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

The Correct Option is D

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