Statement I:Two forces ( P + Q) and ( P - Q) where P Q, when act at an angle θ1 to each other, the magnitude of their resultant is 3 ( P 2+ Q 2), when they act at an angle θ2, the magnitude of their resultant becomes 2 ( P 2+ Q 2). This is possible only when θ1<θ2. Statement II: In the situation given above. θ1=60° and θ2=90° In the light of the above statements, choose the most appropriate answer from the options given below

(A) Statement-I is false but Statement-II is true

(C) Statement-I is true but Statement-II is false

(D) Both Statement-I and Statement-II are false.

JEE Mains · Physics · STD 11 - Vectors

Statement \(I :\)Two forces \((\overrightarrow{{P}}+\overrightarrow{{Q}})\) and \((\overrightarrow{{P}}-\overrightarrow{{Q}})\) where \(\overrightarrow{{P}} \perp \overrightarrow{{Q}}\), when act at an angle \(\theta_{1}\) to each other, the magnitude of their resultant is \(\sqrt{3\left({P}^{2}+{Q}^{2}\right)}\), when they act at an angle \(\theta_{2}\), the magnitude of their resultant becomes \(\sqrt{2\left({P}^{2}+{Q}^{2}\right)}\). This is possible only when \(\theta_{1}<\theta_{2}\). Statement \(II :\) In the situation given above. \(\theta_{1}=60^{\circ} \text { and } \theta_{2}=90^{\circ}\) In the light of the above statements, choose the most appropriate answer from the options given below

Pick an option to check your answer.

Verified Solution

Correct Answer

(B) Both Statement\(-I\) and Statement\(-II\) are true

Detailed explanation

\(\overrightarrow{{A}}=\overrightarrow{{P}}+\overrightarrow{{Q}}\)

\(\overrightarrow{{B}}=\overrightarrow{{P}}-\overrightarrow{{Q}} \quad \overrightarrow{{P}} \perp \overrightarrow{{Q}}\)

\(|\overrightarrow{{A}}|=|\overrightarrow{{B}}|=\sqrt{{P}^{2}+{Q}^{2}}\)

\(|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{2\left({P}^{2}+{Q}^{2}\right)(1+\cos \theta)}\)

\(\text { For }|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{3\left({P}^{2}+{Q}^{2}\right)}\)

\(\theta_{1}=60^{\circ}\)

For \(|\vec{A}+\vec{B}|=\sqrt{2\left(P^{2}+Q^{2}\right)}\)

\(\theta_{2}=90^{\circ}\)

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