When an air bubble of radius `'r'` rises from the bottom to the surface of a lake, its radius becomes `5r//4` (the pressure of the atmosphere is equal to the `10 m` height of water column). If the temperature is constant and the surface tension is neglected, the depth of the lake is

Options

A. 3.53 m
B. 6.53 m
C. 9.53 m
D. 12.53 m

Answer

C

Explanation

as T = constant, (Boyle.s law) `(P_1V_1) = P_2V_2`
`(10+h)rhogxx4/3pir^3=10rhog4/3pi((5r)/4)^3`
`rArr(10+h)r^3=10xx(125)/(64)r^3`
`rArrh=10[(125-64)/(64)]=9.53m`