The terminal velocity \(v\) of a spherical ball of lead of radius \(R\) falling through a viscous liquid…

The terminal velocity \(v\) of a spherical ball of lead of radius \(R\) falling through a viscous liquid varies with \(R\) such that
  1. \(\frac{v}{R}=\) constant
  2. \(v R=\) constant
  3. \(v=\) constant
  4. \(\frac{v}{R^2}=\) constant

Solution

Terminal velocity of spherical ball falling through viscous liquid is given as \(v=\frac{2}{9} \times \frac{R^2(\rho-\sigma) g}{\eta}\) ...(i) where, \(R=\) radius of ball, \(\rho=\) density of ball, \(\sigma=\) density of liquid and \(\eta=\) coefficient of viscosity. From Eq. (i), \(\frac{v}{R^2}=\frac{2(\rho-\sigma) g}{9 \eta}=\text { constant }\)

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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