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 \(\sqrt{3\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}\), when they act at an angle \(\theta_{2}\), the magnitude of their resultant becomes \(\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}\). 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

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Vector

Statement-I: Two forces

(P + Q)

and

(P - Q)

where

P ⊥ Q

, when act at an angle

θ_{1}

to each other, the magnitude of their resultant is

3 (P^{2} + Q^{2})

, when they act at an angle

θ_{2}

, the magnitude of their resultant becomes

2 (P^{2} + Q^{2})

. This is possible only when

θ_{1} < θ_{2}

.
Statement-II: In the situation given above.

θ_{1} = 6 0^{\circ}

and

θ_{2} = 9 0^{\circ}

In the light of the above statements, choose the most appropriate answer from the options given below :-

Answer: (A)

Let

x = ∣ P + Q ∣ = P^{2} + Q^{2}

and

y = ∣ P - Q ∣ = P^{2} + Q^{2}

(since

P ⊥ Q

)

Resultant magnitude:

R = x^{2} + y^{2} + 2 x y cos θ

For

R_{1} = 3 (P^{2} + Q^{2})

3 (P^{2} + Q^{2}) = 2 (P^{2} + Q^{2}) + 2 (P^{2} + Q^{2}) cos θ_{1}

cos θ_{1} = \frac{1}{2}

, so

θ_{1} = 60°

For

R_{2} = 2 (P^{2} + Q^{2})

2 (P^{2} + Q^{2}) = 2 (P^{2} + Q^{2}) + 2 (P^{2} + Q^{2}) cos θ_{2}

cos θ_{2} = 0

, so

θ_{2} = 90°

Since

60° < 90°

, Statement-I is true. Statement-II correctly identifies the angles. Both statements are true.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option