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Q.

If n number of identical droplets of water, each of radius r, coalesce to form a single drop of radius R, the resulting rise in the temperature of water is given by (here ρ is the density of water, s its specific heat and σ its surface tension) 

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a

3σρs1r-1R

b

3σρs1r+1R

c

σρs1r-1R

d

σρs1r+1R

answer is B.

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Detailed Solution

Final energy = 4πR2 σ=πD2σ  where R is the radius of the big drop.

Initial energy = 4πr2nσ=πd2nσ, where d = 2r, is the radius of each tiny drop

Now R3=nr3 or D3 = nd3 or D=n1/3d  or  d=D/n1/3 . 

Therefore, Initial energy= πD2n2/3=πD2σn1/3

 Hence, change in energy = πD2σn1/3-πD2σ=πD2σn1/3-1=4πR2(n1/3-1)σ

Work done is

          W=4πR2n1/3-1σ=4πR2Rr-1σ----- (1)     R=n1/3r

Heat produced  Q=msT=43πR3ρsT    2

Equating (1) and (2), we find that choice (b) gives the correct expression for T. 

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