Fluid Mechanics

JEE Main level, numerical answer, medium

116 of 192

Question 116 of 192

The surface tension of soap solution is 3.5×10−2Nm−13.5\times {10}^{-2}N{m}^{-1}. The amount of work done required to increase the radius of soap bubble from 10cm10cm to 20cm20cm is _____×10−4J\times {10}^{-4}J . (take π=227\pi =\dfrac{22}{7})

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Answer: 264

The amount of work done is equal to the change in the surface energy to increase the radius of the bubble.
The formula to calculate the change in surface energy is given by
ΔU=2(T×ΔA)...(1)\Delta U=2(T\times \Delta A)...\left(1\right)
Substitute the values of the known parameters into equation (1) to calculate the required work done.
∆U=3.5×10−2×4π[(0.2)2−(0.1)2]×2J=3.5×10−2×4×227×0.03×2=2×1.32×10−2=264×10−4J\begin{matrix}∆U & = & 3.5\times {10}^{-2}\times 4\pi \left[(0.2{)}^{2}-(0.1{)}^{2}\right]\times 2J \\ & = & 3.5\times {10}^{-2}\times 4\times \dfrac{22}{7}\times 0.03\times 2 \\ & = & 2\times 1.32\times {10}^{-2} \\ & = & 264\times {10}^{-4}J\end{matrix}

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