[PDF] IIT JEE Main 2015 Question Paper with Solutions

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(Pg. 1)

PAPER - 1 : PHYSICS, CHEMISTRY & MATHEMATICS

Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet.

Important Instructions :

1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use

of pencil is strictly prohibited.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take

out the Answer Sheet and fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are 360.

5. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each

correct response.

6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each

question. 1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No

deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in each question

will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side

1 and Side2

of the Answer Sheet. Use of pencil is strictly prohibited.

9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile

phone, any electronic device, etc., except the Admit Card inside the examination hall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is

given at the bottom of each page and in 3 pages (Pages 21

23) at the end of the booklet.

11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in

the Room / Hall. However, the candidates are allowed to take away this Test Booklet with them.

12. The CODE for this Booklet is A. Make sure that the CODE printed on Side

2 of the Answer Sheet and

also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet.

In case of discrepancy, the candidate should immediately report the matter to the invigilator for

replacement of both the Test Booklet and the Answer Sheet.

13. Do not fold or make any stray marks on the Answer Sheet.

Name of the Candidate (in Capital letters) : ______________________________________________

Roll Number : in figures

: in words ______________________________________________________________

Examination Centre Number :

Name of Examination Centre (in Capital letters) : ____________________________________________ _________________

Test Booklet Code A

JEE-MAIN 2015 : Paper and Solution (2)

(Pg. 2)

Read the following instructions carefully :

1. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with

Blue/Black Ball Point Pen.

2. For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.

3. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the

Test Booklet/Answer Sheet.

4. Out of the four options given for each question, only one option is the correct answer.

5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be

deducted from the total score. No deduction from the total score, however, will be made if no response

is indicated for an item in the Answer Sheet.

6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy

in Test Booklet Code and Answer Sheet Code), another set will be provided.

7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All

calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself,

3 pages (Pages

21 23) at the end of the booklet.

8. On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in

the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9. Each candidate must show on demand his/her Admit Card to the Invigilator.

10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.

11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the

Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the

Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with

as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.

12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.

13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct

in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the

JAB/Board.

14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.

15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager,

mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room. (3) VIDYALANKAR : JEE-MAIN 2015 : Paper and Solution (Pg. 3) .Questions and Solutions.

PART- A : PHYSICS

1. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed

of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time

variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s

2). (The figures are schematic and not drawn to scale)

(1) (2) (3) (4)

1. (3)

For the second stone time required to reach the ground is given by y = ut 21gt2

240 = 40 t

21 10 t2

5t2 40 t 240 = 0

(t 12) (t + 8) = 0 t = 12 s

For the first stone :

240 = 10t

21 10 t2

240 = 10t 5t2

5t

2 10t 240 = 0

(t 8) (t + 6) = 0

T = 8s

During first 8 seconds both the stones are in air : y2 y1 = (u2 u1) t = 30 t graph of (y2 y1) against t is a straight line.

After 8 seconds

y

2 = u2t

21gt 2402

Stones two has acceleration with respect to stone one. Hence graph (3) is the correct description.

JEE-MAIN 2015 : Paper and Solution (4)

(Pg. 4)

2. The period of oscillation of a simple pendulum is T = L2g. Measured value of L is 20.0 cm

known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is (1) 2 % (2) 3 % (3) 1 % (4) 5 %

2. (2)

22LT4g

g = 2 2L4T g L T100 100 2 100g L T

0.1 1100 2 10020 90 = 0.5 + 2.2 = 2.7 3.

3. Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is : (1) 100 N (2) 80 N (3) 120 N (4) 150 N

3. (3)

For Block A :

m

1g = 1F

20 = 0.1 F

F =

20

0.1 = 200 N

Frictional force on block A in upward direction = 1F = 0.1 200 = 20 N Block A exert a frictional force of 20 N on block B in downward direction.

For block B :

2F = m2g + 1F = 100 + 20 = 120 N

4. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass

2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage

loss in the energy during the collision is close to : (1) 44 % (2) 50 % (3) 56 % (4) 62 %

4. (3)

E 1 =

2211m(2v) 2m.v22

221m.4v mv2

= 2mv

2 + mv2

= 3 mv 2

After Collision

3mV =

2.2mv V = 2 2 v 3 A B F 2 mV 3 mV 2 mV (5) VIDYALANKAR : JEE-MAIN 2015 : Paper and Solution (Pg. 5) E2 =

21 2 2 v3m.23

23 8vm.29

= 24v3 E

1 E2 =

2243v v3

229v 4v

3 = 25v3 12 1EE E = 2 25v3
3v = 5 9

Percentage loss =

51009
= 55.6 56.

5. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its

base is R and its height is h the z

0 is equal to :

(1) 2h 4R (2) 3h

4 (3) 5h

8 (4)

23h
8R

5. (2)

6. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :

(A) 2MR 32 2
(2) 2MR 16 2 (3) 24MR
93
(4) 24MR
33

6. (3)

Figure alongside shows a solid sphere of mass M.

The radius of the sphere is R.

The volume of the sphere is

34VR3

The density of the sphere is

33M M 3M

4 V4RR3 From this solid sphere a cube of maximum possible volume is cut.

Therefore

2R 3a, where a is the length of the side of the cube of maximum volume

2Ra3

Mass of the cube is

33

33M 8RMa334R 2M

3

The moment of inertia of the cube is

22a 2M 1 2RM6633 228MR 4MR

18 3 9 3

7. From a solid sphere of mass M and radius R, a spherical portion of

radius R

2 is removed, as shown in the figure. Taking gravitational

potential V = 0 at r = , the potential at the centre of the cavity thus formed is : (G = gravitational constant) (1) GM

2R (2) GM

R (3) 2GM

3R (4) 2GM

R M 2R a

JEE-MAIN 2015 : Paper and Solution (6)

(Pg. 6)

7. (2)

Potential at internal point of solid

V = 2

2GM 3 r

R2 2R at r = R 2 V 1 = 2

2GM 3 R

R2 8R = 11 GM 8R

Because of sphere removed.

V 2 =

MG3 3 GM8

R2 8 R

2

Net potential, V = V1 + V2 =

11 GM 3 GM

8 R 8 R = GM

R

8. A pendulum made of a uniform wire of cross sectional area A has time period T. When an

additional mass M is added to its bob, the time period changes to T

M. If the Young's modulus of

the material of the wire is Y then 1

Y is equal to :

(g = gravitational acceleration) (1) 2

MTA1T Mg

(2) 2

MTMg1TA

(3) 2

MTA1T Mg

(3) 2 MTA 1T Mg

8. (1)

T = 2

12M, T 2gg

T 2 =

2 2 212M4 , T 4gg

2M2 21T
T x

2M 2 2 1

211T11T

1 21Mg
A 221M
1T

1 A A1Mg Mg T

R (7) VIDYALANKAR : JEE-MAIN 2015 : Paper and Solution (Pg. 7)

9. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume u =

4UTV and

pressure P = 1U 3V . If the shell now undergoes an adiabatic expansion the relation between T and R is : (1)

RTe (2) 3RTe (3) 1TR (4) 31TR

9. (3)

41UP T3V

For an ideal gas PV = nRT

P = nR'T

V (R molar gas constant)

4nR'TTV

3nR'TV

31TV
3

31T4R3

T 3 3 1 R T 1 R

10. A solid of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in

two ways :

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat.

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same

amount of heat. In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is : (1) n2, 4 n2 (2) n2, n2 (3) n2, 2 n2 (4) 2 n2, 8 n2

10. (No option matches)

Entropy is a state function so depends on initial temperature and final temperature.

Case I : s =

423 423

373 423dT dT 473

cnT T 373

Case II : s =

385.5 398 473

373 385.5 460.5dT dT dT 473c ...... nT T T 373

Note : In the question temperatures are given in C, but should be taken in kelvin so no option matches.

JEE-MAIN 2015 : Paper and Solution (8)

(Pg. 8)

11. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic

expansion, the average time of collision between molecules increases as V q, where V is the volume of the gas. The value of q is : p vC C (1) 35

6 (2) 35

6 (3) 1

2 (4) 1

2

11. (3)

Average time taken collisions

t = runsv = mean free path 21
N2dV

N = Number of molecular

V = volumes

t = 2 2 1

N2dMVV

3RT 2 N d 3R T

M t = VkT where k is a constant = 2M

2 N d 3R

T 2 2V t

For adiabatic process

TV1 = constant

21

2VV constantt

quotesdbs_dbs46.pdfusesText_46

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