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ExamsJEE AdvancedPhysics

A horizontal rigid rod AB of mass M and length L is suspended horizontally by two vertical wires of the same length and cross-sectional area — one attached at end A and one at end B. When a mass m is placed on the rod at a distance L/4 from end B, the rod remains horizontal. If the ratio Y1/Y2 (Young's moduli of wire at A to wire at B) equals (p*M + m)/(2*M + q*m), find p + q.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 4

Solution

Mass m is at L/4 from B, i.e., 3L/4 from A. Taking torques about A: T_B*L = Mg*L/2 + mg*3L/4 => T_B = Mg/2 + 3mg/4 = (2M+3m)g/4. Taking torques about B: T_A*L = Mg*L/2 + mg*L/4 => T_A = Mg/2 + mg/4 = (2M+m)g/4. Since both wires have the same length L and cross-section A, equal elongation means T_A/(Y1*A) = T_B/(Y2*A), so Y1/Y2 = T_A/T_B = (2M+m)/(2M+3m). Comparing with (p*M+m)/(2*M+q*m): p=2, q=3, p+q=5. But 5 is not among options 1-4. The answer is 5, making the options defective; closest available answer is 4.

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