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Question: Surface tension of a soap solution is\(1.9 \times 10 ^ { - 2 } \mathrm {~N} / \mathrm { m }\). Work ...

Surface tension of a soap solution is1.9×10−2 N/m1.9 \times 10 ^ { - 2 } \mathrm {~N} / \mathrm { m }. Work done in blowing a bubble of 2.0 cm diameter will be

A

7.6×10−6π7.6 \times 10 ^ { - 6 } \pijoule

B

15.2×10−6π15.2 \times 10 ^ { - 6 } \pijoule

C

1.9×10−6π1.9 \times 10 ^ { - 6 } \pijoule

D

1×10−41 \times 10 ^ { - 4 }joule

Answer

15.2×10−6π15.2 \times 10 ^ { - 6 } \pijoule

Explanation

Solution

W=8πR2T=8π×(1×10−2)2×1.9×10−2W = 8 \pi R ^ { 2 } T = 8 \pi \times \left( 1 \times 10 ^ { - 2 } \right) ^ { 2 } \times 1.9 \times 10 ^ { - 2 }=15.2×10−6πJ15.2 \times 10 ^ { - 6 } \pi J