1. Use Hooke’s law for extension of the string

For a light elastic string, the extension is proportional to the tension:

$T = \lambda \frac{x}{L_0}$

where:

• $T$ = tension,
• $\lambda$ = modulus of elasticity,
• $x$ = extension,
• $L_0$ = original length.

So the total length under tension $T$ is:

$L = L_0 + x = L_0 + \frac{T L_0}{\lambda} = L_0\left(1 + \frac{T}{\lambda}\right)$

2. Form equations from the given data

When tension is $5\,\text{N}$, length is $1.4\,\text{m}$:

$1.4 = L_0\left(1 + \frac{5}{\lambda}\right)$

When tension is $7\,\text{N}$, length is $1.56\,\text{m}$:

$1.56 = L_0\left(1 + \frac{7}{\lambda}\right)$

3. Eliminate $\lambda$ by dividing the equations

$\frac{1.56}{1.4} = \frac{1 + 7/\lambda}{1 + 5/\lambda}$

But an easier method is to subtract the two equations.

Subtracting:

$1.56 - 1.4 = L_0\left(\frac{7}{\lambda} - \frac{5}{\lambda}\right)$

$0.16 = L_0\cdot \frac{2}{\lambda}$

$\frac{L_0}{\lambda} = 0.08$

4. Substitute back into one equation

Using:

$1.4 = L_0 + 5\left(\frac{L_0}{\lambda}\right)$

$1.4 = L_0 + 5(0.08)$

$1.4 = L_0 + 0.4$

$L_0 = 1.0\,\text{m}$

5. Final answer

The original length of the string is:

$\boxed{1\,\text{m}}$