Surface tension of a soap bubble is . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take ] [2023]

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Q.
Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

1 $0.72 \times 10^{- 4} J$

2 $5.76 \times 10^{- 4} J$

3 $18.48 \times 10^{- 4} J$

4 $9.24 \times 10^{- 4} J$

Ans.
(3)

$Surface area of soap bubble = 2 \times 4 π R^{2}$

$Work done = change in surface energy \times T_{s}$

$= T_{s} \times 8 π \times (R_{2}^{2} - R_{1}^{2})$

$= 2 \times 10^{- 2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{- 4}$

$= 18.48 \times 10^{- 4} J$

Solution

Q.
Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

1 $0.72 \times 10^{- 4} J$

2 $5.76 \times 10^{- 4} J$

3 $18.48 \times 10^{- 4} J$

4 $9.24 \times 10^{- 4} J$

Ans.
(3)

$Surface area of soap bubble = 2 \times 4 π R^{2}$

$Work done = change in surface energy \times T_{s}$

$= T_{s} \times 8 π \times (R_{2}^{2} - R_{1}^{2})$

$= 2 \times 10^{- 2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{- 4}$

$= 18.48 \times 10^{- 4} J$

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Solution

Q. Surface tension of a soap bubble is 2.0×10-2 Nm-1. Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take π=227] [2023]

1 0.72×10-4 J

2 5.76×10-4 J

3 18.48×10-4 J

4 9.24×10-4 J

Ans. (3) Surface area of soap bubble=2×4πR2 Work done=change in surface energy×Ts =Ts×8π×(R22-R12) =2×10-2×8×227×49×34×10-4 =18.48×10-4 J

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Q.
Surface tension of a soap bubble is $2.0 \times 10^{- 2} {Nm}^{- 1}$ . Work done to increase the radius of the soap bubble from 3.5 cm to 7 cm will be [Take $π = \frac{22}{7}$ ] [2023]

1 $0.72 \times 10^{- 4} J$

2 $5.76 \times 10^{- 4} J$

3 $18.48 \times 10^{- 4} J$

4 $9.24 \times 10^{- 4} J$

Ans.
(3)

$Surface area of soap bubble = 2 \times 4 π R^{2}$

$Work done = change in surface energy \times T_{s}$

$= T_{s} \times 8 π \times (R_{2}^{2} - R_{1}^{2})$

$= 2 \times 10^{- 2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{- 4}$

$= 18.48 \times 10^{- 4} J$