Question

A light rod of length $2m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section $0.1$ ${cm}^2$ and the other of brass of cross-section $0.2{cm}^2$. Along the rod at what distance a weight may be hung to produce equal stresses in both the wires.
$(Y_{steel}=2\times {10}^{11}N{m}^{-2},Y_{brass}=1\times {10}^{11}N{m}^{-2})$

• A. $\frac{4}{3}m$ from steel wire
• B. $\frac{4}{3}m$ from brass wire
• C. $1m$ from steel wire
• D. $\frac{1}{4}m$ from brass wire

Answer 1

The correct option is A $\frac{4}{3}m$ from steel wire

The situation is as shown in the figure
Let a weight $W$ be suspended at a distance $x$ from steel wire. Let $T_S$ and $T_B$ be tensions in the steel and brass wires respectively.
$\therefore$ Stress in steel wire $=\frac{T_S}{A_S}$
Stress in brass wire $=\frac{T_B}{A_B}$
for equal stress in both the wires
$\frac{T_S}{A_S}=\frac{T_B}{A_B}$
$\frac{T_S}{T_B}=\frac{A_S}{A_B}=\frac{0.1{cm}^2}{0.2{cm}^2}=\frac{1}{2}...\left(i\right)$

for the rotational equilibrium of the rod,

$T_{S}x=T_B(2-x)$
$\frac{2-x}{x}=\frac{T_S}{T_B}=\frac{1}{2}$ [Using (i)]
$4-2x=x$ or $3x=4$ or $x=\frac{4}{3}m$