⇒Kinetic Theory of Gases

Correct Option is (B)

From Stefan - Boltzmann's law

$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{e}\left(\sigma {\mathrm{AT}}^4\right)$

Given $\mathrm{A}$ and $\frac{\mathrm{dQ}}{\mathrm{dt}}$ are same for both the bodies

$\begin{array}{rl}& \Rightarrow {\mathrm{e}}_1{\mathrm{T}}_1^4={\mathrm{e}}_2{\mathrm{T}}_2^4\\ \therefore & {\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right)}^4=\frac{{\mathrm{e}}_2}{{\mathrm{e}}_1}=\frac{3}{1}\\ & \Rightarrow \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{\sqrt[4]{3}}{1}=\frac{3^{\frac{1}{4}}}{1}\end{array}$

Correct Option is (B)

Power Radiated by black body, $\mathrm{P}=\sigma {\mathrm{AT}}^4$

$\therefore$ For first black body:

${\mathrm{P}}_1=\sigma 4\pi {\mathrm{r}}_1^2{\mathrm{T}}_1^4$

$\therefore$ For second black body:

${\mathrm{P}}_2=\sigma 4\pi {\mathrm{r}}_2{}^2{\mathrm{T}}_2{}^4$

$\therefore$ The ratio will be:

$\begin{array}{rl}& \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{\sigma 4\pi {\mathrm{r}}_1^2{\mathrm{T}}_1^4}{\sigma 4\pi {\mathrm{r}}_2^2{\mathrm{T}}_2^4}\\ & \frac{1}{2}=\frac{{\mathrm{r}}_1^2{\mathrm{T}}_1^4}{{\mathrm{r}}_2^2{\mathrm{T}}_2^4}\\ & \frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}=\frac{1}{2}\frac{{\mathrm{T}}_2^4}{{\mathrm{T}}_1^4}\\ & \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{1}{\sqrt{2}}{\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^2\end{array}$

Correct Option is (D)

$\begin{array}{rl}& \text{ Given: }{\mathrm{T}}_2={\mathrm{T}}_1+\frac{50}{100}{\mathrm{T}}_1\\ & {\mathrm{T}}_2=1.5{\mathrm{T}}_1\end{array}$

According to Stefan's law,

$\frac{\mathrm{Q}}{\mathrm{At}}=\sigma {\mathrm{T}}^4$

Percentage change in radiation is,

$\begin{array}{rl}& \mathrm{\Delta }E\mathrm{\%}=\frac{T_2^4-T_1^4}{T_1^4}\times 100\\ & \mathrm{\Delta }E\mathrm{\%}=\frac{(1.5{)}^4{T}_1^4-T_1^4}{T_1^4}\times 100\\ & \mathrm{\Delta }E\mathrm{\%}\approx 400\mathrm{\%}\end{array}$

Correct Option is (A)

From Stefan-Boltzmann's law,

$\mathrm{P}=\frac{\mathrm{dQ}}{\mathrm{dt}}={\mathrm{A}\sigma \mathrm{T}}^4$

Also, from Wien's displacement law,

$\begin{array}{rl}& {\lambda }_{max}=\frac{\mathrm{b}}{\mathrm{T}}(\mathrm{b}\to \text{ Wien's constant })\\ & \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda }\end{array}$

$\begin{array}{rl}& \therefore \mathrm{P}=\mathrm{A}\cdot \sigma {\left(\frac{\mathrm{b}}{\lambda }\right)}^4\\ & \Rightarrow \mathrm{P}\propto \frac{1}{(\lambda {)}^4}\end{array}$

$\therefore$ Ratio of power dissipated is $\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}={\left(\frac{{\lambda }_1}{{\lambda }_2}\right)}^4$ Given ${\lambda }_1=\lambda$ and ${\lambda }_2=\frac{2\lambda }{3}$

$\therefore \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{(\lambda {)}^4}{{\left(\frac{2\lambda }{3}\right)}^4}=\frac{81}{16}$

Correct Option is (B)

From Wien's Displacement law,

${\lambda }_{max}=\frac{b}{T}\Rightarrow T=\frac{b}{{\lambda }_{max}}$

From Stefan-Boltzmann law

$E=\sigma T^4=\sigma {\left(\frac{b}{{\lambda }_{max}}\right)}^4$

Let the new emissive power be ${\mathrm{E}}^{\prime }$.

$\begin{array}{rl}\therefore & E^{\prime }=\sigma {\left(\frac{b}{\frac{2{\lambda }_{max}}{3}}\right)}^4\\ & E^{\prime }=\frac{81}{16}E\end{array}$