Question

The length of a metal wire is $l_1$ when the tension in it is $T_1$ and is $I_2$ when the tension is $T_2$. The natural length of the wire is?

• A. $\frac{l_1+l_2}{2}$
• B. $\sqrt{l_1/I_2}$
• C. $\frac{I_1{T}_2-I_2{T}_1}{T_2-T_1}$
• D. $\frac{I_1{T}_2+l_2{T}_1}{T_1+T_2}$

Answer 1

Answer: (C)

$\frac{I_1{T}_2-I_2{T}_1}{T_2-T_1}$

Let $l$ = original length of material – wire.

$A$ = original length of metal – wire.

Change in length in first case = $l_1-l$

change in length in second case = $l_2-l$

Young Modulus= $\frac{\text{Normal Stress}}{\text{Longitudinal strain}}$

$Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$

here $Y=$ Young Modulus

$T=$ Tension

$A=$ Area

$\Delta l$= Change in Length

$l$ Original length

In First Case

$Y=\frac{T_1}{A}\ast \frac{l}{l_1-l}$

In Second case :

$Y=\frac{T_2}{A}\ast \frac{l}{l_2-l}$

Since Young Modulus remains the same

So,

$\frac{T_1}{A}\ast \frac{l}{l_1-l}=\frac{T_2}{A}\ast \frac{l}{l_2-l}$

On solving we get

$l=\frac{I_1{T}_2-I_2{T}_1}{T_2-T_1}$