Paper - I (Solution) Pace

8/13/2019 Paper - I (Solution) Pace

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SOLUTIONS (PHYSICS)

1. (A) As 2 3 p V T V i.e. if temperature increases , volume also increases .Hence w.d. will be positive.

2. (A) Sol.

The electric field due to Q and 2Q have direction opposite to each other, left of Q on the linesegment as shown. Let p be the point where electric field is zero due to the system.

22kQ k2Qr r d

Solving this,r d d 2

As r is positive

r d d 2 Now the potential at that point

k Q k 2QV

d d 2 2d d 2

Solving it will get,V will positive

3. (A) Sol. From graph a,

xE at y 2 slope of line y 2 2V m Similarly, yE at x 1 slope of line x 1 1V m

4. (C) Sol. Initial total energy = Initial kinetic energy + Initial potential energy

2

0 0

1 GMm GMmm 0

2 R R

Total energy, when it reached the surface of earth

21 GMmmv2 R

Applying energy conservation,

2

0

1 GMm GMmmv

2 R R

0

1 1v 2GM

R R


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5. (A)

Sol. Time period T 2k

where 1 21 2

M M M 2M 2MM M M 2M 3

So, time period2M

T 2 3 k

6. (D) Sol. Cons. Of. ang. Momentum about P gives

22M 2LL2MV w

2 12

3Vw

2L

7. (C)

Sol. 1 2

1 1 11

f R R

R 10cm

1 1f ' 3 110 10

10f '

4

eq

1 1 4 1 2 4f 20 10 20 20 10

10feq

3

8. (C)

Sol. 12

rad / hr 1

22

rad / hr 8

32

1 1

2 3

T R

T R

4

2R 4 10 km 41

1

2 R V 2 10 km / hr

1h 422

2 R V 10 km / hr

8h

At closest separation4

rel4

V to line joining 10 km / hr rad/ hr

length of line journing 3 10 km 3


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9. (A)

Sol. 0v

n2

1v

n2 2

2

vn

2 2

beat freq. 1 2n n 1 1

v

02 2 2 2

2 2 8 n4 8 vv

4 2

10. (B)

Sol. p wy

v vx

: as y 0x

; wv 0

pv 0 ; particle moves down

Multiple Choice Questions

11. (B, D)

Sol. Equation of process2P

constant C (1)

Equation of State p R

TM

(2)

From 1 and 2 PT = constant C is false, D is true.

As - changes to2

P changes toP

2

From equation (1) A is false

Hence T changes to 2T . B is true.

12. (B, C)

Sol. No. of collisions per unit area1

time between two collisions area

r m sVn Adistance b w walls A

rm sV T

Distance b/w walls A = volume So

1 2n TA V

If both T and V are halvednA

increases

13. (A, D)

Sol. At R, net0 0

Q QE E E

2A 2A


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At S, net0 0 0

Q Q QE E E

2A 2A A

At T, net0 0

Q QE E E

2A 2A

14. (C) Sol. Since body is floating, Buoyant force is same in both liquids and is equal to the weight of the

body

15. (A, D) Sol. For object at 2P

1 1v 2R R

2R v 4R

2

image is on same side so image will be vertual and upright.

Integer Type Question

16. (0) Sol. As ON MN OM r So, it is equilateral triangle:

Potential at N due to two dipoles;

1 2V v v

oo2 2

kp cos 60kpcos60r r

0

17. (8) Sol. When the ball reaches the robot,

21u cos T aT2

a

u cos T2

Where T is time of flight of ball2usin

g

a 2u sinucos2 g

2a g cot 8m s

18. (2) Sol. Work done by gas

W 200 2.8 300 5.62

W 2560J


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2.56kJ By 1 st law of thermodynamics

U Q W 5.76 2.56 kJ

3.20kJ

19. (2)

Sol. - 4 (1) + 6(1) = 00

q q 2

20. (4)

Sol. 2 2

GM GM g4R 4R R

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Paper - I (Solution) Pace