In this experiment, you will use vernier calipers to measure the external diameter of five cylindrical metal rods labelled P, Q, R, S and T. Each rod is mad...

(a) All five diameter readings are recorded in the table. [1]

(b) The cross-sectional area of each rod is calculated using the formula \( A = \frac{\pi d^2}{4} \): [2]

Working for Rod P:

\[ A = \frac{\pi \times 12.4^2}{4} = \frac{\pi \times 153.76}{4} = \frac{483.1}{4} = 121 \text{ mm}^2 \]

Working for Rod Q: \( A = \frac{\pi \times 8.6^2}{4} = \frac{\pi \times 73.96}{4} = 58.1 \) mm²

Working for Rod R: \( A = \frac{\pi \times 15.2^2}{4} = \frac{\pi \times 231.04}{4} = 181 \) mm²

Working for Rod S: \( A = \frac{\pi \times 6.0^2}{4} = \frac{\pi \times 36}{4} = 28.3 \) mm²

Working for Rod T: \( A = \frac{\pi \times 10.8^2}{4} = \frac{\pi \times 116.64}{4} = 91.6 \) mm²

(c) Hold the vernier calipers so the jaws close gently and perpendicularly around the rod, at right angles to the rod's axis. Do not over-tighten the jaws, as this could deform a soft rod or damage the calipers. The rod should fit snugly between the jaws without any rocking. [1]

(d) The thinnest rod is S with \( d = 6.0 \) mm. The thickest rod is R with \( d = 15.2 \) mm. The difference is:

\[ 15.2 - 6.0 = 9.2 \text{ mm} \]

[1]

(e) The graph of Area / mm² (y-axis) against \( d \) / mm (x-axis) is shown below. The axes are labelled with quantities and units, all five data points are plotted correctly, and a smooth curve is drawn through them: [3]

(f) The graph is a curve with an upward (parabolic) shape, not a straight line. [1] This is because the area depends on the square of the diameter (\( A = \frac{\pi d^2}{4} \)). When one variable is proportional to the square of another, the graph is a parabola. If \( A \) were plotted against \( d^2 \) instead, the result would be a straight line through the origin with gradient \( \frac{\pi}{4} \). [1]