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Physics NEET Questions — Free MCQ Practice

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Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles triangle ABC , right angled at B. The points A and B are maintained at temperatures T and $\sqrt{2} T$ respectively. In the steady state the temperature of the point C is $T_{C}$ . Assuming that only heat conduction takes place, $\frac{T_{C}}{T}$ is equal to -

A.

$\frac{1}{(\sqrt{2} + 1)}$

B.

$\frac{3}{(\sqrt{2} + 1)}$

C.

$\frac{1}{2 (\sqrt{2} - 1)}$

D.

$\frac{1}{\sqrt{3} (\sqrt{2} - 1)}$

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A sphere, a cube and a thin circular plate, all made of the same material and having the same mass are initially heated to a temperature of 1000°C. Which one of these will cool first ?

A.

Plate

B.

Sphere

C.

Cube

D.

None of these

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Explanation

(a) Rate of cooling $\frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc} \Rightarrow \frac{∆ θ}{t} \propto A$ . Since area of plate is largest so it will cool fastest.

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Two identical conducting rods are first connected independently to two vessels, one containing water at 100 $° C$ and the other containing ice at 0 $° C$ . In the second case, the rods are joined end to end and connected to the same vessels. Let $q_{1}$ and $q_{2}$ g / s be the rate of melting of ice in two cases respectively. The ratio of is $q_{1}$ / $q_{2}$

A.

$\frac{1}{2}$

B.

$\frac{2}{1}$

C.

$\frac{4}{1}$

D.

$\frac{1}{4}$

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

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

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

Physics MCQs for NEET — Practice Questions with Answers

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