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NEET PHYSICSLAWS OF MOTIONMedium

Question

A body of weight 2 kg2 \text{ kg} is suspended as shown in the figure. The tension T1T_1 in the horizontal string (in kg wt\text{kg wt}) is:

A

23\frac{2}{\sqrt{3}}

B

32\frac{\sqrt{3}}{2}

C

232\sqrt{3}

D

22

Step-by-Step Solution

  1. Equilibrium Principle: For the knot connecting the strings to remain in equilibrium, the vector sum of all forces acting on it must be zero [NCERT Class 11, Physics Part I, Section 5.8].
  2. Free Body Diagram Analysis:
  • Downward force: Weight W=2 kg wtW = 2 \text{ kg wt}.
  • Horizontal force: Tension T1T_1.
  • Inclined force: Tension T2T_2 in the string attached to the support.
  1. Resolution of Forces: Let the inclined string make an angle θ\theta with the vertical. Resolving T2T_2 into rectangular components:
  • The vertical component balances the weight: T2cos⁡θ=W=2T_2 \cos \theta = W = 2.
  • The horizontal component balances the tension T1T_1: T2sin⁡θ=T1T_2 \sin \theta = T_1.
  1. Calculation: Dividing the horizontal equation by the vertical equation eliminates T2T_2: T12=T2sin⁡θT2cos⁡θ=tan⁡θ\frac{T_1}{2} = \frac{T_2 \sin \theta}{T_2 \cos \theta} = \tan \theta T1=2tan⁡θT_1 = 2 \tan \theta
  2. Deduction: The probable answer 232\sqrt{3} implies that tan⁡θ=3\tan \theta = \sqrt{3}, which corresponds to θ=60∘\theta = 60^\circ (angle with the vertical). Assuming this standard configuration from the missing figure, the calculated tension is 23 kg wt2\sqrt{3} \text{ kg wt}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONweightsuspendedfiguretensionhorizontal

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