Q 11-08-001NEETNEET 2024Top questionEasy
The maximum elongation of a steel wire of $1$ m length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8\ \text{N m}^{-2}$ and $2 \times 10^{11}\ \text{N m}^{-2}$, is :
Answer: (C) $4$ mm
Concept. The largest stress the wire can take without permanent deformation is the elastic limit; Hooke's law then gives the maximum strain, $\text{strain} = \dfrac{\text{stress}}{Y}$.
Step 1. Maximum strain.
$$\frac{\Delta l}{l} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$$
Step 2. Maximum elongation.
$$\Delta l = 4 \times 10^{-3} \times 1\ \text{m} = 4 \times 10^{-3}\ \text{m} = 4\ \text{mm}$$
Answer. $4$ mm (Option C).
Solution by Sreeraj P, M.Sc Physics