A metal wire of length L, area of cross-section A and Young’s modulus Y behaves as a spring. The equivalent spring constant will be

The correct option is YA/L.

To find the equivalent spring constant of a metal wire, we can relate the formula for the elongation of a wire under load to Hooke's Law for a spring.

Let's denote the following parameters for the metal wire:
- Length of the wire = $L$
- Area of cross-section = $A$
- Young's modulus = $Y$
- Applied force (or load) = $F$
- Elongation (change in length) = $\Delta L$

By definition, Young's modulus ($Y$) is the ratio of tensile stress to tensile strain:

$Y=\frac{\text{Stress}}{\text{Strain}}$

Here, tensile stress is given by the force per unit area:

$\text{Stress}=\frac{F}{A}$

And tensile strain is given by the fractional change in length:

$\text{Strain}=\frac{\Delta L}{L}$

Substituting these expressions into the formula for Young's modulus, we get:

$Y=\frac{F/A}{\Delta L/L}=\frac{F·L}{A·\Delta L}$

Now, we can rearrange this equation to express the force $F$ in terms of the elongation $\Delta L$:

$F=\left(\frac{YA}{L}\right)\Delta L$

According to Hooke's Law for a spring, the restoring force $F$ is directly proportional to the displacement (or elongation) $x$:

$F=kx$

where $k$ is the spring constant and $x=\Delta L$. Comparing the two equations:

$F=\left(\frac{YA}{L}\right)\Delta L$ and $F=k\Delta L$

We find that the equivalent spring constant ($k$) of the wire is:

$k=\frac{YA}{L}$