A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \

Question

A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta}$. The value of $g$ is taken to be $9.8 \mathrm{~m} / \mathrm{s}^{2}$, without any significant error, his observation are as following. Then the fractional error in the measurement of $Y$ is\

Professor PC Thomas Classes · Answer

(d) : $Y=\frac{M g L^{3}}{4 b d^{3} \delta}$\ $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ (constant)\ $\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+\frac{3 \Delta L}{L}+\frac{\Delta b}{b}+\frac{3 \Delta d}{d}+\frac{\Delta \delta}{\delta}$\ $=\frac{1}{2000}+\frac{3}{1000}+\frac{0.1}{40}+\frac{0.03}{4}+\frac{0.01}{5}$\ $	herefore \quad \frac{\Delta Y}{Y}=0.0155$\ So, fractional error is measurement of $Y$ is 0.0155 .\