The fuel consumption rate $F$ (in $\text{L/h}$) of an agricultural tractor is directly proportional to the square root of the engine power output $P$ (in $\text{kW}$) and inversely proportional to the specific engine speed $N$ (in hundreds of $\text{rpm}$): $F = k \frac{\sqrt{P}}{N}$. Which of the following statements is/are TRUE?
Options:
(A) If power $P$ is quadrupled ($4P$) while speed $N$ is kept constant, fuel consumption rate $F$ doubles.
(B) If power $P$ is quadrupled ($4P$) and speed $N$ is doubled ($2N$), fuel consumption rate $F$ remains unchanged.
(C) If speed $N$ is halved while power $P$ remains constant, fuel consumption rate $F$ is halved.
(D) To increase $F$ by $50\%$ at constant speed $N$, power $P$ must be increased by $125\%$.
Answer Key:
A, B, D
Step-by-Step Solution & Explanation
**Method 1: Proportional Scaling Analysis**
Given $F = k \frac{\sqrt{P}}{N}$:
- (A) $F_{\text{new}} = k \frac{\sqrt{4P}}{N} = 2k \frac{\sqrt{P}}{N} = 2F$. True.
- (B) $F_{\text{new}} = k \frac{\sqrt{4P}}{2N} = \frac{2\sqrt{P}}{2N} k = F$. True.
- (C) If $N \to N/2$, $F_{\text{new}} = k \frac{\sqrt{P}}{N/2} = 2F$ (doubles, not halved). False.
- (D) If $F \to 1.5F$ at constant $N$, then $\sqrt{P_{\text{new}}} = 1.5\sqrt{P} \implies P_{\text{new}} = (1.5)^2 P = 2.25P$. The percentage increase in power is $(2.25 - 1) \times 100 = 125\%$. True.
Therefore, options **A**, **B**, and **D** are correct.