Question
The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm . The length measured by a scale of least count 0.1 cm is 150 cm . When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm , measured by a micrometer of least count 0.001 cm . The error in the measured Young's modulus is $\alpha \times 10^{9}\text{ }N/m^{2}$. The value of $\alpha$ is $\_\_\_\_$ .
(Ignore the contribution of the load to Young's modulus error calculation)
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Solution
Sol. [Key Concept: Propagate relative errors: $\frac{\Delta\ Y}{Y} = \frac{\Delta\ F}{F} + \frac{\Delta\mathcal{\ l}}{\mathcal{l}} + 2\frac{\Delta\ d}{d} + \frac{\Delta(\Delta\mathcal{l)}}{\Delta\mathcal{l}}$.]
$${Y = \frac{F\mathcal{l}}{A\Delta\mathcal{l}} }{\therefore\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta\mathcal{l}}{\mathcal{l}} + 2\frac{\Delta\text{ }d}{\text{ }d} + \frac{\Delta\left( \Delta\mathcal{l} \right)}{\Delta\mathcal{l}} }$$
$$\left\lbrack A = \pi\frac{d^{2}}{4} \Rightarrow \frac{\Delta\text{ }A}{\text{ }A} = \frac{2\Delta\text{ }d}{\text{ }d} \right\rbrack $$Since F is known exactly $\Rightarrow \Delta F = 0$
$$\frac{\Delta Y}{Y} = \frac{0.1}{150} + \frac{2 \times 0.001}{0.08} + \frac{0.001}{0.5} = 0.0277 $$
$$\therefore\Delta Y = 0.0277Y $$
$$Y = \frac{100 \times 1.5}{\frac{\pi}{4}\left( \frac{0.08}{100} \right)^{2} \times 0.5 \times 10^{- 2}} = 5.97 \times \frac{10^{10}\text{ }N}{m^{2}} $$
$$\therefore\Delta Y = 0.0277 \times 5.97 \times 10^{10}\text{ }N/m^{2} = 1.65 \times 10^{9}\text{ }N/m^{2} $$
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