Question

A solid sphere at a temperature of 400 K radiates a power $P$. If the radius of the sphere is halved and its absolute temperature is doubled, then the power radiated by it is

• A. $\frac{P}{4}$
• B. $\frac{P}{2}$
• C. $2P$
• D. $4P$

Hint:

In the Stefan-Boltzmann law, power depends on $T^4$, so doubling the temperature increases the power by a factor of 16, while halving the radius reduces the surface area by a factor of (4)

Answer 1

The Correct Option is D

The power radiated by a black body (assuming the sphere behaves as one) is given by the Stefan-Boltzmann law: 
\( P = \sigma A T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the absolute temperature. 
For a sphere, \( A = 4\pi r^2 \). 
Initially, \( T_1 = 400 \) K, \( r_1 = r \), and power \( P_1 = P = \sigma (4\pi r^2) (400)^4 \).

Now, the radius is halved (\( r_2 = \frac{r}{2} \)), so the new surface area \( A_2 = 4\pi \left(\frac{r}{2}\right)^2 = \pi r^2 = \frac{A_1}{4} \). 
The temperature is doubled (\( T_2 = 2 \times 400 = 800 \) K). The new power
\[ P_2 = \sigma A_2 T_2^4 = \sigma \left(\frac{4\pi r^2}{4}\right) (800)^4 \]
Since \( 800 = 2 \times 400 \), \( (800)^4 = (2 \times 400)^4 = 16 \times (400)^4 \). Thus:
\[ P_2 = \sigma (\pi r^2) (800)^4 = \sigma (\pi r^2) (16 \times (400)^4) = \left(\sigma (4\pi r^2) (400)^4\right) \times \frac{16}{4} = P \times 4 = 4P \]
The new power radiated is \( 4P \).