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JEE Main 2002
Exam
JEE Main
Year
2002
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2002 · Physics · Mathematics in Physics

Two forces are such that the sum of their magnitudes is 18 ~N and...

Two forces are such that the sum of their magnitudes is 18 N18 \mathrm{~N} and their resultant is 12 N12 \mathrm{~N} which is perpendicular to the smaller force. Then the magnitudes of the forces are

Options

  1. A12 N,6 N12 \mathrm{~N}, 6 \mathrm{~N}
  2. B13 N,5 N13 \mathrm{~N}, 5 \mathrm{~N}
  3. C10 N,8 N10 \mathrm{~N}, 8 \mathrm{~N}
  4. D16 N,2 N16 \mathrm{~N}, 2 \mathrm{~N}

Correct answer

B. 13 N,5 N13 \mathrm{~N}, 5 \mathrm{~N}

Step-by-step solution

Use tan⁡α=Psin⁡θQ+Pcos⁡θ⇒tan⁡90∘=Psin⁡θQ+Pcos⁡θ=∞\tan \alpha=\frac{P \sin \theta}{Q+P \cos \theta} \Rightarrow \tan 90^{\circ}=\frac{P \sin \theta}{Q+P \cos \theta}=\infty ∴Q+Pcos⁡θ=0⇒Pcos⁡θ=−Q\therefore Q+P \cos \theta=0 \Rightarrow P \cos \theta=-Q R=P2+Q2+2PQcos⁡θR=P2+Q2−2Q2\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta} \mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{Q}^2} or R=P2−Q2=12\mathrm{R}=\sqrt{\mathrm{P}^2-\mathrm{Q}^2}=12 144=(P+Q)(P−Q)144=(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q}) or P−Q=144/18=8∴P=13 N\mathrm{P}-\mathrm{Q}=144 / 18=8 \quad \therefore \mathrm{P}=13 \mathrm{~N} and Q=5 N\mathrm{Q}=5 \mathrm{~N}
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