Question

A force of $10^3 N$ stretches the length of a hanging wire by 1 millimeter. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimeter is:
A) ${{4}} \times {{1}}{{{0}}^3}{{ N}}$
B) $16 \times {{1}}{{{0}}^3}{{ N}}$
C) $\dfrac{1}{4} \times {{1}}{{{0}}^3}{{ N}}$
D) $\dfrac{1}{{16}} \times {{1}}{{{0}}^3}{{ N}}$

Answer 1

In this question, we will use Hooke's law, according to the Hooke's law “stress$\left( \sigma \right)$ is directly proportional to the strain, and the Stress is the ratio of the force applied and cross-sectional area. Strain $\left( \varepsilon \right)$ produced by applying force P can be defined as the ratio of the change in length to the original length.

Complete step by step answer:
We will write the expression for stress as follows:
$\sigma = \dfrac{P}{A}$…… (1)
Here, $P$ is the force applied, and $A$ is the cross-sectional area of the wire.
We will write the expression for strain caused by force P as follows:
$\Rightarrow \varepsilon = \dfrac{{\Delta L}}{L}$…… (2)
Here, $\Delta L$ is the change in length of the wire and $L$ is the length of the wire.
As we know the relationship between stress and strain by Hooke's law as follows:
$\Rightarrow \sigma = \varepsilon E$…… (3)
Here, $E$ is the young modulus of the wire’s material.
From equations (1) and (3), we will get

\dfrac{P}{A} = \varepsilon E \\\ \varepsilon = \dfrac{P}{{AE}} \\\ $$…… (4) From equations (2) and (4), we will get

\Rightarrow \dfrac{P}{{AE}} = \dfrac{{\Delta L}}{L} \\
\Rightarrow \Delta L = \dfrac{{PL}}{{AE}} \\
\Rightarrow \Delta L = \dfrac{{PL}}{{\left( {\dfrac{{\pi {d^2}}}{4}} \right)E}} \\
\Rightarrow \Delta L = \dfrac{{4PL}}{{\pi {d^2}E}} \\

$$CONDITION 1:$$

{d_1} = d \\
P = {P_1} \\

From equation (5), we will get $${\left( {\Delta L} \right)_1} = \dfrac{{4{P_1}L}}{{\pi {d^2}E}}$$…… (6) CONDITION 2:

{d_1} = 4d \\
P = {P_2} \\

$$From equation (5), we get$$

\Rightarrow {\left( {\Delta L} \right)_2} = \dfrac{{4{P_2}L}}{{\pi {{\left( {4d} \right)}^2}E}} \\
\Rightarrow {\left( {\Delta L} \right)_2} = \dfrac{{4{P_2}L}}{{16\pi {d^2}E}} \\
\Rightarrow {\left( {\Delta L} \right)_2} = \dfrac{{{P_2}L}}{{4\pi {d^2}E}} \\

$$Divide equations (7) and (6) and obtain,$$

\Rightarrow \dfrac{{{{\left( {\Delta L} \right)}_2}}}{{{{\left( {\Delta L} \right)}_1}}} = \dfrac{{\dfrac{{{P_2}L}}{{4\pi {d^2}E}}}}{{\dfrac{{4{P_1}L}}{{\pi {d^2}E}}}} \\
\Rightarrow \dfrac{{{{\left( {\Delta L} \right)}_2}}}{{{{\left( {\Delta L} \right)}_1}}} = \dfrac{{{P_2}L}}{{4\pi {d^2}E}} \times \dfrac{{\pi {d^2}E}}{{4{P_1}L}} \\
\Rightarrow \dfrac{{{{\left( {\Delta L} \right)}_2}}}{{{{\left( {\Delta L} \right)}_1}}} = \dfrac{1}{{16}} \times \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) \\
\Rightarrow \dfrac{{1{{ mm}}}}{{1{{ mm}}}} = \dfrac{1}{{16}} \times \left( {\dfrac{{{P_2}}}{{{P_1}}}} \right) \\
\Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = 16 \\
\Rightarrow {P_2} = 16{P_1} \\
\Rightarrow {P_2} = 16 \times \left( {{{10}^3}} \right){{ N}} \\
\Rightarrow {P_2} = 16 \times {10^3}{{ N}} \\

So, from the above calculation, the force required is $$16 \times {10^3}{{ N}}$$. **Therefore, the correct option is B.** **Note:** Be careful about the units of the quantities such that all quantities should be in the same unit. Make sure that while calculating the strain we should use the original length of the wire and apply the correct formula.