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Physics Question on elastic moduli

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

A

l1+l22\frac{l_{1}+l_{2}}{2}

B

l1l2\sqrt{l_{1}l_{2}}

C

l1T2−l2T1T2−T1\frac{l_{1}T_{2}-l_{2}T_{1}}{T_{2}-T_{1}}

D

l1T2+l2T1T2−T1\frac{l_{1}T_{2}+l_{2}T_{1}}{T_{2}-T_{1}}

Answer

l1T2−l2T1T2−T1\frac{l_{1}T_{2}-l_{2}T_{1}}{T_{2}-T_{1}}

Explanation

Solution

Let the natural length of the wire will be l
Young?s modulus =TA×l0Δl0=\frac{T}{A}\times\frac{l_{0}}{\Delta l_{0}}
First case:
y=T1A=l0(l1−l0)y=\frac{T_{1}}{A}=\frac{l_{0}}{\left(l_{1}-l_{0}\right)}
Second case:
Y=T2A=l0(l2−l0)Y=\frac{T_{2}}{A}=\frac{l_{0}}{\left(l_{2}-l_{0}\right)}
So, T1A×l0(l2−l0)=T2Al0(l2−l0)\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)}
⇒T1(l2−l0)=T2(l1−l0)\Rightarrow T_{1}\left(l_{2}-l_{0}\right)=T_{2}\left(l_{1}-l_{0}\right)
⇒T1l2−T1l0=T2l1−T2l0\Rightarrow T_{1} l_{2}-T_{1} l_{0}=T_{2}l_{1}-T_{2}l_{0}
⇒l0=(T2l1−T2l2)(T2−T1)\Rightarrow l_{0}=\frac{\left(T_{2}l_{1}-T_{2}l_{2}\right)}{\left(T_{2}-T_{1}\right)}