In searle’s experiment for the determination of young’s modulus of material of the wire, the diameter of wire as measured by a screw gauge of least count 0.001cm is 0.050cm. The length of the wire, measured by a scale of least count 0.1cm is 100.0cm. when a mass of 5.00kg, as measured by a spring balance of least count 0.01kg, is suspended from the wire, the extension is 0.125cm as measured by a micrometer of least count 0.001cm. From the given data, find the maximum error in the value of the young’s modulus. $(g = 9.8 m s^{- 2})$

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$1.02 \times 10^{11} N m^{- 2}$

$10.2 \times 10^{11} N m^{- 2}$

$0.102 \times 10^{11} N m^{- 2}$

$0.0102 \times 10^{11} N m^{- 2}$

answer is A.

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Detailed Solution

$y = \frac{M g L}{(\frac{π d^{2}}{4}) e} y = \frac{5 \times 9.8 \times 1}{π \times \frac{{(5 \times 10^{- 4})}^{2}}{4} 125 \times 10^{- 5}} y = 1.996 \times 10^{11} N m^{- 2}$

For maximum error,

$\frac{Δ y}{y} = \frac{Δ M}{M} + \frac{Δ L}{L} + 2 \frac{Δ d}{d} + \frac{Δ e}{e} \frac{Δ y}{y} = \frac{0.01}{5.00} + \frac{0.1}{100.0} + 2 (\frac{0.001}{0.050}) + \frac{0.001}{0.125} = 0.051 Δ y = 0.051 \times 1.996 \times 10^{11} = 0.102 \times 10^{11} N m^{- 2}$