A tensile stress of 5.5 × 10⁶ Pa is applied to the ends of a round bar with a length of 115 cm - NSC Technical Sciences - Question 5 - 2024 - Paper 1

To find the most appropriate material, we calculate the Young's Modulus (K) using the formula: $K = \frac{\sigma}{\epsilon}$ Substituting the values gives: $K = \frac{5.5 \times 10^6 \text{ Pa}}{2.75 \times 10^{-3}} = 2 \times 10^{10} \text{ Pa}$ Since the calculated Young's Modulus is closest to 2 × 10¹⁰ Pa, the most appropriate material for the bar is wood.

The change in length (ΔL) can be calculated using: $ΔL = \frac{\sigma \cdot L}{K}$ Substituting the values, $ΔL = \frac{5.5 \times 10^6 \cdot 1.15}{2 \times 10^{10}} = 3.15 \times 10^{-4}\, \text{m} = 0.315\, \text{cm}$

Thrust is the force applied perpendicular to the surface of an object, often contributing to the motion or support of the object.

Pascal's law states that in a confined fluid at rest, a change in pressure applied at any point in the fluid is transmitted undiminished to all parts of the fluid.

The area of piston 2 can be calculated using the formula: $A_2 = \pi \left( \frac{D}{2} \right)^2$ Given that the diameter (D) of piston 2 is 24 cm: $A_2 = \pi \left( \frac{24 \times 10^{-2}}{2} \right)^2 \approx 0.0452\, \text{m}^2$

Using the relationship from Pascal's law, the magnitude of F₂ can be calculated as follows: $F_2 = \frac{F_1 \cdot A_2}{A_1}$ Given that F₁ = 5 N, and determining A₁ to be: $A_1 = \pi \left( \frac{12 \times 10^{-2}}{2} \right)^2 \approx 0.0113\, \text{m}^2$ Thus: $F_2 = 5 \cdot \frac{0.0452}{0.0113} \approx 19.99\, N$