In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050cm. The length, measured by a scale of least count 0.1 cm, is 110.0cm. When a weight of 50 N (measured with no error) is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

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$2 .09 \times 10^{10} {N/m}^{2}$

$1 .09 \times 10^{10} {N/m}^{2}$

$3 .09 \times 10^{10} {N/m}^{2}$

$1 .10 \times 10^{10} {N/m}^{2}$

answer is A.

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Detailed Solution

$y = \frac{F}{π r^{2}} \times \frac{L}{e} = \frac{50}{π {(\frac{0.050 \times 10^{- 2}}{2})}^{2}} \times \frac{110}{0.124 \times 10^{- 2}} = 2.24 \times 10^{11} - - - (1)$

$Δ y = 1.095 \times 10^{10} N / m^{2}$

$\frac{Δ y}{y} = 2 \frac{Δ r}{r} + \frac{Δ l}{l} + \frac{Δ e}{e} = 2 \frac{(0.001)}{0.05} + \frac{(0.1)}{110} + \frac{(0.001)}{0.125} = 0.0489$

$Δ y = 0.0489 \times 2.24 \times 10^{11} = 1.09 \times 10^{10} f r o m (1)$

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