Question

A sinker of weight $W_0$ has an apparent weight $W_1$ when weighed in a liquid at a temperature $t_1$ and $W_2$ when weight in the same liquid at temperature $t_2$. The coefficient of cubical expansion of the material of sinker is $\beta$. What is the coefficient of volume expansion of the liquid?

• A. $\frac{W_2-W_1}{\left(W_0-W_2\right)\left(t_2-t_1\right)}+\frac{\beta \left(W_0-W_1\right)}{W_0-W_2}$
• B. $\frac{W_2-W_0}{t_2-t_1}\beta$
• C. $\frac{W_1-W_0}{t_2-t_1}\beta$
• D. $\frac{W_0}{\left(t_2-t_1\right)}\beta +\frac{W_2-W_1}{\left(W_0-W\right)}\beta$

Answer 1

Answer: (A)

$W_0-W_1=V\times d_{\ell }\times g$......(i)
$W_0-W_2=V^{\prime }\times d^{\prime \prime }\times g$......(ii)
Also, $V^{\prime }=V(1+\beta \Delta T)$.......(iii)
and $d_{\ell }=d^{\prime }{}_{\ell }\left(1+{\gamma }_{\ell }\Delta T\right)$.......(iv)
From (ii), (iii) and (iv)
$W_0-W_2=\frac{V(1+\beta \Delta T)\times d_{\ell }}{1+{\gamma }_{\ell }\Delta T}\times g$.......(v)
Dividing (i) and (v), we get
$\frac{W_0-W_1}{W_0-W_2}=\frac{V{d}_{\ell }g\left(1+{\gamma }_{\ell }\Delta T\right)}{V(1+\beta \Delta T)d_{\ell }g}$
$\Rightarrow \frac{W_0-W_1}{W_0-W_2}=\frac{1+{\gamma }_{\ell }\Delta T}{1+\beta \Delta T}$
$\Rightarrow \frac{W_0-W_1}{W_0-W_2}=\frac{1+{\gamma }_{\ell }\left(t_2-t_1\right)}{1+\beta \left(t_2-t_1\right)}$
$\Rightarrow \left(W_0-W_1\right)[1+\beta \left(t_2-t_1\right)=\left(W_0-W_2\right)\left[1+{\gamma }_{\ell }\left(t_2-t_1\right)\right]$
$\Rightarrow {\gamma }_{\ell }=\frac{W_2-W_1}{\left(W_0-W_2\right)\left(t_2-t_1\right)}+\frac{\beta \left(W_0-W_1\right)}{\left(W_0-W_2\right)}$