Fluid Mechanics

JEE Main level, single correct, medium

155 of 192

Question 155 of 192

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

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Answer: (C) 0.4πmJ0.4 \pi \mathrm{mJ}

W=T×ΔA=T×8π(r22−r12)=0.4πmJ\mathrm{W}=\mathrm{T} \times \Delta \mathrm{A}=\mathrm{T} \times 8 \pi\left(\mathrm{r}_2{ }^2-\mathrm{r}_1{ }^2\right)=0.4 \pi \mathrm{mJ}

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