d
$h=0.1 \mathrm{m}$
By force balance
$\mathrm{mg}=0.8\left(\frac{1}{3} \pi\left(\frac{0.05}{3}\right)^{2}\left(\frac{\mathrm{h}}{3}\right)\right)$ $...(i)$
When liquid is added
$\mathrm{Mg}+\rho\left(\frac{1}{3} \pi\left(\frac{0.05}{3}\right)^{2}\left(\frac{\mathrm{h}}{3}\right)\right)$
$=0.8\left(\frac{1}{3} \pi\left(\frac{0.05}{2}\right)^{2}\left(\frac{\mathrm{h}}{2}\right)\right)$ $...(ii)$
By $(i)$ and $(ii)$
$(0.8)\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{3^{3}}\right)+\rho\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{3^{3}}\right)$
$=0.8\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{2^{3}}\right)$
$\frac{0.8+p}{3^{3}}=\frac{0.8}{8}=0.1$
$0.8+\rho=2.7$
$\rho=1.9$
