1. Principle of Steady State

At steady state, the temperature of the spherical black body becomes constant. This occurs when the total rate of energy absorbed by the body is equal to the total rate of energy radiated by it. $P_{\text{absorbed}} = P_{\text{radiated}}$

2. Rate of Energy Absorption ($P_{\text{absorbed}}$)

The body absorbs energy from two sources:

1. Incident parallel light rays: The light rays are incident on the cross-sectional area of the sphere. For a sphere of radius $r$, the cross-sectional area is $A_{\text{cross}} = \pi r^2$. Since it's a black body, it absorbs all incident radiation. The power absorbed from the light is: $P_{\text{abs, light}} = I \times A_{\text{cross}} = I (\pi r^2)$
2. Surroundings: The body is in an environment at temperature $T_s = 300$ K. It absorbs thermal radiation from the surroundings over its entire surface area, $A_{\text{surface}} = 4\pi r^2$. The power absorbed from the surroundings is: $P_{\text{abs, surr}} = \sigma T_s^4 \times A_{\text{surface}} = \sigma T_s^4 (4\pi r^2)$

The total power absorbed is the sum of these two contributions: $P_{\text{absorbed}} = I (\pi r^2) + 4\pi r^2 \sigma T_s^4$

3. Rate of Energy Radiation ($P_{\text{radiated}}$)

According to the Stefan-Boltzmann law, a black body at temperature $T$ radiates energy from its entire surface. The power radiated is: $P_{\text{radiated}} = \sigma T^4 \times A_{\text{surface}} = \sigma T^4 (4\pi r^2)$

4. Energy Balance Equation

At steady state, we equate the total power absorbed and radiated: $I (\pi r^2) + 4\pi r^2 \sigma T_s^4 = 4\pi r^2 \sigma T^4$ We can divide the entire equation by $\pi r^2$ (the radius of the sphere is not needed): $I + 4\sigma T_s^4 = 4\sigma T^4$

An alternative way to write this is to consider the net power exchange with the surroundings. The power gained from the incident light must balance the net power radiated to the surroundings: $I (\pi r^2) = \sigma (4\pi r^2) (T^4 - T_s^4)$ $I = 4\sigma (T^4 - T_s^4)$ Both formulations lead to the same equation for $T$.

5. Calculation

We are given:

• Intensity, $I = 912 \text{ Wm}^{-2}$
• Stefan-Boltzmann constant, $\sigma = 5.7 \times 10^{-8} \text{ Wm}^{-2}\text{K}^{-4}$
• Surrounding temperature, $T_s = 300 \text{ K}$

Rearranging the equation to solve for $T^4$: $T^4 = \frac{I}{4\sigma} + T_s^4$

Let's calculate the terms: $4\sigma = 4 \times (5.7 \times 10^{-8}) = 22.8 \times 10^{-8} \text{ Wm}^{-2}\text{K}^{-4}$ $\frac{I}{4\sigma} = \frac{912}{22.8 \times 10^{-8}} = \frac{9120}{228} \times 10^8 = 40 \times 10^8 \text{ K}^4$ $T_s^4 = (300)^4 = (3 \times 10^2)^4 = 3^4 \times (10^2)^4 = 81 \times 10^8 \text{ K}^4$

Now substitute these values back into the equation for $T^4$: $T^4 = (40 \times 10^8) + (81 \times 10^8) = 121 \times 10^8 \text{ K}^4$

To find $T$, we take the fourth root: $T = (121 \times 10^8)^{1/4} = (121)^{1/4} \times (10^8)^{1/4}$ $T = (11^2)^{1/4} \times 10^2 = 11^{1/2} \times 100 = \sqrt{11} \times 100$

6. Final Answer

We know that $3^2=9$ and $3.5^2=12.25$. So, $\sqrt{11}$ is slightly more than 3. More accurately, $3.3^2 = 10.89$, so $\sqrt{11} \approx 3.317$. $T \approx 3.317 \times 100 = 331.7 \text{ K}$ This value is very close to 330 K. To verify, let's check $T=330$ K: $T^4 = (330)^4 = (3.3 \times 10^2)^4 = (3.3)^4 \times 10^8 = (10.89)^2 \times 10^8 \approx 118.6 \times 10^8 \text{ K}^4$. This is close to our calculated value of $121 \times 10^8 \text{ K}^4$. The difference is due to approximation. Our calculated value of 331.7 K is closest to 330 K among the given options.