Question

A drop of mercury of radius 0.1 cm is broken into 27 droplets of the same size .find the work done if the surface tension of mercury is 540 dyne/cm

Answer 1

Answer:

Explanation:

Here,

T = 0.072 Nm-1

R = 1×10-3 m

Now surface area of the bigger drop

 = 4πR2

 = 4×22/7×(1×10-3)2

 S1 = 0.1257×10-4 m2

Total initial volume = (4/3)πR3

Volume of one of the smaller droplets =  (4/3)πR3/1000

Let r be the volume of one of the smaller droplets

Now, volume of the droplets =  (4/3)πr3    =

Thus, (4/3)πr3 = (4/3)πR3/1000

=> r = R/10

Now surface area of one of the small drop = 4πr2

 = 4π(R/10)2

 = 4×22/7×(1×10-4)2

 = 12.57×10-8 m2

Total surface area of 1000 drops S2 = 1000×12.57×10-8

    = 1.257×10-4 m2

Change in surface area = S2 – S1

   = 1.257×10-4 - 0.1257×10-4

  ΔS = 1.13×10-4

Now work done = change in surface energy = TΔS

 Work done =  0.072 × 1.13×10-4

   = 0.08×10-4

   = 8×10-6 J