Question

A small metal sphere of density $\rho$ is dropped from height $h$ into a jar containing liquid of density $\sigma (\sigma > \rho)$. The maximum depth up to which the sphere sinks is (Neglect damping forces)

• A. $\frac{\rho}{\rho-\sigma}$
• B. $\frac{\text{h}\sigma}{(\rho-\sigma)}$
• C. $\frac{\sigma}{(\rho-\sigma)}$
• D. $\frac{\text{h}\rho}{(\sigma-\rho)}$

Hint:

Work-Energy Theorem: $W_{net} = \Delta KE$. At max depth, $KE = 0$. $W_g + W_b = 0$.

Answer 1

The Correct Option is D

Step 1: Concept
Apply the principle of conservation of energy: the loss in gravitational potential energy equals the work done against the buoyant force.

Step 2: Meaning
Let the maximum depth be $d$. The total vertical distance fallen is $h + d$.

Step 3: Analysis
Loss in P.E. $= mg(h+d) = V\rho g(h+d)$. Work done by buoyant force $= F_B \cdot d = V\sigma gd$. Equating them: $V\rho g(h+d) = V\sigma gd \implies \rho h + \rho d = \sigma d \implies \rho h = (\sigma - \rho)d$.

Step 4: Conclusion
$d = \frac{h\rho}{\sigma - \rho}$. Final Answer: (D)