Question

A two metrelong rod is suspended with the help of two wires of equal length. One wire is of steel and its cross-sectional area is 0.1 cm² and another wire is of brass and its cross-sectional area is 0.2 cm². If a load W is suspended from the rod and stress produced in both the wires is same then the ratio of tensions in them will be

• A. Will depend on the position of W
• B. $T_{1}/T_{2} = 2$
• C. $T_{1}/T_{2} = 1$
• D. $T_{1}/T_{2} = 0.5$

Answer 1

Answer: (D)

$T_{1}/T_{2} = 0.5$

Answer 2

Stress = $\frac{\text{Tension}}{\text{Area of cross} ⥂ - ⥂ \text{section}} =$ constant

∴ $\frac{T_{1}}{A_{1}} = \frac{T_{2}}{A_{2}}$⇒ $\frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{0.1}{0.2} = \frac{1}{2} = 0.5$.