Question

A metal ball weighs $0.096\text{N}$ in air. If suspended in water, it has an apparent weight of $0.071\text{N}$. The density of the metal ball is

• A. $3840{\text{kg/m}}^3$
• B. $2040{\text{kg/m}}^3$
• C. $2540{\text{kg/m}}^3$
• D. $4838{\text{kg/m}}^3$

Answer 1

The correct option is A $3840{\text{kg/m}}^3$

The loss in weight of ball (when suspended in water) is due to the buoyant force acting vertically upwards.
$\Rightarrow \text{Decrease in weight}=\text{Buoyant force}$
$\Rightarrow 0.096-0.071={\rho }_{w}gV$
$\Rightarrow 0.025={\rho }_w\times 10\times \left(\frac{0.0096}{{\rho }_m}\right)...\left(i\right)$

Since $mg=0.096$,
$\Rightarrow m=\frac{0.096}{10}=0.0096\text{kg}\&V=\frac{m}{{\rho }_m}=\frac{0.0096}{{\rho }_m}$

Substituting in Eq.$\left(i\right)$,
${\rho }_m=\frac{1000\times 10\times 0.0096}{0.025}$
$\therefore {\rho }_m=3840{\text{kg/m}}^3$
Density of the metal ball is $3840{\text{kg/m}}^3$