A student performs an experiment to determine the young’s modulus of a wire, exactly 2m long by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8mm with an uncertainty of $\pm$ 0.05mm at load of exactly 1.0kg. The student also measures the diameter of the wire to be 0.4mm with an uncertainty of $\pm$ 0.01mm . Take G = 9.8 m/s² (exact) then Y=?

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(2.0 $\pm$ 0.3) $\times$ 10¹¹N /m²

(2.0 $\pm$ 0.2) $\times$ 10¹¹N /m²

(2.0 $\pm$ 0.1) $\times$ 10¹¹N /m²

(2.0 $\pm$ 0.5) $\times$ 10¹¹N /m²

answer is A.

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Detailed Solution

We know that,

$Y = \frac{4 F l}{{πD}^{2} e}$

taking log,

$\log Y = \log 4 F l - \log π D^{2} e$

now partially differentiating,

$\frac{∆ Y}{Y} = - (\frac{2 ∆ D}{D} + \frac{∆ e}{e}) = - (\frac{2 \times 0.01}{0.4} + \frac{0.05}{0.8}) = - 0.1125$

$A l s o, Y = \frac{F l}{A e} = \frac{9.8 \times 2}{π \times {(0.2)}^{2} \times 0.8} = 194.96 \times 10^{9} \approx 2 \times 10^{11}$

$∆ Y = - Y \times (\frac{2 ∆ r}{r} + \frac{∆ e}{e}) = 0.22 \times 10^{11}$

$Y = (2 \pm 0.2) \times 10^{11} N m^{- 2}$

Hence the correct answer is $(2 \pm 0.2) \times 10^{11} N m^{- 2} .$

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