Question

Find the work required to be done to increase the volume of a bubble of soap solution having a radius 1 mm to 8 times. (Surface tension of soap solution is 30 dyne/cm).

Answer 1

Explanation:

pls refer the attachment

Answer 2

Answer:

$W.d =23760\times 10^{-9}J$

Explanation:

Since we know that the surface tension is the force per unit length.

So, work done in term of surface tension is given as

$W.d= Surface\ Tension\times (Change\ in\ area)$

We had given that the

Surface tension = 30 dyne/cm.

Change in area = $4\pi R^{2} - 4\pi r^{2}$

R= 8 mm r=1 mm.

$Change\ in\ area = 4\pi(63)\times 10^{-6} =792\times 10^{-6} m^{2}$

So,

$W.d = 30\times 10^{-3}\times 792\times 10^{-6} = 23760\times 10^{-9}J$