Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A wire can be broken by applying a load of $200{\text{N}}$. Find the force required to break another wire of the same length and same material, but double in diameter.A) $200{\text{N}}$B) $400{\text{N}}$C) $600{\text{N}}$D) $800{\text{N}}$

Last updated date: 05th Mar 2024
Total views: 21.6k
Views today: 0.21k
Verified
21.6k+ views
Hint: The force applied to break the first wire is proportional to the cross-sectional area of the wire and the fractional change in its length. The second wire is mentioned to have the same length and made out of the same material, but as its diameter becomes twice, the cross-sectional areas of the two wires become different.

Formula used:
The force applied to a wire is given by, $F = \dfrac{{YA\Delta l}}{l}$ where $Y$ is Young’s modulus of the material, $A$ is the cross-sectional area of the wire, $\Delta l$ is the change in length and $l$ is the original length of the wire.

Step 1: Express the relation for the applied force on a wire.
Let ${l_1} = {l_2} = l$ be the length of the two wires and $\Delta {l_1} = \Delta {l_2} = \Delta l$ be the change in their lengths.
Let $Y$ be Young’s modulus of the material of the wire.
Let ${d_1}$ and ${d_2} = 2{d_1}$ be the diameter of the first and second wires. Then their respective radii will be ${r_1} = \dfrac{{{d_1}}}{2}$ and ${r_2} = \dfrac{{{d_2}}}{2} = \dfrac{{2{d_1}}}{2} = {d_1}$ .
The cross-sectional areas of the first wire and second wire will be ${A_1} = \pi {r_1}^2$ and ${A_2} = \pi {r_2}^2$ respectively.
Now the force applied to the first wire to break it can be expressed as
${F_1} = \dfrac{{Y{A_1}\Delta l}}{l} = 200{\text{N}}$ ------- (1)
Similarly, the force applied to the second wire to break it is expressed as
${F_2} = \dfrac{{Y{A_2}\Delta l}}{l}$ ---------- (2)

Step 2: Using equations (1) and (2) obtain the force applied to the second wire to break it.
Since both wires have the same length and the same material from equations (1) and (2) we find that ${F_1} \propto {A_1}$ and ${F_2} \propto {A_2}$ .
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{A_1}}}{{{A_2}}}$
Now substituting for ${A_1} = \pi {r_1}^2$ and ${A_2} = \pi {r_2}^2$ in the above relation we get,
$\Rightarrow$ $\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{\pi {r_1}^2}}{{\pi {r_2}^2}} = \dfrac{{{r_1}^2}}{{{r_2}^2}}$ -------- (3)
Substituting for ${r_1} = \dfrac{{{d_1}}}{2}$ and ${r_2} = {d_1}$ in equation (3) we get, $\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{{\left( {\dfrac{{{d_1}}}{2}} \right)}^2}}}{{{d_1}^2}} = \dfrac{{{d_1}^2}}{{4{d_1}^2}} = \dfrac{1}{4}$
Substituting for ${F_1} = 200{\text{N}}$ in the above ratio we obtain, $\dfrac{{200}}{{{F_2}}} = \dfrac{1}{4} \Rightarrow {F_2} = 4 \times 200 = 800{\text{N}}$
$\therefore$ the force applied to break the second wire is obtained to be ${F_2} = 800{\text{N}}$ .

Hence the correct option is D.

Note: The wire is considered to have a cylindrical shape and so the cross-sectional area of the wire will be equal to the area of a circle with the same diameter. So we have the cross-sectional areas of the two wires as ${A_1} = \pi {r_1}^2$ and ${A_2} = \pi {r_2}^2$ . Young’s modulus is a material property which means that it has a unique value for a particular material. So when we say that both wires are made out of the same material we mean that Young’s modulus will be the same for both wires.