Radiation MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & E = \sigma A{T^4};\,\,A\,\propto \,{R^2} \cr & \therefore \,\,E\,\propto \,{R^2}{T^4} \cr & \therefore \,\,\frac{{{E_2}}}{{{E_1}}} = \frac{{R_2^2\,T_2^4}}{{R_1^2T_1^4}} \cr & \Rightarrow \,\,\frac{{{E_2}}}{{{E_1}}} = \frac{{{{\left( {2\,R} \right)}^2}{{\left( {2\,T} \right)}^4}}}{{{R^2}{T^4}}} \cr & = 64 \cr} $$

$$ - \frac{{dQ}}{{dt}} \propto \left( {\Delta \theta } \right)$$

According to Stefan's law
$$\eqalign{ & \Delta Q = e\sigma A{T^4}\Delta t \cr & {\text{also, }}\Delta Q = mc\,\Delta T \cr & {\text{or, }}mc\,\Delta T = e\sigma A{T^4}\Delta t \cr & {\text{or, }}\frac{{\Delta T}}{{\Delta t}} = \frac{{e\sigma A{T^4}}}{{mc}} \cr & = \frac{{e\sigma {T^4}}}{{mc}}\left[ {\pi {{\left( {\frac{{3\,m}}{{4\,\pi P}}} \right)}^{\frac{2}{3}}}} \right] \cr & = k{\left( {\frac{1}{m}} \right)^{\frac{1}{3}}} \cr & \therefore \,\,\frac{{\frac{{\Delta {T_1}}}{{\Delta {t_1}}}}}{{\frac{{\Delta {T_2}}}{{\Delta {t_2}}}}} = {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{1}{3}} \right)^{\frac{1}{3}}} \cr} $$

According to Wien's law
$${\lambda _m}T = {\text{constant}}\left( {{\text{say}}\,b} \right)$$
where, $${\lambda _m}$$ wavelength corresponding to maximum intensity of radiation and $$T$$ is temperatures of the body in kelvin.
So for two different cases i.e. at two different temperature of body
$$\therefore \frac{{{{\lambda '}_{m'}}}}{{{\lambda _m}}} = \frac{T}{{T'}}$$
Given, $$T = 1227 + 273 = 1500\,K,$$
$$\eqalign{ & T' = 1227 + 1000 + 273 = 2500\,K \cr & {\lambda _m} = 5000\mathop A\limits^ \circ \cr} $$
Hence, $${{\lambda '}_m} = \frac{{1500}}{{2500}} \times 5000 = 3000\mathop A\limits^ \circ $$

In steady state
Energy lost = Energy gained
$$\eqalign{ & \sigma \left( {{T^4} - T_0^4} \right) \times 4\,\pi {R^2} = I\left( {\pi {R^2}} \right) \cr & \therefore \,\,5.7 \times {10^{ - 8}}\left[ {{T^4} - {{\left( {300} \right)}^4}} \right] \times 4 = 912 \cr & \therefore \,\,T = 330\,K \cr} $$

Wein’s law correctly explains the spectrum

Initially at $$t = 0$$
Rate of cooling $$\left( R \right) \propto $$  Fall in temperature of body $$\left( {\theta - {\theta _0}} \right)$$
$$\frac{{{R_1}}}{{{R_2}}} = \frac{{{\theta _1} - {\theta _0}}}{{{\theta _2} - {\theta _0}}} = \frac{{100 - 40}}{{80 - 40}} = \frac{3}{2}$$

According to Wien’s displacement law, the quantity of energy radiated out by a body is not uniformly distributed over all the wavelengths emitted by it. It is maximum for a particular wavelength $$\left( \lambda \right),$$  which is different at different temperatures. As the temperature is increased, the value of the wavelength which carries maximum energy is decreased.
The statement of this law is as follows
“The wavelength corresponding to maximum energy is inversely proportional to the absolute temperature of the body.”
i.e. $${\lambda _m} \propto \frac{1}{T}\,\,{\text{or}}\,\,{\lambda _m}T = {\text{constant}}$$

Given, temperature, $${T_1} = 5760\,K$$
Since, it is given that energy of radiation emitted by the body at wavelength $$250\,nm$$  in $${U_1},$$ at wavelength $$500\,nm$$  is $${U_2}$$ and that at $$1000\,nm$$   is $${U_3}.$$
$$\because $$ According to Wien’s law, we get $${\lambda _m}T = b$$
where, $$b =$$  Wien’s constant $$ = 2.88 \times {10^6}nmK$$
$$\eqalign{ & \Rightarrow {\lambda _m} = \frac{b}{T} \cr & \Rightarrow {\lambda _m} = \frac{{2.88 \times {{10}^6}nmK}}{{5760\,K}} \cr & \Rightarrow {\lambda _m} = 500\,nm \cr} $$
$$\therefore {\lambda _m} = $$    wavelength corresponding to maximum energy, so, $${U_2} > {U_1}.$$

Newton's law of cooling
$$\eqalign{ & \frac{{d\theta }}{{dt}} = - k\left( {\theta - {\theta _0}} \right) \cr & \Rightarrow \,\,\frac{{d\theta }}{{\left( {\theta - {\theta _0}} \right)}} = kdt \cr} $$
Integrating
$$ \Rightarrow \,\,\log \left( {\theta - {\theta _0}} \right) = - kt + c$$
Which represents an equation of straight line.
Thus the option (A) is correct.