Question

Two bodies, a sphere and a cube made of the same material and having the same surface areas are heated to the same temperature and allowed to cool in the same surroundings. The ratio of the amount of radiation emitted in equal time intervals will be

• A. $1:1$
• B. $\frac{4\pi }{3}:1$
• C. ${\left(\frac{\pi }{6}\right)}^{1/3}:1$
• D. $\frac{1}{2}{\left(\frac{4\pi }{3}\right)}^{2/3}:1$

Answer 1

The correct option is C ${\left(\frac{\pi }{6}\right)}^{1/3}:1$

Rate of heat loss is given by,
$\frac{Q}{t}=Ae\sigma (T^4-T_s^4)\left[\begin{array}{c}T=\text{Temperature of body}\\ T_s=\text{Temperature of surroundings}\end{array}\right]$
If two bodies are made of same material, have same surface area and are at the same initital temperature, then
$\frac{Q}{t}\propto A$
$\Rightarrow \frac{Q_{Sphere}}{Q_{Cube}}=\frac{4\pi r^2}{6a^2}......\left(1\right)$
[$T,T_s,\sigma$ and $t$ are same]
Given that, volume of sphere is equal to volume of cube
$\therefore a={\left(\frac{4}{3}\pi \right)}^{1/3}r.......\left(2\right)$
Substituting $\left(2\right)$ in $\left(1\right)$, we get:
$\frac{Q_{Sphere}}{Q_{Cube}}=\frac{4\pi r^2}{6[{\left(\frac{4}{3}\pi \right)}^{2/3}.r^2]}$
$\Rightarrow Q_{Sphere}:Q_{Cube}={\left(\frac{\pi }{6}\right)}^{1/3}:1$
Thus, option (c) is the correct answer.