Question

Identify the ascending order of the work done in the following cases.
a) Work done in increasing the radius of soap bubble from 2 cm to 4 cm (T $=$ 0.04 Nm${}^{-1}$).
b) Work done in increasing the radius of a soap bubble from 2 cm to 4 cm (T $=$0.08 Nm ${}^{-1}$).
c) Work done in breaking a liquid drop of radius 1 cm into 10${}^6$ droplets of same size (T$=$0.08 Nm${}^{-1}$).

• A. b, a & c
• B. c, a & b
• C. a, b & c
• D. c, b & a

Answer 1

The correct option is C a, b & c

a) Work done = $2\times 4\pi \left[\right(\frac{4}{100}{)}^2-\left(\frac{2}{100}{)}^2\right]\times 0.04$
= 0.12 x ${10}^{-2}$ N - m
b) work done = $2\times 4\pi \left[\right(\frac{4}{100}{)}^2-\left(\frac{2}{100}{)}^2\right]\times 0.08$
= 0.24 x ${10}^{-2}$ N - m
c) Volume will be conserved
$\frac{4}{3}\pi (\frac{1}{100}{)}^3=\frac{4}{3}\pi (r{)}^3\times {10}^6$
$\therefore r=\frac{1}{10000}m$
work done = final energy - initial energy
= $\left[4\pi \right(\frac{1}{10000}{)}^2{10}^6-4\pi \left(\frac{1}{100}{)}^2\right]\times 0.08$
= 0.99 x ${10}^{-2}$ N-m
Therefore, ascending order of work done is a < b < c