JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
The diameter of an air bubble which was initially \(2\,mm\), rises steadily through a solution of density \(1750\,kg\,m\,m ^{-3}\) at the rate of \(0.35\,cms ^{-1}\). The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).
- A \(12\)
- B \(11\)
- C \(10\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(11\)
Step-by-step Solution
Detailed explanation
As the bubble is rising steadily the net force acting on it will be zero (Because of density of air the value of mg can be neglected) So \(B = F \Rightarrow \frac{4 \pi}{3} R ^{3} \rho g =6 \pi \eta Rv\) Putting \(R =1\,mm =10^{-3}\,m\) \(\rho=1.75 \times 10^{3}\,kg / m ^{3}\)…
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