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Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

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A constrained steel rod of length l, area of cross-section A Young's modulus Y and coefficient of linear expansion $α$ is heated through $t ° C$ . The work that can be performed by the rod when heated is

(YA $α$ t)(l $α$ t)

$\frac{1}{2}$ (YA $α$ t)(l $α$ t)

$\frac{1}{2}$ (YA $α$ t) $\frac{1}{2}$ (l $α$ t)

2(YA $α$ t)(l $α$ t)

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Explanation

$W = \frac{1}{2} F ∆ L w h e r e F = Y A α ∆ t a n d ∆ L = α L ∆ t$

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A spherical black body with a radius of 12cm radiates 450-watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be:

225

450

1000

1800

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Explanation

(d)

Radiated power of the black body, P = $δ A T^{4}$

where A=surface area of the body

T=temperature of the body

and, $δ$ =Stefan's constant

when radius of the sphere is halved, new area,

$A^{'} = \frac{A}{4}$

$∴$ power radiated.

$P^{'} = δ (\frac{A}{4}) {(2 T)}^{4} = \frac{16}{4} (δ A T^{4})$

=4P= $4 \times 450 = 1800$ Watts.

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Two identical bodies are made of a material for which the heat capacity increases with temperature.One of these is at $100^{°}$ C,while the other one is at $0^{°}$ C. If the two bodies are brought into contact, then assuming no heat loss,the final common temperature is -

$50^{°} C$

more than $50^{°}$ C

less than $50^{°}$ C but greater than $0^{°}$ C

$0 {}_{0}C$

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Explanation

(b) Heat lost by 1st body=heat gained by 2nd body. Body at $100^{°}$ C temperature has greater heat capacity than body at $0^{°}$ C so final temperature will be closer to $100^{°}$ C. So $T_{c} > 50^{°} C .$

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A body cools from a temperature 3T to 2T in10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be -

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$

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Explanation

(b)

$C a s e 1 : T e m p . f a l l s f r o m 3 T t o 2 T \frac{3 T - 2 T}{10} = k (\frac{3 T + 2 T}{2} - T) C a s e 2 : L e t f i n a l t e m p . b e x \frac{2 T - x}{10} = k (\frac{x + 2 T}{2} - T) S o l v i n g b o t h e q u a t i o n s, x = \frac{3 T}{2}$

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A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U₁, at wavelength 500nm is U₂ and that at 1000nm is U₃ . Given Wien's constant $b = 2.88 \times 10^{6} n m - K$ . Which of the following is correct?

U₃=0

U₁>U₂

U₂>U₁

U₁=0

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Explanation

(c) Given, temperature, T₁=5760K Since, it is given that energy of radiation emitted by the body at wavelength 250 nm in U₁, at wavelength 500 nm is U₂and that at 1000 nm is U₃.

∴ According to Wien’s law, we get λ_mT=b

where, b=Wien’s constant=2.88 x 10⁶ nmK

=> λ_m=b/T
=>λ_m=2.88x10⁶ nmK/5760K
=> λ_m=500 nm

∴ λ_m= wavelength corresponding to maximum energy,so, U₂>U₁.

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A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4x10⁵J/Kg and g=10N/Kg]

544 km

136 km

68 km

34 km

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Explanation

(b) According to question as conservation of energy, energy gained by the ice during its fall from height h is given by
E=mgh

As given, only one quarter of its energy is absorbed by the ice.
so, mgh/4=mL_I
=> h=mL_I x 4/mg
=L_I x 4/g
=3.4x10⁵x4/10=13.6x10⁴=13600m=136km

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The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be

44.0 J/s

16.8 J/s

8.0 J/s

4.0 J/s

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Explanation

Here, ΔT₁=110-100=10°C

$\frac{d Q_{1}}{d t}$ =4J/s

ΔT₂=210-200=10°C

$\frac{d Q_{2}}{d t}$ =?

As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. 10°C. So, the same rate of heat will flow in the second case

Hence, $\frac{d Q_{2}}{d t}$ =4 J/s

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Two metal wires of identical dimensions are connected in series. If $σ$ ₁ and $σ$ ₂ are the conductivities of the metal wires respectively, the effective conductivity of the combination is

_{$\frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$}

$\frac{σ_{1} σ_{2}}{2 σ_{1} σ_{2}}$

$\frac{σ_{1} + σ_{2}}{σ_{1} σ_{2}}$

$\frac{σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

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The value of coefficient of volume expansion of glycerin is 5x10^-4 K^-1. The fractional change in the density of glycerin for a rise of 40°C in its temperature is -

0.015

0.020

0.025

0.010

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Explanation

$ρ_{T} = ρ_{0} (1 - γ ∆ T) \frac{ρ_{T}}{ρ_{0}} = (1 - γ ∆ T) 1 - \frac{ρ_{T}}{ρ_{0}} = γ ∆ T \frac{ρ_{0} - ρ_{T}}{ρ_{0}} = γ ∆ T = 5 \times 10^{- 4} \times 40 = 0.020$

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Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be (Take specific heat of water=1 cal g^-1 °C^-1 and latent heat of steam = 540 cal g^-1)

24 g

31.5g

42.5 g

22.5 g

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Explanation

Heat lost by steam = Heat gained by water

Let m' amount of steam converts into water.

m' x L +m's(100-80) =msΔt

m' x 540 + m' X 20 = 20 x 1 x(80-10)

m'=20 x 70/560=2.5g

Now, net water=20+2.5=22.5g

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