In an experiment to determine the Young's modulus of a wire of length exactly $1\ \text{m}$, the extension in the length of the wire is measured as $0.4\ \text{mm}$ with an uncertainty of $\pm0.02\ \text{mm}$ when a load of $1\ \text{kg}$ is applied. The diameter of the wire is measured as $0.4\ \text{mm}$ with an uncertainty of $\pm0.01\ \text{mm}$. The error in the measurement of Young's modulus $(\Delta Y)$ is found to be $x\times10^{10}\ \text{N m}^{-2}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
$Y = \dfrac{4FL}{\pi d^2\,\Delta l}$. Taking $g = 10\ \text{m s}^{-2}$:
$$Y = \frac{4\times10\times1}{\pi(0.4\times10^{-3})^2(0.4\times10^{-3})}\approx2\times10^{11}\ \text{N m}^{-2}$$
$$\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta l)}{\Delta l} = 2\times\frac{0.01}{0.4} + \frac{0.02}{0.4} = 0.1$$
$\Delta Y = 0.1\times2\times10^{11} = 2\times10^{10}\ \text{N m}^{-2}$, so $x = 2$.
Solution by Sreeraj P, M.Sc Physics