A glass ball of radius 4 mm is released into the sea. It starts m… - Vidyadip

Question

A glass ball of radius 4 mm is released into the sea. It starts moving with a velocity of 52.3 $m / sec$. If the density of sea water is $1.015 \times 10^3 kg /$ $m ^3$ then what will be the value of the viscous force acting on the bullet? Find out coefficient of viscosity of sea water $=0.01$ Poise.

✓

Answer

Given : $r=4 mm=4 \times 10^{-3} m$
$\begin{aligned}v & =52.3 m / sec \\\eta & =0.01 \text { Poise }=\frac{0.01}{10} kg / m-sec \\& =10^{-3} kg / m-sec\end{aligned}$
Therefore, viscous force
$F=6 \pi \eta r v$
Putting values,
$\begin{array}{l}F=6 \times 3.14 \times 10^{-3} \times 4 \times 10^{-3} \times 52.3 \\F=9.84 \times 10^{-4} \text { Newton }\end{array}$

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