv+

dv = dx [By separating the variables]

⇒

+− 22

21

ava dv = dx

⇒ ∫ dv.1 – a2 ∫ + 221

avdv = ∫ + cdx

[On integration]

⇒ v – a tan–1

av = x + c

⇒ (x + y) –a tan–1

+

ayx = x + c

5. Let l, m, n be the direction cosines of the given vector r

r(say). Then, its projections on the coordinate

axes are l | rr

|, m | rr

|, n | rr

| ∴ l | r

r| = 6, m | r

r|= –3, n | r

r| = 2 … (i)

⇒ l | rr

|2 + m | rr

|2 + n| rr

|2 = 62 + (–3)2 + (2)2 ⇒ | r

r|2 (l2 +m2 + n2) = 36 + 9 + 4

⇒ | rr

|2 = 49 ⇒ | r

r| = 7 [Q l2 + m2 + n2 =1]

Putting | rr

| = 7 in (i), we obtain that the direction

cosines of rr

are l = 76 , m =

73− , n =

72

6. The angle θ between vectors ar

and br

is given by

cos θ = ||||

.ba

barr

rr

For the angle θ to be obtuse, we must have cos θ < 0

⇒ ||||

.ba

barr

rr

< 0

⇒ ar

. br

< 0 [Q | ar

|, | br

| > 0] ⇒ 14x2 –8x + x < 0 ⇒ 7x (2x –1) < 0 ⇒ x (2x –1) < 0 ⇒ 0 < x < 1/2 Hence, the angle between the given vectors is obtuse

if x ∈ (0, 1, 2)

7. Let the equation of the required plane be

ax +

by +

cz = 1 … (i)

Then, the coordinates of A, B and C are A (a, 0, 0), B(0, b, 0), and C(0, 0, c) respectively. So the centroid of triangle ABC is (a/3, b/3, c/3). But the coordinates of the centroid are (p, q, r)

∴ p = 3a , q =

3b and r =

3c

⇒ a = 3p, b = 3q and c = 3r Substituting the values of a, b and c in (i), we get the

required plane as

px

3+

qy

3+

rz

3= 1 ⇒

px +

qy +

rz = 3

8. a + b = 6, …(i)

ab = 8 ⇒ b = a8 … (ii)

Putting in (i)

a +a8 = 6 ⇒ a = 4, 2

From (ii)

⇒ b =48 ,

28

⇒ b = 2, 4

Hence (a, b) = (4, 2) or (2, 4)

9. 2X =

−++

1906235

⇒ X =

4044

and Y =

9025

–

4044

⇒Y =

−5021

10. ∆ = 1/2 1264121122

kkkk

kk

−−−+−

−= 0

⇒ 8k2 + 4k – 4 = 0 ⇒ k = 1/2, –1

Section B 11. Let A → total of 8 in first throw B → total of 8 in 2nd throw Number of exhaustive cases when a pair of dice is

thrown = 6 × 6 = 36 Cases favourable to a total of 8 in each throw are

(2, 6), (6, 2), (3, 5), (5, 3), (4, 4) Their number = 5 P(A) = P(B) = 5/36

P(A and B) = P(A). P(B) = 365 ×

365 =

129625


12. f(x) = sin2 x + [1 – cos2 (π/3 + x)]

+ 21 2 cos x cos (π/3 + x)

= 1 –[cos2(π/3 + x) – sin2 x]

+21 [cos(π/3+2x) + cos π/3]

= 1 – [cos (π/3+2x) –cos π/3] +21 cos (π/3 +2x) +

41

f(x) = 5/4 g(f(x)) = g(5/4) = 1 → constant function 13. Q f(x) is continuous at x = 2, 4 ∴ f(2) = f(2–) ⇒ 3 × 2 + 2 =

0lim→h

(2 –h)2 + a(2 – h) + b

⇒ 8 = 2a + b + 4 ⇒ 2a + b = 4 … (i) again, f(4) = f(4+) ⇒ 3 × 4 + 2 =

0lim→h

2a (4 + h) + 5b

⇒ 14 = 8a + 5b … (ii) Solve (i) , (ii) a = 3, b = –2

14. y = tan–1

+

xx

cossin1

y = tan–1

−+

2/sin2/cos)2/sin2/(cos

22

2

xxxx

y = tan–1

−+

2/sin2/cos2/sin2/cos

xxxx

y = tan–1[tan (π/4 + x/2)] y = π/4 + x/2

dxdy =

21

15. Let y = tan–1

−++

−−+22

22

11

11

xx

xx , z = cos–1 x2

put x2 = cos 2θ

y = tan–1

θ−+θ+

θ−−θ+

2cos12cos12cos12cos1

y = tan–1

θ+θθ−θ

sincossincos

dxdz =

41

1

x−

− .2x

y = tan–1 [tan (π/4 –θ)]

y = 4π –

21 cos–1 x2

dxdy =

21−

−

− xx

2.1

14

dxdy =

41 x

x

−

∴ dzdy =

dxdzdxdy

// =

4

4

1/2

1/

xx

xx

−−

− = 21−

16. Area A = 21 × b × AD

= 2b .

4

22 bx −

A = 4b 224 bx −

dtdA =

2248 bx

b

− 8x

dtdx

dtdA =

224 bx

bx

−× 3

= 3

dtdx

Q

bxdt

dA

=

=

22

2

4

3

bb

b

− × 3 = 3 b cm2/sec.

OR (i) Since (x – 2) ≥ 0 in [2, 3] so f(x) = 2−x is continuous

(ii) f ´(x) = 22

1−x

exists for all x ∈(2, 3)

∴ f(x) is differentiable in (2, 3) Thus lagrang's mean value theorem is applicable; ∴ There exists at least one real number in (2, 3)

such that

f ´(c) = 23

)2(f)3(f−−

or 22

1−c

= 1

0)1( − ⇒ 22 −c = 1

c = 2 + 41 = 2.25 ∈(2, 3)

⇒ LMV is verified and the req. point is (2.25, 0.5)

17. dxxx

x∫

−− 222 )(1

2

Let x2 = t. Then, d(x2) = dt ⇒ 2x dx = dt ⇒ dx = x

dt2

∴ I = ∫−− 21 tt

dt = ∫−+− 1 2 tt

dt

= ∫−−++− 1

41

41 2 tt

dt

⇒ Ι = ∫

−

+−

45

21 2

t

dt = ∫

+−

2

21

45 t

dt

A

B C

x x

b/2 b/2D


= ∫

+−

22

21

25 t

dt

⇒ I = sin–1

+

2/52/1t + C

= sin–1

+

512t + C

= sin–1

+

512 2x + C

OR

I = ∫ ++

xcos1xsin2 ex/2 dx

= ∫

++

+ xcos1xsin

xcos12 ex/2 dx

= ∫

+

2xcos2

2xcos

2xsin2

2xcos

222

ex/2 dx

= ∫

+

2xtan

2xsec2 ex/2 dx

2 tan2x .ex/2 + c

18. I = dx

xx

x∫+

+

22

2

16

41

⇒ I = dx

xx

x∫+−

+

+

884

41

22

2

⇒ I = dx

xx

x∫+

−

+

84

41

2

2

Let x – x4 = t. Then, d

−

xx 4 = dt

⇒

+ 2

41x

dx = dt

∴ I = ∫ + 22 )22(tdt

⇒ I = 22

1 tan–1

22t + C

⇒ I = 22

1 tan–1

−

22

4x

x+ C

= 22

1 tan–1

−

2242

xx + C

19. We are given that dxdy + (–2)y = cos 3x …(i)

This is a linear differential equation of the form

dxdy + Py = Q, where P = –2 and Q = cos 3x

∴ I.F. = ∫e

Pdx = ∫ −e

dx2 = e–2x

Multiplying both sides of (i) by I.F. = e–2x, we get

e–2x

dxdy –2ye–2x = cos 3x. e–2x

Integrating both sides w.r.t. x, we get ye–2x = ∫ − xe 2 cos 3x dx + C

[Using : y (I.F.) = ∫ + CdxFIQ .).( ]

⇒ ye–2x = I + C, where I = e–2x cos 3x Now, I =

IIIdxxe x∫ − 3cos2

⇒ I = 31 e–2x sin 3x – dxxe x 3sin

3)2( 2−∫

−

⇒ I = 31 e–2x sin 3x +

32 dxxe x 3sin2−∫

⇒ I = 31 e–2x sin 3x

+32

−

−−−

∫ −− dxxexe xx

33cos)2(3cos

31 22

⇒ I = 31 e–2x sin 3x

+32

−− ∫ −− dxxexe xx 3cos

323cos

31 22

⇒ I =31 e–2x sin 3x–

92 e–2x cos 3x –

94 I

⇒

+ II

94 =

9

2xe−

(3 sin 3x –2 cos 3x)

⇒ I = 13

2xe−

(3 sin 3x – 2 cos 3x)

Substituting the value of I in (ii), we get

ye–2x =13

2xe−

(3 sin 3x –2 cos 3x) + C, which is the

required solution.


20. We have, (1 – a

r. br

)2 + | ar

+ br

+( ar

× br

)|2 = 1 – 2( a

r. br

) + ( ar

. br

)2 + ( ar

+ br

+ ar

× br

). ( a

r+ b

r+ a

r× b

r)

= 1 –2 ( ar

. br

) + ( ar

. br

)2 + ( ar

+ br

).( ar

+ br

) + ( a

r+ b

r). ( a

r× b

r)+ ( a

r× b

r).( a

r+ b

r)+ ( a

r× b

r).( a

r× b

r)

= 1 –2 ( ar

. br

) +( ar

. br

)2 + | ar

+ br

|2 + ar

. ( ar

× br

) + b

r. ( a

r× b

r) + ( a

r× b

r) . a

r ( a

r× b

r). b

r+ | a

r× b

r|2

= 1–2 ( ar

. br

) + ( ar

. br

)2 + | ar

+ br

|2 + | ar

× br

|2 [Q a

r⊥( a

r× b

r), b

r⊥( a

r× b

r) ∴ a

r ( a

r× b

r) = b

r.( a

r× b

r)= 0]

= 1 – 2 ( ar

. br

) + ( ar

. br

)2 + | ar

|2 + | br

|2 +2 ( ar

. br

) + | a

r× b

r|2

= 1 + | ar

|2 + | br

|2 + ( ar

. br

)2 + | ar

× br

|2 = 1 + | a

r|2 + | b

r|2 + | a

r|2 | b

r|2

[Q( ar

. br

)2 + | ar

× br

|2 = | ar

|2 | br

|2] = (1 + | a

r|2) (1+| b

r|2)

Hence, (1+ | ar

|2) (1+ | br

|2) = 1 – ( a

r. br

)2 + | ar

+ br

+ ar

× br

|2 21. Let L be the foot of the perpendicular drawn from the

point P (0, 2, 3) to the given line. The coordinates of a general point on

53+x =

21−y =

34+z are given by

53+x =

21−y =

34+z =λ

i.e. x = 5λ –3, y = 2λ + 1, z = 3λ –4 Let the coordinates of L be (5λ –3, 2λ + 1, 3λ – 4) So direction ratios of PL are proportional to 5λ – 3 – 0, 2λ + 1 –2, 3λ –4 – 3 i.e. 5λ –3, 2λ –1, 3λ – 7 Direction ratios of the given line are proportional to

5, 2, 3

P(0, 2, 3)

L (5λ –3, 2λ +1, 3λ –4) 5

3+x = 2

1−y = 3

4+z

Since PL is perpendicular to the given line. ∴ 5(5λ –3) + 2(2λ –1) + 3(3λ –7) = 0 ⇒ λ = 1 Putting λ = 1 in (i), the coordinates of L are (2, 3, –1) ∴Length of the perpendicular from P on the given

line.

= PL = 222 )31()23()02( −−+−+− = 21 units OR

Here integrating factor = ∫ xdxcote = elog sinx = sin x

∴ the solution of differential equation is given by

y.sin x = ∫ + )cot2( 2 xxx sin x dx

= ∫ dxxxsin2 + ∫ dxxx cos2

= ∫ dxxxsin2 + x2 sin x – ∫ dxxxsin2 + c

= x2 sin x + c ...(1) Substituting y = 0 and x = π/2, we get

0 = 4

2π + c or c = –4

2π

∴ (i) ⇒ y sin x = x2 sin x – 4

2π

or y = x2 – 4

2π cosec x

22. xyzzxyyzx

111

= xyz1

xyzzzxyzyyxyzxx

2

2

2

= xyzxyz

1zz1yy1xx

2

2

2

C1 ↔ C3

⇒ 2

2

2

111

zzyyyx

R1 → R1 – R2, R2 → R2 – R3

= 2

22

22

100

zzzyzyyxyx

−−−−

= (x – y) (y – z) 21

1010

zzzyyx

++

= (x – y) (y – z) zyyx

++

11

= (x – y) (y – z) [(y + z) – (x + y)] = (x – y) (y – z) (z – x)

OR

accbbacbbaacbaaccb

+++++++++

= 0

C1 → C1 + C2 + C3

⇒ 2(a + b + c)accbcbbabaac

++++++

111

= 0

R2 → R2 – R1, R3 → R3 – R1

⇒ 2(a + b + c)bcabaccbbaac

−−−−++

001

= 0


⇒ 2(a + b + c) (–a2 – b2 – c2 + ab + bc + ca) = 0 ⇒ –(a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] = 0 ⇒ a + b + c = 0 or a = b = c

Section C

23. Getting 6 → success

p = 61

⇒ q = 1 – p = 65

A can win the game in 1st, 3rd, 5th …..throws. P(A winning) = p + qqp + qqqqp + ……… = p[1 + q2 + q4 + ……..]

= p

− q11 =

116

P(B winning) = 115

24. y = x … (i) y = x3 … (ii) Solving (i) and (ii) O(0, 0), A(1, 1), B(–1, –1) Required area = area BCOB + Area ODAO

A(1, 1) D

y = x y

O(0, 0) x C

B(–1, –1)

•

•

•

Area BCOB = ∫−

−0

1

3 )( dxxx = 0

1

42

42−

−

xx

=

−−

41

21 =

41

Area ODAO = ∫ −1

0

3 )( dxxx = 1

0

42

42

−

xx

= 21 –

41 =

41

∴ required area = 41 +

41 =

21 sq. units.

25. The equations of two given lines are

2

1−x = 3

2−y = 4

3−z … (i)

and 3

2−x = 4

4−y = 5

5−z … (ii)

Line (i) passes through (1, 2, 3) and has direction ratios proportional to 2, 3, 4, so its vector equation is

rr

= 1ar

+ λ 1br

… (iii)

where, 1ar

= i + 2 j + 3 k and 1br

= 2 i + 3 j + 4 k Line (ii) passes through (2, 4, 5) and has direction

ratio proportional to 3, 4, 5. So, its vector equation is r

r= 2a

r+ µ 2b

r … (iv)

where 2ar

= 2 i + 4 j + 5 k and 2br

= 3 i + 4 j + 5 k The shortest distance between the lines (iii) and (iv)

is given by

S.D. = ||

)).((

21

2112

bbbbaa

rr

rrrr

×

×− … (v)

We have, 2ar

– 1ar

= (2 i + 4 j + 5 k ) – ( i + 2 j + 3 k )

= i + 2 j + 2 k

and 1br

× 2br

= 543432

ˆˆˆ kji= – i + 2 j – k

∴ | 1br

× 2br

| = 141 ++ = 16

and,( 2ar

– 1ar

).( 1br

× 2br

) = ( i + 2 j +2 k ).(– i + 2 j – k ) = –1 + 4 –2 = 1 Substituting the values of ( 2a

r– 1a

r) . ( 1b

r× 2b

r) and

| 1br

× 2br

| in (v), we obtain S.D. = 1/ 6

26. A11 = 1231

−= 1 + 6 = 7; A12 = –

1032

= –2

A13 = 20

12−

= –4; A21 = – 1202

−−

= 2

A22 = 1001

= 1; A23 = –2021

−−

= 2

A31 = 3102−

= –6; A32 = –3201

= –3

A33 = 1221 −

= 1 + 4 = 5

∴ adj A =

′

−−

−−

536212427

=

−−−−

524312627

Also |A| = (1) (7) + (–2) (–2) +0(–4) = 7 + 4 = 11

∴ A–1 = |A|Aadj =

111

−−−−

524312627

Again the given system of equations can be written as

AX = B where X =

zyx

, B =

78

10⇒ X = A–1B


i.e.

zyx

=111

−−−−

524312627

78

10

= 111

++−−+−−+

35164021820421670

= 111

−1133

44=

−13

4

∴ x = 4, y = –3, z = 1

27. Let y = sinpθ cosq θ Let z = log y = p log sin θ + q log cos θ

θd

dz = p(cot θ) – q tan θ

For maximum θd

dz = 0

⇒ p cot θ = q tan θ

⇒ tan θ = qp /

Now

2

2

θdzd = – p cosec2 θ – q sec2 θ

= – p(1 + cot2 θ) – q (1 + tan2 θ)

= – p(1 + q/p) – q(1 + p/q) (Q tan θ = qp / )

= – 2(p + q)

2

2

dzd

θ< 0

⇒ at θ = tan–1 qp / , y attains maximum value.

28. ∫b

a

dxxf )( = 0

lim→h

h[f(a) + f(a + h) + f(a + 2h) + ........

…….+ f(a + (n –1) h],

where h = n

ab −

Here a = 1, b = 3, f(x) = x2 + 5x and

h = n

13 − = n2

∴ ∫ +3

1

2 )5( dxxx

=0

lim→h

h [ f(1) + f(1 + h) + f(1 + 2h) + …..........

….+ f(1+(n –1) h)] =

0lim→h

h [(12 + 5 × 1) + (1 + h)2 + 5 (1 + h)

+ (1 + 2h)2 + 5(1 + 2h) + ....... …..+(1 + (n –1)h)2 + 5(1 + (n –1) h)] =

0lim→h

h [(12 + (1 + h)2 + (1 + 2h)2 + ….+

(1 + (n –1)h)2] + 51 + (1 + h) + (1 + 2h) + ........ ….+ (1 +(n –1)h)]

= 0

lim→h

h [n + 2h (1 + 2 + 3 + …..+(n –1) + h2

(12 + 22 + …..+(n –1)2) + 5n + h (1 + 2 + …... ......+ (n –1))]

0

lim→h

h [6n + 7h (1 + 2+ 3+ …..+(n –1) + h2

(12 + 22 + ……+……+ (n –1)2]

= 0

lim→h

h

−−

+−

−+6

)12()1(.2

)1(76 2 nnnhnnhn

= ∞→n

limn2

−−

+−

+6

)12)(1(.42

)1(.146 2nnn

nnn

nn

= ∞→n

lim

−−+

−

+ 2)12)(1(.

6811412

nnn

nn

= ∞→n

lim

−

−+

−+

nnn1211.

34111412

= 12 + 14 + 34 × 2 = 12 + 14 +

38 =

386

OR

∫b

a

dxxf )( = 0

lim→h

h [f(a) + f(a + h) + f(a + 2h) +….....

….+ f(a + (n –1) h)],

where h = n

ab −

Here, a = –1, b = 1, f (x) = ex and h = n

)1(1 −− = n2

∴ ∫−

1

1

dxex

= 0

lim→h

h [f(–1) + f(–1+ h) + f(–1 + 2h) + ...........

……..+ f (–1 + (n –1) h)] =

0lim→h

h [e–1 + e–1+h + e–1+2h+…+ e–1+(n –1)h]

= 0

lim→h

h e–1 [1 + eh + e2h + ….+e(n –1)h]

= 0

lim→h

h e–1

−−1

1)(.1 h

nh

ee

−−

=+++1r1raar.......ar a using

n1–n

= 0

lim→h

e–1

−

−

hee

h 112

[Q h = 2/n ⇒ nh = 2]

= e–1

−1

12e = e – e–1

=

−→

11lim0 h

eh

hQ


29. The given information can be exhibited diagrammatically as follows

Depot A

5 units

Depot C

4 units

Depot B

5 units

Factory Q

6 units

Factory P

8 units

Rs.10 Rs.16

Rs.15 Rs.10

Rs.10 Rs.12 y units (5–y) units

(5–x) units x units

8–(x + y) units 6 – (5 – x + 5 – y) = x + y – 4 units

Let the factory at P transports x units of commodity to depot at A and y units to depot at B. Since the factory at P has the capacity of 8 units of the commodity. Therefore, the left out (8 – x – y) units will be transported to depot at C.

Since the requirements are always non-negative quantities.

Therefore, x ≥ 0, y ≥ 0 and 8 – x – y ≥ 0 ⇒ x ≥ 0, y ≥ 0 and x + y ≤ 8. Since the weekly requirement of the depot at A is 5

units of the commodity and x units are transported from the factory at P. Therefore the remaining (5 – x) units are to be transported from the factory at Q. Similarly, 5 – y units of the commodity will be transported from the factory at Q to the depot at B. But the factory at Q has the capacity of 6 units only, therefore the remaining 6 – (5 – x + 5 – y) = x + y – 4 units will be transported to the depot at C. As the requirements at the depots at A, B and C are always non-negative.

∴ 5 – x ≥ 0, 5 – y ≥ 0 and x + y – 4 ≥ 0 ⇒ x ≤ 5, y ≤ 5 and x + y ≥ 4 The transportation cost from the factory at P to the

depots at A, B and C are respectively Rs.16x, 10y and 15(8 – x –y). Similarly, the transportation cost from the factory at Q to the depots at A, B and C are respectively Rs.10 (5 – x), 12(5 – y) and 10(x + y –4). Therefore, the total transportation cost Z is given by

Z = 16x + 10y + 15(8 – x – y) + 10 (5 – x) + 12 (5 – y) + 10 (x + y – 4) = x –7y + 190

Z = 16x + 10y + 15(8 – x – y) + 10(5 – x) + 12(5 – y) + 10(x + y – 4) = x –7y + 190)

Hence, the above LPP can be stated mathematically as follows, find x & y which Minimize

Z = x – 7y + 190 x + y ≤ 8 x + y ≥ 4 x ≤ 5 y ≤ 5 and x ≥ 0, y ≥ 0

OR Let the depot A transport x thousand bricks to builder

P and y thousand bricks to builder Q. Then the above LPP can be stated mathematically as follows.

Minimize Z = 30x –30y + 1800 Subject to x + y ≤ 30 x ≤ 15 y ≤ 20 x + y ≥ 15 and, x ≥ 0, y ≥ 0 To solve this LPP graphically, we first convert

inequations into equations and then draw the corresponding lines. The feasible region of the LPP is shaded in fig. The coordinates of the corner points of the feasible region A2PQB3B2 are A2 (15, 0), P(15, 15), Q (10, 20), B3 (0, 20) and B2(0, 15). These points have been obtained by solving the corresponding intersecting lines simultaneously.

x = 15 y

y = 20

B (0, 30) Q (10, 20) B3 (0, 20)

P (15, 15) B2 (0, 15)

A2 (15, 0) O

x

x + y = 15

x + y = 30

A (30, 0)

•• •

•

•

The values of the objective function at the corner

points of the feasible region are given in the following table.

Point (x, y) Value of the objective function Z = 30x –30y + 1800

A2(15, 0) Z = 30 × 15 –30 × 0 + 1800 = 2250 P(15,15) Z = 30 × 15 –30 × 15 + 1800 = 1800 Q(10,20) Z = 30 × 10 –30 × 20 + 1800 = 1500 B3(0, 20) Z = 30 × 0 –30 × 20 + 1800 = 1200 B2(0, 15) Z = 30 × 0 –30 × 15 + 1800 = 1350

Clearly, Z is minimum at x = 0, y = 20 and the minimum value of Z is 1200. Thus, the manufacturer should supply 0, 20 and 10 thousand bricks to builders P, Q and R from depot A and 15, 0, and 5 thousand bricks to builders P, Q and R from depot B respectively. In this case the minimum transportation cost will be Rs.1200.


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 A n s D A C A A C B C B B,C Ques 11 12 13 14 15 16 17 18 19 20 A n s A,B,C A,C,D A,B,C,D A,B B A B B D A

21 A → P,Q,R,S B → P,Q,R,S C → P,Q,R D → P,Q,R,S Column Matching 22 A → P B → R C → Q D → S

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s C C A D A D D C A A,C Ques 11 12 13 14 15 16 17 18 19 20 A n s B,C,D A,D B,C,D B,C,D C B D B B C

21 A → Q B → R C → S D → P Column Matching 22 A → Q,S B → R,S C → P,Q D → R,S

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s C D D C B C A C C B,C,D Ques 11 12 13 14 15 16 17 18 19 20 A n s A,D C,D A,B,D A,B,C B C D B A C

21 A → R B → P C → Q D → S Column Matching 22 A → S B → Q C → P D → R

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s C C C D C C B C B A,C Ques 11 12 13 14 15 16 17 18 19 20 A n s A,B,D A,B B, D A,D A B C C A D

21 A →R B → P C → S D → Q Column Matching 22 A → R B → S C → Q D → P

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s D D B A C A C C B D Ques 11 12 13 14 15 16 17 18 19 20 A n s A,B, D A,C A,C B C A A B B C

21 A → P,R,S B → Q C → P,R,S D → S Column Matching 22 A → S B → R C → P D → Q

MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 10 A n s C B B D A B B B A A,B,D Ques 11 12 13 14 15 16 17 18 19 20 A n s A,B,C B,D D A,B,C,D A C B B C C

21 A → S B → R C → Q D → P Column Matching 22 A → P B → R C → Q D → S

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XtraEdge Feb 2011