A drop of radius r is broken into n equa...

A drop of radius r is broken into n equal drips. Calulate the work done if surface tension of water is T.

`r gt ((2 R sigma)/(3 rho g))^(1//3)`

`r gt ((2 sigma)/(3rho g))`

`(2sigma)/(r ) gt ` atmospheric pressure

`r gt ((2R sigma)/(3 rho g))^(2//3)`

Text Solution

Verified by Experts

Weight of drop > (surface tension ) ` xx 2pi R`
` rArr rho (4/3 pi r^3 ) g gt sigma xx 2pi r rArr r^3 gt 3/2 (R sigma)/(rho g)`
` rArr r gt (3/2 (R sigma)/(rho g))^(1//3)`
none of the given options is correct .

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A water drop of radius 1 cm is broken into 1000 equal droplets. If the surface tension of water is 0.075 N/m , then the gain in surface enrgy will be

`8.5xx10^(-4)` J

`7.5xx10^(-4)` J

infinite

A liquid drop of radius R is broken up into n small droplets. The work done is proportional to

`n^(1//3)`

`n^(0)`

`n^(2)`

A water drop of radius R is split into n smaller drops , each of radius r. If T is the surface tension of water ,then the work done in the process is

`4/3piR^(3)T(1/r-1/R)`

`3/4piR^(3)T(1/R-1/r)`

`4piR^(3)T(1/r-1/R)`

`6piR^(2)(1/R-1/r)`