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The volume of the air bubble increases by 10%, when it rises from bottom to the surface of the water pond. If the temperature of the water is constant, find the depth of the pond. ( 1 atm = 10 m of water column)

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> :Volume of the air bubble at the bottom of the <a href="https://interviewquestions.tuteehub.com/tag/pond-592257" style="font-weight:bold;" target="_blank" title="Click to know more about POND">POND</a> `(V_(1) ) = V` <br/> Volume of the air bubble at the surface of the pond`V_(2) = V xx (110)/(100)` <br/> Pressure at the bottom of the pond `(P_(1) )` = (<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>+ H) cm of water column<br/> Where .h. is the <a href="https://interviewquestions.tuteehub.com/tag/depth-948801" style="font-weight:bold;" target="_blank" title="Click to know more about DEPTH">DEPTH</a> of the pond and .H. is the atmospheric pressure <br/> pressure on surface of the pond `(P_(2)) = H ` cm of water column <br/> From Boyle.s law, `P_(1) V_(1) = P_(2) V_(2) ` or <br/> `(h + H) V = H (V xx (110)/(100) ) , "" h + H = H xx (110)/(100)` <br/> h = `H xx (110)/(100) - H = H xx (<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)/(100) = (H)/(10)` <br/> `therefore` Depth of the pond = H/10 cm <br/> `= (1000)/(10) = 100 ` cm = 1m</body></html>


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