Question

A hemispherical portion of radius $4\text{cm}$ is removed from the bottom of a cylinder of radius $4\text{cm}$. The volume of remaining cylinder is $128\pi {\text{cm}}^3$ and it is suspended by a string in a liquid of density $0.8{\text{g/cm}}^3$ where it stays vertical. The upper surface of the cylinder is at a depth $2\text{cm}$ below the liquid surface. The force on the bottom of the cylinder by the liquid is ($g$ is the acceleration due to gravity)

• A. $0.102\pi gN$
• B. $0.160\pi gN$
• C. $0.128\pi gN$
• D. $0.64\pi gN$

Answer 1

The correct option is C $0.128\pi gN$

Given that,
Volume of cylinder, $V=128\pi {\text{cm}}^3=128\pi \times {10}^{-6}{\text{m}}^3$
Density of the liquid, $\rho =0.8{\text{g/cm}}^3=0.8\times {10}^3{\text{kg/m}}^3$
Weight of liquid displaced by cylinder is
$W=\rho Vg=(0.8\times {10}^3\times 128\pi \times {10}^{-6}\times g)\text{N}$
Along vertical direction.


$F_{\text{bottom}}-F_{\text{top}}=$ weight of liquid displaced by the cylinder

$\Rightarrow F_{\text{bottom}}-P\times A=\rho Vg$

$\Rightarrow F_{\text{bottom}}=P\times A+\rho Vg$

Putting all data $\Rightarrow F_{\text{bottom}}=(\rho gh\times \pi R^2)+\rho Vg$

$=0.8\times {10}^3\times g\times 2\times {10}^{-2}\times \pi \times 16\times {10}^{-4}+0.8\times {10}^3\times 128\pi \times {10}^{-6}\times g$

$=0.128\pi g\text{N}$

Tips : According to Archimedes principle, Upthrust = weight of fluid displaced. It explains the concept of apparent weight of object in fluid.