Question

Two mercury droplets of radii 0.2 cm and 0.4 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury 440 xx 10^(-3) Nm^(-1) and g=10ms^(-2).

Answer 1

ps

∴43πR3=43πr31+43πr32

or, R3=r31+r32orR=[r31+r32]1/3

or R=[(2×10−3)3+(4×10−3)3]1/3

=(72×10−9)1/3=4.16×10−3m

DecreHere, r1=0.2c,=0.2×10−2

=2×10−3m,

r2=0.4cm=0.4×10−2m=4×10−3m

S=440×10−3Nm−1

Let R be the radius of big drop formed.

Now volume of big drop = volume of two small droase in surface area

ΔA=4π(r21+r22)−4πR2

=4π[(2×10−3)2+(4×10−3)2−(4.16×10−3)2]

=4π[20×10−6−17.30×10−6]

=4×227×2.70×10−6=33.94×10−6^(2)

Energy released = S×ΔA

=(440×10−3)×933.94×10−6

=14.935×10−6J.