Two wires A and B made of different materials of length 6.0 cm and 5.4 cm , respectively and area of cross sections 3.0 × 10 -5 m 2 and 4.5 × 10 -5 m 2 , respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x : 3 . The value of x is \ \ \ \ 。

Physics · Mechanical Properties of Matter

JEE Main 2026 — 28 January, Morning Shift — Question 34

Two wires $A$ and $B$ made of different materials of length 6.0 cm and 5.4 cm , respectively and area of cross sections $3.0 \times 10^{-5} \mathrm{~m}^{2}$ and $4.5 \times 10^{-5} \mathrm{~m}^{2}$ , respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of $A$ to that of $B$ is $\mathrm{x}: 3$ . The value of x is $\_\_\_\_$ 。

Option A:
1
Option B:
4
Option C:
2
Option D:
5
Correct

Answer: D

Step-by-step solution

$\mathrm{T}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell} \Rightarrow \mathrm{Y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$ $\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{B}}}=\frac{\ell_{\mathrm{A}}}{\ell_{\mathrm{B}}}\left(\frac{\mathrm{A}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{A}}}\right)$ $=\frac{6}{5.4}\left(\frac{4.5 \times 10^{-5}}{3 \times 10^{-5}}\right)=\frac{9}{5.4}=\frac{5}{3} \Rightarrow \frac{\mathrm{x}}{3}=\frac{5}{3}$ $\mathrm{x}=5$

Answer key and solution verified before publishing.

Practise Mechanical Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Stress,Strain and Modulus of Elasticity