Fluid Mechanics

JEE Main level, numerical answer, medium

176 of 192

Question 176 of 192

An air bubble of diameter 6mm6mm rises steadily through a solution of density 1750kgm−31750kg{m}^{-3} at the rate of 0.35cms−10.35cm{s}^{-1}. The co-efficient of viscosity of the solution (neglect density of air) is ________ Pas (given, g=10ms−2g=10m{s}^{-2}).

Show answer and solution

Answer: 10

The bubble is moving with a constant velocity so the net force acting on the body is zero.
Stokes' law defines the drag force which is given by F=6πηrvF=6\pi \eta rv
The buoyant force is given by B=Vρg=43πr3ρgB=V\rho g=\dfrac{4}{3}\pi {r}^{3}\rho g
Hence, it can be written that
43πr3ρg=6πηrv⇒η=2r2ρg9v=2×(3×10−3)2×1750×109×0.35×10−2=10Pas\dfrac{4}{3}\pi {r}^{3}\rho g=6\pi \eta rv \Rightarrow \eta =\dfrac{2{r}^{2}\rho g}{9v}=\dfrac{2\times {\left(3\times {10}^{-3}\right)}^{2}\times 1750\times 10}{9\times 0.35\times {10}^{-2}}=10Pas

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Fluid mechanics simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Fluid Mechanics (Viscosity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.