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 \(\alpha \times 10^9\) N/m\(^2\). The value of \(\alpha\) is _______.
(Ignore the contribution of the load to Young's modulus error calculation)

The formula for Young's modulus is given by:

\(Y = \dfrac{F L_0}{A \Delta L} = \dfrac{4 F L_0}{\pi d^2 \Delta L}\)

Substituting the given values into the formula:

\(Y = \dfrac{4 \times 100 \times 1.5}{\pi \times (0.08 \times 10^{-2})^2 \times (0.5 \times 10^{-2})}\)

\(Y = \dfrac{600}{\pi \times 64 \times 10^{-8} \times 5 \times 10^{-3}} = \dfrac{15}{8\pi} \times 10^{11} \text{ N/m}^2\)

The relative error in Young's modulus is:

\(\dfrac{\Delta Y}{Y} = \dfrac{\Delta L_0}{L_0} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta (\Delta L)}{\Delta L}\)

Substituting the least counts as absolute errors and the measured values:

\(\dfrac{\Delta Y}{Y} = \dfrac{0.1}{150} + 2 \times \dfrac{0.001}{0.08} + \dfrac{0.001}{0.5}\)

\(\dfrac{\Delta Y}{Y} = \dfrac{1}{1500} + \dfrac{1}{40} + \dfrac{1}{500}\)

\(\dfrac{\Delta Y}{Y} = \dfrac{2 + 75 + 6}{3000} = \dfrac{83}{3000}\)

The absolute error in Young's modulus is:

\(\Delta Y = Y \times \dfrac{\Delta Y}{Y} = \left(\dfrac{15}{8\pi} \times 10^{11}\right) \times \dfrac{83}{3000}\)

\(\Delta Y = \dfrac{15 \times 83}{24000 \pi} \times 10^{11} = \dfrac{83}{1600 \pi} \times 10^{11} = \dfrac{83}{16\pi} \times 10^9 \text{ N/m}^2\)

Using \(\pi \approx 3.14\):

\(\Delta Y = \dfrac{83}{16 \times 3.14} \times 10^9 = \dfrac{83}{50.24} \times 10^9 \approx 1.65 \times 10^9 \text{ N/m}^2\)

Comparing this with \(\alpha \times 10^9 \text{ N/m}^2\), we get \(\alpha = 1.65\).

Answer: \(1.65\)