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The loss in weight of a solid when immersed in a liquid at $0^o C$ is $W_0$ and at $t^o C$ is $W$. If cubical coefficient of expansion of the solid and the liquid by $\gamma_s$ and $\gamma_l$ respectively, then $W$ is equal to :
$W_0 [1 + ( \gamma _s -\gamma _l) t]$
$W_0 [1 - (\gamma_s -\gamma_l)t]$
$W_0 [(\gamma_s -\gamma_l) t]$
$W_0t/(\gamma_s -\gamma_l)$
Solution
$U_{0}=V_{0} \sigma_{L}^{0} g=W_{0}$ and $U_{t}=V_{t} \sigma_{L}^{t} g=W$
$\frac{W}{W_{0}}=\frac{V_{t}}{V_{0}} \times \frac{\sigma_{L}^{t}}{\sigma_{L}^{0}}=\frac{\left(1+\gamma_{B} \Delta \theta\right)}{\left(1+\gamma_{L} \Delta \theta\right)}$
$=\left(1+\gamma_{S} \Delta \theta\right)\left(1+\gamma_{L} \Delta \theta\right)^{-1}=1+\gamma_{S} \Delta \theta-\gamma_{L} \Delta \theta$
$\left.=W=W_{0}\left[1+\left(\gamma_{S}-\gamma_{L}\right) \Delta \theta\right)\right]$
$=W_{0}\left[1+\left(\gamma_{S}-\gamma_{L}\right) t\right]$
