The loss in weight of a solid when immersed i | vedclass

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Question

(A) The loss in weight is equal to the buoyant force (upthrust),given by $U = V \rho g$,where $V$ is the volume of the solid and $\rho$ is the density

Vedclass · Accepted Answer

$W_0 [1 + ( \gamma _s -\gamma _l) t]$

Vedclass · Answer

(A) The loss in weight is equal to the buoyant force (upthrust),given by $U = V \rho g$,where $V$ is the volume of the solid and $\rho$ is the density of the liquid.At $0^o C$,$W_0 = V_0 \rho_0 g$.At $t^o C$,$W = V_t \rho_t g$.Using the volume expansion formula $V_t = V_0(1 + \gamma_s t)$ and the density relation $\rho_t = \rho_0 / (1 + \gamma_l t)$,we get:$W = V_0(1 + \gamma_s t) \cdot \frac{\rho_0}{1 + \gamma_l t} \cdot g$$W = W_0 \frac{1 + \gamma_s t}{1 + \gamma_l t} = W_0 (1 + \gamma_s t)(1 + \gamma_l t)^{-1}$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $\gamma t$,we have:$W \approx W_0 (1 + \gamma_s t)(1 - \gamma_l t) \approx W_0 (1 + \gamma_s t - \gamma_l t) = W_0 [1 + (\gamma_s - \gamma_l)t]$.

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 the cubical coefficients of expansion of the solid and the liquid are $\gamma_s$ and $\gamma_l$ respectively,then $W$ is equal to:

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