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JEE Mains · Physics · STD 11 - Surface tension
- JEE Mains 2018
- Hard
- MCQ
When an air bubble of radius \(r\) rises from the bottom to the surface of a lake, its radius becomes \(\frac{{5r}}{4}\).Taking the atmospheric pressure to be equal to \(10\,m\) height of water column, the depth of the lake would approximately be ....... \(m\) (ignore the surface tension and the effect of temperature)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(9.5\)
Step-by-step Solution
Detailed explanation
Using\({P_1}{V_1} = {P_2}{V_2}\)
\(\left( {{P_1}} \right)\frac{4}{3}\pi {r^3} = \left( {{P_2}} \right)\frac{4}{3}\pi \frac{{125{r^3}}}{{64}}\)
\(\frac{{\rho g\left( {10} \right) + \rho gh}}{{\rho g\left( {10} \right)}} = \frac{{125}}{{64}}\)
\(640 + 64\,h = 1250\)
\(On\,solving\,we\,get\,h = 9.5\,m\)
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