In Searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experimental wire is $D = 0.05 cm$ (measured by a scale of least count $0.001 cm$ ) and length is $L = 110 cm$ (measured by a scale of least count $0.1 cm)$ . A weight of $50 N$ causes an extension of $l = 0.125 cm$ (measured by a micrometer of least count $0.001 cm$ ). Find maximum possible error in the value of Young's modulus. Screw gauge and meter scale are free from error.

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$10.09 \times 10^{10} N / m^{2}$

$1.19 \times 10^{10} N / m^{2}$

$1.09 \times 10^{10} N / m^{2}$

$0.09 \times 10^{10} N / m^{2}$

answer is A.

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Detailed Solution

Young's modulus of elasticity is given by

$Y = \frac{stress}{strain}$

$= \frac{F / A}{l / L} = \frac{F L}{L A} = \frac{F L}{l (\frac{π d^{2}}{4})}$

Substituting the values, we get

$Y = \frac{50 \times 1.1 \times 4}{(1.25 \times 10^{- 3}) \times π \times {(5.0 \times 10^{- 4})}^{2}}$

$= 2.24 \times 10^{11} N / m^{2}$

$\frac{Δ Y}{Y} = \frac{Δ L}{L} + \frac{Δ l}{l} + 2 \frac{Δ d}{d}$

$= (\frac{0.1}{110}) + (\frac{0.001}{0.125}) + 2 (\frac{0.001}{0.05})$

=0.0489

$Δ Y = (0.0489) Y$

$= (0.0489) \times (2.24 \times 10^{11}) N / m^{2}$

$= 1.09 \times 10^{10} N / m^{2}$

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