COMEDK · Physics · 10. Mechanical Properties of Solids
A light rod of length 1 m is suspended from ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of material X and is of cross-section \(0.1\ \text{cm}^2\) and the other of material Y of cross-section \(0.3\ \text{cm}^2\). A weight is hung from the wire at a point to produce equal strain in the wires. The ratio of Young's moduli of wires A to B is 3:1. The location of the point from one end of the wire is
- A \(0.2\ \text{m}\)
- B \(0.25\ \text{m}\)
- C \(0.5\ \text{m}\)
- D \(0.75\ \text{m}\)
Answer & Solution
Correct Answer
(C) \(0.5\ \text{m}\)
Step-by-step Solution
Detailed explanation
Let the length of the rod be \(L = 1\ \text{m}\). Let the weight be suspended at a distance \(x\) from the wire of material X. For rotational equilibrium of the rod, the torques about the point of suspension must balance: \(T_X x = T_Y (1 - x)\) The strain in a wire is given by…
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