The figure represents the extension $(\Delta l)$ of a wire of length $1\text{ m}$, suspended from the ceiling of a room at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-5}\text{ m}^2$, then the Young's modulus of the wire is . . . . . . $\text{N/m}^2$.

  • A
    $1.0 \times 10^{11}$
  • B
    $2.0 \times 10^{10}$
  • C
    $1.0 \times 10^{10}$
  • D
    $2.0 \times 10^{11}$

Explore More

Similar Questions

The length of a metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and the tension in the wire is $T_2$ for length $l_2$. Find the original length $l$.

Let $L$ be the length and $d$ be the diameter of the cross-section of a wire. Wires of the same material with different $L$ and $d$ are subjected to the same tension along the length of the wire. In which of the following cases will the extension of the wire be the maximum?

Two wires $A$ and $B$ are stretched by the same force. If,for $A$ and $B$,$Y_A: Y_B = 1: 2$,$r_A: r_B = 3: 1$,and $L_A: L_B = 4: 1$,then the ratio of their extension $\left(\frac{\Delta L_A}{\Delta L_B}\right)$ will be .............

$A$ wire is stretched $1 \ mm$ by a force $F$. If a second wire of same material,same length and $4$ times the diameter of the first wire is stretched by the same force $F$,then the elongation of the second wire is

The speed of a transverse wave passing through a string of length $50 \; cm$ and mass $10 \; g$ is $60 \; ms^{-1}$. The area of cross-section of the wire is $2.0 \; mm^2$ and its Young's modulus is $1.2 \times 10^{11} \; Nm^{-2}$. The extension of the wire over its natural length due to its tension will be $x \times 10^{-5} \; m$. The value of $x$ is $...$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo