Question

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of bigger body is 400 K. If the energy radiate from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding to be negligible):

• A. 256 E
• B. E
• C. 64 E
• D. 16 E

Hint:

The energy radiated is proportional to the surface area and the fourth power of the temperature.

Answer 1

The Correct Option is B

To determine the energy radiated by two spherical bodies, we utilize the Stefan-Boltzmann law, which states that the energy radiated per unit time is \(P = \sigma A T^4\), where \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature. For a sphere, the surface area is given by \(A = 4\pi r^2\).

First, we calculate the energy radiated by the smaller body:

• Radius \(r_1 = 0.2\ m\)
• Temperature \(T_1 = 800\ K\)
• Surface area \(A_1 = 4\pi (0.2)^2 = 0.16\pi\ m^2\)
• Power \(P_1 = \sigma A_1 T_1^4 = \sigma \times 0.16\pi \times 800^4\). This power is denoted as \(E\).

Next, we calculate the energy radiated by the larger body:

• Radius \(r_2 = 0.8\ m\)
• Temperature \(T_2 = 400\ K\)
• Surface area \(A_2 = 4\pi (0.8)^2 = 2.56\pi\ m^2\)
• Power \(P_2 = \sigma A_2 T_2^4 = \sigma \times 2.56\pi \times 400^4\)

To find the relationship between \(P_1\) and \(P_2\), we compare the two powers: \(\frac{P_2}{P_1}=\frac{\sigma \times 2.56\pi \times 400^4}{\sigma \times 0.16\pi \times 800^4}=\frac{2.56}{0.16}\times \left(\frac{400}{800}\right)^4\)

This simplifies to:

• \(\frac{P_2}{P_1}=\frac{2.56}{0.16}\times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1\)

Therefore, \(P_2 = P_1\). This indicates that the energy radiated from the larger body is equal to \(E\).

The final answer is \(E\).