Question

Water is poured into an inverted conical vessel whose radius of the base is 2m and height is 4 m at the rate of 77 litre/minute.the rate at which the water level is rising at the instant when the depth of water is 70 cm is

Answer 1

let the cone is filled till height "y"

so the volume of liquid that has been filled inside the cone is given by

$V = \frac{1}{3}\pi r^2 y$

here "r" is the radius of liquid level when its height is "y"

now the relation between radius(r) and height(y) is is calculated by similar triangle relation

$\frac{r}{y} = \frac{R}{H}$

now we have

$r = \frac{Ry}{H}$

now plug in the value of "r" in above equation

$V = \frac{1}{3}\pi (\frac{Ry}{H})^2y$

now we have

$V = \frac{1}{3}\piR^2\frac{y^3}{H^2}$

now rate of volume is given by

$\frac{dV}{dt} = \frac{1}{3}\piR^2\frac{3y^2}{H^2}\frac{dy}{dt}$

$\frac{dV}{dt} =\frac{ \pi R^2 y^2}{ H^2}\frac{dy}{dt$

now plug in all given values in it

$\frac{77*10^{-3}}{60} = \frac{\pi*2^2*0.7^2}{4^2}\frac{dy}{dt}$

$1.28*10^{-3} = 0.385*\frac{dy}{dt}$

$\frac{dy}{dt} = 3.33*10^{-3} m/s$

so water level is rising at rate of 0.33 cm/s.

Answer 2

R = 2m

H = 4 m

dv/dt  = 77 lit/min

tan θ = r/h = R/H

v = 1/3 π r²h

r = R/H . h

v = 1/3.π . R²/H² . h³

dv/dt = π/3 R²/H² 3h²

h = 70 cm = 0.7 m

77 x 10 ^-3 = n/3 x 4/16  x 3 x 0.7 x 0.7 dh/dt

11/1000 = 7π/4x100 . dh/dt

dh/dt = 44/40π m/min

dh/dt = 44/70π x 100 cm/min

Solve this to get the answer.