Question

20. Twenty seven droplets of water, each
of radius 0.1 mm coalesce into a single
drop. Find the change in surface energy.
Surface tension of water is 0.072 N/m.​

Answer 1

Explanation:

r=0.1mm=0.1×10

−3

m, T=0.072N/m

Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.

∴27×

3

4

πr

3

=

3

4

πR

3

∴27r

3

=R

3

∴3r=R

Surface area of 27 droplets = 27×4πr

2

Surface area of single drop = 4πR

2

∴ Decrease in surface area = 27×4πr

2

−4πR

2

=4π(27r

2

−R

2

)=4π[27r

2

−(3r)

2

]=4π×18r

2

∴ The energy released = surface tension × decrease in surface area = T×4π×18r

2

=0.072×4×3.142×18×(1×10

−4

)

2

=1.628×10

−7

J.

Answer 2

Answer:

r=0.1mm=0.1×10−3m, T=0.072N/m 

Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.

∴27×34πr3=34πR3

∴27r3=R3

∴3r=R

  Surface area of  27 droplets = 27×4πr2Surface

area of single drop = 4πR2

∴ Decrease in surface area = 27×4πr2−4πR2=4π(27r2−R2)=4π[27r2−(3r)2]=4π×18r2

∴ The energy released =

surface tension × decrease in surface area = T×4π×18r2=0.072×4×3.142×18×(1×10−4)2=1.628×10−7J. 

Explanation:

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