Question

Two wires A and B are of the same material. Their lengths are in the ratio 1:2 and diameters are in ratio 2:1. When stretched by force F^A and F^B respectively, they get an equal increase in lengths. Find the ratio of F^A / F^B.

Answer 1

Step 1: Given

Length ratio $\frac{l_1}{l_2}=\frac{1}{2}$

Here, length of first wire is $l_1$ , length of the second wire is $l_2$ .

Diameter ratio $\frac{d_1}{d_2}=\frac{2}{1}$

Here, diameter of the first wire is $d_1$ and diameter of the second wire is $d_2$.

F^A is the force applied on A.

F^B is the force applied on B.

Step 2: Calculating the change in length

Formula for change in length is as follows from the Hook's law.

$d{l}_1=\frac{F^A{l}_1}{A_1{E}_1}$

Here, change in length is $d{l}_1$ , area of the first wire is $A_1$, Young's modulus of first wire is $E_1$.

$d{l}_2=\frac{F^B{l}_2}{A_2{E}_2}$.

Here, change in length is $d{l}_2$ , area of the second wire is $A_2$, Young's modulus of second wire is $E_2$.

Step 3: Calculating the ratio of forces.

Divide $d{l}_1$ to $d{l}_2$.

$\frac{d{l}_1}{d{l}_2}=\frac{{\displaystyle \frac{F^A{l}_1}{A_1{E}_1}}}{{\displaystyle \frac{F^B{l}_2}{A_2{E}_2}}}$

Since change in length and material is same.

$d{l}_1=d{l}_2$ and $E_1=E_2=E$

$$\begin{gathered} \frac{d{l}_1}{d{l}_1}=\frac{F^A{l}_1}{A_1{E}_1}\times \frac{A_2{E}_1}{F^B{l}_2} \\ 1=\frac{F^A}{F^B}\times \frac{l_1}{l_2}\times \frac{{\displaystyle \frac{{\mathrm{\pi d}}_2^2}{4}}}{{\displaystyle \frac{{\mathrm{\pi d}}_1^2}{4}}} \\ 1=\frac{F^A}{F^B}\times \frac{l_1}{l_2}\times \frac{{\displaystyle d_2^2}}{{\displaystyle d_1^2}} \end{gathered}$$

Here, diameter of the first wire is $d_1$, diameter of the second wire is $d_2$.

$$\begin{gathered} 1=\frac{F^A}{F^B}\times \frac{1}{2}\times \frac{1}{2^2} \\ \frac{F^A}{F^B}=2\times 4 \\ \frac{F^A}{F^B}=8 \end{gathered}$$

Hence, the required ratio of the force is $8:1$.