Question

A metal ball immersed in alcohol weight $W_1$ at $0^{\circ }C$ and $W_2$ at ${59}^{\circ }C$. The coefficient 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. $W_1=\left(W_2/2\right)$

Answer 1

Answer: (C)

$W_1<W_2$

hence The upthrust is given by $\frac{4}{3}\times \pi R_t^3\rho g$
hence $R_t^3=R_0^3(1+{\gamma }_{m}t)$ and ${\rho }_t={\rho }_0/(1+{\gamma }_{a}t)$
So, the upthrust at $t^{0c}$ is given by
$=\frac{4}{3}\times \pi R_0^3(1+{\gamma }_{m}t)\times \left[{\rho }_0/(1+{\gamma }_{a}t)\right]g$
As ${\gamma }_m<{\gamma }_a$, hence upthrust at $t^{0C}$< upthrust at $0^{0C}$
So, the upthrust is decreased. Hence wight in liquid gets increased.
hence $W_1<W_2$