Question

In rotary tiller, the total energy requirement for carrying out tillage will decrease if

• A. the bite length is increased
• B. the bite length is decreased
• C. the cone index of soil is higher
• D. forward speed of the machine is reduced

Hint:

In rotary tillage, bite length \(b=\dfrac{V_f}{N z}\). To ease cutting and reduce energy, make \(b\) smaller (e.g., increase rotor speed or decrease travel speed moderately), keeping agronomic tilth acceptable.

Answer 1

The Correct Option is B

Step 1: Define bite length and power/energy.
For a rotary tiller, the bite length (\(b\)) is the distance along the direction of travel between successive cuts made by the same blade path: \[ b \;=\; \frac{V_f}{N\,z} \] where \(V_f\) = forward speed (m/s), \(N\) = rotor speed (rev/s), and \(z\) = number of blade paths (effective blades per revolution per helical track). The average power required is \(P \approx T\omega\) (torque \(\times\) angular speed) or, equivalently from the soil side, \[ P \;\propto\; (\text{soil resistance})\times(\text{volume rate of soil cut}). \] The total energy to cover a given field area \(A\) is \(E=\int P\,dt\).

Step 2: Effect of bite length on soil volume cut and draft.
Smaller \(b\) (shorter bite) means each blade engages a thinner slice of soil per entry, reducing instantaneous soil-chip thickness and the cutting force per blade (draft). For the same tilth depth and width, reducing \(b\) distributes the total work over more, lighter cuts: \[ F_{\text{per blade}} \downarrow \Rightarrow T \downarrow \Rightarrow P \downarrow \Rightarrow E \downarrow. \] Hence decreasing \(b\) reduces the energy requirement.

Step 3: Why other choices do not reduce total energy.
(A) Increase \(b\) \(\Rightarrow\) thicker chip per cut \(\Rightarrow\) higher cutting force and torque \(\Rightarrow\) \(E\uparrow\). ✗
(C) Higher cone index (harder soil) directly raises soil resistance \(\Rightarrow\) \(P\) and \(E\) increase. ✗
(D) Reduce forward speed \(V_f\): Instantaneous power \(P\) may drop (lower mass rate), but operating time increases for the same area. The field energy \(E=\text{(draft)}\times\text{distance}\) remains roughly unchanged; often fuel/energy per area does not improve materially. ✗

Final Answer:
\[ \boxed{\text{B — the bite length is decreased}} \]