$A$ uniform rod of length $L$ has a mass per unit length $\lambda$ and area of cross-section $A$. If the Young's modulus of the rod is $Y$,then the elongation in the rod due to its own weight is:

  • A
    $\frac{2 \lambda g L^2}{A Y}$
  • B
    $\frac{\lambda g L^2}{2 A Y}$
  • C
    $\frac{\lambda g L^2}{4 A Y}$
  • D
    $\frac{\lambda g L^2}{A Y}$

Explore More

Similar Questions

The work done in stretching an elastic wire per unit volume is

$A$ wire of length $25 \ cm$ and radius $2 \ mm$ is fixed at one end. If a torque is applied at the other end to produce an angular displacement of $45^o$,calculate the work done in $J$. (Given: $\eta = 8 \times 10^{10} \ N/m^2$)

Which of the following is true for elastic potential energy density?

$A$ uniform heavy rod of weight $10 \, N$,cross-sectional area $100 \, \text{cm}^2$ and length $20 \, \text{cm}$ is hanging from a fixed support. The Young's modulus of the material of the rod is $2 \times 10^{11} \, \text{N/m}^2$. Neglecting the lateral contraction,find the elongation of the rod due to its own weight. (In $\times 10^{-10} \, \text{m}$)

If the tension on a wire is removed at once,then

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