A two metre long 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

The correct option is T₁ / T₂ = 0.5.

Let us break down the physical concepts and solve the problem step-by-step.
Recall the definition of tensile stress in a wire:
$\mathrm{Stress}=\frac{\mathrm{Force}}{\mathrm{Area}}=\frac{T}{A}$
where $T$ represents the tension (force) in the wire and $A$ represents its cross-sectional area.

Let the steel wire be wire 1 and the brass wire be wire 2.
According to the problem statement:
- Cross-sectional area of the steel wire ($A_1$) = 0.1 cm²
- Cross-sectional area of the brass wire ($A_2$) = 0.2 cm²
- The stress produced in both wires is the same.

Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire.
Equating the stress in both wires:
${\mathrm{Stress}}_1={\mathrm{Stress}}_2$
$\frac{T_1}{A_1}=\frac{T_2}{A_2}$

To find the ratio of tensions $\frac{T_1}{T_2}$, we rearrange the equation:
$\frac{T_1}{T_2}=\frac{A_1}{A_2}$

Substituting the given values of the cross-sectional areas:
$\frac{T_1}{T_2}=\frac{0.1{\mathrm{cm}}^2}{0.2{\mathrm{cm}}^2}=0.5$

Thus, the ratio of the tensions in the two wires is:
$T_1/T_2=0.5$