Question 277: Units , Dimensions and Error Analysis

Question

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long,
by Searle's method. In a particular reading, the student measures the extension in the length of
the wire to be 0.8 mm with an uncertainty of $$\pm0.05\;\mathrm{mm}$$ at a load of exactly 1.0 kg. The student
also measures the diameter of the wire to be 0.4 mm with an uncertainty of $$\pm0.01\;\mathrm{mm}$$. Take
g = 9.8 m/s
2
(exact). The Young's modulus obtained from the reading is

Accepted Answer

Young's modulus $$(Y)=\frac{	ext { Stress }}{	ext { Strain }}=\frac{4 m g l}{\pi d^{2} \Delta l}$$
$$\begin{gathered}
\mathrm{Y}=\frac{4 m g l}{\pi d^{2} l} \\
=\frac{4 	imes 1 	imes 9.8 	imes 2}{3.14 	imes(0.4)^{2} 	imes(0.8) 	imes 10^{-6} 	imes 10^{-3}} \\
=2 	imes 10^{11} \mathrm{~N} / \mathrm{m}^{2}
\end{gathered}
$$
Where, $$\quad$$ Mass $$(m)=1 \mathrm{~kg}$$
$$\begin{aligned}g & =9.8 \mathrm{~m} / \mathrm{s}^{2} \\
	ext { Length of wire } & =2 \mathrm{~m} \\
\pi & =3.14 \\
	ext { Diameter }(d) & =0.4 \mathrm{~mm} 	imes 10^{-3} \mathrm{~m}
\end{aligned}$$
Extension in the length of wire
$$\begin{aligned}(\Delta l) & =0.8 \mathrm{~mm} \\
& =0.8 	imes 10^{-3} \mathrm{~m} \\
\frac{\Delta y}{y} & =\frac{2 \Delta d}{d}+\frac{\Delta l}{l} \\
& =\frac{2 	imes(0.01)}{0.4}+\frac{(0.05)}{0.8} \\
& =0.1125\end{aligned}$$
$$\begin{aligned}\Rightarrow \Delta \mathrm{Y} & =(\mathrm{Y}) 	imes 0.1125 \\
& =2 	imes 10^{11} 	imes 0.1125 \\
& =0.225 	imes 10^{11} \\
& \simeq 0.2 	imes 10^{11} \\
	ext { (b) }\left(2 \pm 0.2 	imes 10^{11} \mathrm{~N/}ight. & \left.\mathrm{m}^{2}ight)
\end{aligned}$$

Answer

$$\left(2.0\;\pm\;0.3ight)	imes10^{11}\;\mathrm N/\mathrm m^2$$

Answer

$$\left(2.0\;\pm\;0.2ight)	imes10^{11}\;\mathrm N/\mathrm m^2$$

Answer

$$\left(2.0\;\pm\;0.1ight)	imes10^{11}\;\mathrm N/\mathrm m^2$$

Answer

$$\left(2.0\;\pm\;0.05ight)	imes10^{11}\;\mathrm N/\mathrm m^2$$

Units , Dimensions and Error Analysis

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