In a Searle's experiment, the diameter of the wire as measured by a...

First, recall the formula of Young's modulus in terms of force applied, cross- sectional area of the wire and strain produced,

$y = \dfrac{mg}{A}\dfrac{l}{\Deltal} = \dfrac{mg}{\pi{r}^{2}}\dfrac{l}{\Deltal}$,

So, the percentage error in Young's modulus,

$\dfrac{\Deltay}{y} \times 100 = \left[\dfrac{2\Deltar}{r} + \dfrac{\Deltal}{l} + \left[\Delta\dfrac{\left(\Deltal\right)}{\Deltal}\right]\right] \times 100$

$= \left[2 \times \dfrac{0 . 001}{0 . 05} + \dfrac{0 . 1}{100} + \dfrac{0 . 001}{0 . 125}\right] \times 100$

$= \left[\dfrac{2}{50} + \dfrac{1}{1000} + \dfrac{1}{125}\right] \times 100$

$= \left[0 . 04 + 0 . 001 + 0 . 008\right] \times 100 = 4 . 90$.