JEE (Main + Advanced) : NURTURE COURSE

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so.

2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

3. Blank spaces are provided within this booklet for rough work.

4. Write your name and form number in the space provided on the back cover of this booklet.

5. After breaking the seal of the booklet, verify that the booklet contains 32 pages and all the 20 questions in each

subject and along with the options are legible.

QUESTION PAPER FORMAT AND MARKING SCHEME :

6. The question paper has three parts : Physics, Chemistry and Mathematics. Each part has two sections.

7. Carefully read the instructions given at the beginning of each section.

8. Section-I :

(i) Section-I(i) contains 8 multiple choice questions with only one correct option.

Marking scheme : +3 for correct answer, 0 if not attempted and –1 in all other cases.

(ii) Section-I(ii) contains 2 ‘paragraph’ type questions. Each paragraph describes an experiment, a situation or a

problem. Two multiple choice questions will be asked based on each paragraph. One or more than one option

can be correct.

Marking scheme : +4 for correct answer, 0 if not attempted and –1 in all other cases.

9. There is no questions in SECTION-II & III

10. Section-IV contains 8 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive)

Marking scheme : +4 for correct answer and 0 in all other cases.

OPTICAL RESPONSE SHEET :

11. The ORS is machine-gradable and will be collected by the invigilator at the end of the examination.

12. Do not tamper with or mutilate the ORS.

13. Write your name, form number and sign with pen in the space provided for this purpose on the original. Do not write

any of these details anywhere else. Darken the appropriate bubble under each digit of your form number.

PAPER – 2

Test Type : SHUFFLING TEST Test Pattern : JEE-Advanced

TEST DATE : 03 - 04 - 2016

JEE (Main + Advanced) : NURTURE COURSE

Paper Code : 1001CT100115033

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

Please see the last page of this booklet for rest of the instructions

Time : 3 Hours Maximum Marks : 216

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SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

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Target : JEE (Main + Advanced) 2017/03-04-2016/Paper-2

1001CT100115033

Space for Rough Work

Nurture Course/03-04-2016/Paper-2

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PART-1 : PHYSICS

SECTION–I(i) : (Maximum Marks : 24)

This section contains EIGHT questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option(s) is correct.

For each question, darken the bubble corresponding to the correct option in the ORS

Marking scheme :

+3 If the bubble corresponding to the correct option is darkened

0 If none of the bubbles is darkened

–1 In all other cases

1. A block of mass m is supported by a massless string passing through a smooth peg as shown in the

figure. Variation of tension in the string T as a function of best represented by (here the total length of

the string is varied)

Smooth peg

(A)

T

mg/2

/2

(B)

T

mg/2

/2

(C)

T

mg/2

/2

(D)

T

mg/2

/2

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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2. A thin wire is bent into the shape of a parabola in the vertical plane as shown. A small bead 'P' can freely slide

along the wire. The coefficient of friction between the bead and the wire is 'µ'. The condition on y so that bead

does not slip on the wire is :

PY

Ox

Y =4ax2

(A) y 2aµ (B) 2a

y

(C) y 2aµ (D) 2a

y

3. Two waveform, 1(x, t) = A sin (kx – t) and

2(x, t) = A sin (kx + t + /3), travelling along the x-axis,

are superposed. The position of the nodes is given by

2k , 2 f

(A) n

1x n

6 2

(B) n

1x n

6 2

(C) nx n2

(D) none of these


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4. The fundamental frequency of a sonometer wire of length is f0. A bridge is now introduced at a distance

from the centre of the wire ( << ). The number of beats heard if both sides of the bridges are set into

vibration in their fundamental modes are :

(A) 08f

(B)

0f

(C)

02f

(D)

04f

5. A hollow sphere of radius R is made of metal whose specific gravity is . The sphere will float in water

if thickness of wall of the sphere is (thickness of the wall << R)

(A) R

3

(B)

2R

3

(C) R

3

(D)

4R

3

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6. A block of mass m is performing SHM with an amplitude A, on a frictionless surface. When it is at extreme

position, a bullet of mass m moving with a velocity v0 collides and gets embedded into it at a time

2 20t 0 mv 2kA then the displacement x at any time t will be :

mv0 m

t=0

extremeposition

meanposition

x=0

A

(A) k

x 2Asin t2m 4

(B)

kx 2Asin t

2m 4

(C)

kx 2A sin t

m 4(D)

kx 2Asin t

m 4


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7. A glass capillary of length and inside radius r(r << ) is submerged vertically into water. The upper end of the

capillary is sealed. The atmospheric pressure is p0. To what length h has the capillary to be submerged to

make the water levels inside and outside the capillary coincide. Assume that temperature of air in the capillary

remains constant. (Given, surface tension of water = T, angle of contact between glass water interface = 0°)

h

(A) 0p r

1T

(B)

0p r1

2T

(C)

0p r1

4T

(D)

02p r1

T

8. Two waves travel down the same string. The waves have the same velocity, frequency (f0), and wavelength

but different phase constants (1 >

2) and amplitudes (A

1 > A

2). According to the principle of superposition,

the resultant wave has an amplitude A such that :

(A) A = A1 + A

2(B) A = A

1 – A

2

(C) A1 A A

2(D) A

1 – A

2 A A

1 + A

2

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SECTION–I(ii) : (Maximum Marks : 16)

This section contains TWO paragraphs.

Based on each paragraph, there will be TWO questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS

Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened



Paragraph for Questions 9 and 10

Snow skiing is one of the more popular winter sports. People who ski tend to be either downhill skiers

or cross-country skiers. In downhill skiing, skiers are taken to the top of the mountain by means of a lift chair,

although some ski resorts still use tow ropes. With two ropes, the skier grabs on to the tow rope and is pulled

up the mountain (figure-1). Once at the top of the mountain, the idea is for the skier to get back down, and

enjoy the ride while doing it. Inexperienced skiers tend to ski straight down the mountain; their motion is much

like that of a block sliding down an inclined plane (figure-2). Experienced skiers know how to control their

rate of descent by making a series of S-shaped turns as they come down the ski run.

Figure-1

Figure-2

Two rope

Cross-country skiing is quite different from downhill skiing. Essentially, specially designed skis allow

the skier to "walk" across the snow. The base of the ski is coated with a surface that glides easily in the

forward direction. This action is different from just strapping on snowshoes, however. Cross-country

skis allow the skier to bend his leg at the ankle. Ideally, the skis should just glide across the snow.


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9. Occasionally, a skier has to "side step" up a hill. She does this by turning her skis perpendicular to the hill, and

then she makes series of small steps up the hill. This enables her to climb the hill. The maximum steepness of

a hill that a skier can side step is determined by :

I. The weight of the skier.

II. The coefficient of static friction between the skis and the snow.

III. The coefficient of kinetic friction between the skies and the snow.

IV. The initial velocity of the skier.

Mark the CORRECT statement(s) :

(A) I (B) II (C) III (D) IV

10. Suppose a skier falls and begins sliding down the hill shaped as shown in figure. Her speed at the

bottom of the hill depends on :

I. Her weight

II. Her speed of at the time of the fall.

III. Length of the hill.

IV. The steepness of the hill.

Mark the CORRECT statement(s) :


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A massless elastic cord (that obeys Hooke's law) will break if the tension in the cord exceeds Tmax

. One end

of the cord is attached to a fixed point, the other is attached to an object of mass 3 m as shown in the figure.

If a second, smaller object of mass m moving at an initial speed v0 strikes the larger mass and the two stick

together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead

the two collide with a perfectly elastic one dimensional collision, the cord will still break and the larger mass

will move off with a final speed of vf. All motion occurs on a horizontal frictionless surface. (assume that

Hooke's law is obeyed through until the cord breaks)

3mm v0

11. Mark the CORRECT statement(s) :

(A) Ratio of f

0

v

v is given by 1

6

(B) Ratio of f

0

v

v is given by 1

2

(C) Impulse offered by bigger block on the smaller block in second case when collision is elastic is 0mv

2

(D) Impulse offered by bigger block on the smaller block in second case when collision is elastic is 03mv

212. Mark the CORRECT statement(s) :-

(A) Ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision andthe cord has broken to the initial kinetic energy of the smaller mass prior to the collision is 3/4.

(B) Ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and thecord has broken to the initial kinetic energy of the smaller mass prior to the collision is 1/4.

(C) In 1st case of inelastic collision, energy stored in spring when velocity of the combined mass (smaller &

larger mass) have half of their maximum speed (maximum speed i.e. speed just after collision) is 20

3mv

32(D) In 1st case of inelastic collision, energy stored in spring when velocity of the combined mass (smaller &

larger mass) have half of their maximum speed (maximum speed i.e. speed just after collision) is 20mv

32

SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type

No question will be asked in section II and III


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SECTION–IV : (Maximum Marks : 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS Marking scheme :

+4 If the bubble corresponding to the answer is darkened0 In all other cases

1. A bead of mass 5 kg is free to slide on a smooth ring of radius r = 20 cm fixed in a vertical plane. Thebead is attached to one end of a massless spring whose other end is fixed to the topmost point O of thespring. Initially the bead is held at rest at a point A of the ring such that OCA = 60°, C being the centreof the ring. The natural length of the spring is also equal to r. After the bead is released and slides downthe ring, the contact force between the bead and the ring becomes zero when it reaches the lowestposition B. If the force constant of the spring (in N/m) equals n × 100, find n.

60°

B

A

2. On a smooth horizontal surface, two particles of mass m each, travelling with a velocity v0 in opposite directions,

strike the ends of a rigid massless rod of length , kept perpendicular to their velocity. The particles stick to

the rod after the collision. If the tension in the rod during subsequent motion is 20nmv

, then find the

value of n.

v0

m

m

v0

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3. During an experiment, an ideal gas is found to obey a condition 2P

= constant. (P-pressure and

-density of the gas). The gas is initially at temperature T, pressure P and density . The gas expands

such that density changes to /2. The temperature of gas changes to T . Find the value of .

4. A massive horizontal platform is moving horizontally with a constant acceleration of 10 m/s2 as shown in the

figure. A particle P of mass m = 1 kg is kept at rest at the smooth surface as shown in the figure. The particle

is hinged at O with the help of a massless rod OP of length 0.9 m. Hinge O is fixed on the platform and the rod

can freely rotate about point O. Now the particle P is imparted a velocity in the opposite direction of the

platform's acceleration such that it is just able to complete the circular motion about point O. Then the

maximum tension appearing in the rod during the motion is 10n. Find the value of n.

O

P 10m/s2


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5. A hollow sphere of outer radius R is allowed to roll down on an incline without slipping and it reaches a speed

v0 at the bottom of the incline. The same incline is then made perfectly smooth by waxing and the sphere is

allowed to slide without rolling and now the speed attained is 0

5v

4. The radius of gyration of the sphere about

an axis passing through its centre is nR

4, find n.

6. An ideal gas undergoes a process shown in the P-V diagram. If the change in heat content of the gas in this

process is 0 0

KP V

2. Then find the value of K.

2P0

P0

P1

2

VV0 2V0

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7. A vessel containing incompressible and non viscous liquid of density is accelerated horizontally with acceleration

10 m/s2 in a gravity free space. A small sphere of density 2 is released from rest with respect to the vessel.

Find the initial acceleration of the sphere with respect to the vessel.

8. A particle of mass m is released in a circular tube of radius R. If a massless spring of constant K is placed in

the tube as shown in figure and angle made by radius vector with vertical when particle will come at an instant

rest is 60°, then the value of K is

2

6mg

R

then find the value of .

m


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PART-2 : CHEMISTRY





Marking scheme :




1. Half litre each of three samples of H2O

2 labelled 10V, 15V and 20V are mixed and then diluted with

equal volume of water. Calculate volume strength of resultant H2O

2 solution -

(A) 3 (B) 11.2 (C) 7.5 (D) 22.4

2. 74 gm of a sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular

formula of the compound may be

(A) C5H12 (B) C4H10O (C) C3H6O2 (D) C3H7O2

[From mole concept sheet Ex-O1, Q.No. 17]

3. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below.

The value of the van der Waals constant a (atm. liter2 mol–2) is

20.1

21.6

23.1

24.6

0 2.0 3.0

1V

(mol liter )–1

PV

(li

ter-

atm

mol

)–1 [Graph not to scale]

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0

[From real gas sheet Jee-Advanced, Q.No. 13, same language with different data.]


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4. Choose the correct statement among the following -

(A) The mean free path () of gaseous molecules is directly proportional to temperature of gas at

constant volume

(B) The mean free path () of gaseous molecules is inversely proportional to pressure of gas at constant

volume

(C) The mean free path () of gaseous molecules is directly proportional to volume of gas at constant T

(D) The mean free path () of gaseous molecules is inversely proportional to volume of gas at constant P

5. Which of the following property does not increase if 'S' is replaced by 'Xe' in SF6 without changing

the oxidation state of central atom.

(A) Total number of lone pair

(B) Number of d-orbitals involved in the hybridisation of central atom

(C) Number of p-orbtials involved in the hybridisation of central atom

(D) A – F bond length (A = S/Xe)

6. Which of the following order of ionization energy is CORRECT ?

(A) In < Tl (B) Sn > Pb (C) Ga < Al (D) O > N


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7. Identify the binary mixtures (s) that can be separated into the individual compounds, by differential

extraction, as shown in the given scheme :

Binary mixture containingCompund 1 and compound 2

NaOH(aq)

NaHCO (aq)3

Compound 1 Compound 2

Compound 1 Compound 2

+

+

(A) Phenol & Benzoic acid (B) Cyclohexanol & Phenol

(C) Benzene & Cyclohexene (D) Toluene & Picric acid

8. How many hydrogen (H) will replaced by deuterium (D) in the enol form of compound (M) when

treated with OD / D O2

– for prolonged time :

O

(M)

(A) 10 (B) 9 (C) 13 (D) 8

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1001CT100115033

SECTION–I(ii) : (Maximum Marks : 16) This section contains TWO paragraphs. Based on each paragraph, there will be TWO questions Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened0 If none of the bubbles is darkened–1 In all other cases


The French physicist Louis de Broglie in 1924 postulated that matter, like radiation, should exhibit adual behaviour. He proposed the following relationship between the wavelength of a material particle,its linear momentum p and planck constant h.

= p

h =

mv

h

The de Broglie relation implies that the wavelength of a particle should decreases as its velocityincreases. It also implies that for a given velocity heavier particles should have shorter wavelengththan lighter particles. The waves associated with particles in motion are called matter waves or deBroglie waves. These waves differ from the electromagnetic waves as they(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the de Broglie relation was obtained when Davisson and Germer, in1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristicproperty of waves, hence the beam of electron behaves as a wave, as proposed by de Broglie.

9. If a photon of wavelength '' strikes on a metallic surface of thershold wavelength ''. If (

> ),

then the select correct statements (s)

(A) Minimum wavelength of ejected photoelectron (e) is given by :

e 0

1 1 1

(B) Minimum wavelength of ejected photoelectron (e) is given by :

e = 0

0

h

2m ( )c

(C) All ejected photoelectron will have same wavelength

(D) Range of wavelength of ejected photoelectron is 0

0

h

2m ( )c

e <

10. Select the correct statements (s) regarding H-atom in ground state -

(A) When 28.6 eV of energy is provided, the debroglie wavelength of ejected electron is 10 Å.

(B) When 10 eV of energy is provided, debroglie wavelength of electron is 15Å

(C) When 10 eV of energy is provided, debroglie wavelength of electron is 150

Å13.6

(D) When 28.6 eV of energy is provided, the ejected electron is of energy 24 × 10–22 kJ



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For the given compounds :OH O

OH

(I)

O

OH(II)

O

O

(III)

O

O

CH O3 OCH3

OCH3

(IV)

O

OHOH

(V)

O

OH

(VI)

O

11. Choose the correctly matched option(s) ?

(A) (I) & (II) = Functional isomers (B) (II) & (VI) = Metamers

(C) (V) & (I) = Chain isomers (D) (VI) & (III) = Functional iomsers

12. How many geometrical isomers are possible for compound (IV) ?

(A) 4 (B) 2 (C) 3 (D) 8




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1001CT100115033

SECTION–IV : (Maximum Marks : 32)


The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

For each question, darken the bubble corresponding to the correct integer in the ORS

Marking scheme :

+4 If the bubble corresponding to the answer is darkened

0 In all other cases

1. 1 mol of CuSO4.5H

2O is introduced in a 2463 L vessel maintained at a constant temperature of 27°C

containing moist air at relative humidity of 25%. Aqueous tension of water at 27°C is 30.4 torr.

CuSO4.5H

2O(s) CuSO

4(s)+ 5H

2O(g), K

p(atm)= 32 × 10–10

Calculate remaining moles of CuSO4.5H2O at equilibrium.

(Fill your answer by multiplying it with 10)

2. A bulb of constant volume is attached to a very thin manometer tube as shown in figure. Gas starts

leaking through a small hole in the bulb causing change in pressure as

dt

dP = – kP

Where k is constant and its value is 7hr–1 P is pressure at any instant.

If initial height difference 'h' was 76 cm, Find time (in minutes) after which level in both limbs were

found to be same. (ln2 = 0.7)

[From Race-32, Q.No. 9, same language with different data.]



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3. During disproportionation of P4 into PH3 and H2PO2

– in basic medium, percentage of phosphorous

which is converted into PH3 is

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

4. A solution containing 4 mmol of An+ ions requires 1.6 mmol of MnO4

- for oxidation of An+ to AO3

-

in acidic medium. The value of n is


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1001CT100115033

5. Find the total number of orbital(s) in iron having |m| 1 and atleast one electron is present in

orbital.

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

6. Find the number of chemical species having p

– d types of bond

XeO2F

2 ; ;

CH3

; C3O

2 ; C

3H

4 ; POCl

3



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7. Find the number of INCORRECT order of bond angle.

OCl2 > OF

2 ; H

2O < OF

2 ; PF

3 < PCl

3 ; SO

2 > SO

3 ; CH

4 > CF

4

8. How many of the following orders are correctly matched ?

(i) > > (Heat of hydrogenation)

(ii)

O

>

O

(Solubility in water)

(iii) C = CCH3

H

H C3

H< C = C

H

CH3

H C3

H(Melting point)

(iv) + >

+

>

+

(Carbocation stability)

(v)

NH2

CH3H3C>

NH2

(Basic strength)

(vi)–

>

O–

(Number of resonating structure involving monoanion)

(vii) –NO2 > –CN > –F > –I (–I effect)

(viii) C = CNO2

H

H C3

H< C = C

H

NO2

H C3

H(Dipole moment)


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PART-3 : MATHEMATICS





Marking scheme :




1. Let A = log10

2 + log10

50, B = log252 – log

213, then-

(A) A = 2B (B) 2A = B (C) A = B (D) 4A = 3B

2. Let p,q are roots of x2 – 10x + 13 = 0. If A is arithmetic mean of p and q, then the value of A is-

(A) 13 (B) 10 (C) 13 (D) 5

3. If N is total number of possible words which are formed by using all letters of 'SATAYPAL' such

that no two alike letters are together and contain word 'PSA' then sum of digits of N is

(A) 9 (B) 10 (C) 12 (D) 3

4. Consider system of equations in x,y,z,

x + y + z = 2,

tanx tany = 3,

tany tanz = 9.

If total number of ordered triplet (x,y,z), where x,y,z [–2,2] is N, then -

(A) N > 20 (B) 16 < N < 19 (C) 11 < N < 15 (D) N < 10



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5. Let y = ƒ(x) is a linear graph, ƒ(x) = Ax + B, A,B R which is passing through (A, 2A – B2) and

(2B + 3, (A + B)2 – 1). If B1,B

2......B

n, n N, are different possible value(s) of B, then the

value of n

rr 1

B is -

(A) 9

10 (B)

18

5 (C)

9

20 (D)

9

5

6. Let x be a set containing 8 elements and P(x) be it's power set. A and B are picked from P(x). If

n(A) = n(B) and A B, then total number of ordered pair (A,B) is

[Note : n(A) represents number of elements in set A]

(A) 12870 (B) 13420 (C) 13124 (D) 12614

7. Let Ai, i = 1,2,3,......10 are vertices of regular decagon P. If area of polygon P is 5sin

5

square units

then the value of (A1A

2)(A

1A

3)(A

1A

4)......(A

1A

10) is-

(A) 10 (B) 9 (C) 45sin5

(D) 45cos5

8. Let ƒ(x) = x2 – 2x + 5, g(x) = –x2 + 4x + 5, h(x) = ax2 + bx + c, a,b,c I. If inequality ƒ(x) < h(x) < g(x)

holds good for all values of x [0,3], then number of all possible ordered pair (a,b) is -

(A) 3 (B) 7 (C) 2 (D) infinite

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SECTION–I(ii) : (Maximum Marks : 16)

This section contains TWO paragraphs.

Based on each paragraph, there will be TWO questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS

Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened




Let ƒ : R R, ƒ(x) = ax2 + (a + 2)x + c, a 0

1 , x 0

g : R R;g x 0 , x 0

1 , x 0

h : (0,2] (0,2], h(x) = cos–1(cos x)

9. Which of the following statement is/are correct-

(A) Function h(x) is many-one and onto function

(B) Function h(x) is into function

(C) Area bounded by y = h(x) and y = g(x) is ( – 1)2

(D) Area bounded by y = h(x) and y = g(x) is 2(– 1)2

10. If equation ƒ(x) = g(x) has 4 solutions, then-

(A) sum of least six integral values of a is divisible by 7.

(B) sum of least six integral values of a is divisible by 3.

(C) function ƒ(x) is onto function

(D) function g(x) is odd function


MATHEM

ATICS

E-27/321001CT100115033



Let C be circle x2 + y2 – 8x – 8y + 16 = 0 and combined equation of lines L1 and L

2 is y2 – 8y + 3 = 0.

Line L1 and L

2 intersect curve C at P,Q,R,S. A is area of quadrilateral formed by tangents at P,Q,R

and S.

11. Which of the following statement(s) is/are correct ?

(A) Circle C touches both axes (B) 92

A39

(C) 62 3

A13

(D) Circle C neither touch nor intersect coordinate axes

12. If point T is (5,3), then-

(A) Value of (PT)2 + (QT)2 + (RT)2 + (ST)2 is 72

(B) Value of (PT)2 + (QT)2 + (RT)2 + (ST)2 is 64

(C) Value of PT + QT + RT + ST is less than 10 3

(D) Value of PT + QT + RT + ST is greater than 5



E-28/32

MATHEM

ATICS


1001CT100115033


SECTION–IV : (Maximum Marks : 32)


The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

For each question, darken the bubble corresponding to the correct integer in the ORS

Marking scheme :

+4 If the bubble corresponding to the answer is darkened

0 In all other cases

1. Value of 1 1 11 3sin cos tan 3

22

is

a

b

, then value of (3a – b) is, (a and b are relatively

prime numbers)

2. With usual notation, if in triangle ABC, a = 7(sin–1x + cos–1x), x (0,1), b = 3and 1

cos A5

. If

sec C = p

q, then the value of (p – q) is (where p and q are coprime numbers)

3. Consider circles A : x2 + y2 – 4x – 3y + 5 = 0 and B : x2 + y2 + 2ax + by – a – 1 = 0; a,b are real

numbers, intersect orthogonally. Let points M and N are centre of circle A and B respectively.

Let tangents PQ and PR are drawn from point P(a,c) on the circle A to the circle B (where P, M and

N are non-collinear points). Q and R lie on the circle B. If line L which is passing through point Q

and R intersect the circle A at point S(), then value of (a – c + ) is (where points P, S and

N are non-collinear points)


MATHEM

ATICS

E-29/321001CT100115033


4. Let (a) and (a) be the roots of the equation 23 3 61 a 1 x 1 1 a x 1 a 1 a , where

a > 0, (a) > (a). If minimum value of expression ƒ(a) = (a) + 12(a) +

2

4,a 0

a 3 a

is

m, then value of [m] is

[Note : [m] denotes the largest integer less than or equal to m]

5. In a triangle RST, let a, b and c be the length of the sides opposite to the angle R,S and T respectively.

If 18b2 – 4a = 12ab + 3b and 12ab – 3b = 8a2, then maximum possible value of 36 is

[Note : dentoes area of triangle RST]

6. Let (3 – sinBsinC)sinA + (3 – sinAsinC)sinB + (3 – sinAsinB)sinC = 0, A,B,C

n

2

, n I.

If 2 2 2 2 2 2

sin A sin B sin C sin A.sin B.sin C

cos A cos B cos C cos A.cos B.cos C , then sum of all possible integral

values of is

E-30/32

MATHEM

ATICS


1001CT100115033

7. Let curve ƒ(x,y) is locus of point (x,y) which satisfy

2x 2y 6 x y 111 2

2 3 2 4 2 3 81 072 2

1 2 1 2 1 2

R. If minimum distance between curve

ƒ(x,y) and line 4x – 3y + 12 = 0 is d, then the value of d is

8. Let (1 + x)15 0 151 2P PP P0 1 2 15a x a x a x .... a x where 0 1 2 3 15a a a a ..... a 0 and

i 1 i i i 1

i 1 i i i 1

P P , if a a 2n 1, n I, i {0,1,2,.....,14}

P P , if a a 2n, n I, i {0,1,2,.....,14}

, then the value of 5 5 11 11

9 9 14 14

a P a P 869

a P P a 32

is



E-31/321001CT100115033



Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Your Hard Work Leads to Strong Foundation 1001CT100115033E-32/32

DARKENING THE BUBBLES ON THE ORS :

14. Use a BLACK BALL POINT PEN to darken the bubbles in the upper sheet.

15. Darken the bubble COMPLETELY.

16. Darken the bubbles ONLY if you are sure of the answer.

17. The correct way of darkening a bubble is as shown here :

18. There is NO way to erase or "un-darken" a darkened bubble.

19. The marking scheme given at the beginning of each section gives details of how darkened and not darkened

bubbles are evaluated.

20. Take g = 10 m/s2 unless otherwise stated.

I HAVE READ ALL THE INSTRUCTIONS

AND SHALL ABIDE BY THEM

____________________________

Signature of the Candidate

I have verified the identity, name and rollnumber of the candidate, and that questionpaper and ORS codes are the same.

____________________________

Signature of the invigilator

NAME OF THE CANDIDATE ................................................................................................

FORM NO. .............................................


1.

2. (ORS)

3.

4.

5. 32 20

6.

7.

8. -I :

(i) -I(i) 8

: +3 0–1

(ii) -I(ii) 2 ‘’

: +4 0–1

9. –II III

10. -IV 8 0 9 ()

: +4 0

11.

12.

13.

PAPER – 2

Test Type : SHUFFLING TEST Test Pattern : JEE-Advanced

TEST DATE : 03 - 04 - 2016

Time : 3 Hours Maximum Marks : 216

JEE (Main + Advanced) : NURTURE COURSE

Paper Code : 1001CT100115033

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)H

IND

I

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

H-2/32


1001CT100115033


PHYSICS

H-3/321001CT100115033

-1 : – I(i) : ( : 24)

(A), (B), (C) (D)

:

+3 0

–1

1. m

T (

)

Smooth peg

(A)

T

mg/2

/2

(B)

T

mg/2

/2

(C)

T

mg/2

/2

(D)

T

mg/2

/2

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

H-4/32

PHYSICS


1001CT100115033

2. 'P'

'µ' y

PY

Ox

Y =4ax2

(A) y 2aµ (B) 2a

y

(C) y 2aµ (D) 2a

y

3. x-1(x, t) = A sin (kx – t)

2(x, t) = A sin (kx + t + /3)

2k , 2 f

(A) n

1x n

6 2

(B) n

1x n

6 2

(C) nx n2

(D)


PHYSICS

H-5/321001CT100115033

4. f0 ( << )

(A) 08f

(B)

0f

(C)

02f

(D)

04f

5. R

(<< R)

(A) R

3

(B)

2R

3

(C)

R

3

(D)

4R

3

H-6/32

PHYSICS


1001CT100115033

6. m A

m v0 2 2

0t 0 mv 2kA

t x

mv0 m

t=0

extremeposition

meanposition

x=0

A

(A) k

x 2A sin t2m 4

(B)

kx 2Asin t

2m 4

(C)

kx 2A sin t

m 4 (D)

kx 2A sin t

m 4


PHYSICS

H-7/321001CT100115033

7. r(r << )

p0 h

(

= T, = 0° )

h

(A) 0p r

1T

(B)

0p r1

2T

(C)

0p r1

4T

(D)

02p r1

T

8. (f0) (

1 >

2)

(A1 > A

2) A

(A) A = A1 + A

2(B) A = A

1 – A

2

(C) A1 A A

2(D) A

1 – A

2 A A

1 + A

2

H-8/32

PHYSICS


1001CT100115033

–I(ii) : ( : 16)

(A), (B), (C) (D)

:

+4

0

–1

9 10

(downhill skiers) (cross-country skiers)

-1

-2

S-

Figure-1

Figure-2

Two rope


PHYSICS

H-9/321001CT100115033

9.

I.

II.

III.

IV.


10.

I.

II.

III.

IV.


H-10/32

PHYSICS


1001CT100115033

11 12 T

max

3 m m v

0

v

f

3mm v0

11.

(A) f

0

v

v 1

6

(B) f

0

v

v 1

2

(C) 0mv

2

(D) 03mv

2

12. (A) 3/4

(B) 1/4

(C)

20

3mv

32

(D)

20mv

32

– II : & – III :

II III


PHYSICS

H-11/321001CT100115033

–IV : ( : 32)

0 9

:

+4

0

1. 5 kg r = 20 cm O A OCA = 60° C r B (N/m n × 100 n

60°

B

A

2. m v0

20nmv

n

v0

m

m

v0

H-12/32

PHYSICS


1001CT100115033

3. 2P

= P--

T, P /2

T

4. 10 m/s2 m = 1 kg

PO OP

0.9 m O O

P O

10n n

O

P 10m/s2


PHYSICS

H-13/321001CT100115033

5. R

v0

yq<+ds fQlyrk gS rFkk blds }kjk izkIr pky dk eku 0

5v

4

nR

4 n

6. P-V 0 0

KP V

2

K

2P0

P0

P1

2

VV0 2V0

H-14/32

PHYSICS


1001CT100115033

7. 10 m/s2

2

8. m R K

60° K

2

6mg

R

m


CHEM

ISTRY

H-15/321001CT100115033

-2 : –I(i) : ( : 24)

(A), (B), (C) (D)

:

+3

0

–1

1. H2O

2 10V, 15V 20V

H2O

2 -

(A) 3 (B) 11.2 (C) 7.5 (D) 22.4

2. 74 gm 132 gm CO2 54 gm H2O

(A) C5H12 (B) C4H10O (C) C3H6O2 (D) C3H7O2

[From mole concept sheet Ex-O1, Q.No. 17]

3. b = 0 T = 300 K PV 1/V

a (atm. liter2 mol–2)

20.1

21.6

23.1

24.6

0 2.0 3.0

1V

(mol liter )–1

PV

(li

ter-

atm

mol

)–1 [Graph not to scale]

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0

[From real gas sheet Jee-Advanced, Q.No. 13, same language with different data.]

H-16/32

CHEM

ISTRY


1001CT100115033

4. -

(A) ()

(B) ()

(C) () T

(D) () P

5. SF6 , 'S' 'Xe'

(A)

(B) d-

(C) p-

(D) A – F (A = S/Xe)

6. (A) In < Tl (B) Sn > Pb (C) Ga < Al (D) O > N


CHEM

ISTRY

H-17/321001CT100115033

7. (binary) (scheme) (differential

extraction)

1 2

NaOH( )

NaHCO ( )3

1 2

1 2

+

+

(A) (B)

(C) (D)

8. (M) OD / D O2

–

:

O

(M)

(A) 10 (B) 9 (C) 13 (D) 8

H-18/32

CHEM

ISTRY


1001CT100115033

–I(ii) : ( : 16)

(A), (B), (C) (D)

:

+4

0

–1

9 10 1924 ) p h

= p

h =

mv

h

(i) (ii) (iii) 1927 ( beam )

9. '' '' (

> )

(A) (e)

e 0

1 1 1

(B) (e)

e = 0

0

h

2m ( )c

(C)

(D) 0

0

h

2m ( )c

e <

10. H- (A) 28.6 eV 10 Å (B) 10 eV 15Å

(C) 10 eV 150

Å13.6

(D) 28.6 eV 24 × 10–22 kJ


CHEM

ISTRY

H-19/321001CT100115033

11 12

OH O

OH

(I)

O

OH(II)

O

O

(III)

O

O

CH O3 OCH3

OCH3

(IV)

O

OHOH

(V)

O

OH

(VI)

O

11.

(A) (I) (II) = (B) (II) (VI) =

(C) (V) (I) = (D) (VI) (III) =

12. (IV)

(A) 4 (B) 2 (C) 3 (D) 8

– II : & – III :

II III

H-20/32

CHEM

ISTRY


1001CT100115033

–IV : ( : 32)

0 9

:

+4

0

1. 1 CuSO4.5H

2O 2463 L 27°C 25%

27°C 30.4 torr.

CuSO4.5H

2O(s)CuSO

4(s)+ 5H

2O(g), K

p(atm)= 32 × 10–10

CuSO4.5H2O

( 10 )

2.

dt

dP = – kP

k 7hr–1 P

'h' = 76 cm

[From Race-32, Q.No. 9, same language with different data.]


CHEM

ISTRY

H-21/321001CT100115033

3. P4 PH3 H2PO2

– PH3

()

4. An+ 4 mmol An+ AO3

–1.6 mmol

MnO4

– n

H-22/32

CHEM

ISTRY


1001CT100115033

5. |m| 1

()

6. p – d

XeO2F

2 ; ;

CH3

; C3O

2 ; C

3H

4 ; POCl

3


CHEM

ISTRY

H-23/321001CT100115033

7.

OCl2 > OF

2 ; H

2O < OF

2 ; PF

3 < PCl

3 ; SO

2 > SO

3 ; CH

4 > CF

4

8.

(i) > >

(ii)

O

>

O

(iii) C = CCH3

H

H C3

H< C = C

H

CH3

H C3

H

(iv) + >

+

>

+

(v)

NH2

CH3H3C>

NH2

(vi)–

>

O–

(vii) –NO2 > –CN > –F > –I –I

(viii) C = CNO2

H

H C3

H< C = C

H

NO2

H C3

H

H-24/32

MATHEM

ATICS


1001CT100115033

-3 : –I(i) : ( : 24)

(A), (B), (C) (D)

:

+3

0

–1

1. A = log10

2 + log10

50, B = log252 – log

213 -

(A) A = 2B (B) 2A = B (C) A = B (D) 4A = 3B

2. x2 – 10x + 13 = 0 p,q p q A A -

(A) 13 (B) 10 (C) 13 (D) 5

3. 'SATAYPAL' N 'PSA' N

(A) 9 (B) 10 (C) 12 (D) 3

4. x,y,z

x + y + z = 2,

tanx tany = 3,

tany tanz = 9

(x,y,z), x,y,z [–2,2] , N -

(A) N > 20 (B) 16 < N < 19 (C) 11 < N < 15 (D) N < 10


MATHEM

ATICS

H-25/321001CT100115033

5. y = ƒ(x) , ƒ(x) = Ax + B, A,B R (A, 2A – B2) (2B + 3, (A + B)2 – 1)

B1,B

2......B

n, n N, B

n

rr 1

B -

(A) 9

10 (B)

18

5 (C)

9

20 (D)

9

5

6. x, 8 P(x) A B, P(x)

n(A) = n(B) A B (A,B) [: n(A), A ]

(A) 12870 (B) 13420 (C) 13124 (D) 12614

7. Ai, i = 1,2,3,......10 P P 5sin

5

(A1A

2)(A

1A

3)(A

1A

4)......(A

1A

10) -

(A) 10 (B) 9 (C) 45sin5

(D) 45cos5

8. ƒ(x) = x2 – 2x + 5, g(x) = –x2 + 4x + 5, h(x) = ax2 + bx + c, a,b,c I x [0,3]

ƒ(x) < h(x) < g(x) (a,b) -

(A) 3 (B) 7 (C) 2 (D)

H-26/32

MATHEM

ATICS


1001CT100115033

–I(ii) : ( : 16)

(A), (B), (C) (D)

:

+4

0

–1

9 10 ƒ : R R, ƒ(x) = ax2 + (a + 2)x + c, a 0

1 , x 0

g : R R;g x 0 , x 0

1 , x 0

h : (0,2] (0,2], h(x) = cos–1(cos x)

9. -

(A) h(x)

(B) h(x)

(C) y = h(x) y = g(x) ( – 1)2

(D) y = h(x) y = g(x) 2(– 1)2

10. ƒ(x) = g(x) 4 -

(A) a 7

(B) a 3

(C) ƒ(x)

(D) g(x)


MATHEM

ATICS

H-27/321001CT100115033

11 12 C : x2 + y2 – 8x – 8y + 16 = 0 L

1 L

2 y2 – 8y + 3 = 0

L1 L

2, C P,Q,R,S P,Q,R S

A

11. (A) C

(B) 92

A39

(C) 62 3

A13

(D) C

12. T, (5,3) -

(A) (PT)2 + (QT)2 + (RT)2 + (ST)2 72

(B) (PT)2 + (QT)2 + (RT)2 + (ST)2 64

(C) PT + QT + RT + ST 10 3

(D) PT + QT + RT + ST 5

– II : & – III :

II III

H-28/32

MATHEM

ATICS


1001CT100115033

–IV : ( : 32)

0 9

:

+4

0

1. 1 1 11 3sin cos tan 3

22

a

b

(3a – b) , (a b )

2. ABC a = 7(sin–1x + cos–1x), x (0,1), b = 3 1cos A

5

sec C =p

q (p – q) ( p q )

3. A : x2 + y2 – 4x – 3y + 5 = 0 B : x2 + y2 + 2ax + by – a – 1 = 0, a,b

M N A B

PQ PR, A P(a,c) B (P, M N ) Q R

B L, Q R A S()

(a – c + ) ( P, S N )


MATHEM

ATICS

H-29/321001CT100115033

4. (a) (a), 23 3 61 a 1 x 1 1 a x 1 a 1 a a > 0,

(a) > (a) ƒ(a) = (a) + 12(a) +

2

4,a 0

a 3 a

m [m]

[ : [m] m ]

5. RST R, S T a, b c 18b2 – 4a = 12ab + 3b

12ab – 3b = 8a2 36

[ : RST ]

6. (3 – sinBsinC)sinA + (3 – sinAsinC)sinB + (3 – sinAsinB)sinC = 0, A,B,C

n

2

, n I

2 2 2 2 2 2

sin A sin B sin C sin A.sin B.sin C

cos A cos B cos C cos A.cos B.cos C

H-30/32

MATHEM

ATICS


1001CT100115033

7. ƒ(x,y), (x,y) 2x 2y 6 x y 1

11 22 3 2 4 2 3 81 0

72 21 2 1 2 1 2

R

ƒ(x,y) 4x – 3y + 12 = 0 d d

8. (1 + x)15 0 151 2P PP P0 1 2 15a x a x a x .... a x 0 1 2 3 15a a a a ..... a 0

i 1 i i i 1

i 1 i i i 1

P P , a a 2n 1, n I, i {0,1,2,...,14}

P P , a a 2n, n I, i {0,1,2, ...,14}

5 5 11 11

9 9 14 14

a P a P 869

a P P a 32


H-31/321001CT100115033


Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Your Hard Work Leads to Strong Foundation 1001CT100115033H-32/32

:

14.

15.

16.

17. :

18.

19.

20. g = 10 m/s2

____________________________

____________________________

................................................................................................

.............................................

Documents

JEE (Main + Advanced) : NURTURE COURSE