Two spheres $A$ and $B$ with masses in the ratio $2 : 3$ and specific heat $2 : 3$ fall freely rest. If the rest in their temperatures on reaching the ground in the ratio $1 : 2$ the ratio of their heights of fall is:-

$\textbf{Given:-}$ Mass ratio of two sphere A and B = 2:3

Specific heat of two sphere A and B = 2:3

Temperature ratio = 1:2

$\textbf{Solution:-}$ Potential energy of sphere = Thermal energy of sphere

$m_{A}gh_{A} = m_{A}S_{A}(\Delta T)_{A}..............(1)$

$m_{B}gh_{B} = m_{B}S_{B}(\Delta T)_{B}..............(2)$

Now divide equation 1 by 2, we get

$\dfrac{h_{A}}{h_{B}} = \left ( \dfrac{S_{A}}{S_{A}} \right )\times \dfrac{\Delta T_{A}}{\Delta T_{B}}$

$\dfrac{h_{A}}{h_{B}}= \dfrac{2}{3}\times \dfrac{1}{2}$

$\Rightarrow\ \dfrac{h_{A}}{h_{B}} =\ \frac{1}{3}$

$\textbf{Hence the correct option is B}$