Question

A spring with natural length$l_0$ has a tension $T_1$ when its length is $l_1$ and the tension is $T_2$ when its length is$l_2$. The natural length of spring will be:

• A. $\frac{T_1{l}_2-T_2{l}_1}{l_1-l_2}$
• B. $\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$
• C. $\frac{T_2{l}_2-T_1{l}_1}{T_1-T_2}$
• D. $\frac{T_2{l}_1-T_1{l}_2}{T_2+T_1}$

Answer 1

The correct option is B

$\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$

Step 1: Given data:

Natural length$=l_0$

Tension $=T_1$ when its length is $l_1$

Tension $=T_2$ when its length is$l_2$

Step 2: Find the natural length of spring

Using hook’s law,

$Y=\frac{TL}{A∆L}$

Where,

$∆L=L–L_0$ (Change in length)

$Y=$Young's modulus

$T=$Tension

$A=$Area

$L=$Original length

For Tension $=T_1$ when its length is $l_1$

$L_1–L_0=\frac{T_1{L}_0}{AY}\dots \left(1\right)$

For Tension $=T_2$ when its length is$l_2$

$L_2–L_0=\frac{T_2{L}_0}{AY}\dots \left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$, we get

$L_0=\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$

Thus, The natural length of spring will be ${\mathit{L}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{T}}_{\mathbf{2}}{\mathbf{l}}_{\mathbf{1}}\mathbf{-}{\mathbf{T}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}\mathbf{-}{\mathbf{T}}_{\mathbf{1}}}$

Hence, the option $\text{'}B\text{'}$ is correct.