Properties of matterMedium
Question
The work done in splitting a drop of water of 1 mm radius in to 106 droplets is (surface tansion of water 72 × 10-3 N/m):
Options
A.5.98 × 10-5
B.10.98 × 10-5
C.16.95 × 10-5
D.8.95 × 10-5
Solution
Radius of new droplet if be r then.
106 ×
πr3 = π(0.001)3
r3 = 10-15 ⇒ r = 10-5
Increase in surface area
[4π×(10-5)2 × 106] - [4π×(10-3)2]
[4π × 10-4]-[4π × 10-6] = 4π10-6[100-1]
= 4π × 10-6 × 99 = 4π × 10-6 × 99
Work done = surface tension × increase in surface area
= 72 × 4p × 99 × 10-6 × 10-3 = 8.95 × 10-5
106 ×
πr3 = π(0.001)3r3 = 10-15 ⇒ r = 10-5
Increase in surface area
[4π×(10-5)2 × 106] - [4π×(10-3)2]
[4π × 10-4]-[4π × 10-6] = 4π10-6[100-1]
= 4π × 10-6 × 99 = 4π × 10-6 × 99
Work done = surface tension × increase in surface area
= 72 × 4p × 99 × 10-6 × 10-3 = 8.95 × 10-5
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