Physics Circle home Student Login

Home›Chapters›Vectors PYQs›Addition & Subtraction of Vectors›Question 21

JEE Main 2021 (31 Aug, Shift 2) · Addition & Subtraction of Vectors

Addition & Subtraction of Vectors question from JEE Main 2021 (31 Aug, Shift 2)

Q21JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Addition & Subtraction of Vectors

Statement I:

Two forces (P⃗+Q⃗)\left(\vec{P} + \vec{Q}\right) and (P⃗−Q⃗)\left(\vec{P} - \vec{Q}\right) where P⃗⊥Q⃗\vec{P} \perp \vec{Q}, when act at an angle θ\theta1 to each other, the magnitude of their resultant is 3(P2+Q2)\sqrt{3(P^2 + Q^2)}, when they act at an angle θ\theta2, the magnitude of their resultant becomes 2(P2+Q2)\sqrt{2(P^2 + Q^2)}. This is possible only when θ1<θ2\theta_1 < \theta_2.

Statement II:

In the situation given above.

θ\theta1 = 60∘{}^\circ and θ\theta2 = 90∘{}^\circ

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

Show answer and solution

Answer: (b)

  1. Since P⃗⊥Q⃗\vec{P} \perp \vec{Q}: ∣P⃗+Q⃗∣=∣P⃗−Q⃗∣=P2+Q2=S|\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2} = S.
  2. Two equal forces SS at angle θ\theta: R2=2S2(1+cos⁡θ)R^2 = 2S^2(1 + \cos\theta).
  3. R2=3S2⇒cos⁡θ1=12, θ1=60∘R^2 = 3S^2 \Rightarrow \cos\theta_1 = \frac{1}{2},\ \theta_1 = 60^\circ; R2=2S2⇒cos⁡θ2=0, θ2=90∘R^2 = 2S^2 \Rightarrow \cos\theta_2 = 0,\ \theta_2 = 90^\circ.
  4. So θ1<θ2\theta_1 < \theta_2 and both statements are true: answer (b).
Report a mistake

More questions on Addition & Subtraction of Vectors