Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of th

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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…

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Given: Two spherical bodies are made of the same material. For the smaller sphere, $r_1 = 0.2 \, \text{m}$ and $T_1 = 800 \, \text{K}$ . For the bigger sphere, $r_2 = 0.8 \, \text{m}$ and $T_2 = 400 \, \text{K}$ . The energy radiated from the smaller body is $E$ .

Find: Energy radiated from the bigger body.

Using Stefan–Boltzmann law, the radiant energy emitted per second is proportional to surface area and the fourth power of absolute temperature:

P = \sigma A T^4

For a sphere,

A = 4\pi r^2

Therefore,

P \propto r^2 T^4

So,

\frac{P_2}{P_1} = \frac{r_2^2 T_2^4}{r_1^2 T_1^4}

Substituting the given values,

\frac{P_2}{P_1} = \frac{(0.8)^2 (400)^4}{(0.2)^2 (800)^4}

= \left(\frac{0.8}{0.2}\right)^2 \left(\frac{400}{800}\right)^4

= 4^2 \times \left(\frac{1}{2}\right)^4

= 16 \times \frac{1}{16} = 1

Hence,

P_2 = P_1

So the energy radiated from the bigger body is $E$ . Therefore, the correct option is B.

Given: Radiation from a sphere varies as $r^2 T^4$ .

Find: Radiation from the bigger sphere in terms of $E$ .

Compare only the changing factors:

\frac{E_2}{E_1} = \left(\frac{r_2}{r_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4

Here,

\frac{r_2}{r_1} = \frac{0.8}{0.2} = 4

and

\frac{T_2}{T_1} = \frac{400}{800} = \frac{1}{2}

Thus,

\frac{E_2}{E_1} = 4^2 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1

So $E_2 = E_1 = E$ . The correct option is B.

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