Question

A uniform rod of Young's modulus $Y$ is stretched by two tension forces$T_1$ and$T_2$ such that the rods get expanded to length $L_1$ and $L_2$ respectively. Find the initial length of the rod.

• A. $\frac{L_1{T}_1-L_2{T}_2}{T_1-T_2}$
• B. $\frac{L_2{T}_1-L_1{T}_2}{T_2-T_1}$
• C. $\frac{L_1{T}_2-L_2{T}_2}{T_2-T_1}$
• D. $\frac{L_1}{T_1}\times \frac{T_2}{L_2}$

Answer 1

The correct option is C

$\frac{L_1{T}_2-L_2{T}_2}{T_2-T_1}$

Step 1. Hooke's Law

Let the initial length be $L_0$ and area be $A$

$length=I,Changeinlength=∆I$

Using Hooke's law, $\frac{F}{A}=Y\frac{∆l}{l}$

Here tension is acting force on the rod,

$\Rightarrow \frac{T}{A}=Y\frac{∆l}{l}$

Step 2. Solving for initial length:

In the first case, tension is $T_1$ and corresponding length of the rod is $L_1$,

$\Rightarrow \frac{T_1}{A}=Y\frac{L_1-L_0}{L_0}$ $\left(1\right)$

In the second case, tension is $T_2$ and corresponding length of the rod is $L_2$,

$\Rightarrow \frac{T_2}{A}=Y\frac{L_2-L_0}{L_0}$ $\left(2\right)$

Dividing equation $\left(1\right)$ with equation $\left(2\right)$, we get

$\Rightarrow \frac{T_1}{T_2}=\frac{L_1-L_0}{L_2-L_0}$

$\Rightarrow T_1\left(L_2-L_0\right)=T_2\left(L_1-L_0\right)$

$\Rightarrow L_0\left(T_2-T_1\right)=L_1{T}_2-L_2{T}_1$

$\Rightarrow L_0=\frac{L_1{T}_2-L_2{T}_1}{\left(T_2-T_1\right)}$

Hence, option C is correct