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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

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प्रश्न

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?

योग
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उत्तर

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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अध्याय 6: Application of Derivatives - Exercise 6.1 [पृष्ठ १९८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.1 | Q 14 | पृष्ठ १९८

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