Question

A 5 × 20 cm seed drill has a ground drive wheel of rolling diameter 0.5 m. While testing under laboratory condition, 320 g of seeds were collected in 20 revolutions of the ground drive wheel. The same seed drill when operated in a 2 ha field, amount of seeds dropped was found to be 185 kg. The variation in the seed dropped between laboratory and field conditions due to skid of ground drive wheel is __________ (Take \( \pi = 3.14 \))

• A. 6.38%
• B. 9.23%
• C. 10.17%
• D. 12.26%

Hint:

The variation in the seed dropped due to the skid of the ground drive wheel can be found by comparing the seed per revolution in laboratory and field conditions.

Answer 1

The Correct Option is B

Step 1: Calculate the seed per revolution in laboratory conditions.
In the laboratory, 320 g of seeds were collected in 20 revolutions, so the amount of seed dropped per revolution is: \[ \frac{320 \, \text{g}}{20} = 16 \, \text{g/rev} \] Step 2: Calculate the total number of revolutions in the field.
The area of the field is 2 ha, which equals \( 2 \times 10^4 \, \text{m}^2 \). The circumference of the ground drive wheel is: \[ \text{Circumference} = \pi \times \text{diameter} = 3.14 \times 0.5 \, \text{m} = 1.57 \, \text{m} \] The number of revolutions in the field is: \[ \frac{2 \times 10^4}{1.57} = 12739.49 \, \text{revs} \] Step 3: Calculate the total seed dropped in the field.
In the field, the total seed dropped is 185 kg or \( 185 \times 1000 = 185000 \, \text{g} \). The seed dropped per revolution in the field is: \[ \frac{185000}{12739.49} = 14.5 \, \text{g/rev} \] Step 4: Calculate the variation.
The variation in the seed dropped is given by: \[ \text{Variation} = \frac{16 - 14.5}{16} \times 100 = 9.23% \] Final Answer: \[ \boxed{9.23%} \]