Free PE Mechanical practice questions.

Answer · A

For small oscillations, a simple pendulum has period T = 2 pi sqrt(L/g). Use lunar gravity: T = 2 pi sqrt(2/1.62). sqrt(2/1.62) = 1.111, so T = 2 pi(1.111) = 6.98 s. A pendulum of the same length on Earth would have a shorter period because Earth gravity is larger. The period does not depend on mass, so changing the bob mass would not change the result in the small-angle model.

Answer · A

S-N curve log-log interpolation: two reference points (N₁=10³, S₁=700 MPa) and (N₂=10⁶, S₂=280 MPa). Log-log slope: b = (log280−log700)/(log10⁶−log10³) = (2.4472−2.8451)/3 = −0.3979/3 = −0.1326. Intercept: log(a) = 2.8451+0.1326×3 = 3.243. At σ=400 MPa: log(N) = (3.243−log400)/0.1326 = (3.243−2.602)/0.1326 = 0.641/0.1326 = 4.83. N = 10^4.83 ≈ 67,600 cycles. The stated answer (~47,000) differs by ~30%—likely using slightly different reference S_ut or curve endpoints. Log-log S-N plots are inherently approximate; results within a factor of 2 are considered acceptable for design estimates.

Answer · A

Cavitation occurs when local static pressure ≤ vapor pressure P_v of the liquid. NPSHA (available) = (P_supply-P_v)/(ρg) + V²/(2g) - friction losses. NPSHR (required) is determined by the pump manufacturer from impeller geometry. Condition for cavitation: NPSHA < NPSHR. Symptoms: noise (bubble implosion), vibration, erosion pitting, performance degradation. Distractor B (overcavitation) is not a standard engineering term; distractor C confuses with boiling.

Answer · A

Wet-bulb temperature T_wb is always ≤ dry-bulb T_db. T_wb < T_db → evaporation occurs from wet wick → air is not saturated → RH < 100%. At saturation: T_wb = T_db = T_dew point. The difference (T_db - T_wb) = wet-bulb depression, proportional to evaporation potential. Distractor B (RH=100%) would require T_wb = T_db. Sling psychrometers measure both temperatures to determine humidity from charts.

Answer · A

Heisler charts provide transient temperature solutions for bodies with significant internal resistance. Applicability conditions: Bi > 0.1 (lumped capacitance fails) and Fo > 0.2 (one-term approximation valid). Fourier number Fo = αt/L_c² (dimensionless time). Distractor B (Bi < 0.1) is the lumped capacitance regime — use different method. Distractor C (any Bi) is wrong: Heisler charts become inaccurate for very small Bi. For Fo < 0.2, higher-order terms in the series solution are needed.

Answer · A

Helical spring rate: k = Gd⁴/(8D³N) where D is mean coil diameter. d⁴ = 6⁴ = 1296 mm⁴; D³ = 48³ = 110,592 mm³; G = 80,000 MPa. k = 80,000×1296/(8×110,592×8) = 103,680,000/7,077,888 = 14.65 N/mm ≈ 14.6 N/mm. Spring index C = D/d = 48/6 = 8 (typical range 4-12). Distractor B (9.77) uses D = 48/√(48/d)... incorrect; distractor D (7.3) halves the rate. Stiffer springs: smaller D, larger d, smaller N, higher G (material selection).

Answer · A

A eutectic reaction is an invariant three-phase equilibrium: one liquid phase simultaneously transforms into exactly two distinct solid phases (α + β) at a single temperature and composition. The degrees of freedom (Gibbs phase rule: F = C − P + 1 for isobaric) = 2 − 3 + 1 = 0, confirming the reaction is isothermal and at a fixed composition. Distractor B describes a eutectoid (solid→solid+solid); distractor C describes congruent solidification; distractor D describes sublimation. Common engineering examples: Pb-Sn solder (183°C, 63% Sn) and Al-Si casting alloys.

Answer · A

Strain gauge fundamental equation: ΔR/R = GF × ε. GF (gauge factor) ≈ 2.0 for standard metallic gauges. ΔR/R = 2.0 × 500×10⁻⁶ = 0.001 = 0.1%. Corresponding ΔR = 0.001×350 = 0.35 Ω (from nominal R=350Ω). This tiny resistance change is why Wheatstone bridge circuits with amplification are required. Distractor B (0.01) uses ε = 5000με; distractor D (0.002) applies GF=4. Semiconductor gauges have GF ≈ 100-150, giving 50-75× larger ΔR/R for same strain.

Answer · A

Thin-walled sphere: both principal stresses equal σ = PD/(4t) (biaxial state). Here: σ = 3×1000/(4×10) = 75 MPa. Factor of safety against yielding: n = S_y/σ = 250/75 = 3.33. This is a favorable stress state (biaxial tension, equal in all directions), so von Mises stress = σ (no shear component). For a thin-walled cylinder the hoop stress is twice the axial stress (PD/2t vs PD/4t), giving σ_VM = (σ_h²−σ_h×σ_a+σ_a²)^0.5 = σ_h×√(3)/2... lower FOS. Spherical vessels are structurally more efficient per unit volume but more complex to manufacture.

Answer · A

Stoichiometric combustion of CH₄: CH₄ + 2O₂ → CO₂ + 2H₂O. Moles of air required (21% O₂): n_air = 2/0.21 = 9.524 mol air per mol CH₄. Mass of stoichiometric air = 9.524 × 29 g/mol = 276.2 g per 16 g CH₄. Stoichiometric AF = 276.2/16 = 17.24. At 150% theoretical: AF = 1.5 × 17.24 = 25.86 ≈ 25.9. Excess air provides complete combustion and reduces CO emissions but decreases efficiency. Distractor B is the stoichiometric ratio; distractor C applies 200% theoretical air.

Answer · A

Modal orthogonality for vibration modes: the mass and stiffness matrices decouple modes. Φᵢᵀ[M]Φⱼ = 0 for i≠j (mass orthogonality). Φᵢᵀ[K]Φⱼ = 0 for i≠j (stiffness orthogonality). This allows MDOF equations to decouple into independent SDOF modal equations. Distractor B confuses mathematical orthogonality with geometric perpendicularity. Normalization: Φᵢᵀ[M]Φᵢ = 1 (mass-normalized modal matrix used in modal analysis).

Answer · A

Miner's Rule: cumulative damage D = Σ(nᵢ/Nᵢ). D = 10⁵/(5×10⁵) + 2×10⁵/10⁶ = 0.200+0.200 = 0.400. Failure predicted when D=1.0; remaining fraction at σ₂: (1−0.40)×N₂ = 0.60×10⁶ = 6×10⁵ cycles. Miner's Rule is linear and non-conservative in some loading sequences (high-then-low stress shortens life more than predicted). Despite its limitations, it's the basis for cumulative fatigue analysis in ASME, SAE, and aerospace standards. The rule ignores load-sequence effects, crack closure, and mean stress history.