Question

Surface tension of a soap bubble is $2.0 \times 10^{-2} Nm ^{-1}$. Work done to increase the radius of soap bubble from $3.5\, cm$ to $7\, cm$ will be: Take $\left[\pi=\frac{22}{7}\right]$

• A. $9.24 \times 10^{-4} J$
• B. $18.48 \times 10^{-4} J$
• C. $0.72 \times 10^{-4} J$
• D. $5.76 \times 10^{-4} J$

Answer 1

Answer: (B)

Surface area of soap bubble $=2 \times 4 \pi R ^2$
Work done $=$ change in surface energy $\times T _{ S }$
$= T _{ S } \times 8 \pi \times\left( R _2^2- R _1^2\right) $
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4} $
$=18.48 \times 10^{-4} J$