The work down in blowing a soap bubble of volume V is W. The work is to be done in blowing it of volume 2V equals to

Options

A. 2W
B. `root(3)(4)W`
C. `root(3)(2)W`
D. `sqrt2W`

Answer

B

Explanation

When radius of the soap bubble is x, then excess pressure inside the bubble,
`p=(4T)/(x)[T = ` surface tension of soap water ]
Volume, `V=4/3pix^3`
If the radius increases by dx, then increase in volume,
`dV=4/3pi. 3x^2dx=4pix^2 dx`
Now, work done for this increase,
`dW=pdV=(4T)/(x). 4pix^2 dx=16piT.xdx`
If the radii of the rubbles of V and 2V volume are R and R. respectively,
`V=4/3 pi R^3 and 2V=4/3pi R^3`
`therefore (2V)/(V)=(R.)/R)^3 or, (R.)/(R ) = 2^(1//3)`
Work done to increase volume form 0 to V,
`W=int dW=16 pi int_0^R xdx=16pi T(x^2)/(2)|_0^R=8piTR^2`
Similarly, work done to increase volume from 0 to 2V,
`W.=8piTR^(12)=8piTR^2((R.)/(R))^2`
`=W(2^(1//3))^2=root(3)(4) W`