Question

A vessel contains oil (density $0.8 \, \, g \text{m/cm}^{3}$ ) over mercury (density $= 13.6 \, \, g \text{m/cm}^{3}$ ). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g \text{m/cm}^{3}$ is

• A. $3.3$
• B. $6.4$
• C. $7.2$
• D. $12.8$

Answer 1

Answer: (C)

As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it

$\text{Weight} \text{of} \text{sphere} = \frac{4}{3} \pi \left(\text{R}\right)^{3} \rho g \text{......} \left(\text{i}\right)$
$\text{Upthrust due to oil and mercury} = \frac{2}{3} \pi \left(\text{R}\right)^{3} \times \left(\sigma \right)_{\text{oil}} g + \frac{2}{3} \pi \left(\text{R}\right)^{3} \left(\sigma \right)_{\text{Hg}} g \text{......} \left(\text{i} \text{i}\right)$
Equating (i) and (ii)
$\frac{4}{3} \pi \text{R}^{3} \rho g = \frac{2}{3} \pi \text{R}^{3} 0 \text{.} 8 g + \frac{2}{3} \pi \text{R}^{3} \times 1 3 \text{.} 6 g$
$\Rightarrow 2 \rho = 0 \text{.} 8 + 1 3 \text{.} 6 = 1 4 \text{.} 4$
$\Rightarrow \rho = 7 \text{.} 2$