Q. A metal ball immersed in alcohol weighs $W_{1}$ at $0℃$ and $W_{2}$ at $59℃$. The coefficient of cubical of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that
A) $W_{1} > W_{2}$
B) $W_{1} < W_{2}$
C) $W_{1} = W_{2}$
D) $W_{1} = 2W_{2}$
Explanation
(b) Let $V_{0},\ V_{t} =$ Volume of the metal ball at $0℃$ and $t℃$ respectively, $\rho_{0},\ p_{t} =$ density of alcohol at $0℃$ and $t℃$ respecitvely. Then
$W_{1} = W_{0} - V_{0}\rho_{0}g$
$W_{2} = W_{t} - V_{t}\rho_{t}g$
Where $V_{t} = V_{0}(1 + \gamma_{m}t)$ and $\rho_{t} = \frac{\rho_{0}}{(1 + \gamma_{a}t)}g = V_{0}\rho_{0}\frac{(1 + \gamma_{m}l)}{(1 + \gamma_{a}l)}$
As $\gamma_{m} < \gamma_{a}$, hence upthrust at $t℃$ is less than at $0℃$. It means upthrust has been decreased with increase in temperature. Due to which the $W_{2} > W_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.