Question

A wire of mild steel as initial length $1 \cdot 5{\text{m}}$ and diameter $0 \cdot 66{\text{mm}}$ gets extended by $6 \cdot 3{\text{mm}}$ when a certain force is applied to it. If Young’s modulus of mild steel is $2 \cdot 1 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$, calculate the applied force.

Answer 1

Hint:Here when a force was applied to the given wire of mild steel, the length of the wire increased. Young’s modulus of the material of wire is found to be proportional to the applied force and original length of the wire and inversely proportional to its cross-sectional area and the change in its length. We can easily derive the formula for the applied force from the known formula for Young’s modulus of the material of the wire.

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

Complete step by step answer.
Step 1: List the parameters of the given wire.
The original length of the wire is given to be $l = 1 \cdot 5{\text{m}}$ .
The diameter of the wire is given to be $d = 0 \cdot 66{\text{mm}}$ and then its radius will be $r = \dfrac{d}{2} = \dfrac{{0 \cdot 66}}{2} = 0 \cdot 33{\text{mm}}$ and its cross-sectional area will be given by, $A = \pi {r^2}$ .
The extension or increase in the length of the wire is given to be $\Delta l = 6 \cdot 3{\text{mm}}$ .
Young’s modulus of mild steel is given to be $Y = 2 \cdot 1 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$ .
Let $F$ be the force applied to the wire which is to be obtained.
Step 2: Express the relation for Young’s modulus of the material of the wire.
Young’s modulus of the material of the given wire can be expressed as $Y = \dfrac{{Fl}}{{A\Delta l}}$ -------- (1).
$ \Rightarrow F = \dfrac{{YA\Delta l}}{l}$ or $F = \dfrac{{Y\left( {\pi {r^2}} \right)\Delta l}}{l}$ --------- (2)
Substituting for $l = 1 \cdot 5{\text{m}}$, $r = 0 \cdot 33 \times {10^{ - 3}}{\text{m}}$, $\Delta l = 6 \cdot 3 \times {10^{ - 3}}{\text{m}}$ and $Y = 2 \cdot 1 \times {10^{11}}{\text{N}}{{\text{m}}^{ - 2}}$ in equation (2) we get, $F = \dfrac{{2 \cdot 1 \times {{10}^{11}} \times \pi \times {{\left( {0 \cdot 3 \times {{10}^{ - 3}}} \right)}^2}6 \cdot 3 \times {{10}^{ - 3}}}}{{1 \cdot 5}} = 250{\text{N}}$
$\therefore $ the applied force is obtained to be $F = 250{\text{N}}$ .

Note:While substituting values in a given equation, make sure that all the quantities are expressed in their respective S.I units. If not, then the necessary conversion of the units must be done. Here the radius and the increase in the length of the wire were expressed in millimetres. So we convert these quantities into meters as $r = 0 \cdot 33 \times {10^{ - 3}}{\text{m}}$ and $\Delta l = 6 \cdot 3 \times {10^{ - 3}}{\text{m}}$ before substituting in equation (2). Also, the term $\dfrac{{\Delta l}}{l}$ in equation (1) refers to the stress of the wire and $\dfrac{F}{A}$ refers to the strain of the wire and so Young’s modulus is essentially the ratio of strain to stress i.e. $Y = \dfrac{{\left( {\dfrac{F}{A}} \right)}}{{\left( {\dfrac{{\Delta l}}{l}} \right)}} = \dfrac{{{\text{strain}}}}{{{\text{stress}}}}$.