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A certain number of spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R` and volume `V`. If `T` is the surface tension of the liquid, thenA. energy=4VT `((1)/(r )-(1)/(R ))` is releasedB. energy = 3 VT `((1)/(r )+(1)/( R))` is absorbedC. energy `=3V T ((1)/(R )-(1)/(r ))` is releasedD. energy is neither released nor absorbed |
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Answer» Correct Answer - C Here, if the surface area changes, it will change the surface energy as well. If the surface area decreases, it means that energyu is released and vice versa. Change in surface energy `Deltaa xx T…..(i)` Let we have n number of drops initially `So, DeltaA=4piR^(2)-n(4pir^(2))...(ii)` Volume is constant `So, n(4)/(3)pir^(3)=(4)/(3)piR^(3)=V...(iii)` From (ii) and (iii) we get `DeltaA=(3)/(R)(4pi)/(3)xxR^(3)-(3)/(r)(n(4pi)/(3r^(3)))=(3)/(R)xxV-(3)/(r)V` `Rightarrow DeltaA=3V ((1)/(R)-(1)/(r))="negative value"` As, `R gt r, "so" DeltaA` is negative. It means that surface areais decreased, so energy must be released. `"Energy released"=DeltaAxxT=-3VT((1)/(r)-(1)/R)` |
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