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

Step 1: Understand Stefan-Boltzmann Law. 

The energy radiated per unit time (power) by a black body 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. 

Step 2: Calculate the surface areas. The surface area of a sphere is given by \(A = 4\pi r^2\). For the smaller body, \(r_1 = 0.2\) m. \(A_1 = 4\pi (0.2)^2 = 4\pi (0.04) = 0.16\pi\) m\(^2\) For the bigger body, \(r_2 = 0.8\) m. \(A_2 = 4\pi (0.8)^2 = 4\pi (0.64) = 2.56\pi\) m\(^2\) 

Step 3: Calculate the radiated energies. For the smaller body, \(T_1 = 800\) K. \(E = P_1 = \sigma A_1 T_1^4 = \sigma (0.16\pi) (800)^4\) For the bigger body, \(T_2 = 400\) K. \(P_2 = \sigma A_2 T_2^4 = \sigma (2.56\pi) (400)^4\) 

Step 4: Find the ratio of radiated energies. \(\frac{P_2}{P_1} = \frac{\sigma (2.56\pi) (400)^4}{\sigma (0.16\pi) (800)^4} = \frac{2.56}{0.16} \times \left(\frac{400}{800}\right)^4\) \(\frac{P_2}{P_1} = 16 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1\) So, \(P_2 = P_1 = E\).