In this page we have *Multiple choice questions on thermal properties of matter for Class 11,JEE and NEET* . Hope you like them and do not forget to like , social share
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There are three Rods A,B,C of equal length L at same temperature .There coefficient of linear expansion are α

a. L

b. L

c. L

d. L

L = L_{0} (1 + αθ)

if L_{0} & θ are same

then

L is directly proportional to α

So answer will be

L_{b} > L_{a} > L_{c}

Three bodies are having temperature

T

T

T

Which body among these is most warm

a, T

b, T

c, T

d, None of these

Let us calculate all body temperature on same scale

Let choose Celsius

C = K - 273.15

C=(5F-160)/9

T_{A}= - 42 °F = (5 x -42 - 160 )/ 9

= -41 °C

T_{B} = - 10 °C

T_{C}= 200 K = 200 - 273

= - 73.1 °C

So most warm body is T_{B}

A sphere of mass m and diameter D is Heated by temperature ΔT , if Coefficient of linear expansion is α what will be the change in the Surface Area

a, π D

b. π D

c. π D

d. π D

Initial Surface Area = 4πR^{2}

= 4 π (D/2)^{2}

= 4 π D^{2} / 4 = π D^{2}

Surface Area at ΔT

= 4π(D^{'}/2)^{2}

Now

D^{'}=D(1+α ΔT)

So Surface Area at ΔT

= 4π(D^{2}/4 )(1+α ΔT)^{2}

= π D^{2} (1+ α^{2}ΔT^{2} + 2α ΔT)

Change in surface Area

= π D^{2} (α^{2}ΔT^{2} + 2α ΔT)

= π D^{2}α ΔT (α ΔT+ 2)

if α is the Coefficient of linear expansion of block and L denotes length, T denotes Temperature then which one of these is true

a, dL - α L dT = 0

b, dL + α L dT = 0

c, αdL - L dT = 0

D, αdL - +L dT = 0

From the definition of linear expansion

α= (1/L)(dL/dT)

αL dT = dL

dL - αL dT = 0

So a is correct

Consider the following statement

A.If body A and body B are in state of thermal equilibrium, B & C are in state of thermal equilibrium then A & C are in Equilibrium

B. If body A and body B are not in equilibrium and A & C are not in thermal equilibrium then B & C may be in thermal equilibrium.

a. Only A is Correct

b. Only b is Correct

c. A & B is Correct

d. Neither is correct

For any material, density ρ , mass m and volume V are related by ρ = m/V and B is coefficient of volume expansion then which one is true

a, B = (1/ρ) (dρ/ dT)

b, B = -(1/ρ) (dρ/ dT)

C, B = (1/ρ) (dρ/ dV)

d, B = (1/ρ) (dρ/ dV)

ρ = m/V

ρV = m

differentiating at w.r.t T

d(ρV)/dT = 0

ρ(dV/dT) = - V(dρ/dT)

dV/VdT= - (1 /ρ) (dρ/ dT)

Now since B =dV/VdT

So B = -(1/ρ) (dρ/ dT)

A constant volume air thermometer works on

a, Pascal law

b, Charles law

c, Boyles law

d, Archimedes

Ans. Charles, Law

10 litres of benzene weight

a, more in summer than in winter

b, more in winter than in Summer

c, equal in winter and summer

d, none of above

Density decrease with increase in temp. So same volume will weigh less with increase in temperature

since summer are hotter than winter

so winter will have more weight

An Aluminium Rod of length L

a, αΔT

B, Zero

c. -αΔT

D. none of the above

Zero Since no tensile stress is there, so strain will be zero

A metal ball immersed in alcohol weight m

a. m

b. m

c. m

d none of the above

Let α_{m} and α_{a} are cubical expansion of metal and alcohol

V be the volume of metal at 0 °C

and M_{a} weight of metal in air

Given α_{m} < α_{a}

m_{1} = M_{a} - Vρ_{a}g

m_{2} = mass in air - Vρ_{a}g [(1+α_{m}150)/1+α_{a}150)]

Now 1+α_{m}150 < 1+α_{a}150

as α_{m} < α_{a}

so m_{2} = mass in air - Vρ_{a}g(Something less than one)

So that means m_{2} > m_{1}

if a is coefficient of Linear expansion, b coefficient of areal expansion, c coefficient of Volume expansion. Which of the following is true

a. b=2a

b. c=3a

c. b=3a

d. a=2b

it is known by formula

b=2a

c=3a

the coefficient of linear expansion of an in homogeneous rod changes linearly from α

a. α

b. 1/2(α

c. √α

d. None of the these

Consider a small element of length dx from one end of the rod. Let L be the length of the rod. Now the increase in the coefficient of linear expansion by unit length of the rod is (α_{2}-α_{1})/L. Therefore the value of α of the element located at x is

α_{x}=α_{1} +x(α_{2}-α_{1})/L.

Therefore increase in length of the element is α_{x}dxΔT where ΔT is the rise in temperature.Therefore the increase in the length of the rod is

L=∫α_{x}dxΔT

integrating between the limits 0 and L = 1/2(α_{1}+α_{2})LΔT

Which of the following devices is used to detect thermal radiation?

a. Thermopile

b. Constant volume air thermometer

c. Liquid thermometer

d. Six Maximum and minimum thermometer

Answer is a

The sprinkling of water reduces the temperature of the closed room

a. The water has large latent heat of vaporization

b. Water is bad conductor of heat

c. Specific heat of water is high

d. the temperature of water is less than that of room

Sprinkled water vaporises by taking the heat from the room.The latent heat of vaporization is very high so it takes large heat to vaporize and room becomes cool

So a is correct

As the temperature is increased, the time period of a pendulum

(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob.

(b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.

(c) increases as its effective length increases due to shifting of centre of mass below the centre of the bob.

(d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

Answer is a

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