230 Heat Success Magnet-Solutions (Part-I)

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Section - B : Multiple Choice Type Questions

1. Answer (2, 3, 4)

5 5

2

20 10(1 10 ) 0.5 1 10

(0.4)h

0.498 mh When man stands:

5 5

2

95 1010 0.5 10 '

(0.4)h

h' = 0.491 m

h = 0.007 m

1 1 2 2

1 2

PV P V

T T (P

1 = P

2)

1 2

1 2

V V

T T

2

2

0.491 0.498297.2

293T

T

Temperature must be raised to 24.2C.When gas is heated, the process is isobaric and when the piston compresses the gas, the process is isothermal.

2. Answer (1, 4)

mC

AT

at

d )(4 03

0 and mC = same

density

1C

3. Answer (2, 3, 4)

60

/ 2 / 2

B C B CBT T T TT

R R R

120 = 5TB 3T

C

10B C CT T T

R R

2 10C BT T

TC = 24.3C

TC = 38.6C

231Success Magnet-Solutions (Part-I) Heat

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

4. Answer (1, 3)

Heat current = kAx

T

x

T

= kA

current Heat

As x

T

is decreasing, so either k or A is increasing as we move away from hot end.

5. Answer (1, 2, 3, 4)

U = 100J Since no force of interaction between gas molecules (for ideal gas) potential energy is zeroNow for diaomic gas

Translation degree of freedom = 3

Rotational degree of freedom = 2

So KEt = 60 J Equipartion energy

KER = 40 J Equipartion energy

6. Answer (1, 2, 3, 4)

For isobaric process 2

3

2 f

W

U

For isotherm process U = 0 i.e. Q = W7. Answer (2, 3)

For n > 1 if volume increase then temperature decreases i.e. internal energy decreases.

8. Answer (2, 4)

PV = Constant

2PV =

8

VP P 6 atm

P 1 T = Constant

TV 1 = Constant

W < 0 compression

9. Answer (2, 3)

A gas does not obey laws of ideal gas at high pressure and low temperature.

10. Answer (2, 4)

E = T4AIndependent of T

0

11. Answer (1, 2, 3, 4)

T = V 2

232 Heat Success Magnet-Solutions (Part-I)

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

PV1 = Constant

C = 2

RCV

C = 2

RCV

So Q = nCdT

TR

W 2

W = 150 R

Q = 600 R12. Answer (1, 4)

Span AB

A B

A B

V V

T T

. 600 K

B

B A

A

VT T

V

Span CD

.( ) 900

CD v D CQ nC T T R Span DA

1ln 2 300 ln 831.6

4

A

DA DA

D

VQ W nRT R R

V

12a. Answer (2) [JEE (Advanced)-2014]

Hint : Wia = 0 W

af = 200 J, U

i = 100 J, U

b = 200 J

W

W

ib

bf

= 50 J

= 100 J

Q

W

iaf

iaf

= 500 J

= 200 J

Uf U

i = 300 J

P

V

Ui = 100 J

b

fa

i

Ub = 200 J

400 JfU

Qib

= Uib + W

ib

= (200 100) + 50 = 150

Qbf

= Ubf

+ Wbf

= (400 200) + 100 = 300

3002

150

bf

ib

Q

Q

233Success Magnet-Solutions (Part-I) Heat

Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

13. Answer (2, 3)

14. Answer (1, 2, 3, 4)

Tm

1 E = T 4

12 4

3mm

So 122

1

3

4 or

4

3TT

T

T

and 122

1

81

256 or

256

81EE

E

E 15. Answer (3, 4)

Rate of cooling capacity Heat

1

Heat capacity of A is greater than that of B

So A cools faster than B

16. Answer (1, 2, 3, 4)

U1

U2

U3

U1 = <

3 2U U

m

= T

b

= 2880

1088.23

= 1 106

= 10,000 1010 m

= 10,000

17. Answer (1, 2, 3)

18. Answer (1, 3)

19. Answer (1, 4)

20. Answer (1, 2, 3, 4)

b = 0 then dU = 0 i.e. T = constant

a = b then dU + W = 0

dQ = 0 i.e. The process is adiabatic

dV = nCv T

W = nR TDividing these two RdU C

vW = 0

2 dU 3W = 0

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