FE Mechanical Mathematics Practice Problems

The current FE Mechanical specification assigns 6-9 of the 110 exam questions to Mathematics. Form A contains 7 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A right circular conical hopper stands apex down, $8$ ft deep with an $8$ ft top diameter. Granular material is fed in at a steady $12$ ft $^{3}$ /min. When the material depth is $6$ ft, the rate at which the surface level rises is most nearly:

(A)0.106 ft/min
(B)0.141 ft/min
(C)0.424 ft/min
(D)1.273 ft/min

Show worked solution

Answer: (C)

Because the free surface at 6 ft depth spans only 28.3 ft $^{2}$ , a 12 ft $^{3}$ /min feed lifts the level about 0.42 ft/min.

Similar triangles fix the surface radius at half the depth, which collapses the cone volume to a function of depth alone.

r = \frac{h}{2}

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

Differentiating in time links the two rates, and the instantaneous depth of 6 ft closes the problem.

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

\frac{d h}{d t} = \frac{4 ( 12 )}{π ( 6 ) ^{2}} = 0.424 ft/min

FE Reference Handbook — Mathematics: Derivatives and Indefinite Integrals; Mensuration of Areas and Volumes, Right Circular Cone

Why the other choices appear

(A)Taking $r = h$ instead of $h /2$ gives $d h / d t = 12/ [π (6)^{2}] = 0.106$ ft/min.
(B)Dropping the $1/3$ from the cone volume gives $V = π h^{3} /4$ and $d h / d t = 48/ [3 π (6)^{2}] = 0.141$ ft/min.
(D)Carrying the $1/3$ into the rate equation as $d V / d t = (π /3) r^{2} d h / d t$ with $r = 3$ ft gives $12/9.425 = 1.273$ ft/min.