Question

Two metallic spheres ${\text{S}}_1$ and ${\text{S}}_2$ are made of the same material and have got identical surface finish. The mass of ${\text{S}}_1$ is thrice that of ${\text{S}}_2$. 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. The ratio of the initial rate of cooling of ${\text{S}}_1$ to that ${\text{S}}_2$ 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 ${\left(\frac{1}{3}\right)}^{\frac{1}{3}}$

Let, mass of ${\text{S}}_1$, $m_{S_1}=3m_1$
so, mass of ${\text{S}}_2$, $m_{S_2}=m_1$

From <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Stefan-Boltzman's law, the rate at which energy leaves the object is
$\frac{\Delta Q}{\Delta t}=ϵ\sigma A(T^4-T_0^4)$
where,
$ϵ$ is the emissivity of the surface,
$A$ surface area,
$\sigma$ is Stefan's constant.

Since, $\Delta Q=mc\Delta T$, we get

$\frac{\Delta T}{\Delta t}=\frac{ϵ\sigma A(T^4-T_0^4)}{mc}$

Where, $c$ is specific heat capacity.

Also, since, $m=\rho \frac{4}{3}\pi r^3$ for a sphere, so we get
$A=4\pi r^2=4\pi {\left(\frac{3m}{4\pi \rho }\right)}^{\frac{2}{3}}$

Hence,
$\frac{\Delta T}{\Delta t}=\frac{ϵ\sigma (T^4-T_0^4)}{mc}\left[4\pi {\left(\frac{3m}{4\pi \rho }\right)}^{\frac{2}{3}}\right]$

$\Rightarrow \frac{\Delta T}{\Delta t}=K{\left(m\right)}^{\frac{-1}{3}}$

where, $K$ is constant for the given material.

For the given two bodies ${\text{S}}_1$ and ${\text{S}}_2$, $K$ is same because they are made of same material and present in the same surroundings. So,

$\frac{{\left(\frac{\Delta T}{\Delta t}\right)}_{S_1}}{{\left(\frac{\Delta T}{\Delta t}\right)}_{S_2}}={\left(\frac{m_{S_{!}}}{m_{S_2}}\right)}^{\frac{-1}{3}}={\left(\frac{1}{3}\right)}^{\frac{1}{3}}$

Hence, option $\left(\text{D}\right)$ is the correct answer.