A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively

(Y_AI = 70 X 10⁹ Nm^-2 and Y_steel = 200 X 10⁹ Nm^-2)

For the wires to have equal stress

Where T_A and T_B is the tension in wires A and B while A_A and A_B are the areas cross – section of the wires.

Let the mass be placed at a distance of x from wire A. As the system is in equilibrium, we can take moments about the point where the mass is hanging.

T_A (x) = T_B (l – x)

Thus, mass (m) should be suspended close to wire B. Option (b) is correct.

Thus, for same strain

Let the mass be placed at a distance of y units from wire A. Taking moment at the point at which mass (m) is attached

Thus, strain will be equal in both wires when mass m is near the wire A.