Question

A cylinder of mass $100kg$ is immersed in the water. It weighs $80kg$ in the water. If the area of cross-section is $8m^2$, then find out the length of cylinder.
Density of water, ${\rho }_w=1000kg{m}^{-3}$
Acceleration due to gravity, $g=10m{s}^{-2}$

Answer 1

Given:
Mass of cylinder, $M=100kg$
Apparent weight of cylinder in water, $M_w=80kg$
Cross-section area of the cylinder, $A=8m^2$
Density of water, ${\rho }_w=1000kg{m}^{-3}$

Let's assume the length of cylinder be $l$.
Volume of the cylinder be $V$.

Apparent loss of weight of cylinder in water:
$\Delta W=(M-M_w)g$

According to the Archimedes' Principle, weight of the displaced water by an object is equal to the buoyant force or upthrust.
The object feels less weight in the water than in the air comparatively because the weight reduces due to the action of upthrust.

$\Delta W=F_{b,water}$
$(M-M_w)g=V{\rho }_{w}g$
$M-M_w=Al{\rho }_w$
$100-80=8\times l\times 1000$
$l=\frac{20}{8000}m=2.5mm$