When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes 5 r 4 . Taking the atmospheric pressure to be equal to 10 ~m height of water column, the depth of the lake would

Correct answer: D. 9.5 ~m Using P_1 ~V_1=P_2 ~V_2 aligned &(P_1) (4)/(3) ^3=(P_2) (4)/(3) π (125 r^3)/(64) &( g(10)+ g h)/( g(10))=(125)/(64) &640+64 h=1250 aligned On solving we get h=9.5 ~m

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Mechanical Properties of Fluids

When an air bubble of radius r rises from the bottom to the surface...

When an air bubble of radius

r

rises from the bottom to the surface of a lake, its radius becomes

\frac{5 \mathrm{r}}{4}

. Taking the atmospheric pressure to be equal to

10 \mathrm{~m}

height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

Options

A $10.5 \mathrm{~m}$
B $8.7 \mathrm{~m}$
C $11.2 \mathrm{~m}$
D $9.5 \mathrm{~m}$

Correct answer

D. $9.5 \mathrm{~m}$

Step-by-step solution

Using

\mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2

\begin{aligned}&\left(P_1\right) \frac{4}{3} \pir^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 \mathrm{~m}

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