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Question: The radius of a soap bubble is increased from \(\frac { 2 } { \sqrt { \pi } }\)cm. If the surface te...

The radius of a soap bubble is increased from 2π\frac { 2 } { \sqrt { \pi } }cm. If the surface tension of water is 30 dynes per cm, then the work done will be

A

180 ergs

B

360 ergs

C

720 ergs

D

960 ergs

Answer

720 ergs

Explanation

Solution

=8πT[(2π)2−(1π)2]= 8 \pi T \left[ \left( \frac { 2 } { \sqrt { \pi } } \right) ^ { 2 } - \left( \frac { 1 } { \sqrt { \pi } } \right) ^ { 2 } \right]

∴ W = 8×π×30×3π8 \times \pi \times 30 \times \frac { 3 } { \pi }