Question

A 2-bottom, 50 cm MB plough is being operated at a speed of 5 km/h. If the time lost in turning is 8 per cent, how many hours will be required to plough 23 ha of land?

• A. 23
• B. 46
• C. 50
• D. 100

Hint:

Work out the effective field capacity in ha per hour first, factoring in the plough width, speed and turning loss, then divide the area by it.

Answer 1

The Correct Option is C

Step 1: Find the total working width of the plough. It has 2 bottoms, each cutting 50 cm, so the width is \(2 \times 0.5 = 1\) m.

Step 2: Compute the theoretical field capacity using \(\text{TFC (ha/h)} = \dfrac{\text{width (m)} \times \text{speed (km/h)}}{10}\). Substituting, \(\text{TFC} = \dfrac{1 \times 5}{10} = 0.5\) ha/h.

Step 3: Apply the field efficiency. Only time lost in turning is given, which is 8%, so field efficiency is \(100\% - 8\% = 92\% = 0.92\).

Step 4: Find the effective field capacity: \(0.5 \times 0.92 = 0.46\) ha/h. This is the actual area the plough covers in one hour of operation, including the turning losses.

Step 5: Divide the total area to be ploughed by the effective field capacity to get the time required: \(\dfrac{23}{0.46} = 50\) hours.

Step 6: Option 1 (23) results from skipping the field efficiency correction entirely along with a width error, option 2 (46) comes from ignoring the field efficiency correction, and option 4 (100) comes from using half the correct width. The value that correctly accounts for width, speed and the 8% turning loss is 50 hours.