Skip to content
dronacharya.ai
← JEE Advanced 2017 · Paper 1
PhysicsInteger+3Question 8 of 54

JEE Advanced 2017 · Paper 1, Physics question 8

A drop of liquid of radius R=10−2R = 10^{-2} m having surface tension S=0.14π Nm−1S = \frac{0.1}{4\pi}\ \mathrm{Nm^{-1}} divides itself into KK identical drops. In this process the total change in the surface energy ΔU=10−3\Delta U = 10^{-3} J. If K=10αK = 10^\alpha then the value of α\alpha is

Answer and worked solution

Official answer: 6

Volume. Volume is conserved: Kr3=R3Kr^3=R^3, so r=RK−1/3r=RK^{-1/3}.

Surface energy.
ΔU=S⋅4π(Kr2−R2)=S⋅4πR2(K1/3−1)=0.14π⋅4π⋅10−4(K1/3−1)=10−5(K1/3−1).\Delta U=S\cdot4\pi\left(Kr^2-R^2\right)=S\cdot4\pi R^2\left(K^{1/3}-1\right)=\frac{0.1}{4\pi}\cdot4\pi\cdot10^{-4}\left(K^{1/3}-1\right)=10^{-5}\left(K^{1/3}-1\right).

Solve. Setting this to 10−310^{-3} gives K1/3−1=100K^{1/3}-1=100, so K1/3≈100K^{1/3}\approx100 and K≈106K\approx10^6. Hence α=6\alpha=6.

Answer: 6

Topics:Surface tension
← Previous questionNext question →