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(D) Let the sphere have mass $m$,volume $V$,and density $\varrho$. Thus,$m = V \varrho$.When the sphere is dropped from height $h$,its velocity $v$ ju

Vedclass · Accepted Answer

$\frac{h \varrho}{(\sigma - \varrho)}$

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(D) Let the sphere have mass $m$,volume $V$,and density $\varrho$. Thus,$m = V \varrho$.When the sphere is dropped from height $h$,its velocity $v$ just before hitting the liquid surface is given by $v^2 = 2gh$.Inside the liquid,the forces acting on the sphere are its weight $mg$ (downward) and the buoyant force $F_B = V \sigma g$ (upward).The net force on the sphere inside the liquid is $F_{net} = F_B - mg = V \sigma g - V \varrho g = Vg(\sigma - \varrho)$.The acceleration $a$ of the sphere inside the liquid is $a = \frac{F_{net}}{m} = \frac{Vg(\sigma - \varrho)}{V \varrho} = g \frac{(\sigma - \varrho)}{\varrho}$.Since the buoyant force is greater than the weight (as $\sigma > \varrho$),the sphere will decelerate.Let $d$ be the maximum depth reached. Using the equation of motion $v_f^2 = v_i^2 + 2ad$,where $v_f = 0$ (at maximum depth),$v_i^2 = 2gh$,and $a = -g \frac{(\sigma - \varrho)}{\varrho}$:$0 = 2gh - 2 \left( g \frac{(\sigma - \varrho)}{\varrho} \right) d$.Solving for $d$: $d = \frac{h \varrho}{(\sigma - \varrho)}$.

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

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