Question

A block of iron contains a hollow cavity as shown below. The block weighs $6000N$ in air and $4000N$ in water. If the density of iron and water are $6g/c{m}^3$ and $1g/c{m}^3$, then the volume of the cavity is
(Assume $g=10m/s^2$ )

• A. $0.05m^3$
• B. $0.5m^3$
• C. $0.25m^3$
• D. $0.1m^3$

Answer 1

Answer: (D)

Weight of block in air, $w_{\text{air }}=6000N$
Weight of block in water, $w_{\text{water }}=4000N$

According to figure given,
volume of cavity,
$=\frac{\text{ weight of water displaced }}{{\rho }_{\text{water }}\times g}-\frac{\text{ weight of iron }}{{\rho }_{\text{iron }}\times g}$
$=\frac{6000-4000}{10^3\times 10}-\frac{6000}{6\times 10^3\times 10}$
$\left[\begin{array}{c}\because p_{\text{iron }}=6g/c{m}^3=6\times 10^{3}kg/m^3\\ {\rho }_{\text{water }}=1g/c{m}^3=1\times 10^{3}kg/m^3\end{array}\right]$
$=0.2-0.1=0.1m^3$