Question

The work done in increasing the radius of a soap bubble from $4cm$ to $5cm$ is ................Joule.(given surface tension of soap water to be $25\times {10}^{-3}N/m$)

• A. $0.5657\times {10}^{-3}$
• B. $5.657\times {10}^{-3}$
• C. $56.5\times {10}^{-3}$
• D. $565\times {10}^{-3}$

Answer 1

Answer: (A)

$0.5657\times {10}^{-3}$

$\Delta A=2\times \frac{4\pi (5^2-4^2)}{10000}$

$\Delta \bigcup =\Delta A\times T$

$=\frac{8\pi }{10000}\times 9\times 25\times {10}^{-3}$

$=5.65\times {10}^{-4}J$