Question

Match List-I with List-II

\[ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; P \text{ and } Q \text{ are two perpendicular forces, acting at a point} & (I) \; R = |P - Q| \\ (B) \; P \text{ and } Q \text{ are equal, forces acting at a point at an angle } \alpha & (II) \; R = P + Q \\ (C) \; P \text{ and } Q \text{ are acting at a point in same direction} & (III) \; R = 2P \cos(\tfrac{\alpha}{2}) \\ (D) \; P \text{ and } Q \text{ are acting at a point in opposite direction} & (IV) \; R = \sqrt{P^2 + Q^2} \\ \hline \end{array} \]
Choose the correct answer from the options given below:

• (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (I), (B) - (IV), (C) - (III), (D) - (II)
• (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Hint:

Memorize the general formula \(R^2 = P^2 + Q^2 + 2PQ\cos\alpha\). All other cases (perpendicular, parallel, anti-parallel, equal forces) are just specializations of this one rule. This is much more efficient than memorizing four separate formulas.

Answer 1

The Correct Option is D

Step 1: Conceptualization: This problem requires applying the parallelogram law of forces to determine the magnitude of the resultant force (R) when two forces, P and Q, act at a point with an angle \(\alpha\) between them. Step 2: Governing Principle: The magnitude of the resultant force is calculated using the formula: \[ R = \sqrt{P^2 + Q^2 + 2PQ\cos\alpha} \] This formula will be applied to the specific scenarios presented in List-I. Step 3: Scenario Analysis:
(A) Perpendicular Forces (P and Q): Here, \(\alpha = 90^\circ\). As \(\cos(90^\circ) = 0\), the formula simplifies to: \[ R = \sqrt{P^2 + Q^2 + 2PQ(0)} = \sqrt{P^2 + Q^2} \] Therefore, (A) corresponds to (IV).
(B) Equal Forces (P=Q) at an Angle \(\alpha\): Substituting Q=P into the general formula yields: \[ R = \sqrt{P^2 + P^2 + 2P(P)\cos\alpha} = \sqrt{2P^2(1 + \cos\alpha)} \] Using the trigonometric identity \(1 + \cos\alpha = 2\cos^2(\alpha/2)\): \[ R = \sqrt{2P^2(2\cos^2(\alpha/2))} = \sqrt{4P^2\cos^2(\alpha/2)} = 2P\cos(\alpha/2) \] Thus, (B) corresponds to (III).
(C) Forces in the Same Direction: This implies \(\alpha = 0^\circ\). With \(\cos(0^\circ) = 1\), the formula becomes: \[ R = \sqrt{P^2 + Q^2 + 2PQ(1)} = \sqrt{(P+Q)^2} = P+Q \] Consequently, (C) corresponds to (II).
(D) Forces in Opposite Directions: This implies \(\alpha = 180^\circ\). Given \(\cos(180^\circ) = -1\), the formula simplifies to: \[ R = \sqrt{P^2 + Q^2 + 2PQ(-1)} = \sqrt{P^2 + Q^2 - 2PQ} = \sqrt{(P-Q)^2} = |P-Q| \] Hence, (D) corresponds to (I).
Step 4: Conclusion: The correct pairings are (A)-(IV), (B)-(III), (C)-(II), and (D)-(I), which aligns with option (D).