Question

The energy required to break a mercury drop of $1cm$ radius into $1000$ small drops of equal size is $.....J$ (given the surface tension of mercury is $0.4N.m^{-1}$).

• A. $144\pi \times {10}^{-2}$
• B. $1.44\pi$
• C. $14.4\pi \times {10}^{-4}$
• D. $144\pi$

Answer 1

The correct option is C $14.4\pi \times {10}^{-4}$

$T=0.4N/m$

As we know volume of 1000 drops =Volume of big drop

So $\frac{1000\times 4\pi r^3}{3}=\frac{4\pi 1^3}{3}$

$\Rightarrow r^3=\frac{1}{1000}{cm}^3\Rightarrow r=\frac{1}{10}=0.1cm$

Now $W=T(A_2-A_1)=T[(1000\times 4\pi r^2)-4\pi R^2]$

by putting value of T, R and r

we get,

$W=0.4\{(1000\times 4\pi {\left(\frac{0.1}{100}cm\right)}^2)-4\pi {\left(\frac{1}{100}cm\right)}^2\}W=14.4\pi \times {10}^{-4}J$

So $W=14.4\pi \times {10}^{-4}J$

Option C is correct.