Question

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

• A. $0.05\, m ^{3}$
• B. $0.5 \,m ^{3}$
• C. $0.25 \,m ^{3}$
• D. $0.1 \,m ^{3}$

Answer 1

Answer: (D)

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

According to figure given,
volume of cavity,
$=\frac{\text { weight of water displaced }}{\rho_{\text {water }} \times g}-\frac{\text { weight of iron }}{\rho_{\text {iron }} \times g}$
$=\frac{6000-4000}{10^{3} \times 10}-\frac{6000}{6 \times 10^{3} \times 10}$
$\begin{bmatrix}\because p_{\text {iron }}=6\, g / cm ^{3}=6 \times 10^{3} \,kg / m ^{3} \\ \rho_{\text {water }}=1 \,g / cm ^{3}=1 \times 10^{3} \,kg / m ^{3}\end{bmatrix}$
$=0.2-0.1=0.1 \,m ^{3}$