Step 1: Find the cross-sectional area of the trapezoidal furrow cut by the plough, using the trapezoid area formula \(\text{Area} = \dfrac{(a+b)}{2} \times h\), where \(a\) = top width, \(b\) = bottom width and \(h\) = depth.
\[\text{Area} = \frac{(16+4)}{2} \times 15 = \frac{20}{2} \times 15 = 10 \times 15 = 150\ \text{cm}^2\]
Step 2: Multiply the cross-sectional area by the average soil resistance (specific draft) to get the horizontal draft force resisting the plough's forward motion.
\[\text{Draft} = 0.71 \times 150 = 106.5\ \text{kgf}\]
Step 3: This 106.5 kgf is the horizontal component of the resisting force, since soil resistance to cutting acts along the direction of forward travel. But the bullocks do not pull horizontally, they pull along the line of draft, which makes a 45° angle with the horizontal (because of the yoke height and plough beam geometry). The horizontal component of the bullocks' pull must balance the 106.5 kgf draft, so if \(P\) is the actual pull along the 45° line, then \(P\cos 45^\circ = 106.5\).
Step 4: Solve for \(P\):
\[P = \frac{106.5}{\cos 45^\circ} = \frac{106.5}{0.7071} \approx 150.6\ \text{kgf}\]
Step 5: Rounding, the pull exerted by the bullocks is about 150 kgf. Option 1 (107) is simply the horizontal draft before correcting for the 45° angle, option 3 (211) would come from multiplying instead of dividing by cos 45°, and option 4 (300) is far too high and does not match either calculation. The value that correctly accounts for the trapezoidal area, soil resistance, and the pull angle is 150 kgf.