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

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1.65

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

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