First slide

Fluid statics

Question

A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is Just immersed in water is

Difficult

A

$\frac{5}{12} {πr}^{4} ρg$
B

$0 .5 ρrg$
C

$\frac{4}{3} {πr}^{3} ρg$
D

$\frac{2}{3} {πr}^{4} ρg$

Solution

When the ball is pushed down, the water gains P.E whereas the ball losses P.E. Hence gain in P.E. of water
$= Vρrg - \frac{V}{2} ρ (\frac{3}{8} r) g = \frac{13}{12} {πr}^{4} ρg (∵ V = \frac{4}{3} {πr}^{3}) Loss in P.E. of ball = {Vρ}^{'} rg = \frac{4}{3} {πr}^{4} ρ^{'} g Work done = \frac{13}{12} {πr}^{4} ρg - \frac{4 π}{3} r^{4} ρ^{'} g = {πr}^{4} ρg [\frac{13}{12} - \frac{4}{3} \frac{ρ^{'}}{ρ}] = {πr}^{4} ρg [\frac{13}{12} - \frac{4}{3} \times 0 \cdot 5] = \frac{5}{12} {πr}^{4} ρg$

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