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InterviewSolution
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The volume of sphericalballoon being inflated changes at a constant rate. If initially its radius is3 units and after 3 seconds it is 6 units. Find the radius of balloon after tseconds. |
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Answer» Correct Answer - `r=(63t+27)^(1//3)` The volume of a spherical balloon of radius r is given by `V=(4)/(3) pi r^(3).` Now, `(dV)/(dt) = -k, " where " k gt 0 " " ` [note that V is decreasing] `rArr (d)/(dt)((4)/(3)pi r^(3))= -k rArr (4pr^(2))(dr)/(dt) = -k` `rArr int (4pir^(2))dr=int (-k)dt` `rArr (4)/(3) pi r^(3) = -kt +C " " `...(i), where C is an arbitrary constant. Putting t = 0 and r = 3 in (i), we get C = 36 `pi`. `therefore (4)/(3)pir^(3)= -alpha t +36 pi. " " `...(ii) It is being given that when t = 3, then r = 6. Putting t = 3 and r = 6 in (ii), we get k = -84`pi`. Putting k = -84`pi`. in (ii) , we get `r^(3)=(63t +27) rArr r = (63t +27)^(1//3).` |
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