Water is flowing into a right circular conical vessel, 45 cm deep and 27 cm in diameter at rate of 11 cc per minute. How fast is water level rising when water is 30 cm deep?

Let ?r? be the radius and ?h? be the height of the water level at time t.
Then the volume of water level ?V? at time t is given by,
V = 1/3 (πr²h) ???.(1)
Given the radius of the base of the cone is = (OA)? = 27/2 cm and its height = (OB)? = 45cm.
Again at time t, the radius of the water level = r = (CD)? and its height = h = (CB)?
Clearly, the triangle OAB and CBD are similar.
Therefore, (CD)?/(CB)? = (OA)?/(OB)?
Or, r/h = (22/7)/45 = 3/10
or, r = 3h/10
Thus, from (1) we get,
V = 1/3 π (3h/10)² h = (3π/100)h³
Thus, dV/dt = (3π/100)*3h²(dh/dt)
= (9πh²/100)(dh/dt)
When, h = 30cm, then,
11 = (9π/100) (30)² (dh/dt) [as for all the values of t we have, dV/dt = 11]
Or, dh/dt = (11/(9π*9)) = 0.043
Thus, the rising rate of rising is 0.043 cm/minute.