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MHT CET 2025 (26 Apr, Shift 2) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from MHT CET 2025 (26 Apr, Shift 2)

Q5MHT CET 2025 · 26 APR · SHIFT 2Medium

Equilibrium, Third Law & Frames

A mass ' M ' is suspended by a rope from a rigid support at point ' P ' as shown in figure. Another rope is tied at end ' Q ' and pulled horizontally with a force ' FF '. If the rope makes an angle ' θ\theta ' with vertical then the tension in the string ' PQ ' is

Figure

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Answer: (b)

  1. At Q the rope PQ pulls along QP, at θ\theta to the vertical.
  2. Horizontal balance: Tsin⁡θ=FT\sin\theta = F (and vertically Tcos⁡θ=MgT\cos\theta = Mg).
  3. T=Fsin⁡θT = \frac{F}{\sin\theta}
  4. So the answer is (b) Fsin⁡θ\frac{F}{\sin\theta}.
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