A metal ball immersed in alcohol weighs w₁ at 0°C and w₂ at 50°C. Assuming that the density of the metal is large compared to that of alcohol, then..

Apparent weight = true weight - upthrust

$w_1=w_0-V\rho g$

$w_2=w_0-\left[V\left(1+{\gamma }_s∆\theta \right)\right]\left(\frac{\rho }{1+{\gamma }_l∆\theta }\right)g=w_0-V\rho g\frac{\left(1+{\gamma }_s∆\theta \right)}{1+{\gamma }_l∆\theta }$

Alcohol has greater value of $\gamma$ compared to that of metal

Hence, $w_2>w_1$