a drop of mercury radius 0.2 cm is broken into 8 droplets of the same size find the work done if surface tension of mercury is 435.5 dyne/cm
Answers
work done would be 2.2 J
It is given that a drop of mercury radius 0.2 cm is broken into 8 droplets of the same size.
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 πR³ = 8 × 4/3 πr³
⇒R = 2r
⇒r = R/2 = 0.2/2 = 0.1 cm
so, change in area, ∆A = 8 × 4πr² - 4πR²
= 4π [8r² - R²]
= 4π[8r² - (2r)²]
= 4π × 4r²
= 16πr²
= 16π(0.1)²
= 16π × 0.01
= 0.16π cm²
now workdone = T∆A
= 435.5 Dyne/cm × 0.16 π cm²
= 435.5 × 0.16 × 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
Answer:
volume of the drop = volume of 8 droplets
⇒4/3 πR³ = 8 × 4/3 πr³
⇒R = 2r
⇒r = R/2 = 0.2/2 = 0.1 cm
so, change in area, ∆A = 8 × 4πr² - 4πR²
= 4π [8r² - R²]
= 4π[8r² - (2r)²]
= 4π × 4r²
= 16πr²
= 16π(0.1)²
= 16π × 0.01
= 0.16π cm²
now workdone = T∆A
= 435.5 Dyne/cm × 0.16 π cm²
= 435.5 × 0.16 × 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