6. Calculate the energy required to split a water drop of radius 1 × 10−3 m into one
thousand million droplets of same size. Surface tension of water = 0.072 N m−1.
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Answer:
ven, Diameter=2mm,droplets=10
9
,Tension=0.073N/m
Since the liquid drop breaks into 10
9
droplets of equal mass so the volume of the big drop is equal to the volume of each droplet multiplied by 10
9
R
3
=
3
4
πR
3
=10
9
×
3
4
πr
3
⇒R
3
=(10
3
r)
3
=10
r
The initial energy of the liquid drop E
1
=TA where T is the tension and A is the area
E
1
=T×4πR
2
Final energy of the 10
9
droplets E
2
=10
9
Ta
Where T is the tension and a is the area.
E
2
=10
9
×T×4πr
2
NOw, Change in energy E
2
−E
1
=10
9
T×4πr
2
−T4πR
2
=T×4π[10
9
r
2
−R
2
]
⇒T×4π[10
9
(
10
3
R
)
2
−R
2
]=T×4π[10
3
R
2
−R
2
]=3996πTR
2
Now put the value of T and R
3996×3.14×(1×10
−3
)
2
×0.073=915×10
−6
J/m
2
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