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An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 2.5 m and the temperature at the surface is 40^(@)C. What is the temperatue at the bottom of the lake ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let `P_(1), V_(1), T_(1)" and "P_(2), V_(2), T_(2)` are the parameters of the air bubble at the bottom and surface of the lake respectively. <br/> `P_(1)=` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>/<a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> + Pressure of water at the bottom <br/> `""=(76 times 13.6 times <a href="https://interviewquestions.tuteehub.com/tag/980-342886" style="font-weight:bold;" target="_blank" title="Click to know more about 980">980</a>)+(250 times 1 times 980)=1283.6 times 980" dyne "cm^(-2)` <br/> `P_(2)="Atm. pressure"=76 times 13.6 times 980=1033.6 times 980" dyne "cm^(-2)` <br/> `V_(1)=4/3pi(0.36/2)^(3)=0.007776 times <a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a> cm^(3) rArr V_(2)=4/3pi(0.4/2)^(3)=0.01067 times picm^(3)` <br/> `T_(2)=273+40=313K` <br/> From the equation, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` <br/> `rArr T_(1)=P_(1)/P_(2) times V_(1)/V_(2) times T_(2)=(1283.6 times 980)/(1033.6 times 980) times (0.00776pi)/(0.01067pi) times 313=1.242 times 0.7288 times 313=283.3K` <br/> Temperature at the bottom of the lake = 283.3 - 273 = `10.3^(@)C`</body></html>


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