Question

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 $0.1c{m}^2$ and another wire is of brass and its cross-sectional area is $0.2c{m}^2$. 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. $\frac{T_1}{T_2}=2$
• C. $\frac{T_1}{T_2}=1$
• D. $\frac{T_1}{T_2}=0.5$

Answer 1

The correct option is D $\frac{T_1}{T_2}=0.5$

Sress = $\frac{\text{Tension}}{\text{Area of cross section}}=\text{constant}$
$\therefore \frac{T_1}{A_1}=\frac{T_2}{A_2}\Rightarrow \frac{T_1}{A_2}=\frac{A_1}{A_2}=\frac{0.1}{0.2}=\frac{1}{2}=\mathrm{0.5.}$