AB is an iron wire and CD is a copper wire of same length and same cross-section. BD is a rod of length 0.8 m. A load G = 2kg-wt is suspended from the rod. At what distance x from point B should the load be suspended for the rod to remain in a horizontal position ( Y꜀ᵤ = 11.8 x 10¹⁰ N / m², Yբₑ = 19.6 x 10¹⁰ N/ m² )

The correct answer is 0.3 m.

Step-by-step Explanation:
Let us analyze the system consisting of the two wires and the horizontal rod:
- The iron wire $\mathrm{AB}$ is connected to the rod at point $B$. Let the tension in this wire be $T_1$.
- The copper wire $\mathrm{CD}$ is connected to the rod at point $D$. Let the tension in this wire be $T_2$.
- The length of the rod $\mathrm{BD}$ is $L=0.8\text{m}$.
- A load $G=2\text{kg-wt}$ is suspended at a distance $x$ from point $B$ (so its distance from point $D$ is $0.8-x$).

1. Condition for the rod to remain horizontal:
For the rod to remain horizontal, the elongation (extension) in both the iron and copper wires must be equal:
$\Delta L_{\mathrm{Fe}}=\Delta L_{\mathrm{Cu}}$

We know the relation for Young's modulus:
$Y=\frac{T·l}{A·\Delta L}\Rightarrow \Delta L=\frac{T·l}{A·Y}$

Since both wires have the same initial length ($l$) and same area of cross-section ($A$), the condition for equal elongation simplifies to:
$\frac{T_1}{Y_{\mathrm{Fe}}}=\frac{T_2}{Y_{\mathrm{Cu}}}$

Rearranging the terms, we get:
$\frac{T_1}{T_2}=\frac{Y_{\mathrm{Fe}}}{Y_{\mathrm{Cu}}}$

Substituting the given values of Young's moduli ($Y_{\mathrm{Fe}}=19.6\times {10}^{10}{\text{N/m}}^2$ and $Y_{\mathrm{Cu}}=11.8\times {10}^{10}{\text{N/m}}^2$):
$\frac{T_1}{T_2}=\frac{19.6\times {10}^{10}}{11.8\times {10}^{10}}=\frac{19.6}{11.8}\approx 1.66$

2. Rotational Equilibrium of the Rod:
For the rod to remain horizontal, it must be in rotational equilibrium. Taking torque about the point where the load $G$ is suspended:
$T_1·x=T_2·(0.8-x)$

This gives:
$\frac{T_1}{T_2}=\frac{0.8-x}{x}$

3. Solving for x:
Equating the two expressions for the ratio of tensions:
$\frac{0.8-x}{x}=1.66$

$0.8-x=1.66x$
$0.8=2.66x$
$x=\frac{0.8}{2.66}\approx 0.3\text{m}$

Thus, the load must be suspended at a distance of 0.3 m from point B.