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

Using the formula for Young's modulus, we can calculate it as follows:

$K = \frac{\sigma}{\epsilon}$

Where:

• $\sigma = 5.5 \times 10^6 \, \text{Pa}$
• $\epsilon = 2.75 \times 10^{-3}$

Calculating this gives:

$K = \frac{5.5 \times 10^6}{2.75 \times 10^{-3}} = 2 \times 10^{10} \, \text{Pa}$

Thus, the most appropriate material for the bar is wood, which has a Young's modulus of $2 \times 10^{10} \, \text{Pa}$.

The change in length can be calculated using:

$\Delta L = \frac{\sigma L}{K}$

Where:

• $\sigma = 5.5 \times 10^6 \, \text{Pa}$
• $L = 1.15 \, \text{m}$ (length of the bar)
• $K = 2 \times 10^{10} \, \text{Pa}$

Calculating gives:

$\Delta L = \frac{5.5 \times 10^6 \times 1.15}{2 \times 10^{10}} = 3.16 \times 10^{-5} \, \text{m}$

Thrust is defined as the force exerted by a fluid on a surface that is in contact with it.

Pascal's law states that in a continuous liquid at rest, the pressure applied at any point is transmitted equally in all directions throughout the liquid.

The diameter of piston 1 is 12 cm, thus the diameter of piston 2 is 24 cm. The area of piston 2 can be calculated using:

$A_2 = \pi \left( \frac{D}{2} \right)^2$

Substituting the diameter value gives:

$A_2 = \pi \left( \frac{24 \times 10^{-2}}{2} \right)^2 = \pi \left( 12 \times 10^{-2} \right)^2 \approx 0.452 \, \text{m}^2$

Using the principle that force is proportional to area:

$F_2 = F_1 \times \frac{A_2}{A_1}$

Where:

• $F_1 = 5 \, \text{N}$
• $A_1 = \pi \left( \frac{12 \times 10^{-2}}{2} \right)^2$

Calculating gives:

$F_2 = 5 \times \frac{0.452}{0.113} = 19.99 \, \text{N}$