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Q.

A soap bubble (surface tension $30 \times 10^{- 3} N / m)$ has radius 2 cm. The work done in doubling the radius is

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a

zero

b

$1 \cdot 1355 \times 10^{- 4} J$

c

$2 \cdot 261 \times 10^{- 4} J$

d

$9.04 \times 10^{- 4} J$

answer is D.

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Detailed Solution

work done
$= 4 π [{(4 \times 10^{- 2})}^{2} - {(2 \times 10^{- 2})}^{2}] (30 \times 10^{- 3}) \times 2 = 9.04 \times 10^{- 4} J$

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