Properties Of Matter Ques 66
- A hemispherical portion of radius $R$ is removed from the bottom of a cylinder of radius $R$. The volume of the remaining cylinder is $V$ and mass $M$. It is suspended by a string in a liquid of density $\rho$, where it stays vertical.The
upper surface of the cylinder is at a depth $h$ below the liquid surface. The force on the bottom of the cylinder by the liquid is
$(2001,2\ \text{M})$
(a) $M g$
(b) $M g-V \rho g$
(c) $M g+\pi R^{2} h \rho g$
(d) $\rho g\left(V+\pi R^{2} h\right)$
Show Answer
Answer:
Correct Answer: 66.(a)
Solution:
Formula:
- $F_2 - F_1 =$ upthrust
$\therefore \quad F _2=F _1+$ upthrust
$F _2=\left(p _0+\rho g h\right) \pi R^{2}+\rho g V$
$$ =p _0 \pi R^{2}+\rho g\left(\pi R^{2} h+V\right) $$
$\therefore$ Most appropriate option is (d).
BETA
