Young's modulus is determined by the equation given by $Y = 49000\dfrac{M}{l}\ \dfrac{\text{dyn}}{\text{cm}^2}$, where $M$ is the mass and $l$ is the extension of the wire used in the experiment. Now the error in Young's modulus ($Y$) is estimated by taking data from the $M$–$l$ plot in graph paper. The smallest scale divisions are $5\ \text{g}$ and $0.02\ \text{cm}$ along the load axis and extension axis respectively. If the values of $M$ and $l$ are $500\ \text{g}$ and $2\ \text{cm}$ respectively, then the percentage error of $Y$ is:
Answer: (B) $2\%$
The least counts give the errors $\Delta M = 5\ \text{g}$ and $\Delta l = 0.02\ \text{cm}$.
$$\frac{\Delta Y}{Y}\times100 = \left(\frac{5}{500} + \frac{0.02}{2}\right)\times100 = 1\% + 1\% = 2\%$$
Solution by Sreeraj P, M.Sc Physics