Ramsay Test Mechanical Aptitude 6

Question 1

A compound gear train has three meshing gears: Gear A (12 teeth) drives Gear B (36 teeth), which is fixed to the same shaft as Gear C (9 teeth), which drives Gear D (27 teeth). If Gear A rotates at 720 RPM, what is the output speed of Gear D?

Accepted Answer

80 RPM

Answer

In a compound gear train, the overall ratio multiplies stage by stage. Stage 1: A→B ratio = 12/36 = 1/3, so B (and C on the same shaft) spin at 720 × (1/3) = 240 RPM. Stage 2: C→D ratio = 9/27 = 1/3, so D spins at 240 × (1/3) = 80 RPM. Overall reduction = 1/9.

Question 2

A hydraulic press has a small piston with a 2 cm² cross-sectional area and a large piston with a 50 cm² cross-sectional area. If a 40 N force is applied to the small piston, what is the pressure transmitted through the fluid, and how far must the small piston move to raise the large piston by 1 cm?

Accepted Answer

20 N/cm²; 25 cm

Answer

Pressure = Force/Area = 40 N ÷ 2 cm² = 20 N/cm². By conservation of volume, the displaced fluid volume must be equal: small piston area × small piston displacement = large piston area × large piston displacement → 2 × d = 50 × 1 → d = 25 cm. The mechanical advantage (50/2 = 25) means the small piston must travel 25× farther.

Question 3

A steel rod 4 meters long is fixed between two rigid walls at 20°C with no initial stress. The coefficient of linear thermal expansion for steel is 12 × 10⁻⁶ /°C and its modulus of elasticity is 200 GPa. If the temperature rises to 70°C, what is the magnitude of the compressive stress induced in the rod?

Accepted Answer

120 MPa

Answer

When a rod is fully constrained, the thermal strain (α × ΔT) is entirely converted to mechanical stress. Thermal strain = 12×10⁻⁶ × (70−20) = 12×10⁻⁶ × 50 = 6×10⁻⁴. Compressive stress = E × thermal strain = 200×10³ MPa × 6×10⁻⁴ = 120 MPa. The rod length is irrelevant since both walls are perfectly rigid.

Question 4

A first-class lever has its fulcrum positioned such that the effort arm is 3 times the length of the resistance arm. A 90 kg load is placed on the resistance side. Ignoring the lever's own weight, how much effort force is needed, and in which direction does the effort move relative to the load?

Accepted Answer

30 kgf, opposite direction to the load's movement

Answer

In a first-class lever (fulcrum between effort and load), the mechanical advantage = effort arm / resistance arm = 3, so effort = 90/3 = 30 kgf. Because the fulcrum is between the two forces, when the effort side goes down the load side goes up — they move in opposite directions. This distinguishes first-class from second- and third-class levers.

Question 5

Two V-belt pulleys are connected: a driver pulley with a 150 mm diameter running at 1,200 RPM, and a driven pulley. The required output speed is 300 RPM. Due to belt slip of 4%, what diameter must the driven pulley actually be to achieve the target 300 RPM output?

Accepted Answer

576 mm

Answer

Without slip, the speed ratio requires D_driven = D_driver × (N_driver/N_driven) = 150 × (1200/300) = 600 mm. However, belt slip means the driven pulley receives only 96% of the peripheral velocity. To compensate, the driven pulley must be smaller so the same lower effective speed still yields 300 RPM: D_driven = 600 × 0.96 = 576 mm. A smaller driven pulley at reduced belt velocity hits the target speed.

Ramsay Test Practice Test

Ramsay Test Practice Test

Ramsay Test Mechanical Aptitude 6