Question

Two spheres $S_1$ and $S_2$ have radii $R$ and $3R$, temperature $T$ K and $\dfrac{T{\sqrt{3}}$ K respectively. If they are coated with a material of same emissivity, rate of radiation of $S_1$ is $E$ then rate of radiation of $S_2$ is}

• A. $\dfrac{E}{6}$
• B. $\dfrac{E}{3}$
• C. $\dfrac{E}{9}$
• D. $\dfrac{E}{12}$

Hint:

Radiation depends on surface area $(R^2)$ and temperature to the power four $(T^4)$.

Answer 1

The Correct Option is C

Step 1: Using Stefan–Boltzmann law.
The rate of radiation from a body is given by: \[ E = \varepsilon \sigma A T^4 \] where $\varepsilon$ is emissivity, $\sigma$ is Stefan constant, $A$ is surface area and $T$ is absolute temperature.

Step 2: Radiation from sphere $S_1$.
For $S_1$, \[ A_1 = 4\pi R^2,\quad T_1 = T \] \[ E_1 = \varepsilon \sigma (4\pi R^2)T^4 = E \]
Step 3: Radiation from sphere $S_2$.
For $S_2$, \[ A_2 = 4\pi (3R)^2 = 36\pi R^2,\quad T_2 = \frac{T}{\sqrt{3}} \] \[ E_2 = \varepsilon \sigma (36\pi R^2)\left(\frac{T}{\sqrt{3}}\right)^4 \] \[ E_2 = \varepsilon \sigma (36\pi R^2)\frac{T^4}{9} \]
Step 4: Comparing with $E_1$.
\[ E_2 = \frac{1}{9}\varepsilon \sigma (4\pi R^2)T^4 = \frac{E}{9} \]
Step 5: Conclusion.
The rate of radiation of sphere $S_2$ is $\dfrac{E}{9}$.