The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section,one of steel and another of brass,are suspended from the same roof. If we want the lower ends of the wires to be at the same level,then the weights added to the steel and brass wires must be in the ratio of

  • A
    $2:1$
  • B
    $1:2$
  • C
    $1:1$
  • D
    $4:1$

Explore More

Similar Questions

What is bending? How can bending problems be prevented,and what is buckling?

Difficult
View Solution

At $40\,^oC$,a brass wire of radius $0.5\, mm$ (diameter $1\, mm$) is hung from the ceiling. $A$ small mass $M$ is hung from the free end of the wire. When the wire is cooled down from $40\,^oC$ to $20\,^oC$,it regains its original length of $0.2\, m$. The value of $M$ is close to ........$kg$. (Coefficient of linear expansion $\alpha = 10^{-5}/^oC$ and Young's modulus $Y = 10^{11}\, N/m^2$; $g = 10\, ms^{-2}$)

The cross-sectional area of all wires is $10^{-4} \ m^2$. What is the displacement of point $D$?

Difficult
View Solution

Which material has a higher Young's modulus: copper or steel?

Wires $A$ and $B$ are connected with blocks $P$ and $Q$ as shown. The ratio of lengths,radii,and Young's modulus of wires $A$ and $B$ are $r, 2r$,and $3r$ respectively ($r$ is a constant). Find the mass of block $P$ if the ratio of the increase in their corresponding lengths is $1/(6r^2)$. The mass of block $Q$ is $3M$.

Difficult
View Solution

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