Question

Work done in increasing the size of a soap bubble from a radius of $3cm$ to $5cm$ is nearly (surface tension of soap solution $=0.03N{m}^{-1}$ ).

• A. $4\pi mJ$
• B. $0.4\pi mJ$
• C. $0.2\pi mJ$
• D. $2\pi mJ$

Answer 1

Answer: (B)

Given, $R_1=3cm=3\times 10^{-2}m$
$R_2=5cm=5\times 10^{-2}m$ and $T=0.03N/m$
Original surface area $=2\times 4\pi R_1$
For second bubble $=2\times 4\pi R_2$
Work done $=$ Surface tension $\times$ extension in area
$=T\times \Delta A$
$=0.03\times 2[4\pi R_2^2-4\pi R_1^2]$
$=0.03\times 8\pi ((5{)}^2-(3{)}^2]\times 10^{-4}$
$=0.03\times 8\pi \times 16\times 10^{-4}$
$=0.384\pi \times 10^{-3}J$
$\simeq 0.4\pi mJ$