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Chapter: 11th Physics : Heat and Thermodynamics

Multiple Choice Questions - Physics: Heat and Thermodynamics

1 mark question and answers, Multiple Choice Questions - Physics: Heat and Thermodynamics - Physics : Heat and Thermodynamics

Heat and Thermodynamics (Physics)

Multiple choice questions:

 

1. In hot summer after a bath, the body’s

a) internal energy decreases

b) internal energy increases

c) heat decreases

d) no change in internal energy and heat

Answer : a) internal energy decreases

Solution : Temperature decreases, So internal energy decreases.

 

2. The graph between volume and temperature in Charles’ law is

a) an ellipse

b) a circle

c) a straight line

d) a parabola

Answer : c) a straight line

Solution: P - constant, VT

 

3. When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is

a) isothermal

b) adiabatic

c) isobaric

d) isochoric

Answer : b) adiabatic

 

4. An ideal gas passes from one equilibrium state (P1V1T1N) to another equilibrium state (2P1, 3V1T2N). Then

a) T1=T2

b) T1=T2/6

c) T1=6T2

d) T1=3T2

Answer: b) T1 = T2 / 6

Solution:

T1 = T2 / 6

2P1 3V1 = NKT2

6P1V1 = NKT2

P1V1 = NKT2 / 6

NKT1 = NK T2 / 6

P1V1 = NKT1

NKT1 = NK T2 / 6

T1 = T2 / 6

 

5. When a uniform rod is heated, which of the following quantity of the rod will increase

a) mass

b) weight

c) center of mass

d) moment of inertia

Answer : d) moment of inertia

Solution:

I = mr2

Due to thermal expansion, if T increases I also increases.

 

6. When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process

a) Q > 0, W > 0,

b) Q < 0, W > 0,

c) Q > 0, W < 0,

d) Q < 0, W < 0,

Answer: a) Q>0,W>0,

Solution:

Heat is given to the system

System gains heat → Q > 0

Lid Pushes upwards → Work done

by the system → W > 0

 

7. When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true?

a) U > 0, W > 0,

b) U < 0, W > 0,

c) U < 0, W < 0,

d) U = 0, W > 0,

Answer : b) ΔU < 0, W > 0,

Solution:

Workdone by the system, W > 0

Heat flows outward → ΔU ↓

 

8. A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true

a) U > 0, Q = 0

b) U > 0, W < 0

c) U > 0, Q > 0

d) U = 0, Q > 0

Answer : c) ΔU > 0, Q > 0

 

9. An ideal gas is taken from (Pi,Vi) to (Pf,Vf) in three different ways. Identify the process in which the work done on the gas the most.


a) Process A

b)Process B

c) Process C

d) Equal work is done in Process A,B &C

Answer : b) Process B

 

10. The V-T diagram of an ideal gas which goes through a reversible cycle A→B→C→D is shown below. (Processes D→A and B→C are adiabatic)


The corresponding PV diagram for the process is (all figures are schematic)


Ans: b

 

11. A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is

a) 8280 K

b) 5000K

c) 7260 K

d) 9044 K

Answer: a) 8280 K

Solution:

λm = b / T

T = b / λm

= 2.898×10-3 / 350×10-9

λm = 8280 K

 

12. Identify the state variables given here?

a) Q, T, W

b) P, T, U

c) Q, W      

d) P, T, Q

Answer : b) P, T, U

 

13. In an isochoric process, we have

a) W = 0

b) Q = 0

c) ∆U = 0

d) ∆T = 0

Answer: a) W = 0

 

14. The efficiency of a heat engine working between the freezing point and boiling point of water is

a) 6.25%

b) 20%

c) 26.8%

d) 12.5%

Answer: c) 26.8%

Solution:

Tn = 100°C

Tn = 373

TL = 0°C

TL = 273 K

 [ TH − TL ] / TH = n

= ( [373 – 273] / 373 ) × 100

= ( 100 / 373 ) × 100

n = 26.8%,

 

15. An ideal refrigerator has a freezer at temperature −12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is

(a) 50°C     

(b) 45.2°C

(c) 40.2°C

(d)37.5°C

Answer : c) 40.2°C

Solution:

TL = −12°C + 278 K

TL = 261 K,

B=5,

TH = ?

β = TL / [TH – TL ]

TH = [ TL β ] + TL

= [261/5] + 261

= 525 + 261

= 313.2K

TH = 313.2 K − 273°C

TH = 40.2°C

 

Answers:

 1) a     2) c    3) b   4) b

  5) d   6) a    7) b   8) c

  9) b 10) b 11) a  12) b

13) a  14) b 15) c

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11th Physics : Heat and Thermodynamics : Multiple Choice Questions - Physics: Heat and Thermodynamics |


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