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Question

Physics Question on potential energy

The potential energy possessed by a soap bubble, having surface tension equal to 0.04 N/m0.04 \,N/m of diameter 1 cm1\, cm is

A

6π×10−6J6 \pi \times 10^{-6} J

B

2π×10−6J2 \pi \times 10^{-6} J

C

8π×10−6J8 \pi \times 10^{-6} J

D

4π×10−6J4\pi \times 10^{-6} J

Answer

8π×10−6J8 \pi \times 10^{-6} J

Explanation

Solution

Given: Surface tension in soap bubble (T)(T) =0.04 N/m=0.04\, N / m and diameter of soap bubble (d)=1 cm(d)=1\, cm =0.01 m=0.01\, m or radius (r)=0.005 m(r)=0.005 \,m. We know that potential energy == Surface tension ×\times Surface area. We also know that for a soap bubble, there are two surfaces. Therefore total area =2×4πr2.=2 \times 4 \pi r^{2} . Thus potential energy =(0.04)×[2×4π×(0.005)2]=8π×10−6 J.=(0.04) \times\left[2 \times 4 \pi \times(0.005)^{2}\right]=8 \pi \times 10^{-6}\, J .