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Q.

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

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a

3TJ1r−1R

b

3TrJ

c

2TJ1r−1R

d

2TrJ

answer is A.

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Detailed Solution

Conserving volume,  N43πr3=43πR3 ⇒R3=Nr3Loss of surface energy =  Ui-Uf=TAf-TAi=N4πr2T−4πR2T ⇒ heat produced  =4πTJNr2−R2⇒ heat produced per unit volume =4πTNr2−R243πR3 J =3TJNr2R3−1R=3TJ1r−1R
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