Question

An iron casting containing a number of cavities weighs $16000\text{N}$ in air and $12000\text{N}$ in water. Volume of the cavities in the casting is -
[ take density of iron $=8000{\text{kg/m}}^3$ & $g=10{\text{m/s}}^2$]

• A. $0.2{\text{m}}^3$
• B. $0.3{\text{m}}^3$
• C. $0.1{\text{m}}^3$
• D. $0.4{\text{m}}^3$

Answer 1

The correct option is A $0.2{\text{m}}^3$

Given:
Weight of iron casting in air, $w_a=16000\text{N}$
Weight of iron casting in water, $w_w=12000\text{N}$
Density of iron, ${\rho }_i=8000{\text{kg/m}}^3$

Let us suppose the total volume of cavities is $V$.
Now, volume of the iron, $V_0=\frac{\text{mass}}{\text{density}}=\frac{16000/10}{8000}=0.2{\text{m}}^3...\left(1\right)$
[ weight = mass $\times g$ ]
We know that,
Decrease in weight $=$ upthrust on iron casting
$\Rightarrow w_a-w_w=(V_0+V){\rho }_{w}g$
[ ${\rho }_w$ is density of water ]
$\Rightarrow 16000-12000=(0.2+V)\times 1000\times 10$
[ from (1) ]
$\Rightarrow 0.2+V=0.4$
$\Rightarrow V=0.2{\text{m}}^3$ is the volume of cavities.