1.

A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)A. `4.35 xx 10^(-2) J`B. `4.35 xx 10^(-3) J`C. `4.35 xx 10^(-6) J`D. `4.35 xx 10^(-8) J`

Answer» Correct Answer - A
If r is the radius of smaller droplet and R is the radius of bigger drop, then according to question,
`(4)/(3) pi R^(3) = 10^(6) xx (4)/(3) pi r^(3) rArr r = (R)/(100) = 0.01 R`
`0.01 xx 10^(-2) m = 10^(-4) m`
`:.` Work done = Surface tension `xx` increase in area
`= 35 xx 10^(-2) xx [(10^(6) xx 4pi xx (10^(-4))^(2) - 4pi xx (10^(-3))^(2)]`
`= 4.35 xx 10^(-2) J`


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