Question

A block of silver of mass $4\, kg$ hanging from a string is immersed in a liquid of relative density $0.72$. If relative density of silver is $10$, then tension in the string will be

• A. 37.12 N
• B. 42 N
• C. 73 N
• D. 21 N

Answer 1

Answer: (A)

Let $\rho_{S}$ and $\rho_{L}$ be the densities of silver and liquid, respectively, and $m$ and $V$ be the mass and volume, respectively, of the silver block. Therefore,
Tension in the string $=m g$ - buoyant force
$\Rightarrow T=\rho_{S} V g-\rho_{L} V g=\left(\rho_{S}-\rho_{L}\right) V g$
Also, $V=\frac{m}{\rho_{S}}$
$\therefore T=\left(\frac{\rho_{S}-\rho_{L}}{\rho_{S}}\right) m g$
$=\frac{(10-0.72) \times 10^{3}}{10 \times 10^{3}} \times 4 \times 10$
$=37.12\, N$