A student performs an experiment to determine 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 $\pm 0.05 mm$ at a load of $1.0 kg$ (exact). The student also measures the diameter of the wire to be $0.4 mm$ with an uncertainty of $\pm 0.01 mm$ . Take $g = 9.8 m / s^{2}$ (exact). Find Young's modulus of elasticity with limits of error.

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$0.44 \times 10^{11} N / m^{2}$

$0.33 \times 10^{11} N / m^{2}$

$0.22 \times 10^{11} N / m^{2}$

$0.11 \times 10^{11} N / m^{2}$

answer is B.

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Detailed Solution

$Y = \frac{4 F L}{π d^{2} l} = \frac{4 \times 1.0 \times 9.8 \times 2}{(π) {(0.4 \times 10^{- 3})}^{2} (0.8 \times 10^{- 3})}$

Further, $Y = \frac{4 M g L}{π d^{2} l} \Rightarrow \frac{Δ Y}{Y} = 2 (\frac{Δ d}{d}) + \frac{Δ l}{l}$

$Δ Y = [2 (\frac{Δ d}{d}) + (\frac{Δ l}{l})] \times Y$

$= [2 \times (\frac{0.01}{0.4}) + (\frac{0.05}{0.8})] \times 1.94 \times 10^{11}$

$= 0.22 \times 10^{11} N / m^{2}$

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