Question

A mercury drop of radius $1cm$ is broken into ${10}^6$ droplets of equal size. The work done is $\rho =35\times {10}^{-2}N/m$

• A. $4.35\times {10}^{-2}$
• B. $4.35\times {10}^{-3}J$
• C. $4.35\times {10}^{-6}J$
• D. $4.35\times {10}^{-8}J$

Answer 1

The correct option is A $4.35\times {10}^{-2}$

If $r$ is the radius of a small droplet and $R$ is the radius of a big drop, then according to question:

$\frac{4}{3}\pi R^3={10}^6\times \frac{4}{3}\pi R^3$

So, $r=0.01R={10}^{-4}m$

Now, work done $=S.A$

$=35\times {10}^{-2}[{10}^6\times 4\pi \times ({10}^{-4}{)}^2-4\pi \times \left({10}^{-3}{)}^2\right]$

$=4.35\times {10}^{-2}J$