A small metal sphere of density ρ is dropped from height h into a jar containing liquid of density σ (σ > ρ). The maximum depth up to which the sphere sinks is (Neglect damping forces) (a) (ρ)/(ρ - σ) (b) (hσ)/((ρ - σ)) (c) (σ)/((ρ - σ)) (d) (hρ)/((σ - ρ))

Answer: (d). Speed on reaching the liquid: v² = 2gh. In the liquid, net upward force = V(σ - ρ)g, so retardation a = ((σ - ρ)g)/(ρ). Maximum depth = v²/2a = (2gh ρ)/(2(σ - ρ)g) = (hρ)/(σ - ρ) So the answer is (d).

A small metal sphere of density ρ is dropped from height h into a jar… | MHT CET 2025 (19 Apr, Shift 2)

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Archimedes' Principle & Floatation

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)

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Answer: (d)

Speed on reaching the liquid: $v^2 = 2gh$ .
In the liquid, net upward force $= V(\sigma - \rho)g$ , so retardation $a = \dfrac{(\sigma - \rho)g}{\rho}$ .
Maximum depth $= \dfrac{v^2}{2a} = \dfrac{2gh\,\rho}{2(\sigma - \rho)g} = \dfrac{h\rho}{\sigma - \rho}$
So the answer is (d).

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Archimedes' Principle & Floatation question from MHT CET 2025 (19 Apr, Shift 2)

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