==> $R \propto \frac{A}{m} \propto \frac{{{\rm{Area}}}}{{{\rm{volume}}}} \propto \frac{{{r^2}}}{{{r^3}}} \propto \frac{1}{r}$
==> Rate $(R) \propto \frac{1}{r} \propto \frac{1}{{{m^{1/3}}}}$ $\left[ {\because \,\;m = \rho \times \frac{4}{3}\pi {r^3} \Rightarrow r \propto {m^{1/3}}} \right]$
==> $\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)^{1/3}} = {\left( {\frac{1}{3}} \right)^{1/3}}$
$A.$ When small temperature difference between a liquid and its surrounding is doubled the rate of loss of heat of the liquid becomes twice.
$B.$ Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ}\,C$ and $20^{\circ}\,C$. The thermal radiation emitted in a given time by $P$ and $Q$ are in the ratio $1: 1.15$
$C.$ A carnot Engine working between $100\,K$ and $400\,K$ has an efficiency of $75 \%$
$D.$ When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below :

