Question

A sphere and a cube made of the same metal have equal volumes, identical surface characteristics and are at the same temperature. If they are allowed to cool in identical surroundings, the ratio of their rates of loss of heat will be :

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

Answer 1

Answer: (D)

${\left(\frac{\pi }{6}\right)}^{\frac{1}{3}}:1$

The radiation emitted

$Q=\sigma AT(T^4-T_0^4)$

Here, for sphere and cube , $\sigma ,t,T$ and $T_0$ are same.

$\therefore \frac{Q_{\text{sphere}}}{Q_{\text{cube}}}=\frac{A_{\text{sphere}}}{A_{\text{cube}}}=\frac{4\pi r^2}{6a^2}$ ........(i)

Given Volume of sphere = Volume of Cube

$\frac{4}{3}\pi r^3=a^3$

$a={\left(\frac{4}{3}\pi \right)}^{1/3}\cdot r$ ........(ii)

Substituting value of Eqs. (ii) in Eq. (i) , we get

$\frac{Q_{\text{sphere}}}{Q_{\text{cube}}}=\frac{4\pi r^2}{6{\{{\left(\frac{4}{3}\pi \right)}^{1/3}\cdot r\}}^2}={\left(\frac{\pi }{6}\right)}^{\frac{1}{3}}:1$