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A hollow cone floats with its axis vertical upto one-third of its height in a liquid of relative density $0.8$ and with its vertex submerged. When another liquid of relative density $\rho$ is filled in it upto one-third of its height, the cone floats upto half its vertical height. The height of the cone is $0.10$ $m$ and the radius of the circular base is $0.05$ $m$. The specific gravity $\rho$ is given by
$1.0$
$1.5$
$2.1$
$1.9$
Solution
$h=0.1 \mathrm{m}$
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 $(ii)$
$(0.8)\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{3^{3}}\right)+\rho\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{3^{3}}\right)$
$=0.8\left(\frac{1}{3} \pi \frac{(0.05)^{2} h}{2^{3}}\right)$
$\frac{0.8+p}{3^{3}}=\frac{0.8}{8}=0.1$
$0.8+\rho=2.7$
$\rho=1.9$
