The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm

Question

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young’s modulus is $α \times 10^{9} N / m^{2}$ . The value of $α$ is ______.

(Ignore the contribution of the load to Young’s modulus error calculation) [2026]

Smart Achievers · Accepted Answer

(2)

$Y = \frac{F ℓ}{A Δ ℓ}$

$∴$ $\frac{Δ Y}{Y} = \frac{Δ F}{F} + \frac{Δ ℓ}{ℓ} + \frac{2 Δ d}{d} + \frac{Δ (Δ ℓ)}{Δ ℓ}$ $[A = π \frac{d^{2}}{4} \Rightarrow \frac{Δ A}{A} = \frac{2 Δ d}{d}]$

$Since F is known exactly \Rightarrow Δ F = 0$

$\frac{Δ Y}{Y} = \frac{0.1}{150} + \frac{2 \times 0.001}{0.08} + \frac{0.001}{0.5} = 0.0277$

$∴$ $Δ Y = 0.0277 Y$

$Y = \frac{100 \times 1.5}{\frac{π}{4} {(\frac{0.08}{100})}^{2} \times 0.5 \times 10^{- 2}} = 5.97 \times 10^{10} {N/m}^{2}$

$∴$ $Δ Y = 0.0277 \times 5.97 \times 10^{10} {N/m}^{2}$

$= 1.65 \times 10^{9} {N/m}^{2}$

Solution

1 1.3

2 1.65

3 0.13

4 0.25

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