A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. [2007]

Question

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of $\pm 0.05 mm$ at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of $\pm 0.01 mm$ . Take $g = 9.8 {m/s}^{2}$ (exact). The Young's modulus obtained from the reading is [2007]

Smart Achievers · Accepted Answer

(2)

$Y = \frac{F / A}{ℓ / L} = \frac{4 m g L}{π D^{2} ℓ} = \frac{4 \times 1 \times 9.8 \times 2}{π {(0.4 \times 10^{- 3})}^{2} \times (0.8 \times 10^{- 3})}$

$= 2.0 \times 10^{11} {N/m}^{2}$

$Now \frac{Δ Y}{Y} = \frac{2 Δ D}{D} + \frac{Δ ℓ}{ℓ}$ $[∵ the value of m, g and L are exact]$

$= 2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8} = 2 \times 0.025 + 0.0625$

$= 0.05 + 0.0625 = 0.1125$

$\Rightarrow Δ Y = Y \times 0.1125 = 2 \times 10^{11} \times 0.1125 = 0.225 \times 10^{11}$

$= 0.2 \times 10^{11} {N/m}^{2}$

$∴$ $Reading of Young's modulus = (Y \pm Δ Y)$

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

Solution

1 $(2.0 \pm 0.3) \times 10^{11} {N/m}^{2}$

2 $(2.0 \pm 0.2) \times 10^{11} {N/m}^{2}$

3 $(2.0 \pm 0.1) \times 10^{11} {N/m}^{2}$

4 $(2.0 \pm 0.05) \times 10^{11} {N/m}^{2}$

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Solution

1 (2.0±0.3)×1011 N/m2

2 (2.0±0.2)×1011 N/m2

3 (2.0±0.1)×1011 N/m2

4 (2.0±0.05)×1011 N/m2

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1 $(2.0 \pm 0.3) \times 10^{11} {N/m}^{2}$

2 $(2.0 \pm 0.2) \times 10^{11} {N/m}^{2}$

3 $(2.0 \pm 0.1) \times 10^{11} {N/m}^{2}$

4 $(2.0 \pm 0.05) \times 10^{11} {N/m}^{2}$