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JEE Main 2019 (10 Jan, Shift 2) · Addition & Subtraction of Vectors

Addition & Subtraction of Vectors question from JEE Main 2019 (10 Jan, Shift 2)

Q30JEE Main 2019 · 10 JAN · SHIFT 2Medium

Addition & Subtraction of Vectors

Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle θ\theta with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ\theta is -

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Answer: (d)

  1. R2=P2+Q2+2PQcos⁡θR^2 = P^2 + Q^2 + 2PQ\cos\theta with P=2FP = 2F, Q=3FQ = 3F: R2=F2(13+12cos⁡θ)R^2 = F^2(13 + 12\cos\theta).
  2. With Q→6FQ \to 6F: R′2=F2(40+24cos⁡θ)R'^2 = F^2(40 + 24\cos\theta), and R′=2RR' = 2R means R′2=4R2R'^2 = 4R^2.
  3. 40+24cos⁡θ=52+48cos⁡θ⇒cos⁡θ=−1240 + 24\cos\theta = 52 + 48\cos\theta \Rightarrow \cos\theta = -\frac{1}{2}
  4. So the answer is (d) 120∘120^\circ.
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