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A ball of radius $r $ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\eta$, the value of $h$ is given by
$\frac{2}{9}{r^2}\left( {\frac{{1 - \rho }}{\eta }} \right)\,g$
$\frac{2}{{81}}{r^2}\left( {\frac{{\rho - 1}}{\eta }} \right)\,g$
$\frac{2}{{81}}{r^4}{\left( {\frac{{\rho - 1}}{\eta }} \right)^2}g$
$\frac{2}{9}{r^4}{\left( {\frac{{\rho - 1}}{\eta }} \right)^2}g$
Solution
(c)Velocity of ball when it strikes the water surface $v = \sqrt {2gh} $ …(i)
Terminal velocity of ball inside the water
$v = \frac{2}{9}{r^2}g\frac{{\left( {\rho – 1} \right)}}{\eta }$ …(ii)
Equating (i) and (ii) we get
$\sqrt {2gh} = \frac{2}{9}\frac{{{r^2}g}}{\eta }(\rho – 1)$
==> $h = \frac{2}{{81}}{r^4}{\left( {\frac{{\rho – 1}}{\eta }} \right)^2}g$
