1. Xtra edge June 2011 - SMAROSA · 2018. 7. 23. · XtraEdge for IIT-JEE 1 JUNE 2011 Dear Students, Motivate Yourself One of the greatest virtues of human beings is their ability

XtraEdge for IIT-JEE 1 JUNE 2011

Dear Students,

Motivate Yourself One of the greatest virtues of human beings is their ability to think and act accordingly. The emergence of the techno savvy human from the tree swinging ape has really been a long journey. This transition has taken a span of countless centuries and lots of thinking caps have been involved. Inquisitiveness and aspiration to come out with the best have been the pillars for man's quest for development. Self-motivation is the sheer force, which pulled him apart and distinguished him from his primitive ancestors. Many times, in our life, when we are reviving old memories we get into a phase of nostalgia. We feel that we could have done better than what we had achieved. But thinking back won’t rewind the tireless worker called time. All we can do is promise ourselves that we will give our very best in the future. But do we really keep up to our mental commitments? I can guess that 90% answers are in the negative. This is because of that creepy careless attitude which is slowly, but surely entering into our mind. We easily forget the pains of yesterday to relish the joys of today. This is the only time in our life, when we can control our fate, by controlling our mind. So it is time to pull up our socks and really motivate ourselves so that we can give our best shot in the future. Self-motivation is the need of the hour. Only we can control and restrict ourselves. It’s up to us, how we use our mental capabilities to the best of our abilities. Here are some Funda's for self-motivation. Don't just read them digest each one of them and apply them and I bet it will make a better YOU. • The ultimate motivator is defeat. Once you are defeated, you have

nowhere to go except the top. • Then only thing stopping you is yourself. • There is no guarantee that tomorrow will come. So do it today. • Intentions don't count, but action's do. • Don't let who you are, stunt what you want to be. • Success is the greatest motivator. • Your goals must be clear, but the guidelines must be flexible. Try to include these one liners in your scrapbook or on your favorite poster. You will be sub-consciously tuned to achieve what you want. Also do keep in mind that nothing can control your destiny but you! With Best Wishes for Your Future.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Volume-6 Issue-12 June, 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.

Xtra Edge Test Series for JEE- 2012 & 2013

S

Success Tips for the Months

• If you haven nothing else to do, look about you and see if there is not something close at hand that you can improve !

• He has achieved success who has worked well, laughed often, and loved much.

• You always pass failure on the way to success.

• A journey of a thousand miles begins with a single step.

• Your success will be largely determined by your ability to concentrates single-mindedly on one thing at a time.

• Success is a journey, not a destination.

• Success comes in "Cans". Failure comes in "Can't".

• Success seems to be largely a matter of hanging on after others have let go.

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 • IIT-B launches 'clean' energy laboratory • Tech tycoon Shiv Nadar to set up research-led university in Greater Noida

IITian ON THE PATH OF SUCCESS 5 Mr. Krishnamurthy Rengarajan

KNOW IIT-JEE 6 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 46

Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper AIEEE- 2011 Examination Paper with Solution 61

Regulars ..........

DYNAMIC PHYSICS 13

8-Challenging Problems [Set # 2] Students’ Forum Physics Fundamentals • Electrostatics-2 • Newton's Law of motion

CATALYSE CHEMISTRY 28

Key Concept • Nomenclature & isomerism • Electro Chemistry Understanding : Inorganic Chemistry

DICEY MATHS 36

Mathematical Challenges Students’ Forum Key Concept

• Inverse Trigonometric Function • Quadratic Equation

Study Time........

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


IIT-B launches ‘clean’ energy laboratory Mumbai: The Indian Institute of Technology-Bombay on Thursday launched a laboratory, set up in association with Applied Materials Inc, a global supplier for manufacturing solutions for semi-conductor and solar industries.

The scope of the Applied Materials Chemistry Laboratory for Energy and Nanoelectronics (CLEAN) will include developing materials that could be used for electronic and renewable energy focused applications, such as next generation solar cells.

Under the first phase of collaboration, Applied Materials supplied equipment worth $7.5 million to IIT-B for a semiconductor fabrication facility. IIT-B director Devang Khakhar said, “CLEAN will begin a new phase of the collaboration in areas related to renewable energy.”

“This is a great example of the kind of university and corporate collaboration that is helping to advance technology by enabling world-class research, innovation and workforce development,” said Mike Splinter, chairman of the board of Applied Materials.

The inauguration marked the renewal of its five-year relationship, during which it endowed the institute with $12 million as research grants.

Officials said as part of the Indo-US collaboration in the sector, Applied Materials will work with the Jawaharlal Nehru National Solar Mission (JNNSM). “Our unique collaboration will allow us to turn material inventions into innovations,” said Omkaram Nalamasu, chief technology officer, Applied Materials.

Tech tycoon Shiv Nadar to set up research-led university in Greater Noida Shiv Nadar, the founder chairman of major HCL Technologies, announced on Monday that he would set up a research-led multidisciplinary university at Greater Noida, a satellite town of Delhi. The Shiv Nadar University, named after the IT billionaire, marks the germination of an idea that was incubated five years ago in the mind of Nadar, who thought of the concept in his zest to make a creative difference in India’s educational system.

“The University will be endowment-led, research-driven with multidisciplinary and interdisciplinary approach,” Nadar told. The total endowment would be R1,000 crore. Education is not a virgin territory for Nadar. The Shiv Nadar Foundation runs the VidyaGyan School at Bulandshahar in Uttar Pradesh that aims at nurturing talent and leadership amongst the highly gifted rural poor children. A visibly happy Nadar said with the setting up of this University, he feels the same excitement as was when HCL was first formed in 1976. Nadar had already committed a 10% of his wealth for philanthropic causes. His total worth has been estimated at around R15,000 crore. Nadar said an undergraduate student spends close to $150,000-$200,000 seeking degrees on foreign shores and a good chunk comes from North India. “At undergraduate levels one doesn’t get scholarships. The idea to create a world class University at Greater Noida was born out of this need.”

It will offer courses in engineering, social sciences, natural sciences, schools of business and other professional courses in due time. “We will start admissions from this academic session itself and the total intake this year would be around 200-

300 students,” said Dr Nikhil Sinha, the founding vice-chancellor of the University. Sinha has served as associate dean for academic affairs at the University of Texas.

Nuclear plans get makeover SAFETY FIRST Independent regulator, reviews by external experts on govt to-do list Stunned by Fukushima and hit by unprecedented anti-nuclear protests in Jaitapur, the government announced a series of dramatic reforms to the country's nuclear review mechanism that are aimed at rebuilding public confidence in India's ambitious nuclear programme.

Soon, a one-shot cure for all cancers Universal Vaccine Uses Body’s Own Defences To Stop All Tumours In Their Tracks London: Coming soon. Universal cancer jab, say scientists who are developing the vaccine which they claim would stop all tumours in their tracks. According to the scientists, the jab is part of a new generation of drugs that use the body’s own defences to fight cancers like pancreatic, prostate and breast, stopping tumours in their tracks. The jabs would hit the market in just two years’ time, say the scientists.

Although vaccines usually prevent disease, the Telo-Vac jab is designed as a treatment. Rather than attacking cancer cells, like many existing drugs, it harnesses the power of the immune system to fight the tumours, they say. It works by encouraging the immune system to seek out and destroy an enzyme called telomerase. Found at high levels in many cancer cells, telomerase effectively makes


them fully immortal, allowing them to live on when healthy cells would die — easing the growth and spread of the tumour.

In the largest trial of its kind in the UK, more than 1,000 men and women in the late stages of pancreatic cancer are either being given the vaccine alongside their normal drugs or treated as usual, the ‘Daily Mail’ reported.

The results from the 53 hospitals taking part will not be available until next year but, anecdotally, some patients credit their participation in the trial with giving them an extra year or two of life.

John Neoptolemos, who is co-ordinating the large-scale British trial, said: “When you have got pancreatic cancer, it is like a timebomb in people.” Pancreatic cancer cells are normally invisible to the immune system but the vaccine “spots” the telomerase spilling out from them and kickstarts the fight back.

Dr Jay Sangjae Kim, the founder of GemVax, the Korean company developing the TeloVac vaccine, said: “We s

trongly believe this has the potential to overcome the limits of other current cancer vaccines and become part of the standard of care not only for pancreatic cancer but for various other types of cancers.”

New engine sends shock waves through auto industry Prototype could potentially decrease auto emissions up to 90 percent Despite shifting into higher gear within the consumer's green conscience, hybrid vehicles are still tethered to the gas pump via a fuel-thirsty 100-year-old invention: the internal combustion engine.

However, researchers at Michigan State University have built a prototype gasoline engine that requires no transmission, crankshaft, pistons, valves, fuel compression, cooling systems or fluids. Their so-called Wave Disk Generator could greatly improve the efficiency of gas-electric hybrid automobiles and potentially

decrease auto emissions up to 90 percent when compared with conventional combustion engines.

The engine has a rotor that's equipped with wave-like channels that trap and mix oxygen and fuel as the rotor spins. These central inlets are blocked off, building pressure within the chamber, causing a shock wave that ignites the compressed air and fuel to transmit energy.

The Wave Disk Generator uses 60 percent of its fuel for propulsion; standard car engines use just 15 percent. As a result, the generator is 3.5 times more fuel efficient than typical combustion engines.

Researchers estimate the new model could shave almost 1,000 pounds off a car's weight currently taken up by conventional engine systems.

Michigan State's team of engineers hope to have a car-sized 25-kilowatt version of the prototype ready by the end of the year.

IITian from Sikar leaves Fortune 500 company to take up rural welfare From a government school at a nondescript village in Sikar district to IIT-Kgp, this youngster has come a long way. Born to a humble farmer's family, Rohit Garhwal now cherishes a dream of developing an international platform for the traditional agriculture art & craft forms and agriculture and minimise the exodus of rural talent to cities. To achieve his dream, the 25-year-old has quit plush jobs at an MNC and an NGO to study at the IIM-A.

"IIM-A offers incubation centre for social ventures. Here, I will like to construct a model of branding products made by villagers and taking them in international markets," said Rohit, who has secured a seat in the most prestigious management school, the result of which was announced on Friday afternoon.

Leaving behind the luxuries of an MNC job earned after receiving a degree in aerospace engineering from the most sought after IIT at Kharagpur, Rohit

has taken the first big leap towards translating his dream into reality by securing the seat.

He had already started the groundwork travelling daily from one dusty village to another on the state roadways bus to the hinterland of Sikar, presiding over small public gatherings, explaining to the villagers the need for education and healthcare to alleviate their social and economic status.

Rohit now plans to take his work to a higher level after completing his IIM-A course.

An alumnus of the IIT-Kgp 2008 batch, Rohit was employed by a financial firm listed in Fortune 500 companies. He called it quits a year later worn out by the daily rigours of explaining and devising business models.

Back home, the IITian, being from a farming background and a strong advocator of inclusive growth, ventured in social entrepreneurship. He joined as consultant of an NGO Helplads.

Using his technical skill, Rohit organised workshops on themes ranging from the basics of a computer to robotics for village children. "I plan fund-raising activities along with developing business models for small-time social entrepreneurs in this region," Rohit said.

• Inspired by other IIM students who have turned entrepreneurs, he said he was determined to find a way to help the rural people. "I am very worried of the migration of villagers to small towns and cities abandoning their agricultural and traditional art and craft practices. I would like to use my technical expertise and management tools to create a self-reliant environment for them," Rohit added.


Knowledge is indeed wealth. Who better exemplifies it than Krishnamurthy Rengarajan, IIT-B gold medallist (B Tech dual-degree course). Krishnamurthy's story is that of hard work, sheer grit and determination. His undying passion for learning and excellence has paid off. Coming from a lower middle class background , Krishnamurthy has made his parents proud when he passed with flying colours. His father, who works as a typist at Bharatiya Vidya Bhavan is overwhelmed by his son's achievement. Rengarajan, who hails from Tamil Nadu, came to Mumbai 28 years ago and settled down in a distant Mumbai suburb of Dombivli. Though the family went through a lot of hardships initially, he made sure that his children were well educated. "My son always wanted to join the IIT. When people asked him what if you don't get through the entrance examinations, he used to say, `there is no question of me not clearing the test,'" says his proud father. And, of course, he did top all the five years at IIT, a result of sheer hard work and brilliance, says his mother, barely able to control her excitement. "I am very happy for him," says Radha Rengarajan. Krishnamurthy did his schooling at the Kidland School in Dombivli and pre-degree from V G Vaze College at Mulund. His favourite subject being mathematics it was obvious that he would pursue a degree in engineering. He won the Rakesh Mathur award of Rs 1 lakh (Rs 100,000) during his third year and other scholarships throughout the four years. Here's what Krishnamurthy had to say on his IIT experience. My IIT experience The five years I spent at IIT were the best in my life. I will cherish each and every moment here. I loved everything here: the professors are the best one can ever get, the facilities to study and the extra-curricular activities are

excellent. I made best of friends and thoroughly enjoyed my college life. I don't think I will ever get this experience anywhere else. On studies Before joining IIT, I used to study for 7 to 8 hours daily. After joining IIT, I used to spend about a couple of hours. I did not go for anything coaching classes. I learnt through Brilliant Tutorial correspondence course and my preparation began after I finished my 10th standard. Why IIT IIT is one of the premier institutes in India. I always wanted to get good higher education, so I opted for IIT. My mantra for success There is no short cut to success. One has to work very hard, put in a lot of effort, should have a problem-solving mentality and a right approach to every problem. My parents always stood by me, their support has been invaluable and am overwhelmed. Advice to IIT aspirants Work hard. You have to spend a lot of time preparing as exams are getting more and more competitive. You must also have problem-solving skills. Next move "I have been selected for the scholarship programme at Stanford University for a PhD in operations research. I would like to research on optimising computer networks and operation systems. Will you come back to India? I have been good in studies from my childhood. But preparation for JEE is altogether different from earlier studies. It is all about conceptual clarity rather than cramming. So I always focussed on concepts and supplemented it with required no. of problems. My advice to all the IIT is aspirants out there is: Believe in your capabilities and never under estimate yourself.

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

Mr. Krishnamurthy Rengarajan IIT-B Gold Medallist


PHYSICS

1. Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless, in extensible string and a frictionless pulley as shown in figure. The wedge is inclined at 45º to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the system A and B is released from rest, find

[IIT-1997]

A

m 45º 45º

2m

B

(i) the acceleration of A, (ii) tension in the string, and (ii) the magnitude and direction of the force of

friction acting on A

Sol. µAW = 32 ; µBW = 3

1

A

mg 45º 45º

2mg2

NB B

NA T T

2mg2

2mg

2mg

2mg

Diagram shows the various forces acting on the

masses and their resolution in the direction of motion.

Let us consider the two masses to be a system. The forces trying to move the system, such that A moves upwards and B moves downwards

= 2

mg2 – 2

mg = 2

mg

The forces trying to stop this motion (maximum frictional force)

= fA + fB = µA NA + µB NB

= 32 ×

2mg +

31 ×

2mg2 =

2mg

34

Since the stopping force is more therefore the mass system will not move.

⇒ acceleration of the system is zero. (ii) Taking equilibrium of B into consideration

Since, 2

mg2 > µBNB

×=µ

2mg2

31NBB

∴ The difference of this force is provided by the force of tension

T = 2

mg2 – 31

2mg2 =

32 ×

2mg2 =

3mg22

(iii) T = 3mg22 in upwards direction; weight

component 2

mg in downward direction.

Since T > 2

mg

∴ The frictional force will act in downward direction

f = T – 2

mg = 3

22 mg – 2

mg = 233–4 mg =

2mg

2. A uniform thin rod of mass M and length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin(0, 0). A slight disturbance at t = 0 causes the lower end to slip on the smooth surface along the positive x-axis, and the rod starts falling. [IIT-1993]

(i) What is the path followed by the centre of mass of the rod during its fall?

(ii) Find the equation of the trajectory of a point on the rod located at a distance r from the lower end. What is the shaper of the path of this point?

Sol. (i) Let the position of the rod at any instant be as shown in the figure at any instant. Let the end point of the rod have dimensions(x, 0) and (0, y) and rod making an angle θ with the x-axis as shown in the figure.

B(0,y)

N C

M O (x, 0)A

L/2

L/2

θ

∆CBN is similar to ∆ABO

∴ CBCN =

ABAO ⇒

2/LCN =

Lx

KNOW IIT-JEE By Previous Exam Questions


⇒ CN = L/2 Also ∆ACM is similar to ∆ABO

∴ CACM =

BABO ⇒

2/LCM =

2y

∴ The co-ordinates of C (centre of mass) are

2y,

2x

Also x2 + y2 = L2 [in ∆ABO] using pythagorous theorem]

∴ 22

2y

2x

+

=

2

2L

⇒ Motion of c.m. is a circular path of radius 2L

and the centre of the circle is at origin. (ii) Let us consider a point P(x, y) at a distance r

from A and q be the inclination of rod AB with x-axis at any arbitraty instant of time t.

cos θ = r–L

x (In ∆BPN)

sin θ = ry (in ∆APM)

But cos2θ + sin2θ = 1

⇒ 2

R–Lx

+

2

ry

= 1

3. Three particles A, B and C, each of mass m, are

connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictioness table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity ω.

[IIT-2002]

C B

→F

l

A x

y

ω

(a) Find the magnitude of the horizontal force

exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the

x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time. T.

Sol. The mass B is moving in a circular path centred at A. The centripetal force (mlω2) F'. Therefore a force F' acts on a (the hinge) which is equal to mlω2. The same is the case for mass C. Therefore the net force in the hinge is

Fnet = º60cos'F'F2'F'F 22 ++

Fnet = 21'F2'F2'F2 222 ××+ = 'F3 = 3 mlω2

CB l

A X

y

60º F'

F'

Fnet l l

(b) The force F acting on B will provide a torque to

the system. This torque is

F × 2

3l = Iα

F × 23l = (2ml2)α

⇒ a = lm

F43

The total force acting on the system along x-direction is F + (Fnet)x

This force is responsible for giving an acceleration ax to the system.

c.m 23

l

F Therefore F + (Fnet)x = 3m (ax) c.m.

= 3mm4F Q ax = αr = 4

3lm

F × 3l =

4F

= 4F3

∴ (Fnet)x = – 4F

(Fnet)y remains the same as before = 3 mlω2. 4. Two fixed equal, positive charges, each of magnitude

5 × 10–5 coul. are located at points A and B separated by a distance of 5 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. [IIT-1985]


A +q

– q C

B +q

O D

The moving charge, when it reaches the point C at a

distance of 4 m from O, has a kinetic energy of 4 joules. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

Sol. Total energy of the system of three charges when the charge – q is at C

= P.E + K.E.

= 45

)q(–Kq5

)q)(–q(K6

qKq+

++

× ...(i)

D C

+q

+q

3m 3m 4m

5m

–q

5m

B

A

x

22 3x +

22 3x +

Final energy of the system of three charges when – q

is at D and momentarily at rest = P.E. + K.E.

=

++

++

×2222 3x

)q(–Kq

3x

)q(–Kq6

qKq

= 6

qKq× = 22 3x

)q(–Kq2

+ ...(ii)

By the principle of conservation of energy from (i) and (ii) we get

6

qkq × + 5

)q(–kq2 = 4 = 6

qkq × + 22 3x

)q(–kq2

+

2 = kq2

+ 22 3x

1–51

5–5–9 1051051092

××××× =

51 –

22 3x

1

+

559

20××

= 51 –

22 3x

1

+

22 3x

1

+ =

51 –

454 =

454–9 =

455 =

91

∴ x2 + 9 = 91 ∴ x + 8.48 m

5. A circular ring of radius R with uniform positive charge density λ per unit length is located in the y-z plane with its centre at the origin O, A particle of mass m and positive charge q is projected from the point P(R 3 , 0,0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. [IIT-1993]

Sol. Let the particle be at a distance x from the origin at a point P at any instant. The electric field at P due to charge on the circular ring is

Px

q v A X

)0,0,R3(

Z

Y

E = 04

1πε 2/322 )xR(

xR2+

λπ from P to X

∴ Force acting on q at P F = qE Small amount of work done in moving the particle

toward O for an infinitsinally small distance dx dW = Fdx

= 2/3220 )xR(

xqdxR24

1+λπ

πε

The total work done

W = 04qR2

πελπ ∫ +

0

R32/322 )xR(

xdx ...(i)

Now, ∫ +0

R32/322 )xR(

xdx

Let R2 + x2 = t ∴ 2xdx = dt

⇒ xdx = 2dt

∴ ∫ 2/3t2dt =

t2–

21 = –

t1

= 0

R322 xR

1–

+

= – R1 +

22 R3R

1

+ = –

R1 +

R21 = –

R21

Substituting in (i) we get

W = 04qR2

πελπ ×

R21 =

04qε

λ


This work done needs to be supplied by the kinetic energy given to the charge initially.

∴ 21 mv2 =

04qε

λ ⇒ v = m2

q

0ελ

This is the minimum velocity through which the charge q must be projected to reach the origin.

CHEMISTRY

6. The equilibrium constant Kp of the reaction 2SO2(g) + O2(g) 2SO3(g) is 900 atm–1 at 800

K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989]

Sol. Since to start with SO2 is not present, it is expected that some of SO3 will decompose to give SO2 and O2 at equilibrium. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have

2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x

Hence, Kp = )p()p(

)p(

22

3

O2

SO

2SO =

)xatm2()x2()x2atm1(

2

2

+−

= 900 atm–1 Assuming x


(iv) The gas mixture (B) and (C) is passed into dichromate solution. The solution turns green.

(v) The green solution from step (iv) gives a white ppt. (E) with a solution of Ba(NO3)2.

(vi) Residue (D) from (v) is heated on charcoal in reducing flame. It gives a magnetic substance. Identify compounds (A) to (E) and predict all the equations. [IIT-1980]

Sol. The fore said observations may be briefly summarised as follows :

(a) solidgreen Light

A → 42SOH.Dil )A(

ofSolution

→ 4KMnO Pink colour disappears

(b) A →∆ gas

smelling PungentCB + +

residueBrownD

(c) Solution

722 OCrK →+ CB Green solution

→ 23 )NO(Ba .pptWhite

E

(d) residueBrown

D∆

→ coalChar a magnetic substance

From the last step, one may conclude that brown residue (D) (hence also compound (A)) must be a salt of iron. Since (A) decolourises KMnO4 solution hence it should be a salt of Fe (II). The reactions involved are given below.

MnO4– + 8H+ + 5e– → Mn2+ + 4H2O [Fe2+ → Fe3+ + e–] × 5

Pink4MnO− +

Green

2Fe5 + + 8H+ → Colourless

2Mn + + 5Fe3+ + 4H2O

From observations of (b) and (c), one concludes that compound (A) should be FeSO4 as on heating, it gives pungent gases SO2 and SO3.

)A(

4FeSO2 →∆

)Brown()D(

32OFe + )B(2SO +

)C(3SO

SO2, gas turns dichromate solution green due to formation of green coloured sulphate of chromium (III), the different equations are,

Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O [SO2 + 2H2O → 4H+ + SO42– + 2e–] × 3 Cr2O72– + 3SO2 + 2H+ → 2Cr3+ + 3SO42– + H2O White ppt. (E) is of BaSO4

Green342 )SO(Cr + 3Ba(NO3)2 →

)E(4BaSO3 ↓ + 2Cr(NO3)3

Hence, (A) is FeSO4 (B) is SO2 (C) is SO3 (D) is Fe2O3 and (E) is BaSO4

9. 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 CH32,3-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)

H2Ni

CH3 – CH – CH – CH3

CH3 CH3

CH3 – CH – C = CH2

CH3 CH3(Z)

H2Ni

CH3 – CH – CH – CH3

CH3 CH3

Both, Y and Z can be obtained from following alkyl halide :

CH3 – C – CH – CH3

CH3 CH32-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


10. 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 O

Base

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)

MATHEMATICS

11. If one of the vertices of the square circumscribing the

circle |z – 1| = 2 is 2 + i3 . Find the other vertices of square. [IIT-2005]

Sol. Here, centre of circle is (1, 0) is also the mid-point of diagonals of square

z1(2, 3 )

(1, 0) x

z4

z2

z0

z3

y

⇒ 2

21 zz + = z0

⇒ z2 = – i3 , (where z0 = 1 + 0 i)

and 1–1–

1

3

zz

= e± iπ/2

⇒ z3 = 1 + (1 + i3 ).

π±

π2

sin2

cos i ,

as z1 = 2 + i3

= 1 ± i(1 + i3 )

= (1 m 3 ) ± i

z3 = (1 – 3 ) + i and z4 = (1 + 3 ) – i 12. If M is a 3 × 3 matrix, where MT M = I and det (M) = 1,

then prove that det (M – I) = 0. [IIT-2004] Sol. As, MT M = I and |M| = 1 ⇒ |MT M| = | I | ⇒ |MT M| = |M| {as | I | = 1 = |M|} ⇒ |MT| |M| – |M| = 0 ⇒ |M| (|MT| – 1) = 0 ⇒ |M| = 0 or |MT| = 1 {neglecting |M| = 0} ⇒ |MT| = 1 ...(i) ∴ |M – I| = |M – I| |MT| = |MMT –MT| = |I – MT| = – |MT – I| = – |M – I|T ⇒ |M – I| + |M – I| = 0 ⇒ |M – I| = 0 13. A function f : R → R satisfies the equation

f (x + y) = f (x) f (y) for all x, y in R and f (x) ≠ 0 for any x in R. Let the function be differentiable at x = 0 and f ' (0) = 2. Show that f '(x) = 2 f (x) for all x in R. Hence, determine f (x). [IIT-1990]

Sol. We have, f(x + y) = f (x). f (y) for all x, y ∈ R. ∴ f (0) = f (0). f (0) ⇒ f (0) {f (0) – 1} = 0 ⇒ f (0) = 1 {Q f (0) ≠ 0}

Now, f '(0) = 2 ⇒ h

fhfh

)0(–)0(lim0

+→

= 2

⇒ hhf

h

1–)(lim0→

= 2 {Q f (0) = 1}

⇒ f '(x) = h

xfhxfh

)(–)(lim0

+→

= h

xfhfxfh

)(–)().(lim0→

,

{using f (x + y) = f (x). f(y)}

= f (x) .

→ hhf

h

1–)(lim0

= 2 f (x) ∴ f ' (x) = 2 f (x)


or )()('

xfxf = 2 ; integrating both sides between 0 to x,

we get

∫x

dxxfxf

0)()(' = 2x

⇒ log |f (x)| – log |f (0)| = 2x ⇒ loge|f (x)| = 2x,

{as, f (0) = 1 ⇒ log|f (0)| = 0} ⇒ f (x) = e2x

14. Let O(0, 0), A(2, 0) and B(1, 3

1 ) be the vertices of a

triangle. Let R be the region consisting of all those points P inside ∆OAB which satisfy d(P, OA) ≥ min {d(P, OB), d(P, AB)}, when d denotes the distance from the point to the corresponding line. Sketch the region R and find its area. [IIT-1997]

Sol. Let the coordinate of P be (x ,y)

O

y

(0, 0)

P

A(2, 0) x

B(1, 1/ 3 )

Equation of line OA be y = 0

Equation of line OB be 3 y = x

Equation of line AB be 3 y = 2 – x d(P, OA) = distance of P from line OA = y

d(P, OB) distance of P from line OB = 2

|–3| xy

d(P, AB) = distance of P from line

AB = 2

|2–3| xy +

Given : d(P, OA) ≤ min. {d(P, OB), d(P, AB)}

y ≤ min.

+

22–3|,

2|–3| xyxy

⇒ y ≤ 2

|–3| xy and y ≤ 2

|2–3| xy +

Case I : y ≤ 2

|–3| xy , {since 3 y – x < 0}

⇒ y ≤ 2

3– yx

⇒ (2 + 3 ) y ≤ x. ⇒ y ≤ x tan 15º

Case II : If y ≤ 2

|2–3| xy +

⇒ 2y ≤ 2 – x – 3 y [i.e. 3 y + x – 2 < 0]

⇒ (2 + 3 )y ≤ 2 – x ⇒ y ≤ – tan 15º. (2 – x) from above discussion P moves in side the ∆ as

shown,

O

y

(0, 0)

Q

A(2, 0) x

B(1, 1/ 3 )

C (1, 0)

⇒ Area of shaded region = area of ∆OQA

= 21 (base) × height

= 21 (2) (tan 15º) = tan 15º

= (2 – 3 ) sq. units

15. Circle with radii 3, 4 and 5 touch each other externally if P is the point of intersection of tangents to these circles at their points of contact. Find the distance of P from the point of contact. [IIT-2005]

Sol. As the circles with radii 3, 4 and 5 touch each other externally and P is the point of intersection of tangents

R

C3 3 3

4

4C2 C1 5

5

A

⇒ P is incentre of ∆C1C2C3. Thus distance of point P from the points of contact

= In radius (r) of ∆C1C2C3.

i.e., r = s∆ =

scsbsass )–)(–)(–(

where 2s = 7 + 8 + 9 ∴ s = 12

Hence, r = 12

)9–12)(8–12)(7–12( =

123.4.5 = 5


1. Initially the capacitor is uncharged, the switch is put to position 1. t1 sec later charge on the capacitor is

C9µ . After steady state is reached the switch is put to position 2. t1 sec. later charge on the capacitor is

C3µ . The capacitance of the capacitor is –

C

S 1 2

4 V (A) 1.5 µF (B) 3 µF

(C) 6 µF (D) 3.37 µF

2. A pure inductor is connected across a battery with no internal resistance. The graph of current (i) Vs time (t) is best represented by –

(A)

i

t

(B)

i

t

(C)

i

t

(D)

i

t

3. When two identical voltmeters are connected so as to measure the potential difference across the resistors of Ω3 and Ω8 respectively, their readings are same. The resistance of each voltmeter is –

2Ω R 8Ω

5Ω

3Ω

(A) 72 Ω (B) 76 Ω

(C) 80 Ω (D) 88 Ω

4. A circular loop is placed near a long straight current carrying wire

i

Column – I Column – II

(A) Induced current in (P) i is increased loop is in clockwise direction shown direction

(B) Induced current in (Q) wire with constant loop is in anticlockwise i moved away direction from loop (C) wire will repel loop (R) i is decreased in the direction shown (D) wire will attract loop (S) loop is moved towards the wire

(T) Direction of current i is reversed and then it is further decreased.

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 # 2


Passage # (Q. No. 5 to Q. No. 7) A loudspeaker system uses alternating current to amplify sound of certain frequencies. It consists of 2 speakers.

C

v0 sin(ωt) ~

woofer L

Tweeter

Tweeter-which has smaller diameter produces high

frequency sounds. Woofer which has large diameter produces low frequency sound. For purpose of circuit analysis, we can take both speakers to be of equal resistance R. The equivalent circuit is show in the figure. The 2 speakers are connected to the amplifier via capacitance and inductance respectively. The capacitor in tweeter branch blocks the low frequency sound but passes the high frequency. The inductor in woofer branch does the opposite.

5. Which plot correctly represents r.m.s. current against frequency –

(A)

Woofer

Tweeter f

I

(B)

Woofer

Tweeterf

I

(C)

Woofer

Tweeter f

I (D) Woofer

Tweeter

f

I

6. What is the frequency which is sounded equally loudly by both speakers –

(A) LC1

LR

21

2

2−

π (B)

LC1

LR4

21

2

2−

π

(C) 22

L4R

LC1

21

−π

(D) LC21π

7. For a combination of L, R and C the current in

woofer and tweeter are always found to have a phase

difference of 2π . What is the relation between L, R

and C –

(A) L = 2R2C (B) CR2L 2=

(C) L = R2C (D) 2CRL

2=

8. The four terminal network shown in the figure, consists of four equal resistors and is a part of a larger circuit. The points A, B and C are at same potential. The potential difference between A and D is 40V. Find the potential difference between O and D.

A

B D

C

O r

r

r

r

Regents Physics You Should Know Electricity

1. A coulomb is charge, an amp is current [coulomb/sec] and a volt is potential difference [joule/coulomb].

2. Short fat cold wires make the best conductors. 3. Electrons and protons have equal amounts of

charge (1.6 x 10-19 coulombs each). 4. Adding a resistor in parallel decreases the total

resistance of a circuit. 5. Adding a resistor in series increases the total

resistance of a circuit.

6. All resistors in series have equal current (I).

7. All resistors in parallel have equal voltage (V). 8. If two charged spheres touch each other add the

charges and divide by two to find the final charge on each sphere.

9. Insulators contain no free electrons. 10. Ionized gases conduct electric current using

positive ions, negative ions and electrons.

11. Electric fields all point in the direction of the force on a positive test charge.

12. Electric fields between two parallel plates are uniform in strength except at the edges.

13. Millikan determined the charge on a single electron using his famous oil-drop experiment.

14. All charge changes result from the movement of electrons not protons (an object becomes positive by losing electrons)


1. Dipole moment = 2Lλ And torque on dipole WxLE =λ= Option [C] is correct

2. Conceptual Option [B] is correct 3. Option [A] is correct

4. Final current is RV that distributes in inverse ratio

to inductance. Option [C] is correct

5. Option [B] is correct

6. Option [B] is correct Transitions emitting photons of energy more than

work function W = 2.3eV can result in a photoelectric.

max

minmK2

hmK2h

ph

=λ⇒==λ

Kmax = 13.6 – 0.85

7. Option [C] is correct 8. Option [C] is correct

Acceleration = masspulled

forcepullingNet

2s/m10gmqE

mmgqE

=−=−

=

at 0.5 sec v = 5 m/s

and RvBmgqE

dtdvm

22 l−−=

a = 20 – 10 – 10 = 0

dtv210

dvv210dtdV

=−

⇒−=

]e31[5v )4t2( −−+=⇒

Solution Physics Challenging Problems

Set # 1

8 Questions were Published in May Issue

THE BRAND NEW EMOTIONAL ROBOT

Most of the people always think, the robots are only the machines that didn’t have any feeling. It’s definitely impossible for the scientists to apply emotion on the robots.

Well, please don’t be so sure about your thought yet, as the scientists at Georgia Tech had decided to test our ability to interpret a robot’s emotion. The research group discovered that older adults showed some unexpected differences in the way they read a robot’s face from the way younger adult did.

Jenay Beer, a graduate student in Georgia Tech’s School of Psychology described that the home-based assistive robots have the potential to help older adults, as they can be used to keep the older adults independent longer. As a result, it reduces healthcare needs and provides everyday assistance to the elders.

Based on the previous research, the robot found out that older adults are less accurate in recognizing anger, fear and happiness. Furthermore, the older adults have problem recognizing the happy robot compared with their success in recognizing happy people.

Another interesting fact about the experiment was the researchers discovered that neither the young nor old could easily distinguish the emotion disgust on the virtual iCat. It might be due to the difficulty in programming a robot to show the emotion!


1. Two blocks of mass m1 and m2 are attached at the

ends of an ideal spring of force constant K and natural length l0. The system rests on a smooth horizontal plane. Blocks having mass m1 and m2 are pulled apart by applying force F1 and F2 respectively as shown in figure. Calculate maximum elongation of the spring.

F2F1 m1 m2 K

Sol. Let at an instant, displacement of m2 be x2 (rightwards) and that of m1 be x1 (rightwards) and let x2 > x1.

Elongation of spring at this instant is x = (x2 – x1) But (x2 – x1) is displacement of m2 relative to m1 . Let rightward acceleration of two blocks at this

instant be a1 and a2 , Considering the free body diagrams, as shown in

figure

m1.a1

Kx

N1

F1

m1g

m2.a2

F2

N2

F1

m2g

Considering horizontal forces,

Kx – F1 = m1a1 or a1 = 1

1

mF–Kx

F2 – Kx = m2a2 or a2 = 2

2

mKx–F

Acceleration of m2 relative to m1 is a = a2 – a1

or a = 2

2

mKx–F –

1

1

mF–Kx

= 21

1221

m.mFmFm + – Kx

+

21

21

mmmm

But a = dxdyv where v = velocity of m2 relative to m1.

At initial moment, elongation x = 0 and v = 0. At the instant of maximum elongation velocities of two blocks are identical or at x = xmax, v = 0.

∴ ∫=

=

0v

0v

duv = ∫=

++maxx

0x 21

21

21

1221 xKm.mmm–

m.mFmFm dx

or 0= 21

1221

m.mFmFm + (xmax – 0)

–

+

21

21

m.mmm )0–x(

2K 2

max

or xmax = K2 .

++

21

121

mmFmFm Ans.

2. One end of an ideal spring is fixed to a wall at origin O and axis of spring is parallel to x-axis. A block of mass m = 1 kg is attached to free end of the spring and its is performing SHM. Equation of position of the block in co-ordinate system shown in figure is x = 10 + 3. sin (10.t),t is in second and x in cm.

Another block of mass M = 3 kg, moving towards the origin with velocity 30 cm/sec collides with the block performing SHM at t = 0 and gets stuck to it Calculate

3 Kg1 Kg

xO

(i) new amplitude of oscillations, (ii) new equation for position of the combined body

and (iii) loss of energy during collision. Neglect friction. Sol. Natural length of the spring is 10 cm and force

constant of the spring is K = 100 Nm–1 . Just before collision, velocities of 1 kg block and

3 kg block are (+ 0.30 ms–1) and (– 0.30 ms–1) respectively. Let velocity of combined body just after collision be v, then, according to law of conservation of momentum, (1 + 3) v = 1 (0.30) + 3 ( – 0.30)

or v = – 0.15 ms–1. Negative sign indicates that combined body starts to

move leftward. But at the instant of collision spring is in its natural length or combined body is in equilibrium position. Hence at t = 0, phase of combined body becomes equal to π.

Now angular frequency of oscillations of combined body is

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

Students' ForumPHYSICS


ω' = Mm

K+

= 31

100+

= 5 rad sec–1 .

∴ New amplitude of oscillations is

a' = '|v|

ω =

515.0 = 0.3 m or 3 cm Ans. (i)

∴ Equation for position x of combined body is given by

x = l0 + a' sin(ω't + π) or x = 10 + 3 sin (5t + π) cm or x = 10 – 3 sin (5t) cm Ans. (ii) Kinetic energy of two blocks (Just before collision)

= 21 m(0.3)3 +

21 M(0.3)2 = 0.18 joule

Kinetic energy of combined body (just after collision)

= 21 (m + M) v2 = 0.045 Joule

∴ Loss of energy, during collision = 0.18 – 0.045 joule = 0.135 joule Ans.(iii) 3. A small sphere is charged uniformly and placed at

point A(u, v) so that at point B(8, 7) electric field

strength is →E = (54

î + 72

^j ) NC–1 and potential is +

900 volt. Calculate (i) magnitude of charge, (ii) co-ordinates of point A, and (iii) if di-electric strength of air is 3 × 106 Vm–1,

minimum possible radius of the sphere. Sol. Since, potential due to sphere is positive, therefore, it

is positively charged. Let magnitude of charge on sphere be q and let distance AB be equal to r.

Potential at B, V = 9 × 109 2rq = 900 volt ...(i)

Magnitude of Electric field at B is E

= 9 × 109 2rq = 22 7254 + NC–1

or 9 × 109 2rq = 90 NC–1 ...(ii)

Dividing equation (i) by (ii) r = 10 m Substituting this value in equation (i), q = 10–6 coulomb or 1µC Ans. (i) Since q is a positive charge, therefore

→

AB||→E or

|AB|

AB→

→

= |E|

E→

→

or r

j)v–7(i)u–8(^^

+ = 90

j72i54^^

+

or (8 – u)î + (7 – v)

^j = 6

î + 8

^j

∴ u = 2 and v = –1 or Co-ordinates of point A are (2, –1) Ans. (ii) Since, minimum radius of sphere corresponds to

electric field strength at surface of sphere to be equal to dielectric strength of air.

Therefore radius R of sphere is given by,

9 × 109 2Rq = 3 × 106 Vm–1

∴ R = m103 3–× = 30 cm = 5.48 cm Ans. (iii) 4. A steady beam of α-particles travelling with kinetic

energy E = 83.5 keV carries current of I = 0.2 µA. (i) If this beam strikes a plane surface at an angle

θ = 30º with normal to the surface, how many α-particles strike the surface in t = 4 second?

(ii) How many α-particles are there in length l = 20 cm of the beam?

(iii Calculate power of the source used to accelerate these α-particles from rest.

(Mass of α-particle = 6.68 × 10–27 kg) Sol. Since, current is rate of flow of charge through a

section, there fore, a current I = 0.2 µA means that a charge 0.2 µC is flowing per second.

Charge of an α-particle θ = 2e = 3.2 × 10–19C ∴ Rate of flow of α-particles,

n = qI = 6.25 × 1011 per second

∴ Number of α-particles striking against a surface in t = 4 second

= n × t = 6.25 ×1011 × 4 = 2.5 × 1012 Ans.(i) (Note : There is no significance of angle θ for

calculation of number of α-particles striking the surface.)

Kinetic energy of each α-particle is E = 83.5 ke V or E = (83.5 × 103) (1.6 × 10–19) J

But E = 21 mv2 where m = 6.68 × 10–27 kg

∴ Velocity of α-particles is v = 2 × 106ms–1 . It means a beam of length v = 2 × 106 m crosses a

section in one second. But number of α-particles passing through a section in one second is n = 6.25 × 1011

∴ Number of α-particles in unit length of the beam

= vn = 3.125 × 105 per m.

∴ Number of α-particle in length l of beam = vn

l

= 6.25 × 104 Ans.(ii)


Let potential difference of the source be V volt. Kinetic energy of α-particles accelerated by this

source,

E = qV or V = qE = 41.75 kV

Power supplied by the source to accelerate α-particles,

P = VI = 8.35 × 10–3 watt Ans. (iii) 5. A two way switch S is used in the circuit shown in

Fig. First, the capacitor is charged by putting the switch in position 1.

2

0.1F 1

S

60V+ – 10Ω

4Ω

6Ω

3Ω Calculate heat generated across each resistor when

switch is in position 2.

Sol. Initially the switch was in position 1. Therefore, initially potential difference across capacitor was equal to emf of the battery i.e. 60 volt.

∴ Initially energy stored in the capacitor was

U =21 CV2 =

21 × 0.1 × 602 J

= 180 J q

4Ω

+ –

i 6Ω

3Ω

i1

i

i2

When switch is shifted to position 2, capacitor begins to discharge and energy stored in it is dissipated in the form of heat across resistances. Let at some instant discharging current through the capacitor be i as shown in Fig.

According to Kirchhoff's laws, i1 + i2 = i … (1) 6i1 – 3i2 = 0 or i2 = 2i1 … (2) From above two equations,

i1 = 3i and i2 = 3

2 i

But thermal power generated in a resistance R is P = i2R where i is current flowing through it. Therefore, heat generated P1, P2 and P3 across 4Ω, 6Ω and 3Ω resistances is in ration 22

21

2 i3:i6:i4

or P1 : P2 : P3 = 4 : 32 :

34 = 6 : 1 : 2

But total heat generated is P1 + P2 + P3 = U ∴ Heat generated across 4Ω is P1 = 120 J Ans. Heat generated across 6Ω is P2 = 20 J Ans.

Heat generated across 3Ω is P3 = 40 J Ans. Since, during discharging, no current flows through

10Ω, therefore heat generated across it is equal to zero. Ans.

Regents Physics You Should Know Mechanics

1. Weight (force of gravity) decreases as you move away from the earth by distance squared.

2. Mass and inertia are the same thing.

3. Constant velocity and zero velocity means the net force is zero and acceleration is zero.

4. Weight (in newtons) is mass x acceleration (w = mg). Mass is not weight!

5. Velocity, displacement [s], momentum, force and acceleration are vectors.

6. Speed, distance [d], time, and energy (joules) are scalar quantities.

7. The slope of the velocity-time graph is acceleration.

8. At zero (0) degrees two vectors have a resultant equal to their sum. At 180 degrees two vectors have a resultant equal to their difference. From the difference to the sum is the total range of possible resultants.


Electric Potential Energy: If a point charge q1 is present in an electric field

where potential is V, by definition V = (U/q1) i.e., U = q1V And if the field is produced by a point charge q2

which is at a distance r12 from q1,

U =

πε 12

2

01 r

q4

1q = 12

21

0 rqq

41πε

…(1)

So in case of discrete distribution of charges

++

πε= .......

rqq

rqq

41U

23

32

12

21

0 =

ji ij

ji

0 rqq

41

21

≠∑πε

Here it is worth noting

(a) Electrical potential energy is not localised but is distributed all over the field

(b) If the electric potential energy of a system in one configuration is U1 and in the other UF, work done in changing the configuration will be

WIF = – UIF = – (U1 – UF) = UF – U1 And as potential energy at infinity is zero, work done

in assembling or disassembling a given charge distribution will be respectively,

W = UF [as U1 = 0] and W = – U1 [as UF = 0] (Assembling) (Disassembling) Motion of a Charged Particle in an Electric Field In case of motion of a charged particle in an electric

field :

(a) As by definition of electric intensity →E ,

→F = q

→E , a

point charge always experiences a force either at rest or in motion.

(b) The direction of force is parallel to the field if the charge is positive and opposite to the field if charge is negative.

E

→→= EqF

+

E

→→= Eq–F

–

(A) (B)

(c) Electric field is conservative in an electric field work is path independent and work done in moving a point

charge q between two fixed points having potential difference V is equal to,

WAB = – UAB = q(VB – VA)qV …(1) And hence in moving a charged particle in an electric

field work is always done unless the points are at same potential as shown in figure. [However, in magnetic field as force is always perpendicular to motion, work done is always zero.]

B

A

III

+q

+Q L L

B A

I

II

B

III +Q

A

(A) (B) (C)

(d) When a charged particle is accelerated by an electric field (uniform or non-uniform) by Work energy theorem, i.e., ∆KE = W, we have

22 mu21–m

21

υ = qV [as from Eq. (1) W = qV]

or υ =

+mqV2

u 2 …(2)

And if the charged particle is initially at rest, i.e., u = 0

υ = mqV2 …(3)

And if the field is uniform, i.e., E = (V/d)

υ = mqEd2 …(4)

(e) In case of motion of a charged particle in a uniform

electric field if force of gravity does not exist (or is balanced by some other force

→a =

mF→

= mEq→

= constant [as →F = q

→E ] …(5)

So equation of motion are valid. Now there are two possibilities:

(i) If the particle is initially at rest From Eq. υ = u + at, we get

υ = at = tmqE

==

mqEaand0uas …(6)

Electrostatics-2

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


And from Eq. s = ut + 2at21

s = 2at21 = 2t

mqE

21 …(7)

i.e., the motion is accelerated translatory with a ∝ t0 ; υ ∝ t and s ∝ t2 Further more in this situation :

W = ∆KE = 21 mυ2 =

21 m

2

tmqE

=υ

mqEt)6.(fromEqas

+ + +

+ +

E

+

– – –

– –

–

d P D = V

+q F

Which in the light of Eq. (7) with s = d, gives W = qEd = qV [as E = V/d] …(8)

(ii) If the particle is projected perpendicular to the field with an initial velocity υ0

From Eq. υ = u + at and s = ut +21 at2 respectively

for motion along x-axis as u = υ0 and a = 0, υx = υ0 = constt. and x = υ0t …(9)

While for motion along y-axis as u = 0 and a = (qE/m),

υy = tmqE

and y = 2t

mqE

21

…(10)

+ + + + + + + +

– – – – – – –

θ

y

L D

–q V0

Y

So eliminating t between equation for x and y,

we have

y = 2

0

xm2

qE

υ

= 20m2

qEυ

…(11)

i.e., the path is a parabola. [However, under same conditions in magnetic

field path is a circle.] Electric Dipole

Definition: If two equal and opposite point charges are separated by a distance 2l such that the distance of field point r >>2l, the system is called a dipole.

(a) Field of a Dipole Potential due to dipole at a point (r, θ) as shown

in Figure will be

V = V1 + V2 =

πε 210 r

q–rq

41

=

πε 21

21

0 rrr–r

4q

2lcosθ

r2 r r1

P

+q–ql l

θ

Now as , r >> 2l r1 × r2 = r2 and r2 – r1 = 2l cosθ

So, V = 20 r

)cos2(q4

1 θπε

l

V = 20 r

cosP.4

1 θπε

[as p = q × 2l] ] …(1)

(b) Potential will be minimum when |cos θ| = min = 0, i.e., θ = 90º. So for broad on, equitorial or tan B position, potential is minimum and is zero, i.e.,

Vmin = 0 This all is shown Figure

+q

BA

–q

O P

r

Vmax = 2

0 rP

41πε

Vmax = 3

0 rP2

41πε

End-on, Axial or tan A position

(A)

+q

BA

–qO

P

Vmax = 0

Vmax = 30 r

P4

1πε

Broad on, Equitorialor tan B position

(B) (c) Now as electric field,

→E = –

→n

dxdV

So component of electric intensity in the direction of r :

Er =

θπε 20 r

cosp4

1drd– = 3

0 rcosp2

41 θπε

…(i)

And perpendicular to r,

Eθ =

θπεθ 20 r

cosp4

1rdd– = 3

0 rsinp

41 θπε


…(ii)

So that, tanφ = 21

EE

r=θ tan θ

and, E = 22r EE θ+ = )cos31(rP

41 2

30

θ+πε

…(2) From this it is clear that : (1) Intensity due to a dipole varies as (1/r3) and can

never be zero unless r → ∞ or p → 0. (2) E will be maximum when cos2θ = max = 1, i.e.,

θ = 0º, i.e., for end on, axial or tan A position E is maximum and is,

Emax = 30 r

p24

1πε

(3) E will be minimum when cos2 θ = min = 0, i.e., θ = 90º, i.e., for broad on, equitorial or tan B position E is minimum and is,

Emin = 30 r

p4

1πε

Conductor in Electrostatics The substances such as metals which allow the

charge to flow freely through them are called conductors. In metals conduction envolves the movement of free electrons. In case of conductors in electrostatics, it is worth noting that :

(a) In charging a conductor electrons are removed, conductor becomes positively charged and its potential increases and if added, it becomes negatively charged and its potential decreases.

(b) When a conductor is charges by induction, induced charge (Which is free to move) is equal and opposite to the inducing charge, i.e.,q' = –q

(c) Charge resides on the outer surface of a conductor. However, distribution of charge on the surface is generally not uniform and surface density of charge varies inversly as the radius of curvature of that part of the conductor, i.e.,

σ ∝ (1/R) (d) The dielectric constant of conductors in

electrostatics is infinite, i.e., K = ∞ (e) Electric intensity inside a conductor is zero

while outside (near its surface) is (σ/ε0), i.e., Ein = 0 and Eout = (σ/ε0)

(f) Conductor is an equipotential surface, i.e., potential at its surface or inside everywhere is same, i.e., for a conductor

V = constt. (g) Electric field and hence lines of force are

normal to the surface of a conductor. (h) The field in a cavity inside a conductor is zero

resulting in 'electrostatic shielding.'

Vs = vin = constt.

++

++ + + + +

+ + +

+

0

2

2dsdF

εσ

=

0outE ε

σ=

E0 = 0Ein = 0

σ1(>σ2) σ2(A) (B)

1. Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the position where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the position of the charge q due to the other two charges ?

Sol. As potential energy of two point charges separated by a distance r is given by U(=q1q2/4πε0r), so to have minimum potential energy the charges of greater value should be farthest, i.e., q must be between 2q and 8q. Let q be at a distance x from 2q, then potential energy of the system will be

U =

×++

πε dq2q8

)x–d(qq8

xqq2

41

0

For U to be minimum (dU/dx) = 0 ]

i.e., 22

2

2

)x–d(q8

xq2– + = 0

2q q 8qx (d–x)

d i.e., 2x = (d – x) or x = (d/3) = (9/3) = 3 cm So to have minimum potential energy, the charge q

must be placed at a distance of 3 cm from 2q between 2q and 8q on the line joining the charges. In this situation, Field at q

E = 0)6(q8–

)3(q2

41

220

=

πε Ans.

2. Three point charges 1C, 2C and 3C are placed at the corners of an equilateral triangle of side 1m. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m as shown in Figure (A)

B C

A

CB

A 1

(A) 3

Solved Examples


Sol. As potential energy of two charges separated by a distance r is given by U = [q1q2/4πε0r], the initial and final potential energy of the system will be

(US)I =

×+

×+

×πε 1

131

321

214

1

0

= 9 × 109 × 11 = 9.9 × 1010 J

(US)F =

×+

×+

×πε 5.0

135.032

5.021

41

0

= 9 × 109 × 22 = 19.8 × 1010J So work done in changing the configuration of the

system: W = (US)F – (US)I = (19.8 – 9.9) × 1010 = 9.9 × 1010J Ans. 3. A particle of mass 40 µg and carrying a charge

5 × 10–9 C is moving directly towards fixed positive point charge of magnitude 10–8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest ? Is the acceleration constant during motion ?

Sol. If the particle comes to rest momentarily at a distance r from the fixed charge, then from 'conservation of energy' we have

2mu21 +

aQq

41

0πε =

rQq

41

0πε

Substituting the given date, we get :

21 × 40 × 10–6 ×

21 ×

21 = 9 × 109 × 10–8 × 5 × 10–9

10–

r1

or r1 – 10 = 8–

6–

1059105××

× = 9

190

or r1 =

9100 + 10 =

9100 m

i.e., r = 4.7 × 10–2 m

As here, F = 20 r

qQ4

1πε

So acc. = mF ∝ 2r

1

i.e., acceleration is not constant during motion. 4. A very small sphere of mass 80 g having a charge q

is held at a height 9 m vertically above the centre of a fixed conducting sphere of radius 1m, carrying an equal charge q. When released it falls until it is repelled just before it comes in contact with the sphere. Calculate the charge q. [g = 9.8 m/s2]

+ + + + +

+++ ++

+ + +

+

+qA

9m

B

1m

Sol. Keeping in mind that here both electric and gravitational potential energy are changing and for external point a charged sphere behaves as whole of its charge were concentrated at its applying conservation of energy between initial and final position, we have

9

qq4

1

0πε + mg × 9 =

1q

41 2

0πε + mg × 1

or q2 = 93–

108.91080 ××

q = 28 µC Ans. 5. The distance between the two plates of a cathode-ray

oscillograph is 1 cm and potential difference between them is 1200 volt. If an electron of energy 2000 eV enters at right-angles to the field, what will be its deflection if the plates be 1.5 cm long ?

Sol. As distance between the plates is 1 cm and potential difference 1200 V, the field between the plates

E = dV = 2–101

1200×

= 1.2 × 105 mV …(1)

+ + + + + +

– – – – – –

y

L

–eV0 E

x d

a

y

So the electron will experience a force Fe = eE

opposite to the field as shown in Figure and hence acceleration of electron along y-axis:

a = mF =

meE

= constt. ...(2) So from equation of motion,

s = ut + 2at21

Along x-axis, L = υ0t [as a = 0] …(3)

and along y-axis, y = 2at21 [as u = 0] …(4)

Eliminating t, between Eqs. (3) and (4)

y = 2

0uLa

21

= 2

0

2

mueEL

21

=

meEa)2(.Eqfromas

or y = 2LKeE

41

υ= 20m2

1Kas

Substituting the given data and value of E from Eq. (1),

y = 19–22–519–

106.1200)105.1()102.1)(106.1(

41

××××××

= 3.375 mm Ans.


Newton's First Law (or Law of Inertia) : A body continues to maintain its state of equilibrium

till disturbed by an unbalanced force i.e. it continues to maintain its state of rest or of uniform motion till an unbalanced external force disturbs it. This law is also called Galileo Law or Law of Inertia.

Newton's second Law : The rate of change of momentum is equal to the force

applied on the body and this change takes place in the

direction of force applied i.e. →F =

dtpdr

Newton's third law (Action-reaction law) : To every action there is equal and opposite reaction

and both act on two different bodies.

Mathematically ABF→

= BAF→

or mAaA = mBaB (in magnitude) i.e. for the same force acting on two bodies the

massive body has less acceleration than a light body. Impulse : If two bodies moving along a straight line collide,

then the collision is small and the force experienced during collision on any of the two bodies varies with time and has a large value. In such cases the net effect of force can be measured with the help of a physical quantity called Impulse.

since →F =

dtpdr

⇒ pdr

= →F dt = Id

r

⇒ Ir

= ∫2

1

p

p

pdr

= dtF2

1

t

t∫

r

⇒ Ir

= dtF2

1

t

t∫

r = 2p – 1p

So, Impulse = total change in momentum Also called Impulse-Momentum theorem Important :

Newton's second law is the real law of motion as the First law and Third law can be derived form it.

The concept of impulse must be applied at those places where an extremely large force acts on a body for a very small time interval. Then, impulse is just

defined as the product of extremely large force with the very small time value.

Impulse is a vector quantity having direction along the force.

Equilibrium : A body is said to be in the equilibrium state when (a) no net force acts on the body 0F

rr=∑ (Condition for translational equilibrium)

⇒ ∑ xF = 0 ∑ yF = 0 ∑ zF = 0 (b) no net torque acts on the body : 0p

rr=∑ (condition for rotational equilibrium)

⇒ ∑τx = 0 ∑τy = 0 ∑τz = 0 This statement is none other than law of conservation

of moments according to which the above condition can be restated as

∑

momentus

clockwise total = ∑

momentus

iseanticlockw total

Important : For Rotational Equilibrium

∑

momentus

clockwise total = ∑

momentus

iseanticlockw total

Frames of Reference : The system/co-ordinate system/a platform w.r.t.

which the position or the motion of a body is determined is called a frame of reference. The simplest frame of reference having all the properties of a frame is the Cartesian co-ordinate frame/system.

Frame of reference are of two types : S.N. Inertial frame Non-Inertial frame 1. Newton's laws are valid

in the inertial frames. Newton's Laws are not valid in the non-inertial frame. They are to be modified by introducingthe concept of pseudo force.

2. All non-accelerated frames (frames at rest or frames moving with uniform velocity) are inertial frames.

All accelerated frames are non-inertial frames.

Newton's Law of motion PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


3. A particle moves with uniform velocity in the absence of an external force.

In this frame of reference the particle doesn't move with uniform velocity.

4. A frame of reference moving with constant velocity with respect to an inertial frame is also inertial.

A rotating frame of reference is a non-inertial frames and an example for this is the earth.

Concept of Pseudo force : If a body of mass m is placed in a non-inertial frame

having acceleration 0ar

then it experiences a pseudo force m 0a

racting in a direction opposite to the

direction of 0ar

(the acceleration of non-inertial frame). So,

pseudoFr

= – m 0ar

where, negative sign indicates the pseudo force is always directed in a direction opposite to the direction of the acceleration of the frame.

While drawing free Body Diagrams (FBDs) in which pseudo force is involved, we must first see the acceleration of the non internal frame and then in the FBD, plot the pseudo force with a value ma in a direction opposite to the acceleration of non-inertial frame.

y

xm Fpseudo = ma0

a0

"No negative sign has then to be applied to the value

of pseudo force as its direction in the indicated correctly".

Problem Solving strategy Newton's First Law : Equilibrium of a Particle Step 1 : Identify the relevant concepts : You must

use Newton's first law for any problem that involves forces acting on a body in equilibrium. Remember that "equilibrium" means that the body either remains at rest or moves with constant velocity. For example, a car is in equilibrium when it's parked, but also when it's driving down a straight road at a steady speed.

If the problem involves more than one body, and the bodies interact with each other, you'll also need to use Newton's third law. This law allows you to relate the force that one body exerts on a second body to the force that the second body exerts on the first one.

Be certain that you identify the target variable(s). Common target variables in equilibrium problems

include the magnitude of one of the forces, the components of a force, or the direction of a force.

Step 2: Set up the problem using the following steps: Draw a very simple sketch of the physical

situation, showing dimensions and angles. You don't have to be an artist!

For each body that is in equilibrium, draw a free-body diagram of this body. For the present, we consider the body as a particle, so a large dot will do to represent it. In your free-body diagram, do not include the other bodies that interact with it, such as surface it may be resting on, or a rope pulling on it.

Now ask yourself what is interacting with the body by touching it or in any other way. On your free-body diagram, draw a force vector for each interaction. If you know the angle at which a force is directed, draw the angle accurately and label it. A surface in contact with the body exerts a normal force perpendicular to the surface and possibly a friction force parallel to the surface. Remember that a rope or chain can't push on a body, but can only pull in a direction along its length. Be sure to include the body's weight, except in case where the body has negligible mass (and hence negligible weight). If the mass is given, use w = mg to find the weight. Label each force with a symbol representing the magnitude of the force.

Do not show in the free-body diagram any forces exerted by the body on any other body. The sums in Eq. ∑F

r = 0 (particle in equilibrium, vector

form) and ∑ xF = 0; ∑ yF = 0 (particle in equilibrium, component form) include only forces that act on the body. Make sure you can answer the question "What other body causes that force?" for each force. If you can't answer that question, you may be imagining a force that isn't there.

Choose a set of coordinate axes and include them in your free-body diagram. (If there is more than one body in the problem, you'll need to choose axes for each body separately.) Make sure you label the positive direction for each axis. This will be crucially important when you take components of the force vectors as part of your solution. Often you can simplify the problem by your choice of coordinate axes. For example, when a body rests or sides on a plane surface, it's usually simplest to take the axes in the directions parallel and perpendicular to this surface, even when the plane is tilted.

Step 3 : Execute the solution as follows : Find the components of each force along of the

body's coordinate axes. Draw a wiggly line through each force vector that has been replaced


by its components, so you don't count it twice. Keep in mind that while the magnitude of a force is always positive, the component of a force along a particular direction may be positive or negative.

Set the algebraic sum of all x-components of force equal to zero. In a separate equation, set the algebraic sum of all y-components equal to zero. (Never add x-and y-components in a single equations.) You can then solve these equations for up to two unknown quantities, which may be force magnitudes, components, or angles.

If there are two or more bodies, repeat all of the above steps for each body. If the bodies interact with each other, use Newton's third law to relate the forces they exert on each other.

Make sure that you have as many independent equations as the number of unknown quantities. Then solve these equations to obtain the target variables. This part is algebra, not physics, but it's an essential step.

Step 4 : Evaluate your answer : Look at your results and ask whether they make sense. When the result is a symbolic expression or formula, try to think of special cases (particular values or extreme cases for the various quantities) for which you can guess what the results ought to be. Check to see that your formula works in these particular cases.

Newton's Second Law : Dynamics of Particles Step 1: Identify the relevant concepts : You have to

use Newton's second law for any problem that involves forces acting on an accelerating body.

As with any problem, identify the target variable – usually an acceleration or a force. If the target variable is something else, you'll need to identify another concept to use. For example, suppose the problem asks you to find how fast a sled is moving when it reaches the bottom of a hill. This means your target variable is the sled's final velocity. To find this, you'll first need to use Newton's second law to find the sled's acceleration. Then you'll also have to use the constant –acceleration relationships to find velocity from acceleration.

Step 2: Set up the problem using the following steps: Draw a simple sketch of the situation. Identify

one or more moving bodies to which you'll apply Newton's second law.

For each body you identified, draw a free-body diagram that shows all the forces acting on the body. (Don't try to be fancy–just represent the object by a point.) Be careful not to include any forces exerted by the object on some other object. Remember, the acceleration of a body is determined by the forces that act on it, not by the forces that it exerts on anything else. Make sure you can answer the question "What other body is applying this force ?" for each force in your

diagram. Furthermore, never include the quantity amr

in your free-body diagram; it's not a force! Label each force with an algebraic symbol for the

force's magnitude, as well as a numerical value of the magnitude if it's given in the problem. (Remember that magnitudes are always positive. Minus signs show up later when you take components of the forces.) Usually, one of the forces will be the body's weight; it's usually best to label this as w = mg. If a numerical value of mass is given, you can compute the corresponding weight.

Choose your x-and y-coordinate axes for each object, and show them explicitly in each free body diagram. Be sure to indicate which is the positive direction for each axis. If you know the direction of the acceleration, it usually simplifies things to take one positive axis along that direction. Note that if your problem involves more than one object and the objects accelerate in different directions, you can use a different set of axes for each object.

In addition to Newton's second law, ∑Fr

= amr

, identify any other equations you might need. (You need one equation for each target variable.) For example, you might need one or more of the equations for motion with constant acceleration. If more than one body is involved, there may be relationships among their motions; for example, they may be connected by a rope. Express any such relationships as equations relating the accelerations of the various bodies.

Step 3 : Execute the solution as follows : For each object, determine the components of the

forces along each of the object's coordinate axes. When you represent a force in terms of its components, draw a wiggly line through the original force vector to remind you not to include it twice.

For each object, write a separate equation for each component of Newton's second law, as in Eq. ΣFx = max; ΣFy = may (Newton's second law, component form)

Make a list of all the known and unknown quantities. In your list, identify the target variable or variables.

Check that you have as many equations as there are unknowns. If you have too few equations, go back to step 5 of "Set up the problem." If you have too many equations, perhaps there is an unknown quantity that you haven't identified as such.

Do the easy part – the math! Solve the equations to find the target variable(s).


Step 4 : Evaluate your answer : Does your answer have the correct units ? (When appropriate, use the conversion 1N = 1kg . m/s2) Does it have the correct algebraic sign ? (If the problem is about a sled sliding downhill, you probably took the positive x-axis to point down the hill. If you then find that the sled has a negative acceleration – that is, the acceleration is uphill – then something went wrong in your calculations.) When possible, consider particular values or extreme cases of quantities and compare the results with your intuitive expectations. Ask, "Does this result make sense ?"

1. A balloon is descending with a constant acceleration

a. The mass of the balloon and its contents is M. What mass m of its contents should be released so that the balloon starts ascending with the same acceleration a ? Assume that the volume of the balloon remains the same when the mass m is released.

Sol. The forces acting on the balloon are its weight and the upthrust U due to air. Since the volume of the balloon remains the same the upthrust is the same in both the cases. We have, According to Newton's second law,

Mg – U = Ma and U – (M – m)g = (M – m)a Solving these, we get

m = ga

a2+

M

U

Mg

a

U

(M – m)g

a

(a) (b) 2. A frictionless cart of mass M carries two other

frictionless carts having masses m1 and m2 connected by a string passing over a pulley, as shown in the figure. What horizontal force F must be applied on M so that m1 and m2 do not move relative to it ?

Sol. Since m1 and m2 are in accelerating frame, we can assume that inertial force m1a and m2a act on them, respectively, a being the acceleration of the system. Clearly,

a = 21 mmM

F++

...(1)

N1

m1 Tm1a

m1g

M N2m2

T

m2a F

m2g

The forces acting on m1 and m2 are shown in fig. We have

For m1 : T = m1a For m2 : T = m2g ⇒ m1a = m2g

or a = 1

2

mm g ...(2)

Eqs. (1) & (2) give

F = (M + m1 + m2) 1

2

mm g

3. The total mass of an elevator with a 80 kg man in it is

1000 kg. This elevator moving upward with a speed of 8 m/sec, is brought to rest over a distance of 16m. Calculate (a) the tension T in the cables supporting the elevator and (b) the force exerted on the man by the elevator floor.

Sol. (a) The elevator having an initial upward speed of 8 m/sec is brought to rest within a distance of 16 m. Hence,

0 = (8)2 + 2a(16)

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1. Xtra edge June 2011 - SMAROSA · 2018. 7. 23. · XtraEdge for IIT-JEE 1 JUNE 2011 Dear Students, Motivate Yourself One of the greatest virtues of human beings is their ability