Question

A steel rope has length $L$, area of cross-section $A$, Young's modulus $Y$. [$Density=d$]. If the steel rope is vertical and moving with the force acting vertically up at the upper end, find the strain at a point $\frac{L}{3}$ from lower end.

• A. $\left(dgL\right)/2Y$
• B. $\left(dgL\right)/4Y$
• C. $\left(dgL\right)/6Y$
• D. $\left(dgL\right)/8Y$

Answer 1

The correct option is C $\left(dgL\right)/6Y$

Since steel rope has mass therefore, tension in it will not remain constant
and it will proportional to the length of it from the end with which it is tied up.
$T\alpha l$

Therefore tension at $l=L/3$ will be .

$T_{l=L/3}=T/3$

$T_{l=L/3}=\left[dALg\right]/6$

We know that,

$stress=F/A=T/A$

$stress=\left[dALg\right]/6A$

$strain=stress/Y$

$strain=\left[dALg\right]/6AY$

Therefore, option(C) is correct