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Physics Question on mechanical properties of fluid

Work done in increasing the size of a soap bubble from a radius of 3 cm3 \,cm to 5 cm5 \,cm is nearly (surface tension of soap solution =0.03 Nm−1= 0.03\, Nm^{-1} ).

A

4π mJ4\pi \,mJ

B

0.4π mJ0.4 \pi \,mJ

C

0.2π mJ0.2 \pi \,mJ

D

2π mJ2 \pi \,mJ

Answer

0.4π mJ0.4 \pi \,mJ

Explanation

Solution

Given, R1=3 cm=3×10−2 mR_1 =3\, cm = 3 \times 10^{-2}\, m
R2=5 cm=5×10−2 mR_2 = 5 \,cm = 5 \times 10^{-2}\, m
and T=0.03 N/mT = 0.03\, N/m
Original surface area =2×4 πR1= 2 \times 4 \,\pi R_1
For second bubble =2×4 πR2= 2 \times 4 \,\pi R_2
Work done == Surface tension ×\times extension in area
=T×ΔA= T \times \Delta A
=0.03×2[4 πR22−4 πR12]= 0.03 \times 2[4\,\pi R_2^2 - 4 \,\pi R_1^2]
=0.03×8π((5)2−(3)2]×10−4= 0.03 \times 8 \pi ((5)^2 - (3)^2] \times 10^{-4}
=0.03×8π×16×10−4= 0.03 \times 8 \pi \times 16 \times 10^{-4}
=0.384 π×10−3 J= 0.384 \,\pi \times 10^{-3}\, J
≃0.4 π mJ\simeq 0.4\,\pi \,mJ