Question

Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of the same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a:b, then \[\dfrac{a}{b}\]can be expressed as \[\dfrac{1}{x}\], then find the value of x.

Answer 1

Hint:Before we start addressing the problem, we need to know about the stress and strain. Stress of a material is defined as the ratio of force applied on the material per unit area. Strain is the ratio of change in length of the material to its original length.

Formula Used:
The formula for young’s modulus is given by,
\[Y = \dfrac{\text{Stress}}{\text{Strain}}\]
Where,
Stress is force per unit area.
Strain is changed in length of material to the original length.

Complete step by step solution:
Consider that, we have two wires A and B. If the wire stretches this wire, then the length will become \[\Delta {l_1} = 2mm\]and \[\Delta {l_2} = 4mm\]. The material used in both wires is the same. Then the force of 2N is applied and the radius of wire B is four times of radius of A that is, \[{r_A} = {r_A}\]and \[{r_B} = 4{r_A}\]. If the length of the wires A and B are in the ratio of a:b, then \[\dfrac{a}{b}\]can be expressed as \[\dfrac{1}{x}\]. We need to find the value of x.

We know that from the formula,
\[Y = \dfrac{\text{Stress}}{\text{Strain}}\]
Stress is the ratio of force per unit area that is, \[\dfrac{F}{A}\]and strain is the ratio of change in length to the original length that is, \[\dfrac{{\Delta l}}{l}\]so, substitute these values in above equation we get,
\[Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}}\]
Now, rearrange the equation for l, we get,
\[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{\left( {\dfrac{{Y{A_1}\Delta {l_1}}}{F}} \right)}}{{\left( {\dfrac{{Y{A_2}\Delta {l_2}}}{F}} \right)}}\]

Since we have two wires, the area and length will be different but, both the wires are made up of the same material, the young’s modulus will be the same and cancel out each other.Then, above equation can be written as,
\[\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{\pi {r_A}^2\Delta {l_1}}}{{\pi {r_B}^2\Delta {l_2}}}\]
\[\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{r_A}^2 \times 2}}{{{{\left( {4{r_B}} \right)}^2} \times 4}}\]
\[\Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{r_A}^2}}{{16{r_A}^2 \times 2}}\]
\[ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{{16 \times 2}}\]
\[ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{{32}}\]
The length of the wires is in the ratio a:b that is \[\dfrac{a}{b} = \dfrac{1}{x}\]
\[ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{32}}\]
\[ \therefore x = 32\]

Therefore, the value of x is 32.

Note:Remember that, since we have two separate wires taken and the radius of both the wires are different, we need to take the ratios of length, area etc.