A 5 m long aluminum wire ( Y = 7 x 10¹⁰ N / m²) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire ( Y = 12 x 10¹⁰ N / m² ) of the same length under the same weight, the diameter should now be, in mm

The correct option is 2.3.

Step-by-Step Explanation:

Let us analyze the relationship between the elongation of a wire, its physical properties, and the applied force using Hooke's Law and the definition of Young's Modulus ($Y$).

Young's Modulus is defined as:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$

Here:
- $F$ is the applied force (weight of the supported mass, $F=mg$).
- $A$ is the cross-sectional area of the wire, given by $A=\frac{\pi }{4}{d}^2$ where $d$ is the diameter.
- $L$ is the original length of the wire.
- $\Delta L$ is the elongation of the wire.

Rearranging the formula for elongation ($\Delta L$), we get:
$\Delta L=\frac{FL}{AY}=\frac{4FL}{\pi d^{2}Y}$

According to the problem statement, we have two wires: one made of aluminum (Al) and one made of copper (Cu). The following conditions are identical for both wires:
1. Same length: $L_{\text{Al}}=L_{\text{Cu}}=L$
2. Same weight (applied force): $F_{\text{Al}}=F_{\text{Cu}}=F$
3. Same elongation: $\Delta L_{\text{Al}}=\Delta L_{\text{Cu}}$

Equating the expressions for elongation for both wires:
$\frac{4FL}{\pi d_{\text{Al}}^2{Y}_{\text{Al}}}=\frac{4FL}{\pi d_{\text{Cu}}^2{Y}_{\text{Cu}}}$

Since $F$, $L$, $4$, and $\pi$ are common on both sides, they cancel out, leaving:
$d_{\text{Al}}^2{Y}_{\text{Al}}=d_{\text{Cu}}^2{Y}_{\text{Cu}}$

Now, let us solve for the diameter of the copper wire ($d_{\text{Cu}}$):
$d_{\text{Cu}}^2=d_{\text{Al}}^2\cdot \frac{Y_{\text{Al}}}{Y_{\text{Cu}}}$

Taking the square root on both sides:
$d_{\text{Cu}}=d_{\text{Al}}\cdot \sqrt{\frac{Y_{\text{Al}}}{Y_{\text{Cu}}}}$

Given data:
- Young's modulus of aluminum, $Y_{\text{Al}}=7\times {10}^{10}{\text{N/m}}^2$
- Young's modulus of copper, $Y_{\text{Cu}}=12\times {10}^{10}{\text{N/m}}^2$
- Diameter of aluminum wire, $d_{\text{Al}}=3\text{mm}$

Substitute these values into the equation:
$d_{\text{Cu}}=3\cdot \sqrt{\frac{7\times {10}^{10}}{12\times {10}^{10}}}$

$d_{\text{Cu}}=3\cdot \sqrt{\frac{7}{12}}$

$d_{\text{Cu}}\approx 3\cdot \sqrt{0.5833}$

$d_{\text{Cu}}\approx 3\cdot 0.7638\approx 2.29\text{mm}$

Rounding to one decimal place, we obtain:
$d_{\text{Cu}}\approx 2.3\text{mm}$