Mechanical Properties of Solids

JEE Main level, single correct, medium

16 of 118

Question 16 of 118

The diameter of a wire measured by a screw gauge of least count 0.0010.001 cm is 0.080.08 cm. The length measured by a scale of least count 0.10.1 cm is 150150 cm. When a weight of 100100 N is applied to the wire, the extension in length is 0.50.5 cm, measured by a micrometer of least count 0.0010.001 cm. The error in the measured Young's modulus is α×109\alpha \times 10^9 N/m2^2. The value of α\alpha is _______.
(Ignore the contribution of the load to Young's modulus error calculation)

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Answer: (B) 1.651.65

The formula for Young's modulus is given by:

Y=FL0AΔL=4FL0πd2ΔLY = \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=4×100×1.5π×(0.08×10−2)2×(0.5×10−2)Y = \dfrac{4 \times 100 \times 1.5}{\pi \times (0.08 \times 10^{-2})^2 \times (0.5 \times 10^{-2})}

Y=600π×64×10−8×5×10−3=158π×1011 N/m2Y = \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:

ΔYY=ΔL0L0+2Δdd+Δ(ΔL)ΔL\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:

ΔYY=0.1150+2×0.0010.08+0.0010.5\dfrac{\Delta Y}{Y} = \dfrac{0.1}{150} + 2 \times \dfrac{0.001}{0.08} + \dfrac{0.001}{0.5}

ΔYY=11500+140+1500\dfrac{\Delta Y}{Y} = \dfrac{1}{1500} + \dfrac{1}{40} + \dfrac{1}{500}

ΔYY=2+75+63000=833000\dfrac{\Delta Y}{Y} = \dfrac{2 + 75 + 6}{3000} = \dfrac{83}{3000}

The absolute error in Young's modulus is:

ΔY=Y×ΔYY=(158π×1011)×833000\Delta Y = Y \times \dfrac{\Delta Y}{Y} = \left(\dfrac{15}{8\pi} \times 10^{11}\right) \times \dfrac{83}{3000}

ΔY=15×8324000π×1011=831600π×1011=8316π×109 N/m2\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 π≈3.14\pi \approx 3.14:

ΔY=8316×3.14×109=8350.24×109≈1.65×109 N/m2\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 α×109 N/m2\alpha \times 10^9 \text{ N/m}^2, we get α=1.65\alpha = 1.65.

Answer: 1.651.65

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