Question

Calculate the worldene or energy evolved
when 8 small droplets of water of
radius 0.5 mm combines to form
One big drop. Given Surface tension of
water is 0.072 N/m.​

Answer 1

Explanation:

Let the number of droplets be = n (Given)

Surface tension of water = 0.072N (Given)

Radius of each droplet r = 1/2 = 5 × 10`4mm (Given)

Since, droplets are combining to form into a bigger droplet, the total energy released will be -

Eloss = 4π ( n - n`2/3) r²T

= 4π ( 8 - 8`2/3) × 5 × 10`4 × 0.072

= 9.0478 × 10`7

Therefore, the energy evolved during the combining into a single drop is 9.0478 × 10`7.

Answer 2

Explanation:

Let R be the radius of the big drop formed by the combination of 8 small droplet.

$\sf Volume_{Big \: drop}=Volume _{8\: small \: droplets}$

$\sf \dfrac{4}{3}πR^3=8 \bigg( \dfrac{4}{3}πr^3 \bigg)$

$\sf R=2r=10^{-3}\: m$

$\sf Surface \: area_{big\:drop}=4πR^2$

$\sf = 4π×10^{-6}\:m^2$

Surface area of 8 small droplets:-

$\sf = 8×4π(5×10^{-4})^2=8π×10^{-6}\: m^2$

Decrease in surface area:-

$\sf = 8π×10^{-6}\:m^2-4π×10^{-6}\: m^2=4π×10^{-6}\: m^2$

Energy evolved:-

$\sf W = \Delta A$

$\sf = 72×10^{-3}×4π×10^{-6}$

$\sf = \bf 9×10^{-7}\:J$