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Standard VIII

Physics

Viscosity

A small spher...

Question

A small spherical ball of radius r falls freely under gravity through a distance h before entering a tank of water. If, after entering the water, the velocity of the ball does not change, then h is proportional

r^{2}

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r^{3}

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r^{4}

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r^{5}

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Solution

The correct option is C $r^{4}$
Velocity acquired on falling a height $h$ will be $v = \sqrt[2]{2 g h}$
No change in velocity means the net force is $z e r o$ i.e the $v i s c o u s$ force is balancing the $w e i g h t$ so $6 π r η v = m g$
or putting $m = ρ V = \frac{ρ 4 π r^{3}}{3}$ and $v = \sqrt[2]{2 g h}$
so we get $6 π r η \sqrt[2]{2 g h} = \frac{ρ 4 π r^{3}}{3} g$
squaring both sides we get $h \propto r^{4}$

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A small spherical ball of radius r falls freely under gravity through a distance h before entering a tank of water. If, after entering the water, the velocity of the ball does not change, then h is proportional

The correct option is C r4Velocity acquired on falling a height h will be v=2√2ghNo change in velocity means the net force is zero i.e the viscous force is balancing the weight so 6πrηv=mgor putting m=ρV=ρ4πr33 and v=2√2ghso we get 6πrη2√2gh=ρ4πr33gsquaring both sides we get h∝r4