"Solution 1 ---- General Solution of Calculus"

This is a differential application problem (Rate of change with time)

Set the depth of the water, h, to be the radius of the plane at any time t to be r.

Volume of the water at time t: V = (1/3)πr?h

According to the similarity triangle: height of cone/depth of water = radius of cone mouth/radius of water surface

8/h = 4/r, h = 2r

∴V = (1/3)πr?h = (1/3)π(h/2)?h

= (1/12)πh?

dV/dt = (1/4)πh?dh/dt

dh/dt = 4(dV/dt)/(πh?)

= 4*4/(25π)

= 16/25π

≈ 0.204 (m/min)

"Solution 2 ---- Elementary Algebraic Solutions"

Velocity at which the water surface rises at a given instant = Volume increase of the water at that instant/area of the water at that instant

= 4/(πr^2) = 4/(π2.5^2) = 16/25π (m/min)

[Note: Based on the similarity ratio: height of cone/depth of water = radius of cone's mouth/radius of water's surface, we get: r=2.5m)