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 $0.2{cm}^2$. At which distance a weight may be hung along the rod, in order to produce equal strains 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

Answer: (C)

$1m$ from steel wire

As, $Strain=\frac{Stress}{Youn{g}^{\prime }sModulus}$

$\therefore$ Strain in steel wire $=\frac{T_S}{A_S{Y}_S}$

Strain in brass wire $=\frac{T_B}{A_B{Y}_B}$

For equal strain in both the wires,

$\frac{T_S}{A_S{Y}_S}=\frac{T_B}{A_B{Y}_B}$

$\therefore \frac{T_S}{T_B}=\frac{A_S}{A_B}\frac{Y_S}{Y_B}=\frac{0.1{cm}^2}{0.2{cm}^2}\times \frac{(2\times {10}^{11}N{m}^{-2})}{1\times {10}^{11}N{m}^{-2}}=\mathrm{1...}\left(i\right)$

For the rotational equilibrium of the rod,

$T_{B}x=T_B(2-x)$

$\frac{2-x}{x}=\frac{T_S}{T_B}=1$ [Using (i)]

$2-x-x$ or $x=1m$