$\rho = 300\frac{{kg}}{{{m^3}}}$
$m + {\ell ^3}\rho = {\ell ^3}{\rho _\omega }$
$M = {\ell ^3}\left( {{\rho _w} - \rho } \right) = {\left( 0.5 \right)^3}\left\{ {1000 - 300} \right\}$
$ = 700 \times {\left( 0.5 \right)^3}$
$ = 87.5\,kg$


[Given: Surface tension of the liquid is $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]


