Q58 / 75 · JEE Main 2026 (06 April Shift 2) · Physics
Figure represents the extension ($\Delta l$) of a wire of length $1$ meter, suspended from the ceiling of the room at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5}$ m$^2$ then the Young's modulus of the wire is __________ N/m$^2$.
- A$1.0 \times 10^{11}$
- B$2.0 \times 10^{10}$
- Correct answerC$1.0 \times 10^{10}$
- D$2.0 \times 10^{11}$
Solution
From the given graph, we can observe that for a load $W = 20$ N, the extension is $\Delta l = 2 \times 10^{-4}$ m.
The formula for Young's modulus $Y$ is:
$Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{W / A}{\Delta l / L} = \dfrac{W L}{A \Delta l}$
Given:
Length of the wire, $L = 1$ m
Cross-sectional area, $A = 10^{-5}$ m$^2$
Substituting the values into the formula:
$Y = \dfrac{20 \times 1}{10^{-5} \times 2 \times 10^{-4}}$
$Y = \dfrac{20}{2 \times 10^{-9}}$
$Y = 10 \times 10^9 = 1.0 \times 10^{10}$ N/m$^2$
Answer: $1.0 \times 10^{10}$
The formula for Young's modulus $Y$ is:
$Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{W / A}{\Delta l / L} = \dfrac{W L}{A \Delta l}$
Given:
Length of the wire, $L = 1$ m
Cross-sectional area, $A = 10^{-5}$ m$^2$
Substituting the values into the formula:
$Y = \dfrac{20 \times 1}{10^{-5} \times 2 \times 10^{-4}}$
$Y = \dfrac{20}{2 \times 10^{-9}}$
$Y = 10 \times 10^9 = 1.0 \times 10^{10}$ N/m$^2$
Answer: $1.0 \times 10^{10}$
Hence, the correct answer is option: 'C'.