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A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time the velocity of the ball attains a constant value known as terminal velocity $v_T$. The terminal velocity depends on (i) the mass of the ball $m$, (ii) $\eta$, (iii) $r$ and (iv) acceleration due to gravity $g$. Which of the following relations is dimensionally correct?
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The correct answer is:
$v_T \propto \frac{m g}{\eta r}$
By substituting dimension of each quantity in R.H.S. of option
(a) we get $\left[\frac{m g}{\eta r}\right]=\left[\frac{M \times L T^{-2}}{M {L}^{-1} T^{-1} \times L}\right]=\left[L T^{-1}\right]$.
This option gives the dimension of velocity.
(a) we get $\left[\frac{m g}{\eta r}\right]=\left[\frac{M \times L T^{-2}}{M {L}^{-1} T^{-1} \times L}\right]=\left[L T^{-1}\right]$.
This option gives the dimension of velocity.
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