Young's modulus is determined by the equation given by $Y=49000 \frac{M}{l} \frac{ ext { dyne }}{\m

Question

Young's modulus is determined by the equation given by $Y=49000 \frac{M}{l} \frac{	ext { dyne }}{\mathrm{cm}^{2}}$, where $M$ is the mass and $l$ is the extension of wire used in the experiment. Now error in Young's modulus ( $Y$ ) is estimated by taking data from M-l plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis, respectively. If the value of $M$ and $l$ are 500 g and 2 cm , respectively then percentage error of $Y$ is\

Professor PC Thomas Classes · Answer

(d) : Given, $Y=49000 	imes \frac{M}{l}$ dyne $/ \mathrm{cm}$\ $\left(\frac{d Y}{Y}ight) \%=\left(\frac{d M}{M}+\frac{d l}{l}ight) 	imes 100 \%$\ $=\left(\frac{5}{500}+\frac{0.02}{2}ight) 	imes 100=2 \%$\