11. ⇒ (MHT CET 2023 10th May Morning Shift )

Heat energy is incident on the surface at the rate of X J/min . If ' $a$ ' and ' $r$ ' represent coefficient of absorption and reflection respectively then the heat energy transmitted by the surface in ' $t$ ' minutes is

A. $(a + r) x t$

B. $\frac{(a + r)}{xt}$

C. $- (a + r) x t$

D. $\frac{xt}{(a + r)}$

Correct Option is (A)

$We know 1 = a + r + t_{r}$

$\Rightarrow t_{r} = 1 - (a + r)$

$Heat transmitted = t_{r} \times time$

$∴ Q = X \times (1 - a + r) \times t \dots (Substituting for t_{r})$

$Q = X (1 - a + r) \times t$

12. ⇒ (MHT CET 2021 21th September Morning Shift )

A black body has maximum wavelength ' $λ_{m}$ ' at temperature $2000 K$ . Its corresponding wavelength at temperature $3000 K$ will be

A. $\frac{4 λ_{m}}{9}$

B. $\frac{2 λ_{m}}{9}$

C. $\frac{3 λ_{m}}{2}$

D. $\frac{9}{4} λ_{m}$

Correct Option is (B)

By Wien's displacement law $λ T =$ constant

$\begin{aligned} ∴ λ \propto \frac{1}{T} \\ ∴ \frac{λ_{2}}{λ_{1}} = \frac{T_{1}}{T_{1}} = \frac{2000}{3000} = \frac{2}{3} \\ ∴ λ_{2} = \frac{2}{3} λ_{1} \end{aligned}$

13. ⇒ (MHT CET 2021 21th September Morning Shift )

A black reactangular surface of area ' $a$ ' emits energy ' $E$ ' per second at $27^{\circ} C$ . If length and breadth is reduced to ${(\frac{1}{3})}^{rd}$ of initial value and temperature is raised to $327^{\circ} C$ then energy emitted per second becomes

A. $\frac{16 E}{9}$

B. $\frac{8 E}{9}$

C. $\frac{4 E}{9}$

D. $\frac{12 E}{9}$

Correct Option is (A)

$E = e σ \cdot A (T^{4} - T_{0}^{4}) and A = ℓ b$

When $ℓ$ and $b$ change to $\frac{ℓ}{3}$ and $\frac{b}{3}$

$\begin{aligned} A^{'} = \frac{A}{9} \\ \frac{E^{'}}{E} = \frac{A^{'}}{A} \frac{(327 + 273)^{4}}{(27 + 273)^{4}}; \frac{E^{'}}{E} = \frac{1}{9} {(\frac{600}{300})}^{4} \\ ∴ E^{'} = \frac{1}{9} \times (2)^{4} \times E \Rightarrow E^{'} = \frac{16 E}{9} \end{aligned}$

14. ⇒ (MHT CET 2021 20th September Evening Shift )

A black rectangular surface of area ' $A$ ' emits energy ' $E$ ' per second at $27^{\circ} C$ . If length and breadth is reduced to $(1 / 3)^{rd}$ of its initial value and temperature is raised to $327^{\circ} C$ then energy emitted per second becomes

A. $\frac{20 E}{9}$

B. $\frac{8 E}{9}$

C. $\frac{16 E}{9}$

D. $\frac{4 E}{9}$

Correct Option is (C)

The energy emitted by a black body per unit area per unit time (or power per unit area) is given by Stefan-Boltzmann Law :

$σ T^{4}$

Where :

$σ$ is the Stefan-Boltzmann constant.
$T$ is the absolute temperature of the body.

Given :

Initial energy emitted $E$ for area $A$ at temperature $27^{\circ} C = 273 + 27 = 300 K$

$E = A σ (300^{4})$

Now, the area is reduced to $(1 / 3)^{2}$ of its original value :

$A^{'} = {(\frac{1}{3})}^{2} A = \frac{A}{9}$

The temperature is raised to $327^{\circ} C = 273 + 327 = 600 K$

The new energy emitted $E^{'}$ will be :

$E^{'} = A^{'} σ (600^{4})$

Dividing the two equations :

$\frac{E^{'}}{E} = \frac{A^{'} σ (600^{4})}{A σ (300^{4})}$

Given $A^{'} = \frac{A}{9}$ and using $(600^{4}) / (300^{4}) = 2^{4} = 16$ :

$\frac{E^{'}}{E} = \frac{16}{9}$

So, $E^{'} = \frac{16 E}{9}$

The correct answer is :

Option C

$\frac{16 E}{9}$

15. ⇒ (MHT CET 2021 20th September Evening Shift )

Two stars 'P' and 'Q' emit yellow and blue light respectively. The relation between their temperatures $(T_{P}$ and $T_{Q})$ is

A. $T_{P} = T_{Q}$

B. $T_{P} = \frac{T_{Q}}{2}$

C. $T_{P} > T_{Q}$

D. $T_{P} < T_{Q}$

Correct Option is (D)

According to Wien's law, $λ$ T = constant

$∴ T \propto \frac{1}{λ}$

Wavelength of blue light is less than that of yellow light. Hence, temperature of Q is greater than temperature of P.

⇒Kinetic Theory of Gases