Q169 / 200 · TS EAMCET 2001 · Physics

When an air bubble of radius $r$ rises from the bottom to the surface of a lake. Its radius becomes $\frac{5 r}{4}$ (the pressure of the atmosphere is equal to the $10 \mathrm{~m}$ height to water column). If the temperature is constant and the surface tension is neglected the depth of the lake is

  • A
    $5.53 \mathrm{~m}$
  • B
    $6.53 \mathrm{~m}$
  • Correct answer
    C
    $9.53 \mathrm{~m}$
  • D
    $12.53 \mathrm{~m}$

Solution

Initial pressure $p_1=$ atmospheric pressure + pressure of liquid column $=h d g+h_1 d g$ $h=10 \mathrm{~m}$ of water, $h_1=$ depth of lake $p_1=d g\left(h+h_1\right)$ From, $p_1 V_1=p_2 V_2$ $\begin{aligned} d g(10+h) \times \frac{4}{3} \pi r_1^3 & =h d g \times \frac{4}{3} \pi r_2^3 \\ (10+h) \times r^3 & =10 \times\left(\frac{5}{4} r\right)^3 \\ (10+h) & =10 \times \frac{125}{64} \\ h & =\frac{1250}{64}-10 \\ \Rightarrow \quad h & =\frac{610}{64}=9.53 \mathrm{~m}\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions