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 solve the problem, we need to find the energy radiated by two spherical bodies. The energy radiated by a body per unit time 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 temperature. For a sphere, \(A = 4\pi r^2\).

First, let's 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 given as \(E\).

Now, calculate the energy radiated by the bigger 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 relation between \(P_1\) and \(P_2\), 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:

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

Thus, \(P_2 = P_1\), which means the energy radiated from the bigger body is equal to \(E\).

Therefore, the answer is \(E\).