Free FE Practice Test# PrepFE™

## Free FE Other Disciplines Example Practice Problems

We've selected 10 diverse practice problems from our question bank that you can use to review for the Other Disciplines engineering FE exam and give you an idea about some of the content we provide.

◯ A.

1 1 1

◯ B.

1 1 2

◯ C.

1 2 1

◯ D.

1 2 2

$R_1 = \SI{10}{\Omega}$, $R_2 = \SI{3}{\Omega}$, $R_3 = \SI{8}{\Omega}$

$V_1 = \SI{5}{V}$, $V_2 = \SI{8}{V}$

◯ A.

$1.1 \si W$

◯ B.

$3.0 \si W$

◯ C.

$7.1 \si W$

◯ D.

$21 \si W$

◯ A.

0.51

◯ B.

2.18

◯ C.

1.93

◯ D.

1.20

◯ A.

$ (- \frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

◯ B.

$ (- \frac{1}{5},\frac{2}{5}), (1,1), (\frac{1}{7}, \frac{2}{7}) $

◯ C.

$ (- \frac{1}{5},-\frac{2}{5}), (0,1), (\frac{1}{7}, \frac{2}{7}) $

◯ D.

$ (\frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

$$ \begin{array}{c|c|c} \text{Bids Won} &0&1&2\\ \hline \text{\$ Earned} & -\$12,000 & \$88,000 & \$188,000 \\P(x) & 0.81 & 0.18 &0.01 \\ \end{array} $$

◯ A.

\$8,000

◯ B.

\$22,000

◯ C.

\$12,000

◯ D.

\$0

◯ A.

3.1 m/s

◯ B.

6.1 m/s

◯ C.

7.0 m/s

◯ D.

11 m/s

◯ A.

elastic behavior

◯ B.

Young's modulus

◯ C.

ultimate strength

◯ D.

Necking behavior

◯ E.

plastic behavior

◯ A.

$\SI{49}{\micro\meter}$

◯ B.

0.31 mm

◯ C.

$\SI{265}{\micro\meter}$

◯ D.

230 nm

◯ A.

$\SI{240}{\kelvin}$

◯ B.

$\SI{234}{\kelvin}$

◯ C.

$\SI{108}{\kelvin}$

◯ D.

$\SI{124}{\kelvin}$

A.1 1 1

B.1 1 2

C.1 2 1

D.1 2 2

The correct answer is B.

Note that you could have solved for other variables besides b and still gotten to the same final solution.

$R_1 = \SI{10}{\Omega}$, $R_2 = \SI{3}{\Omega}$, $R_3 = \SI{8}{\Omega}$

$V_1 = \SI{5}{V}$, $V_2 = \SI{8}{V}$

A.$1.1 \si W$

B.$3.0 \si W$

C.$7.1 \si W$

D.$21 \si W$

The correct answer is B.

Use the equation, $$ P = \frac{V^2}{R} $$ The total power dissipated in the circuit is the sum of the powers consumed by each individual resistor. Write an expression for the power consumed on each individual resistor and sum them together, $$ P_{\text{total}} = \frac{(V_2-V_1)^2}{R_3} + \frac{(V_1)^2}{R_1+R_2} $$ $$ = \frac{(\SI{8}{V}-\SI{5}{V})^2}{\SI{8}{\Omega}} + \frac{(\SI{5}{V})^2}{\SI{13}{\Omega}} $$ $$ \approx \SI{3.05}{W} $$ Remember that the $V$ in the equation $\frac{V^2}{R}$ refers to the voltage drop across the resistor. For the resistor, $R_3$, this is the difference between $V_2$ and $V_1$.

For the resistors $R_1$ and $R_2$, their total power consumption can be found by adding them together and dividing them into the voltage drop across both of them, which is simply the voltage $V_1$.

Another way to look at this is that the resistors $R_1$ and $R_2$ are together in series which are then in parallel with the voltage $V_1$. The voltage drop across the series combination of resistors is $V_1 - 0 = V_1$ because all the voltage sources share a common reference. Thus, the term $V_1$ is all that ends up in that particular term of the total power equation.

A.0.51

B.2.18

C.1.93

D.1.20

The correct answer is C.

The population is 50, but the sample is 9. Since we only have pressure values for 9 measurements out of the 50 total measurements, we must use the sample standard deviation formula. Ultimately, we must solve for: $$ s = \sqrt{ \frac{1}{(n-1)} \sum_{i=1}^n \left(X_i-\bar{X}\right)^2 } $$ Solve for the unknowns.

The total number of samples is $n=9$.

Calculate the arithmetic mean, $\bar{X}.$ $$ \bar{X}=(1/n)(X_1+X_2+...+X_n) \\ \bar{X}=(1/9)(20+20+20+22+ \\ 22+22+22+25+25) $$ $$\bar{X} = 22$$ Now that we have all the unknowns, solve for standard deviation. $$ s = \sqrt{ \frac{1}{9-1} \left[ 3(20-22)^2 + 4(22-22)^2 + 2(25-22)^2 \right] } $$ $$ \approx 1.93 $$

Note: You can use your TI-36X pro calculator's standard deviation function to solve this problem more quickly.

A.$ (- \frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

B.$ (- \frac{1}{5},\frac{2}{5}), (1,1), (\frac{1}{7}, \frac{2}{7}) $

C.$ (- \frac{1}{5},-\frac{2}{5}), (0,1), (\frac{1}{7}, \frac{2}{7}) $

D.$ (\frac{1}{5},\frac{2}{5}), (0,0), (\frac{1}{7}, \frac{2}{7}) $

The correct answer is A.

$$ \begin{array}{c|c|c} \text{Bids Won} &0&1&2\\ \hline \text{\$ Earned} & -\$12,000 & \$88,000 & \$188,000 \\P(x) & 0.81 & 0.18 &0.01 \\ \end{array} $$

A.\$8,000

B.\$22,000

C.\$12,000

D.\$0

The correct answer is A.

First, what are the chances that we lose both bids ($x=0$). If winning has a 10% chance, then losing has a 90% chance. $$ P(0) = 0.9 \cdot 0.9 = 0.81 $$ What about when one bid loses and one bid wins? $$ P = 0.9 \cdot 0.1 = 0.09 $$

Once you work through the example, you see it's easier to understand than what the formal definition of

The expected value is also known as the expectation, mathematical expectation, mean, or first moment. It is the probability-weighted average of a discrete random variable's possible values.

A.3.1 m/s

B.6.1 m/s

C.7.0 m/s

D.11 m/s

The correct answer is A.

Ultimately, we must solve for the velocity at the outlet of the system, $v_2$. Use Bernoulli's equation to solve for $v_2. $ $$ \frac{p_2}{\gamma} +\frac{v_2^2} {2g} +z_2 = \frac{p_1}{\gamma} + \frac{v_1^2} {2g} +z_1 $$ Calculate the unknowns:

$p_1=0 \si{Pa}$ because the water is exposed to the atmosphere, which is 0 gauge pressure or 101.3 kPa absolute pressure

$\gamma_{water}=9.81\si{kN/m^3}$

$z_1=0.5\si{m}$

$g=9.81\si{m/s^2}$

$v_1=0$ since the water at $x_1$ is just sitting there, not moving

$v_2=?$

$p_2=0 \si{Pa}$ because the water is released to the atmosphere, which is 0 gauge pressure or 101.3 kPa absolute pressure

$z_2=0$ $$ \frac{0}{9.81\si{kN/m^3}} +\frac{v_2^2} {2(9.81\si{m/s^2})} +0 = \frac{0}{9.81\si{kN/m^3}} \\ + \frac{0^2} {2(9.81\si{m/s^2})} +0.5\si{m}; \text{solve for }v_2 $$ $$ v_2=3.1\si{m/s} $$

A. elastic behavior

B. Young's modulus

C. ultimate strength

D. Necking behavior

E. plastic behavior

The correct answers are A and B.

From the origin until Point B, the material behaves elastically to tensile force. In elastic behavior, the material deformation is not permanent. The material will go back to its original shape once the tensile stress is removed. Additionally, the slope of A is used to find Young's modulus.

Plastic behavior occurs from Point B to Point D. The material's deformation is permanent and the material will not go back to its original shape if the tensile force is removed. Necking behavior occurs within the plastic behavior region. Necking occurs after the tensile force is so strong that it surpasses the ultimate strength of the material (Point C). Necking refers to when the material's cross-sectional area gets smaller and smaller until the material fractures (Point D).

A.$\SI{49}{\micro\meter}$

B.0.31 mm

C.$\SI{265}{\micro\meter}$

D.230 nm

The correct answer is C.

Ultimately, we must determine the increase in rod diameter after the rod is compressed. We can determine the increase in length of a member from its engineering strain. $$ \epsilon=\Delta L/L_0 $$ Which can be interpreted as $$ \epsilon_{lateral}=\Delta L_{lateral}/L_{0 lateral}=\Delta d/d_0 $$ Solve for the unknowns. Ultimately, we must solve for $\Delta d$.

$ d_0=12\si{mm}=0.012\si{m} \\ \Delta d=? \\ \epsilon_{lateral}=? $

Side note:

$ E= 70\times 10^9\si{Pa} \\ A=\pi/4\cdot d^2=\pi/4\cdot (0.012\si{m})^2=0.000113097 \si{m^2} \\ P=\text{internal loading}=500\si{kN}=500,000\si{N} $ $$ 70\times 10^9\si{Pa}=(-500,000\si{N}/0.000113097 \si{m^2})/\epsilon_{longitudinal} \\ \therefore \epsilon_{longitudinal}=-0.0631569 $$ We can go back and solve for lateral strain using Poisson's ratio. $$ 0.35=\frac{-\text{lateral strain}}{\text{longitudinal strain}}=\frac{-\epsilon_{lateral}}{-0.0631569} $$ $$ \therefore \epsilon_{lateral}=0.022104915 $$ We can go back to the first equation and solve for the change in diameter. $$ \epsilon_{lateral}=\Delta d/d_0 \\ 0.022104915=\Delta d/0.012\si{m};\text{solve for } \Delta d \\ \Delta d = 0.000265259\si{m}=\SI{265}{\micro\meter} $$ Some of the takeaways from this problem are:

Strain and deformation to a rod can occur longitudinally (along the length of the rod) or laterally (along the cross section diameter length). Therefore, you can use any of the Hooke's Law equations and modify them to be in terms of lateral deformation or longitudinal deformation. In this problem, the rod got shorter and the diameter got larger as the two 500 kN pushed into the rod axially.

A.$\SI{240}{\kelvin}$

B.$\SI{234}{\kelvin}$

C.$\SI{108}{\kelvin}$

D.$\SI{124}{\kelvin}$

The correct answer is D.