1000 drops of water, all of same size join together to form a single drop and the energy released raises the temperature of the drop. Given that ? is the surface tension of water, r the radius of each small drop, p the density of liquid, / the mechanical equivalent of heat. What is the rise in temperature ?

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T/Jr

10TJ/

100T/r

none of the above

answer is D.

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Detailed Solution

43πR3=n×43πr3 R=(n)1/3r=10r decrease in surface alea, ΔA=n4πr2−4πR2 ΔA=10004πr2−4π(10r)2 =4πr2[1000−100]=900×4πr2 Energy released =ΔA×T =900×4πr2T/J calorie Let △θ be the rise in temperature, then 1000×43πr3×1×△θ=900×4πr2T/J △θ=[2⋅7T/Jr]

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