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Question
Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
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Solution
Let at any instant of time t, the radius of the base of the cone be r, its height be h and the volume of the sand cone be V, then
`h = 1/6 r`
= r = 6h
and `V = 1/3 pir^2 h = 1/3 pi (6h)^2 h = 12 pi h^3` ....(i)
Differentiating (i) w.r.t. t, we get,
`(dV)/dt = (12 pi) (3h^2 (dh)/dt)`
= 12 cm3 /sec
`= 36 pi (4 cm)^2 (dh)/dt`
(∵ h = 4 cm and `(dV)/dt = 12` cm2/sec)
`= (dh)/dt = 12/ (36 pixx 16)` cm /sec
`= 1/(48 pi)` cm/sec
∴ Rate of increase of the height of the sand cone `1/(48 pi)` cm/sec.
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