Question

Statement I : Two forces \((\vec{p} + \vec{q})\) and \((\vec{p} - \vec{q})\) where \(\vec{p} \perp \vec{q}\), when act at an angle \(\theta_1\) to each other, the magnitude of their resultant is \(\sqrt{3(p^2 + q^2)}\), when they act at an angle \(\theta_2\), the magnitude of their resultant becomes \(\sqrt{2(p^2 + q^2)}\). This is possible only when \(\theta_1<\theta_2\).
Statement II : In the situation given above, \(\theta_1 = 60^{\circ}\) and \(\theta_2 = 90^{\circ}\).
In the light of the above statements, choose the most appropriate answer from the options given below :

• A. Both Statement I and Statement II are true.
• B. Both Statement I and Statement II are false.
• C. Statement I is true but Statement II is false.
• D. Statement I is false but Statement II is true.

Hint:

If two vectors have equal magnitude \(A\), their resultant is \(2A \cos(\theta/2)\). In this problem, both vectors have magnitude \(\sqrt{p^2+q^2}\) because they are perpendicular diagonals of a rectangle formed by \(p\) and \(q\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The magnitude of the resultant \(R\) of two vectors \(\vec{A}\) and \(\vec{B}\) acting at an angle \(\theta\) is given by \(R = \sqrt{A^2 + B^2 + 2AB \cos \theta}\).
Step 2: Key Formula or Approach:
Let \(\vec{A} = \vec{p} + \vec{q}\) and \(\vec{B} = \vec{p} - \vec{q}\).
Since \(\vec{p} \perp \vec{q}\), then \(\vec{p} \cdot \vec{q} = 0\).
The squares of the magnitudes are:
\[ A^2 = |\vec{p} + \vec{q}|^2 = p^2 + q^2 + 2\vec{p} \cdot \vec{q} = p^2 + q^2 \]
\[ B^2 = |\vec{p} - \vec{q}|^2 = p^2 + q^2 - 2\vec{p} \cdot \vec{q} = p^2 + q^2 \]
Note that \(A = B = \sqrt{p^2 + q^2}\). Let this be \(R_0\).
Step 3: Detailed Explanation:
The resultant \(R\) is:
\[ R^2 = A^2 + B^2 + 2AB \cos \theta = R_0^2 + R_0^2 + 2R_0^2 \cos \theta = 2R_0^2(1 + \cos \theta) \]
Case 1: Resultant \(R_1 = \sqrt{3(p^2 + q^2) = \sqrt{3} R_0\)}
\[ (\sqrt{3} R_0)^2 = 2R_0^2(1 + \cos \theta_1) \]
\[ 3 R_0^2 = 2R_0^2(1 + \cos \theta_1) \implies 3 = 2 + 2\cos \theta_1 \]
\[ 2\cos \theta_1 = 1 \implies \cos \theta_1 = \frac{1}{2} \implies \theta_1 = 60^{\circ} \]
Case 2: Resultant \(R_2 = \sqrt{2(p^2 + q^2) = \sqrt{2} R_0\)}
\[ (\sqrt{2} R_0)^2 = 2R_0^2(1 + \cos \theta_2) \]
\[ 2 R_0^2 = 2R_0^2(1 + \cos \theta_2) \implies 1 = 1 + \cos \theta_2 \]
\[ \cos \theta_2 = 0 \implies \theta_2 = 90^{\circ} \]
Since \(60^{\circ}<90^{\circ}\), Statement I (\(\theta_1<\theta_2\)) is true.
Statement II (\(\theta_1 = 60^{\circ}, \theta_2 = 90^{\circ}\)) is also true.
Step 4: Final Answer:
Both Statement I and Statement II are true.