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Eight identical drops of mercury each of radius ‘r’ coalesce to form a single drop. If surface tension of mercury is ‘T’, the heat generated is

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answer is 2.

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Detailed Solution

43πR3=8×43πr3⇒R=2r ∴heat   generated  =Lost  energy                                                                    =Ui−Uf                                                                     =8×4πr2T−4πR2T                                                                      = 16πr2T
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Eight identical drops of mercury each of radius ‘r’ coalesce to form a single drop. If surface tension of mercury is ‘T’, the heat generated is