Question

A mercury drop of a cm radius is broken into ${10}^6$ small drops. The energy used will be (surface tension of mercury is $35\times {10}^{-3}N/cm)$

• A. $4.4\times {10}^{-3}J$
• B. $2.2\times {10}^{-4}J$
• C. $8.8\times {10}^{-4}J$
• D. ${10}^{-4}J$

Answer 1

The correct option is A $4.4\times {10}^{-3}J$

Given,

(surface tension of mercury is $35\times {10}^{-3}N/cm)$

split into =${10}^6$ small drops

$E=4\pi R^{2}T(n^{1/3}-1)$

$=4\ast 3.14\ast {10}^{-4}\ast 35\ast {10}^{-1}({10}^{6/3}-1)=4.4\ast {10}^{-3}J$