The volume of an air bubble increases by `x%` as it raises from the bottom of a water lake to its surface. If the water barometer reads `H`, the depth of the lake is
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`P_(1)V_(1) = P_(2)V_(2), (H + h) dgV_(1) = HdgV^(2)`
`(H + h)V_(1) = HV_(2) , H + h = (HV_(2))/(V_(1))`
`h = H((V_(2) – V_(1))/(V_(1))), h = H((DeltaV)/(V))`
Here `(DeltaV)/(V) xx 100 = (x)/(100) :. h = (Hx)/(100)`