A metal sphere immersed in water weighs $w_1$ at $0^{\circ} \mathrm{C}$ and $w_2$ at $50^{\circ} \mathrm{C}$…
A metal sphere immersed in water weighs $w_1$ at $0^{\circ} \mathrm{C}$ and $w_2$ at $50^{\circ} \mathrm{C}$. The coefficient of cubical expansion of the metal is less than that of water. Then
$W_1>W_2$
$W_1 < W_2$
$W_1=W_2$
$W_1=2 W_2$
Solution
Apparent weight of object in water,
$
\begin{aligned}
& w \propto \frac{\text { density of object (sphere) }}{\text { density of water }} \\
& w \propto \frac{d_s}{d_w}
\end{aligned}
$
As, cubical expansion $(Y)$ of metal is less than water, thats why ratio of density of sphere and water will greater at $50^{\circ} \mathrm{C}$ than $0^{\circ} \mathrm{C}$.
$
\begin{aligned}
& \left(\frac{d_s}{d_w}\right)_{50^{\circ} \mathrm{C}}>\left(\frac{d_s}{d_w}\right)_{0{ }^{\circ} \mathrm{C}} \\
& w_2>w_1
\end{aligned}
$