Question

1. A 20m steel wire is stretched to 20.0m by

Answer 1

Answer:

Force to stretch the wire is 250 N

Explanation:

As we know that modulus of elasticity will remain the same for the wire if the applied stretch to the wire is within elastic limit

So we will have

$\frac{F L}{\Delta L A} = constant$

now we have

$\frac{F_1 L}{\Delta L_1 A} = \frac{F_2 L}{\Delta L_2 A}$

so we can write it as

$F_2 = \frac{F_1 L_2}{L_1}$

$F_2 = \frac{50 (0.05)}{0.01}$

$F_2 = 250 N$