A sphere and a cube, both of copper have equal volumes and are black. They are allowed to cool at same temperature and in same atmosphere. The ratio of their rate of loss of heat will be

Volumes of the cube and the sphere are equal
\(
\begin{array}{ll}
\therefore & \mathrm{a}^3=\frac{4}{3} \pi \mathrm{r}^3 \\
\therefore & \mathrm{a}=\left(\frac{4}{3} \pi \mathrm{r}^3\right)^{\frac{1}{3}}
\end{array}
\)
Rate of loss of radiation is proportional to the surface area of the object.
\(
\begin{aligned}
& \therefore \quad \frac{Q_{\text {sphere }}}{Q_{\text {cube }}}=\frac{4 \pi \mathrm{r}^2}{6 \mathrm{a}^2} \\
& \therefore \quad \frac{\mathrm{Q}_{\text {sphere }}}{\mathrm{Q}_{\text {cube }}}=\frac{4 \pi \mathrm{r}^2}{6\left(\frac{4}{3} \pi \mathrm{r}^3\right)^{\frac{2}{3}}} \\
& \therefore \quad \frac{\mathrm{Q}_{\text {sphere }}}{\mathrm{Q}_{\text {cube }}}=\left(\frac{\pi}{6}\right)^{\left(\frac{1}{3}\right)}
\end{aligned}
\)