JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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 \(g=10\,m / s ^{2}\) ]
- A \(24.125\)
- B \(23.125\)
- C \(25.125\)
- D \(22.125\)
Answer & Solution
Correct Answer
(C) \(25.125\)
Step-by-step Solution
Detailed explanation
\(F _{ V }+ F _{ B }= mg ( v =\) constant \()\) \(F _{ V }= mg - F _{ B }\) \(=\rho_{ B } Vg -\rho_{ L } Vg\) \(=\left(\rho_{ B }-\rho_{ L }\right) Vg\) \(=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^{3}} \times 10\)…
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