JEE Main 2022 · 28 Jun · Shift 1 · Physics Q36

← JEE Main 2022 · 28 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 36 of 90

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

2.8\times10^{-4}

1.5\times10^{-3}

1.9\times10^{-4}

9.4\times10^{-5}

Answer and worked solution

Official answer: A

Droplet radius. $R=1$ cm, $r=\frac R4=0.25$ cm (volume shared 64 ways).

Extra area. $4\pi(64r^2-R^2)=4\pi(4-1)=12\pi$ cm $^2=12\pi\times10^{-4}$ m $^2$ .

Energy. $0.075\times12\pi\times10^{-4}\approx2.8\times10^{-4}$ J.

(A)

Topics:Surface tension

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JEE Main 2022 · 28 Jun · Shift 1, Physics question 36