MCQ
A air bubble rises from bottom of a lake to surface. If its radius increases by $200 \%$ and atmospheric pressure is equal to water coloumn of height $H$. then depth of lake is ..... $H$
- A$21$
- B$8$
- C$9$
- ✓$26$
✓
Answer
Correct option: D.
$26$
d
(d)
(d)
Let initial radius be $=r$
Final radius $=r+200 \%$ of $r$
$=3 r$
Atmospheric pressure $=\rho g H$
Let depth of the lake be $h$
So, pressure at the bottom of lake $=\rho g H+\rho g h$
Using $P_1 V_1=P_2 V_2$
$\rho g H \times \frac{4}{3} \pi(3 r)^3=(\rho g H+\rho g h) \times \frac{4}{3} \pi r^3$
$\rho g H \times \frac{4}{3} \pi \times 27 r^3=(\rho g H) \frac{4}{3} \pi r^3+\rho g h \times \frac{4}{3} \pi r^3$
Solving this equation we get
$26 H=h$
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