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JEE Mains · Physics · STD 11 - Fluid mechanics
  • JEE Mains 2022
  • Medium
  • MCQ

The velocity of a small ball of mass \(0.3\,g\) and density \(8\,g / cc\) when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(1.3\,g / cc\), then the value of viscous force acting on the ball will be \(x \times 10^{-4}\,N\), the value of \(x\) is [use \(g=10\,m / s ^{2}\) ] 

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Answer & Solution

Correct Answer

(C) \(25.125\)

Step-by-step Solution

Detailed explanation

\(F _{ V }+ F _{ B }= mg ( v =\) constant \()\)

\(F _{ V }= mg - F _{ B }\)

\(=\rho_{ B } Vg -\rho_{ L } Vg\)

\(=\left(\rho_{ B }-\rho_{ L }\right) Vg\)

\(=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^{3}} \times 10\)

\(=\frac{6.7 \times 0.3}{8} \times 10^{-2} \quad(g=10)\)

\(=\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}\)

Ans. \(25.125\)

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