Question

A ball of steel whose volume is $100\text{cc}$ was thrown into a vessel filled with water upto its brim. Assuming the vessel is cylindrical with a height of $30\text{cm}$ and a base radius of $5\text{cm}$, find the volume of water displaced, the buoyant force on the ball and reduction in the height of water column in the vessel after the ball is removed. (Relative density of steel is $7.7$)

• A. $100\text{cc},1\text{N},1.27\text{cm}$
• B. $100\text{cc},0.1\text{N},28.73\text{cm}$
• C. $100{\text{m}}^3,10\text{N},1.27\text{cm}$
• D. $100{\text{m}}^3,1\text{N},28.73\text{cm}$

Answer 1

The correct option is A $100\text{cc},1\text{N},1.27\text{cm}$

Force of buoyancy acting on the steel ball $\left({\text{F}}_{\text{B}}\right)=$ the weight of water displaced by it
${\text{F}}_{\text{B}}={\text{V}}_{\text{liq disp}}\times {\text{d}}_{\text{liq}}$
where, ${\text{V}}_{\text{liq disp}}$ is the volume of liquid displaced by the ball,
${\text{d}}_{\text{liq}}$ is the density of the liquid.

${\text{V}}_{\text{liq disp}}={\text{V}}_{\text{ball}}$

Given,
Radius of the base, $r=5\text{cm}$
${\text{V}}_{\text{ball}}=100\text{cc}$
${\text{d}}_{\text{liq}}=1\text{g / cc}$
$\Rightarrow {\text{F}}_{\text{B}}=100\times 1=100\text{g}\approx 1\text{N}$

Height of the water column lost, $h=\frac{\text{Volume lost}}{\pi r^2}$

$\Rightarrow h=\frac{100}{\pi \times 5^2}=1.27\text{cm}$

Therefore,
the volume of water displaced $=100\text{cc}$
the buoyant force, ${\text{F}}_{\text{B}}\approx 1\text{N}$
the height of water column lost $=1.27\text{cm}$