Two wires each of radius 0 . 2cm and negligible mass, one made of...

The force acting on steel wire $F$ is \left(2 + 1 . 14\right)g = 3 . 14g$</p><p>From the relation of Young's modulus, it can be written</p><p>$\dfrac{F}{A} = Y\left(\dfrac{\DeltaL}{L}\right)$</p><p>$⇒ \DeltaL = \left(\dfrac{FL}{AY}\right) = \dfrac{3 . 14 \times g \times 1 . 6}{\pi \times {\left(0 . 2\right)}^{2} \times 2 \times {10}^{11} \times {10}^{- 4}}$</p><p>$⇒ ∆ L = 2 \times {10}^{- 5}$</p><p>$⇒ ∆ L = 20 \times {10}^{- 6}m