Question

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

(a) 0.033cm/minute

(b) 0.043cm/minute

(c) 0.053cm/minute

(d) 0.045cm/minute

I got this question during an interview for a job.

I'm obligated to ask this question of Application of Derivative in division Application of Derivatives of Mathematics – Class 12

Answer 1

Right answer is (b) 0.043cm/minute

Best explanation: 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 (πr^2h) ……….(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)h^3

Thus, dV/dt = (3π/100)*3h^2(dh/dt)

= (9πh^2/100)(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.