Let the volume of the truncated cone = V

Then V = (1/3)*pi*(r1^2+r1*r2+r2^2)h

Where r1=lower radius ,r2=upper radius and h=height

The angle of the cone surface = tan^(-1)(4-2)/(6) = 18.43 deg.

Lower radius is constant always when filling the cup and only upper radius changing with height.

Then we can say r1=2 and r2=r

So at any point of the cone if the upper radius is r then;

h= (r-2)/tan18.43 = 3(r-2)

When cup is half filled h= h/2 = 3

Then 3 = 3(r-2) and we get r=2

V = (1/3)*pi*(r1^2+r1*r2+r2^2)h

V = (1/3)*pi*(2^2+2*r+r^2)* 3(r-2)

V = pi*(r^2+2*r+4)(r-2)

The rate of pouring of coffee = dV/dr

= pi*[*(r^2+2*r+4)+(r-2)(2r+2)]

The rate at h=h/2 is (dV/dr) at r=2

dV/dr = pi*[*(2^2+2*2+4)+(2-2)(2*2+2)]

= 12*pi

= 37.68 cm^3/s

**The rate of coffee pour at half way fill is 37.68 cm^3/s**

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