Surface tension of a soap solution is 1⋅9×10−2N/m Work done in blowing a bubble of 2 .0 cm diameter will be

Work done = [change in surface area) x T=2×4×π×10−221⋅9×10−2 =15⋅2×10−6π joule

Surface tension of a soap solution is 1⋅9×10−2N/m Work done in blowing a bubble of 2 .0 cm diameter will be

Work done = [change in surface area) x T=2×4×π×10−221⋅9×10−2 =15⋅2×10−6π joule

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Surface tension of a soap solution is $1 \cdot 9 \times 10^{- 2} N / m$ Work done in blowing a bubble of 2 .0 cm diameter will be

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$(7 \cdot 6 \times 10^{- 6} π) J$

$(15 \cdot 2 \times 10^{- 6} π) J$

$(1 \cdot 9 \times 10^{- 6} π) J$

$(1 \times 10^{- 4} π) J$

answer is B.

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Detailed Solution

Work done = [change in surface area) x T

$= [2 \times 4 \times π \times {(10^{- 2})}^{2}] 1 \cdot 9 \times 10^{- 2} = (15 \cdot 2 \times 10^{- 6} π) joule$

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Surface tension of a soap solution is 1⋅9×10−2N/m Work done in blowing a bubble of 2 .0 cm diameter will be

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Surface tension of a soap solution is $1 \cdot 9 \times 10^{- 2} N / m$ Work done in blowing a bubble of 2 .0 cm diameter will be