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A metal ball immersed in alcohol weighs ${W_1}$ at $0°C$ and ${W_2}$ at $59°C.$ The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol, it can be shown that
${W_1} > {W_2}$
${W_1} = {W_2}$
${W_1} < {W_2}$
${W_2} = ({W_1}/2)$
Solution
(c) As the coefficient of cubical expansion of metal is less as compared to the coefficient of cubical expansion of liquid, we may neglect the expansion of metal ball. So when the ball is immersed in alcohol at $0°C,$ it displaces some volume $V$ of alcohol at $0°C$ and has weight $W_1$.
$W_1 = W_0 -V\rho_0g$
where $W_0 $= weight of ball in air
Similarly, $W_2 = W0 -V\rho_{50}g$
where $\rho_{0} $= density of alcohol at $0°C$
and $\rho_{50} $= density of alcohol at $50°C$
As $\rho_{50} < \rho_{0}, ==> W_2 > W_1 \,or\, W_1 < W_2$
