Question

Two metallic spheres A and B are made of the same material and have identical surface finish. The mass of A is eight times the mass of B. Both the spheres are heated to the same temperature and placed in the same room having lower temperature but thermally insulated from each other. The ratio of initial rate of cooling of A to that of B is –

& \text{A) }\dfrac{1}{2} \\\ & \text{B) }\dfrac{1}{4} \\\ & \text{C) }\dfrac{4}{1} \\\ & \text{D) }\dfrac{1}{8} \\\ \end{aligned}$$

Answer 1

We need to understand the factors which affect the heating of a material such as the mass, surface area or volume. These quantities can be different for the same material and can cause a relative difference in getting heated which is required here.

Complete Solution:
We are given two metallic spheres that are made of the same material but have different masses. We know that this condition can result in the surface area and the volume of the spheres to be different as the density is a constant for the same material.

The ratio of the radius of the two spheres can be found from this information. The volume of each of them can be equated as –

& \rho =\dfrac{\text{Mass}}{\text{Volume}} \\\ & {{\rho }_{A}}={{\rho }_{B}} \\\ & \Rightarrow \dfrac{8m}{\dfrac{4}{3}\pi {{r}_{A}}^{3}}=\dfrac{m}{\dfrac{4}{3}\pi {{r}_{B}}^{3}} \\\ & \Rightarrow 8{{r}_{B}}^{3}={{r}_{A}}^{3} \\\ & \therefore {{r}_{A}}=2{{r}_{B}} \\\ \end{aligned}$$ Now, we know that eh ratio of the surface areas of these spheres can be found as – $$\begin{aligned} & {{S}_{A}}=4\pi {{r}_{A}}^{2}=4\pi {{(2{{r}_{B}})}^{2}} \\\ & \therefore {{S}_{A}}=16\pi {{r}_{B}}^{2} \\\ & {{S}_{B}}=4\pi {{r}_{B}}^{2} \\\ & \text{So,} \\\ & \dfrac{{{S}_{A}}}{{{S}_{B}}}=\dfrac{4}{1} \\\ & \therefore {{S}_{A}}=4{{S}_{B}} \\\ \end{aligned}$$ Now, we can find the rate of heating as – $$\begin{aligned} & P=(e\sigma {{T}^{4}})A \\\ & \Rightarrow P=kA \\\ & \Rightarrow {{P}_{A}}=k(4{{S}_{B}}) \\\ & \text{and,} \\\ & \Rightarrow {{\text{P}}_{B}}=k{{S}_{B}} \\\ & \therefore {{P}_{A}}=4{{P}_{B}} \\\ \end{aligned}$$ The initial rate of heating is inversely proportional to the mass of the two bodies also. So, we can find the initial rate of heating as – $$\begin{aligned} & H\propto \dfrac{P}{m} \\\ & \Rightarrow \dfrac{{{H}_{A}}}{{{H}_{B}}}=\dfrac{{{P}_{A}}}{{{m}_{A}}}\times \dfrac{{{m}_{B}}}{{{P}_{B}}} \\\ & \Rightarrow \dfrac{{{H}_{A}}}{{{H}_{B}}}=\dfrac{4}{8} \\\ & \therefore \dfrac{{{H}_{A}}}{{{H}_{B}}}=\dfrac{1}{2} \\\ \end{aligned}$$ This is the required solution. **The correct answer is option A.** **Note:** The initial rate of heating is highly dependent on the surface area and the mass of the spheres made of the same material. The whole bulk of the material takes more time as compared to a smaller sphere, but the further heating is proportional to the surface area.