Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

There are two wires of the same material. Their radii and lengths are both in the ratio $1:2$ . If the extensions produced are equal, what is the ratio of the loads?
A. $1:2$
B. $2:1$
C. $1:4$
D. $4:1$

Answer
VerifiedVerified
554.7k+ views
Hint: There are two wires of the same material, therefore, the young’s modulus of the wires will be the same. Also, the extensions of the wire will be the same. Here, we will use the value of young’s modulus to calculate the loads on the wire.

Complete step by step answer:
Consider a wire of the same material. Now, the radii and lengths of the wire are in the ratio $1:2$ .Now, if ${r_1}$ and ${r_2}$ are the radii of both the wires. Therefore, the ratio of the radii of the wires is given by,
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{1}{2}$
And, if ${l_1}$ and ${l_2}$ are the lengths of both the wires. Therefore, the ratio of the lengths of both the wires is given by,
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{2}$
Now, as the material of the wire is the same, therefore, the extension produced in the wires are given by
$\Delta {l_1} = \Delta {l_2}$
Now, as the material of both the wires is the same. Therefore, Young’s modulus of both the wires are the same and is shown below
${Y_1} = {Y_2}$
Now, if ${A_1}$ and ${A_2}$ are the areas of both the wires, therefore, the areas of both the wires are given by
${A_1} = \pi r_1^2$ and ${A_2} = \pi r_2^2$
Therefore, their ratio is given by
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi r_1^2}}{{\pi r_2^2}} = \dfrac{{r_1^2}}{{r_2^2}}$
$ \Rightarrow \,\dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}$
$ \Rightarrow \,\dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$
Now, let ${F_1}$ and ${F_2}$ are the loads on the wire, therefore, their ratio is given by
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{Y_1}{A_1}\Delta {l_1}{l_2}}}{{{Y_2}{A_2}\Delta {l_2}{l_1}}}$
$ \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{4} \times \dfrac{2}{1}$
$ \therefore \,\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{2}$
Therefore, the ratio of loads of the wire is $1:2$ .

Hence, option A is the correct option.

Note:Young’s Modulus (also referred to as the Elastic Modulus or Tensile Modulus), is a measure of mechanical properties of linear elastic solids like rods, wires, and such. There are other numbers that give us a measure of elastic properties of a material, like Bulk modulus and shear modulus, but the value of Young’s Modulus is most commonly used. This is because it gives us information about the tensile elasticity of a material (ability to deform along an axis).