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A spherical rain drop, falling in a constant gravitational field, grows by absorption of moisture from the surroundings at a rate proportional to its surface area. If it starts with zero radius, find its acceleration. |
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Answer» SOLUTION :`(dm)/(dt) = K.4pir^(2)` Where K is a constant and .r. is the radius of the DROP at any instant . But ` m=(4)/(m) PIR^(3) RHO=(4)/(3) pi r^(3)` since ` rho`= density of water is 1 gram/c.c `(dm)/(dt) =(4)/(3) pi. 3R^(2) (dr)/(dt) = 4pir^(2) (dr)/(dt) = K. 4 pir^(3)"":. K=(dr)/(dt)` `:. ` r=Kt ( since it starts with zero radius ) Since ` m(dv)/(dt) + v(dm)/(dt) = mg ` `(4)/(3) pir^(3) (dv)/(dt) + v.k 4pir^(2)=(4)/(3) pir^(2) g , (dv)/(dt) + vK(3)/(r) =g ` But `(dv)/(dt) =a , v = at ` `:. a + at. K. (3)/(Kt) =g ` becomes `4a=g,a =` acceleration of drop =g/4 |
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