Question

A metal ball is immersed in alcohol weighing $W_1$ at $0°C$ and $W_2$ at $50°C.$ The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that density of metal is large compared to that of alcohol, it can be shown

• A. $W_1>W_2$
• B. $W_1=W_2$
• C. $W_1<W_2$
• D. Relation between $W_1\&W_2$ can not be established.

Answer 1

The correct option is A $W_1>W_2$

Let volume and density of ball at $0°C$ be $V_1\&{\rho }_1$ respectively and at $50°C$be $V_2\&{\rho }_2$respectively.
As in both the case, ball is completely immersed in alcohol , so volume displaced of liquid will be same as the volume of ball.
By Law of Buoyancy,
$W_1=V_1{\rho }_{1}g$ & $W_2=V_2{\rho }_{2}g$
$\Rightarrow \frac{W_1}{W2}=\frac{V_1{\rho }_1}{V_2{\rho }_2}$
Now, ${\rho }_{1>}{\rho }_2$
as volume of ball increases. Because of expansion of alcohol, volume of alcohol increases while mass remain same so density of alcohol decreases. Though $V_2>V_1$ , but as the coefficient of cubical expansion of the metal is less than that of alcohol, its effect can be neglected compared to effect of liquid and it can be considered as $V_2\simeq V_1$.
Therefore $\frac{W_1}{W_2}>1\Rightarrow W_1>W_2$