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Question

In Searle's experiment to find Young's modulus, the diameter of wire is measured as $$D = 0.05cm$$, the length of wire is $$L = 125 cm, $$ and when a weight, $$m = 20 kg$$ is the put, extension in wire was found to be $$0.100 cm$$. Find the maximum permissible error in Young's modulus (Y).


Solution

$$\dfrac{mg}{\pi d^2/4} = Y \left(\dfrac{x}{l}\right) \Rightarrow Y = \dfrac{mgl}{(\pi/4)d^2 x}$$
$$\left(\dfrac{dY}{Y}\right)_{max} = \dfrac{\triangle m}{m} + \dfrac{\triangle l}{l} + 2 \dfrac{\triangle d}{d} + \dfrac{\triangle x}{x}$$
$$m = 20.0 kg   \Rightarrow \triangle m = 0.1 kg$$
$$l = 125 m    \Rightarrow  \triangle l = 1 cm$$
$$d = 0.050 cm    \Rightarrow  \triangle d = 0.001 cm$$
$$x = 0.100 cm \Rightarrow   \triangle x = 0.001 cm$$
$$\left(\dfrac{dY}{Y}\right)_{max} = \left(\dfrac{0.1kg}{20.0kg} + \dfrac{1cm}{125 cm}+ \dfrac{0.001 cm}{0.05cm} + \dfrac{0.001cm}{0.100cm}\right) \times 100\% = 4.3\%$$

Physics

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