JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2022
- Hard
- Numerical
A small spherical ball of radius \(0.1 \,mm\) and density \(10^{4} \,kg m ^{-3}\) falls freely under gravity through a a distance \(h\) before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of \(h\) wil be \(m\). (Given \(g =10 \,ms ^{-2}\), viscosity of water \(=1.0 \times 10^{-5} \,N - sm ^{-2}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
Speed after falling through height \(h\) Should be equal to terminal velocity
\(\sqrt{2 gh }=\frac{2}{9} \frac{ r ^{2}( d -\rho) g }{\eta}\)
\(\sqrt{2 gh }=\frac{2}{9} \frac{10^{-8}(10000-1000) \times 10}{10^{-5}}\)
\(=\frac{2}{9} \times 10^{-8} \frac{9 \times 10^{4}}{10^{-5}}=20\)
\(2 \times 10 \times h =400\)
\(h =20 \,m\)
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