Question

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

Physical Quantity	Least count of the Equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm

Then the fractional error in the measurement of $Y$ is :

• A. 0.0083
• B. 0.0155
• C. 0.155
• D. 0.083

Answer 1

Answer: (B)

$y=\frac{Mg{L}^3}{4b{d}^3\delta }$
$\frac{\Delta y}{y}=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+\frac{3\Delta d}{d}+\frac{\Delta \delta }{\delta }$
$\frac{\Delta y}{y}=\frac{10^{-3}}{2}+\frac{3\times 10^{-3}}{1}+\frac{10^{-2}}{4}+\frac{3\times 10^{-2}}{4}+\frac{10^{-2}}{5}$
$=10^{-3}[0.5+3+2.5+7.5+2]$
$=0.0155$