FE Other Disciplines Mathematics Practice Problems

The current FE Other Disciplines specification assigns 8-12 of the 110 exam questions to Mathematics. Form A contains 9 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

Water drains from an inverted right circular cone, apex down, having a top radius of 3.0 m and a height of 6.0 m. The discharge is steady at 0.25 m $^{3}$ /s. When the water depth is 4.0 m, the rate of fall of the free surface is most nearly:

(A)0.0050 m/s
(B)0.0088 m/s
(C)0.0199 m/s
(D)0.0597 m/s

Show worked solution

Answer: (C)

The free surface falls about 0.0199 m/s, since at a depth of 4.0 m its area is only $4 π$ m $^{2}$ .

Geometric similarity ties the surface radius to the depth, reducing the volume to a single variable.

r = \frac{h}{2}

V = \frac{1}{3} π r^{2} h = \frac{π h ^{3}}{12}

Differentiating volume with respect to depth produces the instantaneous free-surface area, the only geometry the drain rate responds to.

\frac{d V}{d t} = \frac{π h ^{2}}{4} \cdot \frac{d h}{d t}

At 4.0 m depth the surface radius is 2.0 m and the surface area is $4 π$ m $^{2}$ , so the level drops at

\frac{d h}{d t} = \frac{0.25}{4 π} = 0.0199 m/s

FE Reference Handbook — Mathematics: Mensuration of Areas and Volumes, Right Circular Cone; Differential Calculus, The Chain Rule

Why the other choices appear

(A)Sets the surface radius equal to the depth, $r = h = 4.0$ m, giving $0.25/ (16 π) = 0.0050$ m/s.
(B)Uses the fixed 3.0 m top radius as the surface radius, giving $0.25/ (9 π) = 0.0088$ m/s.
(D)Divides the volume by the depth instead of differentiating, taking $d V / d h = π h^{2} /12 = 4.19$ m $^{2}$ and giving 0.0597 m/s.