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

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$4πmJ$

$0.2πmJ$

$2πmJ$

$0.4 π mJ$

answer is D.

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Detailed Solution

w = $σΔA$ $= σ \times [8 π (r_{2}^{2} - r_{1}^{2})]$ $= 0.03 \times [8 π (5^{2} - 3^{2}) \times 10^{- 4}]$

$= {0.384π×10}^{-3} J = 0.4πmJ$

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