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Thermal Properties of Matter NEET Questions — Free MCQ Practice

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A solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature 120 $° C$ , then -

A.

Both the cube and the sphere cool down at the same rate

B.

The cube cools down faster than the sphere

C.

The sphere cools down faster than the cube

D.

Whichever is having more mass will cool down faster

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Explanation

(b) Rate of cooling of a body

$R = \frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc} \Rightarrow R \propto \frac{A}{m} \propto \frac{Area}{Volume}$

$\Rightarrow$ For the same surface area. $R \propto \frac{1}{Volume}$

$∵$ Volume of cube < Volume of sphere

$\Rightarrow R_{Cube} > R_{Sphere}$ i.e. cube cools down with faster rate.

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Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength $λ_{B}$ corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 $μm$ . If the temperature of A is 5802 K -

A.

The temperature of B is 1964 K

B.

$λ_{B}$ =1.5 $μm$

C.

The temperature of B is 11604 K

D.

The temperature of B is 2901 K

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Explanation

(b) According to Stefan’s law

$E = {eAσT}^{4} \Rightarrow E_{1} = e_{1} {AσT}_{1}^{4} and E_{2} = e_{2} {AσT}_{2}^{4} ∵ E_{1} = E_{2} ∴ e_{1} T_{1}^{4} = e_{2} T_{2}^{4} \Rightarrow T_{2} = {(\frac{e_{1}}{e_{2}} T_{1}^{4})}^{1 / 4} = {(\frac{1}{81} \times {(5802)}^{4})}^{1 / 4} \Rightarrow T_{B} = 1934 K$

And, from Wein's law $λ_{A} \times T_{A} = λ_{B} \times T_{B}$

$\Rightarrow \frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}} \Rightarrow \frac{λ_{B} - λ_{A}}{λ_{B}} = \frac{T_{A} - T_{B}}{T_{A}} \Rightarrow \frac{1}{λ_{B}} = \frac{5802 - 1934}{5802} = \frac{3968}{5802} \Rightarrow λ_{B} = 1.5 μm$

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A black metal foil is warmed by radiation from a small sphere at temperature T and at a distance d. It is found that the power received by the foil is `P'. If both the temperature and the distance are doubled, the power received by the foil will be

A.

16P

B.

4P

C.

2P

D.

P

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Explanation

(b) Energy received per second i.e., power

$P \propto (T^{4} - T_{0}^{4}) \Rightarrow P \propto T^{4} (∵ T_{0} < < T)$

Also energy received per sec (p) ∝ $\frac{1}{d^{2}}$

(inverse square law)

$\Rightarrow P \propto \frac{T^{4}}{d^{2}} \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{2} \times {(\frac{d_{2}}{d_{1}})}^{2} \Rightarrow \frac{P}{P_{2}} = {(\frac{T}{2 T})}^{2} \times {(\frac{2 d}{d})}^{2} = \frac{1}{4} \Rightarrow P_{2} = 4 P$

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Two metallic spheres $S_{1}$ and $S_{2}$ are made of the same material and have identical surface finish. The mass of $S_{1}$ is three times that of $S_{2}$ . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_{1}$ to that of $S_{2}$ is

A.

$1 / 3$

B.

${(1 / 3)}^{1 / 3}$

C.

$1 / \sqrt{3}$

D.

$\sqrt{3} / 1$

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Explanation

(b) Rate of cooling (R) $= \frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc}$

$\Rightarrow R \propto \frac{A}{m} \propto \frac{Area}{volume} \propto \frac{r^{2}}{r^{3}} \propto \frac{1}{r} \Rightarrow (R) \propto \frac{1}{r} \propto \frac{1}{m^{1 / 3}} [∵ m = ρ \times \frac{4}{3} {πr}^{3} \Rightarrow r \propto m^{1 / 3}] \Rightarrow \frac{R_{1}}{R_{2}} = {(\frac{m_{2}}{m_{1}})}^{1 / 3} = {(\frac{1}{3})}^{1 / 3}$

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Three discs A, B and C having radii 2m, 4m, and 6m respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are $Q_{a}$ , $Q_{b}$ , and $Q_{c}$ respectively

A.

$Q_{a}$ is maximum

B.

$Q_{b}$ is maximum

C.

$Q_{c}$ is maximum

D.

$Q_{a} = Q_{b} = Q_{c}$

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Explanation

(b) Radiated power $P = {AεσT}^{4} \Rightarrow P \propto {AT}^{4}$
From Wein’s law, $λ_{m} T = constant \Rightarrow T \propto \frac{1}{λ_{m}}$

$∴ P \propto \frac{A}{{(λ_{m})}^{4}} \propto \frac{r^{2}}{{(λ_{m})}^{4}} \Rightarrow Q_{A} : Q_{B} : Q_{C} = \frac{2^{2}}{{(300)}^{4}} : \frac{4^{2}}{{(400)}^{4}} : \frac{6^{2}}{{(500)}^{4}}$

$∴$ $Q_{B}$ will be maximum.

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A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is same , then

A.

The hollow sphere will cool at a faster rate for all values of T

B.

The solid sphere will cool at a faster rate for all values of T

C.

Both spheres will cool at the same rate for all values of T

D.

Both spheres will cool at the same rate only for small values of T

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Explanation

(a) Rate of cooling $\frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc}$
As surface area, material and temperature difference are same, so rate of loss of heat is same in both the spheres. Now in this case rate of cooling depends on mass.

$\Rightarrow$ Rate of cooling $\frac{∆ θ}{t} \propto \frac{1}{m}$

$∵ m_{solid} > m_{hollow}$ . Hence hollow sphere will cool fast.

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A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100 $° C$ to 99 $° C$ in 100 s . Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from 100 $° C$ to 99 $° C$ will be approximately -

A.

25 s

B.

50 s

C.

200 s

D.

400 s

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Explanation

(c) Rate of cooling $\frac{∆ θ}{t} = \frac{Aeσ (T^{4} - T_{0}^{4})}{mc}$

$\Rightarrow t \propto \frac{m}{A} [∵ ∆ θ, t, σ (T^{4} - T_{0}^{4} are constant)]$

$\Rightarrow t \propto \frac{m}{A} \propto \frac{Volume}{Area} \propto \frac{a^{3}}{a^{2}} \Rightarrow t \propto a \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{1}}{a_{2}} \Rightarrow \frac{100}{t_{2}} = \frac{1}{2} \Rightarrow t_{2} = 200 \sec$

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A body initially at 80 $° C$ cools to 64 $° C$ in 5 minutes and to 52 $° C$ in 10 minutes. The temperature of the body after 15 minutes will be

A.

42.7 $° C$

B.

35 $° C$

C.

47 $° C$

D.

40 $° C$

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Four identical rods of same material are joined end to end to form a square. If the temperature difference between the ends of a diagonal is 100 $° C$ , then the temperature difference between the ends of other diagonal will be -

A.

0 $° C$

B.

$\frac{100}{l} ° C$ ; where l is the length of each rod

C.

$\frac{100}{2 l} ° C$

D.

100 $° C$

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A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 gm of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is $\frac{1}{4}$ that of first, the rate at which ice melts in gm/sec will be -

A.

3.2

B.

1.6

C.

0.2

D.

0.1

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Explanation

(c) $\frac{Q}{t} = \frac{KA ∆ θ}{l} \Rightarrow \frac{mL}{t} = \frac{K ({πr}^{2}) ∆ θ}{l}$

$\Rightarrow$ Rate of melting of ice $(\frac{m}{t}) \propto \frac{{Kr}^{2}}{l}$

Since for second rod K becomes $\frac{1}{4} th$ , r becomes double and length becomes half, so rate of melting will be twice i.e.

${(\frac{m}{t})}_{2} = 2 {(\frac{m}{t})}_{1} = 2 \times 0.1 = 0.2 gm / \sec$

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

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