Therefore, the volume of water flowing in at time t is q * t.

Make a plane parallel to the bottom and intersecting the cone, and let the distance from the plane to the fixed point of the cone be x.

If the radius of the cross section is set to r, then by the property of similar triangles,

R/H=r/x

Where r and h are the radius and height of the bottom of the cone, respectively.

therefore

r=xR/H

So the bottom area

S=πr^2=π(R/H)^2 * x^2

Therefore, the volume of the cone surrounded by the cross section and the conical surface is

S*x/3=π(R/H)^2 * x^3/2

Let the water depth at time t be x, then

q*t = π(R/H)^2 * x^3/2 ①

Derive t at both ends of the above formula at the same time, and get

q = 3π(R/H)^2 * x^2 *v/2 ②

Where v =dx/dt represents the rising speed of the water surface.

The above formula is the basic equation of this problem.

(1) Distance from the spire 12 cm, that is, x =12cm3.

Substitute (* *) to get t = 1728π/ 1000 ④.

Substituting ④ into ②, we can get V =125/54 π cm/s.

(2) There are two solutions to this problem.

(i) A cone surrounded by a visible section and a conical surface is similar to the original cone.

Then the ratio of the height x of the former to the height h of the latter is the similarity ratio, and the volume ratio is the cube of the similarity ratio.

Therefore, x/h = (1/4) (1/3) = 2 (-2/3).

Then x = 2 (-2/3) * h = 2 (1/3) * 20cm.

Substituting the above formula into ② formula, v = 2 (1/3) * 5/12 π cm/s is obtained.

(ii) If the cone is filled with water, the time required.

T = πR^2*H/3q = 128π/3

Because the flow rate of water is constant, the corresponding time when 1/4 is full is

t = T/4 =32π/3

Substituting it into ① formula also gives x = 2 (1/3) * 20 cm.

The following steps are the same as (i).