Question

A light rod of length $2\text{m}$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to it ends. One of the wire is made of steel and it's cross section is ${10}^{-3}{\text{m}}^2$ and the other is of brass of cross section $2\times {10}^{-3}{\text{m}}^2$. The position along the rod at (from the end where brass wire is tied) which a weight may be hung to produce equal stress on both wires $(Y_{steel}=2\times {10}^{11}{\text{Nm}}^{-2};Y_{brass}={10}^{11}{\text{Nm}}^{-2})$

• A. $1\text{m}$
• B. $\frac{4}{3}\text{m}$
• C. $\frac{2}{3}\text{m}$
• D. $\frac{1}{3}\text{m}$

Answer 1

The correct option is C $\frac{2}{3}\text{m}$

Let the weight is hung at distance $x$ from point $\text{B}$ as shown in figure.


Moment of force about the weight is
$T_1(2-x)=T_{2}x.....\left(1\right)$

According to problem the stresses induced are same in both the wires.
$\frac{T_1}{A_1}=\frac{T_2}{A_2}$

Where,
$A_1$ is cross-sectional area of steel, $A_1={10}^{-3}{\text{m}}^2$
And $A_2$ is cross-sectional area of brass,
$A_2=2\times {10}^{-3}{\text{m}}^2$

Substituting the values,
$\frac{T_1}{{10}^{-3}}=\frac{T_2}{2\times {10}^{-3}}$

$\Rightarrow T_2=2T_1.....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$T_1(2-x)=2T_{1}x$

$\Rightarrow 2=3x$

$\therefore x=\frac{2}{3}\text{m}$, from the brass wire.

Hence, option (c) is correct answer.