Question:

The length of a metal wire is l When the tension in it is T and is l when the tension is T. The natural length of the wire is

Updated On: Jun 7, 2022
  • $\frac{l_{1}+l_{2}}{2}$
  • $\sqrt{l_{1}l_{2}}$
  • $\frac{l_{1}T_{2}-l_{2}T_{1}}{T_{2}-T_{1}}$
  • $\frac{l_{1}T_{2}+l_{2}T_{1}}{T_{2}-T_{1}}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Let the natural length of the wire will be l
Young?s modulus $=\frac{T}{A}\times\frac{l_{0}}{\Delta l_{0}}$
First case:
$y=\frac{T_{1}}{A}=\frac{l_{0}}{\left(l_{1}-l_{0}\right)}$
Second case:
$Y=\frac{T_{2}}{A}=\frac{l_{0}}{\left(l_{2}-l_{0}\right)}$
So, $\frac{T_{1}}{A}\times\frac{l_{0}}{\left(l_{2}-l_{0}\right)}=\frac{T_{2}}{A} \frac{l_{0}}{\left(l_{2}-l_{0}\right)}$
$\Rightarrow T_{1}\left(l_{2}-l_{0}\right)=T_{2}\left(l_{1}-l_{0}\right)$
$\Rightarrow T_{1} l_{2}-T_{1} l_{0}=T_{2}l_{1}-T_{2}l_{0}$
$\Rightarrow l_{0}=\frac{\left(T_{2}l_{1}-T_{2}l_{2}\right)}{\left(T_{2}-T_{1}\right)}$
Was this answer helpful?
0
0