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# Engineering

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)
L Lovekush Kumar Saini

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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