Question

A mercury drop of radius 1 cm is broken into 106 droplets of equal size. The work done is

$(T=35{10}^{-2}N/m)$

• A. $1.615\times {10}^{-3}J$
• B. $4.35\times {10}^{-4}J$
• C. $1.615\times {10}^{-6}J$
• D. $4.35\times {10}^{-8}J$

Answer 1

The correct option is A $1.615\times {10}^{-3}J$

Let the radius of smaller drop be r, so

$\frac{4\pi (1{)}^2}{3}=106\frac{4\pi r^3}{3}$

$r=0.21cm=0.21\times {10}^{-2}m$

work done is $W=T\left(dA\right)=T(106\times 4\pi (0.21\times {10}^{-2}{)}^2-4\pi (1\times {10}^{-2}{)}^2)$

here $T=35\times {10}^{-2}$

$W=1615\times {10}^{-6}J$