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n droplets of equal size and each of rad...

n droplets of equal size and each of radius r coalesce to form a bigger drop of radius R. the energy liberated is equal to

A

`4piR^(2)T(n^(1//3)-1)`

B

`4pir^(2)T(n^(1//3)-1)`

C

`4piR^(2)T(n^(2//3)-1)`

D

`4pir^(2)T(n^(2//3)-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Volume of n droplets = Volume of the big drop
` :. n xx 4/3 pir^(3) = 4/3 piR^(3)`
` :. R^(3) = nr^(3) " " :. R = n^(1//3) r`
` :. r = R/(n^(1//3)) and r^(2) = (R^(2))/(n^(2//3))`
` :. ` Change in area ` = 4pir^(2)n - 4piR^(2)`
` = 4pi(nr^(2)-R^(2))`
` dA = 4pi[(nR^(2))/(n^(2//3)) -R^(2)]= 4piR^(2)[n^(1//3)-1]`
` :' T = (dW)/(dA) or (dE)/(dA)`
` :. ` energy liberated dE = TdA
` = 4 piR^(2) T [n^(1//3) -1]`
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