Question

The area of cross-section of steel wire is $0.1c{m^2}$ and the young modulus of the steel wire is $2 \times {10^{11}}N{m^{ - 2}}$. The force required to stretch by $0.1\% $ of its length is then
(A) $1000\;N$
(B) $2000\;N$
(C) $3000\;N$
(D) $4000\;N$

Answer 1

Hint: To solve these questions we first need to understand the term Young modulus given. Young modulus can be defined as the ability of a material to resist changes in its length which can occur due to compression and tension in its length. It is also defined as the ratio of stress to strain.
Formula used:
Young modulus formula
$Y = \dfrac{{stress}}{{strain}} = \dfrac{{F/A}}{{\Delta l/l}}$
where $F$ is forced, $A$ is area and $l$ is length, $\Delta l$ is the change in length.

Complete step-by-step solution:
Here given that the young modulus of the steel wire is $2 \times {10^{11}}N{m^{ - 2}}$, and cross-section area is also given as
$A = 0.1c{m^2} = 0.1 \times {10^{ - 4}}{m^2}$ ………. $(1)$
Here also given that the wire is stretch by $0.1\% $ which is the $\dfrac{{\Delta l}}{l}$,
$\dfrac{{\Delta l}}{l} = 0.1\% $
$ \Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{{0.1}}{{100}} = 0.1 \times {10^{ - 2}}$ ………. $(2)$
Now we know that the young modulus $Y$ is defined as the ratio of stretch and strain. Where stress is defined as the ratio of the force $F$ per unit area $A$ acting on the material while a strain is defined as the ratio of change is length $\Delta l$ to the length $l$ of the martial given.
Hence applying the definition the young modulus is given as
$Y = \dfrac{{stress}}{{strain}} = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$
Substituting the values of young modulus given as $Y = 2 \times {10^{11}}N{m^{ - 2}}$and the values from the equation $(1)$ and $(2)$, we get
$2 \times {10^{11}}N{m^{ - 2}} = \dfrac{F}{{0.1 \times {{10}^{ - 4}}{m^2} \times 0.1 \times {{10}^{ - 2}}}}$
$ \Rightarrow F = 2 \times {10^{11}}N{m^{ - 2}} \times 0.1 \times {10^{ - 4}}{m^2} \times 0.1 \times {10^{ - 2}}$
$\therefore F = 2 \times {10^3}N = 2000N$.
Hence the force required to stretch by $0.1\% $ of its length if the area of cross-section of steel wire is $0.1c{m^2}$ and young modulus of the steel wire is $2 \times {10^{11}}N{m^{ - 2}}$ given by $F = 2000N$.

Therefore option (B) is the correct answer.

Note: While solving this type of question one should ensure about the units used. Here in this question, the area was given in $c{m^2}$ which has to be converted to ${m^2}$ which is a SI unit as young modulus is given in $N{m^{ - 2}}$ which is in the SI unit system.