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

 

Answers (1)

As we learned

Change in Pressure of bubble in the air -

\Delta P=\left ( \frac{2T}{R} \right )\times 2= \frac{4T}{R}

- wherein

T- Temperature

R- Radius

 

 At bottom surface

\rho _{1}=\rho _{a}+\rho gh+\frac{4s}{r}

p_{2}-p _{a}=\frac{4s}{\left ( \frac{5r}{4} \right )}=\left ( \frac{16s}{5r} \right )

Also,P_{1}v_{1}=P_{2}v_{2}

P_{1}.\frac{4\Pi }{3}r^{3}=P_{2}.\frac{4\Pi }{3}*\frac{125}{64}.r^{3}

P_{1}= \frac{125}{64}P_{2}\: \: or\: \: \frac{P_{1}}{P_{2}}=\frac{125}{64}

\therefore \frac{P_{a}+\rho gh+\frac{4s}{r}}{P_{a}+\frac{16s}{5r}}= \frac{125}{64}

Ignoring Surface Tension

\therefore \frac{P_{a}+\rho gh}{P_{a}}=\frac{125}{64} \: \: or\: \: 1+\frac{\rho gh}{P_{a}}=\frac{125}{64}

\therefore \frac{\rho gh}{P_{a}}=\frac{61}{64} \: \: or\: \: \rho gh=\left ( \frac{61}{64} \right )\rho g*10

h=9.5m

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lovekush

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