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Calculate the energy released when 1000 small water drops each of same radius `10^(-7)m` coalesce to form one large drop. The surface tension of water is `7.0xx10^(-2)N//m`. |
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Answer» Correct Answer - A::B Let r be the radius drops and `R` of bigger one. Equanting the initial and final volume, we have `(4)/(3) piR^(3)=(1000)((4)/(3) pi r^(3))` or, `R=10r=(10)(10^(-7))m` or, `R=10^(-6)m` Further, the water drops have only one free surface. Therefore, `Delta A=4pi R^(2)-(1000)(4 pi r^(2))` `=4pi [(10^(-6))^(2)-(10^(3))(10^(-7))^(2)]` `=-36pi (10^(-12))m^(2)` Here, negative sign implies that surface area is decreasing. Hence, energy released in the process. `U=T|Delta A|=(7xx10^(-2))(36 pi xx10^(-12))J` `=7.9xx10^(-12) J`. |
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