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. `4.35xx10^-2J`B. `4.35xx10^-3J`C. `4.35xx10^-6J`D. `4.35xx10^-8J`

Answer» Correct Answer - A
Initial radius of drop,
`R=1cm1xx10^-2m`
suppose, `r` is the radius of droplets
since, volume will be remain constant.
`(4)/(3)piR^3=10^6xx(4)/(3)pir^3`
[since, number of droplets`=10^6`]
`r=(1xx10^-2)/(10^2)`
Increase in surface area,
`triangleS=10^6xx4pir^2-4piR^2`
`=10^6xx4xxpixx10^-8-4pixx10^-4`
`=4xx9.9xxpixx10^-3`
Work done `=`increase in surface area `xx` Surface tension.
`=4xx9.9xxpixx10^-3xx35xx10^-2`
`=4.35xx10^-2J`


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