Explanation

(b) Density at 0$℃$, $\rho_{0}$=1.0127
Density at 100$℃$, $\rho_{100}$=1
Coefficient of real expansion of liquid
$\gamma_{real} = \frac{\rho_{0} - \rho_{100}}{\rho_{100} \times \mathrm{\Delta}t}$
= $\frac{1.0127 - 1}{1 \times 100}$ = 0.0127$\times 10^{- 2}$
=$1.27 \times 10^{- 4}$
$\gamma_{real} = \gamma_{app} + \gamma_{g}$
$\gamma_{g}$=coefficient of volume expansion of glass=3$\alpha$
$\therefore$ $1.27 \times 10^{- 4} = \gamma_{app} + 3\alpha$
$1.27 \times 10^{- 4} = \gamma_{app} + 3 \times 9 \times 10^{- 6}$
$\gamma_{app} = 1.27 \times 10^{- 4} - 27 \times 10^{- 6}$
= $1.27 \times 10^{- 4} - 0.27 \times 10^{- 4}$
= 1$\times 10^{- 4}$
$\therefore$ $\gamma_{app} = \ \frac{Mass\ expelled}{Remaining\ mass \times \mathrm{\Delta}t}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \times 10^{- 4} = \ \frac{m_{1} - m_{2}}{m_{2} \times 100}\ $
$\left( \frac{m_{1}}{m_{2}} - 1 \right) = 1 \times 10^{- 4} \times 100 = 10^{- 2}$
$\frac{m_{1}}{m_{2}}$=1+$10^{- 2} = 1.01$
$m_{2} = \frac{m_{1}}{1.01} = \frac{300}{1.01}$
Mass expelled=$m_{1} - m_{2}$
$= \ 300\ - \ \ \frac{300}{1.01} = \frac{3}{1.01}$