A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)

Option 1 - <p>8.5 × 10<sup>-4</sup> J</p>
Option 2 - <p>8.2 × 10<sup>-4</sup> J&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p>
Option 3 - <p>7.5 × 10<sup>-4</sup> J</p>
Option 4 - <p>5.3 × 10<sup>-4</sup> J</p>
13 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 3
Detailed Solution:

 43πR3=72943πr3

R3 = 729r3

R= (729)13 (r) (13)

R = 9r

Δu=T (4πr2)*729T*4πR2

=T*4π*8R2

=75.39*105JΔu=7.5*104J

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