Fluid Mechanics

JEE Main level, single correct, medium

18 of 192

Question 18 of 192

When an air bubble of radius rr rises from the bottom to the surface of a lake, its radius becomes 5r4\frac{5 \mathrm{r}}{4}. Taking the atmospheric pressure to be equal to 10 m10 \mathrm{~m} height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

Show answer and solution

Answer: (D) 9.5 m9.5 \mathrm{~m}

Using P1 V1=P2 V2\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2
(P1)43πr3=(P2)43π125r364ρg(10)+ρghρg(10)=12564640+64h=1250\begin{aligned} &\left(P_1\right) \frac{4}{3} \pi r^3=\left(P_2\right) \frac{4}{3} \pi \frac{125 r^3}{64} \\ &\frac{\rho g(10)+\rho g h}{\rho g(10)}=\frac{125}{64} \\ &640+64 h=1250 \end{aligned}
On solving we get h=9.5 mh=9.5 \mathrm{~m}

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 (Pressure and thrust), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.