Question

a drop of mercury of radius 0.2 cm is broken into 8 droplets of the same size. find the work done if the surface tension of mercury is 435.5 dyne/cm​

Answer 1

Explanation:

we have to find workdone if surface

tension of mercury is 435.5 Dyne/cm. volume of the drop = volume of 8 droplets

4/3 TR3 = 8 × 4/3 Tr

⇒R=2r

⇒ r = R/2 = 0.2/2 = 0.1 cm

so, change in area, AA = 8 × 4πr² - 4πR²

= 4T [8r² - R²]

= 4π[8r² - (2r)²]

= 4T X 4r²

= 16πr²

= 16T(0.1)²

= 16T X 0.01

= 0.16π cm²

now workdone = TAA

= 435.5 Dyne/cm x 0.16 π cm²

= 435.5 × 0.16 x 3.14 Dyne. cm

= 218.7952 (10^-5 N) ( 10^-2 m)

= 218.7952 × 10^-7 Nm

= 2.187952 × 10^-5 J= 2.2 × 10^-5 J