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JEE Main 2019 (9 Jan, Shift 1) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2019 (9 Jan, Shift 1)

Q35JEE Main 2019 · 9 JAN · SHIFT 1Medium

Torque, Couple & Equilibrium of Rigid Bodies

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is θ\theta, then:

Figure

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Answer: (d)

  1. In equilibrium the centre of mass lies vertically below A. Let AB=BC=lAB = BC = l, each of mass mm.
  2. With A as origin, AB points along (−sin⁡θ,−cos⁡θ)(-\sin\theta, -\cos\theta) and BC (perpendicular) along (cos⁡θ,−sin⁡θ)(\cos\theta, -\sin\theta).
  3. Horizontal positions: centre of AB at −l2sin⁡θ-\frac{l}{2}\sin\theta; centre of BC at −lsin⁡θ+l2cos⁡θ-l\sin\theta + \frac{l}{2}\cos\theta.
  4. Net horizontal CM =0= 0: −12sin⁡θ−sin⁡θ+12cos⁡θ=0⇒3sin⁡θ=cos⁡θ-\frac{1}{2}\sin\theta - \sin\theta + \frac{1}{2}\cos\theta = 0 \Rightarrow 3\sin\theta = \cos\theta
  5. So the answer is (d) tan⁡θ=13\tan\theta = \frac{1}{3}.
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