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NEET PHYSICSMedium

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}.
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