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Question: The work done in blowing a soap bubble of surface tension \(0.06 \mathrm {~N} \mathrm {~m} ^ { - 1 }...

The work done in blowing a soap bubble of surface tension 0.06 N m−10.06 \mathrm {~N} \mathrm {~m} ^ { - 1 }from 2 cm radius to 5 cm radius is

A

3.1 mJ

B

1.25 mj

C

2.51 mJ

D

4.55 mJ

Answer

3.1 mJ

Explanation

Solution

Here, S=0.06Nm−1,r1=2 cm=2×10−2 m\mathrm { S } = 0.06 \mathrm { Nm } ^ { - 1 } , \mathrm { r } _ { 1 } = 2 \mathrm {~cm} = 2 \times 10 ^ { - 2 } \mathrm {~m}

Since bubble has two surfaces initial surface area of the bubble

=32π×10−4 m2= 32 \pi \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }

Final surface area of the bubble

=200π×10−4 m2= 200 \pi \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }

Increase is surface area

= 200π×10−4−32π×10−4200 \pi \times 10 ^ { - 4 } - 32 \pi \times 10 ^ { - 4 }

=168π×10−4 m2= 168 \pi \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }

∴\therefore work done = Surface tension × increase is surface area

=0.06×168π×10−4= 0.06 \times 168 \pi \times 10 ^ { - 4 }

=3.12×10−3 J=3.12 mJ= 3.12 \times 10 ^ { - 3 } \mathrm {~J} = 3.12 \mathrm {~mJ}