MHT CET 2017 Physics Solved Paper

(d) Let radius of big drop = R
and of small drop = r
Volume of big drop = n(Volume of small drop)

4

3

πR3=n⋅

4

3

πr3
R3=nr3 ⇒ R=n1∕3r
Surface energy of n drops,
E2=n×4πr2×T
Surface energy of big drop, E1=4πR2T
∴

E2

E1

=

nr2

R2

=

nr2

(n 1 3 ⋅r)2

=

nr2

n2∕3⋅r2

=n1∕3 [∵R=n1∕3⋅r]
or, ratio of energy, E2:E1=[3]√n:1