Surface tension of a soap bubble is 2.0 × 10-2 Nm -1. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be: Take [π=(22/7)]

Question

Surface tension of a soap bubble is 2.0 × 10-2 Nm -1. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be: Take [π=(22/7)] (A) 9.24 × 10-4 J (B) 18.48 × 10-4 J (C) 0.72 × 10-4 J (D) 5.76 × 10-4 J

Tardigrade Admin · Accepted Answer

Surface area of soap bubble =2 × 4 π R 2 Work done = change in surface energy × T S = T S × 8 π ×( R 22- R 12) =2 × 10-2 × 8 × (22/7) × 49 × (3/4) × 10-4 =18.48 × 10-4 J

Q. Surface tension of a soap bubble is $2.0 \times 1 0^{- 2} N m^{- 1}$ . Work done to increase the radius of soap bubble from $3.5 c m$ to $7 c m$ will be: Take $[π = \frac{22}{7}]$

$9.24 \times 1 0^{- 4} J$

$18.48 \times 1 0^{- 4} J$

$0.72 \times 1 0^{- 4} J$

$5.76 \times 1 0^{- 4} J$

Surface area of soap bubble $= 2 \times 4 π R^{2}$
Work done $=$ change in surface energy $\times T_{S}$
$= T_{S} \times 8 π \times (R_{2}^{2} - R_{1}^{2})$
$= 2 \times 1 0^{- 2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 1 0^{- 4}$
$= 18.48 \times 1 0^{- 4} J$

Q. Surface tension of a soap bubble is 2.0×10−2Nm−1. Work done to increase the radius of soap bubble from 3.5cm to 7cm will be: Take [π=722​]

9.24×10−4J

18.48×10−4J

0.72×10−4J

5.76×10−4J