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A small spherical body of radius ' R ' and mass ' m ' is dropped in a viscous fluid at \mathrm{t}=0. The time taken by the body to reach 99 \% of its terminal velocity is:

Option: 1

\mathrm{\frac{2.3 m}{3 \pi \eta R}}


Option: 2

\mathrm{\frac{m}{6 \pi \eta R}}


Option: 3

\mathrm{\frac{2.3 \mathrm{~m}}{4 \pi \eta \mathrm{R}}}


Option: 4

\mathrm{\text { Can't be calculated }}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{V}=\mathrm{V}_{\mathrm{T}}\left\{1-\mathrm{e}^{-\mathrm{t} / \tau}\right\} \\\\ & \mathrm{T}=\frac{\mathrm{m}}{6 \pi \eta \mathrm{R}} \\\\ & 0.99 \mathrm{~V}_{\mathrm{T}}=\mathrm{V}_{\mathrm{T}}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{T}}\right) \\\\ & -\mathrm{t} / \mathrm{T}=\ln 10^{-10}=-2 \times 2.3 \\\\ & \mathrm{t}=4.6 \mathrm{~T} \\\\ & =\frac{4.6 \mathrm{~m}}{6 \pi \eta \mathrm{R}} \\\\ & =\frac{2.3 \mathrm{~m}}{3 \pi \eta R} \end{aligned}}

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