KEAM 2010 Physics and Chemistry Paper

Volume of big drop =n× volume of small drop
‌

4

3

πR3=n×‌

4

3

πr3
R=n1/3r
Capacitance of small drop, C=4πε0r
Capacitance of big drop, C′=4πε0R
=4πε0n1/3r
C′=n1/3C
The potential of small drop V=‌

q

C

=‌

q

4πε0r

The potential of big drop V′=‌

nq

(4πε0)n1∕3r

V′=n2∕3V
∴ Energy of small drop =‌

1

2

CV2
Energy of big drop =‌

1

2

C′V′2
=‌

1

2

n1/3C(n2/3V)2
=n5/3‌

1

2

CV2
∴‌

‌ Energy ‌(‌big drop ‌)

‌ Energy ‌‌(small drop) ‌

=‌

n5/3

1