Question

What is the percentage increase in the length of a wire of diameter $2.5$mm stretched by a force of $100$ kg weight. Young’s modulus of the wire is$12.5 \times {10^{11}}dyn/c{m^2}$.

Answer 1

To solve this question we have to know the formula of the change in length in percentage. So, here we are considering L is the length of the wire. $\Delta L$ is the change in length. Now we are going to use the $\Delta L/L \times 100$ formula.

Complete step by step answer:
We know that in this question it s given that, Young’s modulus y of the wire is $12.5 \times {10^{11}}dyn/c{m^2}$which is equal to $12.5 \times {10^{10}}N/{m^2}$.According to the question we write, the diameter of the wire D is equal to $2.5$ mm, which is equal to $2.5 \times {10^{ - 3}}$ m. Now, we are going to assume the force is equal to F.
And according to the question we can write, $F = 100$kg f . which is equal to $100 \times 9.8N = 980N$.Here, we are going to use the formula of percentage increase in length is,
$\Delta L/L \times 100$. We are assuming that l is the length. And A is the area of the wire.
So, we can write A is equal to $\pi {r^2} = \pi (1.25 \times {10^{ - 3}}){m^2}$
Now, we know that the Young’s modulus Y is equal to $FL/A\Delta L$
Now, we can write,
$\Delta L/L \times 100 = (F/AY) \times 100 = (F/\pi {r^2}Y) \times 100$
$\Rightarrow\Delta L/L \times 100 = (980/3.142 \times {(1.25 \times {10^{ - 3}})^2} \times \times {10^{10}}) \times 100 \\\ \therefore\Delta L/L \times 100 = 15.96 \times {10^{ - 2}} = 0.16\% \\\$
So the percentage increase in the length of wire is $0.16\%$.

Note: We can forget to convert the units into one system but if we do not do that the answer will not be correct. We have to convert $dyn/c{m^2}$ into $N/{m^2}$. And also kg f to N. we know that N is equal to Newton here.