In Searle’s experiment, which is used to find Young’s Modulus of elasticity, the diameter of experimental wire is (measured by a scale of least count 0.001 cm) and length is (measured by a scale of least count 0.1 cm). [2004]

Question

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 $X = 0.125 cm$ (measured by a micrometer of least count $0.001 cm$ ). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error. [2004]

Smart Achievers · Accepted Answer

(1.09)

$Young's modulus, Y = \frac{\frac{F}{A}}{\frac{X}{L}} = \frac{F}{\frac{π D^{2}}{4}} \times \frac{L}{X}$

$Maximum error in Y$

${(\frac{Δ Y}{Y})}_{\max} = 2 (\frac{Δ D}{D}) + \frac{Δ X}{X} + \frac{Δ L}{L}$

$= 2 (\frac{0.001}{0.05}) + (\frac{0.001}{0.125}) + (\frac{0.1}{110}) = 0.0489$

$Since, W = 50 N; D = 0.05 cm = 0.05 \times 10^{- 2} m;$

$X = 0.125 cm = 0.125 \times 10^{- 2} m;$

$L = 110 cm = 110 \times 10^{- 2} m$

$∴$ $Y = \frac{50 \times 4 \times 110 \times 10^{- 2}}{3.14 (0.05 \times 10^{- 2}) \times (0.125 \times 10^{- 2})} = 2.24 \times 10^{11} {N/m}^{2}$

$∴$ $Maximum possible error in the measurement of Y$

$Δ Y = (0.0489) Y = 0.0489 \times 2.24 \times 10^{11} = 1.09 \times 10^{10} {N/m}^{2}$

Solution

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