JEE Main 2025 Apr 2 Shift 2, Physics Q24: Properties of Solids

Treat the string as an elastic body obeying Hooke's law in terms of its extension from natural length ${\ell }_0$:

$T=K(\ell -{\ell }_0)$.

The two data points give:

$5=K(1.4-{\ell }_0)$ and $7=K(1.56-{\ell }_0)$.

Eliminate $K$ by dividing:

$\frac{5}{1.4-{\ell }_0}=\frac{7}{1.56-{\ell }_0}$.

Cross-multiplying: $5(1.56-{\ell }_0)=7(1.4-{\ell }_0)\Rightarrow 7.8-5{\ell }_0=9.8-7{\ell }_0\Rightarrow 2{\ell }_0=2\Rightarrow {\ell }_0=1\text{m}$.