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

Hint: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.