In an experiment of determine the Young's modulus of wire of a...

Given here,

Length, $L = 1m$ and extension in length, $\DeltaL = 0 . 4 \times {10}^{- 3}m$

Mass of load, $m = 1kg$ and diameter of wire, $d = 0 . 4 \times {10}^{- 3}m$

Young's modulus is given by

$Y = \dfrac{FL}{A\DeltaL} = \dfrac{\left(mg\right) \times \left(1\right)}{\left(\dfrac{\pi{d}^{2}}{4}\right)0 . 4 \times {10}^{- 3}}$

$⇒ Y = \dfrac{10 \times 4}{\pi{\left(0 . 4 \times {10}^{- 3}\right)}^{2} \times 0 . 4 \times {10}^{- 3}}$

$⇒ Y = \dfrac{40 \times 7}{22 \times 64 \times {10}^{- 3} \times {10}^{- 9}}$

$⇒ Y = 0 . 199 \times {10}^{- 12}N{m}^{- 2}$

Error in measurement of Young's modulus,

$\dfrac{\DeltaY}{Y} = \dfrac{\DeltaF}{F} + \dfrac{\DeltaL}{L} + \dfrac{\DeltaA}{A} + \dfrac{\Delta\left(\DeltaL\right)}{\left(\DeltaL\right)}$

$= \dfrac{0 . 02}{0 . 4} + 2\dfrac{\Deltad}{d} = \dfrac{0 . 2}{4} + 2 \times \dfrac{0 . 01}{0 . 4}$

$= \dfrac{0 . 1}{2} + \dfrac{0 . 1}{2} = 0 . 1$

$= 0 . 199 \times {10}^{11} \approx 2 \times {10}^{10}$

Hence, the value of $x = 2$.