A Spherical ball of radius \(1 mm\) and density \(10.5 g / cc\) is dropped in glycerine of coefficient of viscosity \(9.8\) poise and density \(1.5 g / cc\). Viscous force on the ball when it attains constant velocity is \(3696 \times 10^{-x} N\). The value of \(x\) is \(\text { (Given, } g =9.8 m / s ^2 \text { and } \pi=\frac{22}{7} \text { ) }\)

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Answer & Solution

Correct Answer

(C) \(7\)

Detailed explanation

When the ball attain terminal velocity

\(F _r=\left( mg - F _{ B }\right)(\because a =0)\)

\(= V \sigma_{ b } g - V \rho_{\ell} g\)

\(= Vg \left(\sigma_{ b }-\rho_{\ell}\right)\)

\(=\frac{4}{3} \pi\left(10^{-3}\right)^3 \times 9.8(10.5-1.5) \times 10^3\)

\(=3696 \times 10^{-7}\,N\)

So, \(x=7\)

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