Question

Two metallic spheres $S_1$ and $S_2$ are made of the same material and have identical surface finish. The mass of $S_1$ is three times that of $S_2$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature, but are thermally insulated from each other. This ratio of the initial rate of cooling of $S_1$ to that of $S_2$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{1}$
• D. ${\left(\frac{1}{3}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is D ${\left(\frac{1}{3}\right)}^{\frac{1}{3}}$

Given that,
$m_1=3m_2$ where $m_1$ and $m_2$ are masses of sphere $S_1$ and $S_2$ respectively.
$\Rightarrow \rho V_1=3\rho V_2$
$\Rightarrow \frac{V_1}{V_2}=3$
$\Rightarrow \frac{\frac{4\pi r_1^3}{3}}{\frac{4\pi r_2^3}{3}}=3$
$\Rightarrow \frac{r_1}{r_2}=3^{\frac{1}{3}}$ ...(i)
$[\because m=\frac{4}{3}\pi r^3\rho \Rightarrow r\propto m^{\frac{1}{3}}]$

The rate of loss of heat is given by:
$\frac{dQ}{dt}=P=e\sigma A{T}^4$
$\Rightarrow mS\left(\frac{dT}{dt}\right)=eA\sigma T^4$ $\Rightarrow \frac{dT}{dt}mS=e\left(4\pi r^2\right)\sigma T^4$ (rate of cooling)
$\Rightarrow \frac{d{T}_1}{dt}\propto \frac{r_1^2}{m_{1}S}$
and $\frac{d{T}_2}{dt}\propto \frac{r_2^2}{m_{2}S}$
$\Rightarrow \frac{\frac{d{T}_1}{dt}}{\frac{d{T}_2}{dt}}=\frac{\frac{r_1^2}{m_1}}{\frac{r_2^2}{m_2}}\Rightarrow \frac{\frac{d{T}_1}{dt}}{\frac{d{T}_2}{dt}}=\frac{r_1^2}{r_2^2}\times \frac{m_2}{m_1}$

Substituting from equation (i),
$\frac{\frac{d{T}_1}{dt}}{\frac{d{T}_2}{dt}}=(3^{1/3}{)}^2\times \frac{1}{3}={\frac{1}{3}}^{\frac{1}{3}}$