In an experiment of determine the Young’s modulus of wire of a length exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of ± 0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ± 0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ± 0.01 mm. The error in the measurement of Young’s modulus (ΔY) is found to be x × 10¹⁰ Nm^–2. The value of x is ___.
(Take g = 10 ms^–2)

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Detailed Solution

$\begin{array}{l} \frac{F / A}{I / L} = Y, A = {πD}^{2} \\ \begin{aligned} \frac{ΔY}{Y} = \frac{ΔF}{F} + \frac{2 ΔD}{D} + \frac{ΔI}{e} + \frac{ΔL}{L} \\ = 2 \times \frac{0 .01}{0 .4} + \frac{0 .02}{0 .4} \\ = \frac{0 .04}{0 .4} = \frac{1}{10} \\ Y = \frac{FI}{AΔI} \\ = \frac{10}{π (0 .1 mm)^{2} \times 0 .4 mm} \\ = 1 .988 \times 10^{11} \end{aligned} \\ \approx 2 \times 10^{11} \\ \frac{Δy}{y} = \frac{1}{10} \\ Δy = \frac{y}{10} = 2 \times 10^{10} \end{array}$

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