The length of a light string is 1.4 m when the tension on it is 5 N. If the tension increases to 7 N, the length of the string is 1.56 m. The original length of the string is _________ m.

The extension of an ideal string that obeys Hooke’s law is directly proportional to the tension applied: $$\Delta L \propto T$$.
For a given string, the ratio $$\dfrac{\Delta L}{T}$$ remains constant.

Let the natural (unstretched) length of the string be $$L_0$$.

Tension $$T_1 = 5 \text{ N}$$ produces total length $$L_1 = 1.40 \text{ m}$$.
Extension, $$\Delta L_1 = L_1 - L_0 = 1.40 - L_0$$

Tension $$T_2 = 7 \text{ N}$$ produces total length $$L_2 = 1.56 \text{ m}$$.
Extension, $$\Delta L_2 = L_2 - L_0 = 1.56 - L_0$$

Using $$\dfrac{\Delta L_1}{T_1} = \dfrac{\Delta L_2}{T_2}$$:

$$\frac{1.40 - L_0}{5} = \frac{1.56 - L_0}{7}$$

Cross-multiplying gives:
$$7(1.40 - L_0) = 5(1.56 - L_0)$$

Simplify each side:
$$9.8 - 7L_0 = 7.8 - 5L_0$$

Rearrange terms:
$$9.8 - 7.8 = 7L_0 - 5L_0$$
$$2.0 = 2L_0$$

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

Therefore, the original (unstretched) length of the string is 1 m.