Question

A liquid drop having surface energy ‘E’ is spread into 216 droplets of the same size. The final surface energy of the droplets is __.

• A. 3E
• B. 8E
• C. 2E
• D. 6E

Answer 1

Answer: (D)

6E

Answer 2

Surface area of drop, A1 = 4πR2
Surface area of droplets, A2 = 216(4πr2)
Volume of drops = n x (Volume of droplets)
$\frac {4}{3}$ πR3 = 216 ($\frac {4}{3}$ πr3)
R = 6r
Now A2 = 216(4π($\frac {R}{6}$)2)
A2 = 6(4πR2)
E ∝ A
$\frac {E_1}{E_2 }$= $\frac {A_1}{A_2}$
$\frac {E_1}{E_2 }$ = $\frac {4πR^2}{6(4πR^2)}$
$\frac {E_1}{E_2 }$ = $\frac {1}{6}$
6E1 = E2
E2 = 6E1
E2 = 6E {E1 = E}
Therefore, the correct option is (D) 6E