Question:
A thin 1 m long rod has a radius of 5 mm. A force of \(50\pi\) kN is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?
A thin 1 m long rod has a radius of 5 mm. A force of \(50\pi\) kN is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?
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Use \(Y = \frac{Fl}{\pi r^2 \Delta l}\), find the largest measurable \(Y\) by putting \(\Delta l\) equal to the least count, and compare with \(10^{14}\).
Updated On: Nov 11, 2026
- \(\frac{\Delta Y}{Y}\) gets minimum contribution from the uncertainty in the length.
- The figure of merit is the largest for the length of the rod.
- The maximum value of \(Y\) that can be determined is \(10^{14}\ \mathrm{N/m^2}\).
- \(\frac{\Delta Y}{Y}\) gets its maximum contribution from the uncertainty in strain.
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The Correct Option is C
Solution and Explanation
\(Y = \frac{F l}{A\,\Delta l} = \frac{Fl}{\pi r^2\Delta l}\) with \(F = 50\pi\times10^3\) N, \(l = 1\) m, \(r = 5\times10^{-3}\) m.
\(Y = \frac{50\pi\times10^3\times 1}{\pi\times25\times10^{-6}\,\Delta l} = \frac{2\times10^{9}}{\Delta l}\)
The smallest measurable extension is the least count, \(\Delta l = 0.01\) mm \(= 10^{-5}\) m, so \(Y_{max} = 2\times10^{14}\ \mathrm{N/m^2}\), not \(10^{14}\ \mathrm{N/m^2}\). Hence statement (3) is false.
Check the others: \(\frac{\Delta Y}{Y} = \frac{\Delta l_{rod}}{l} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta l)}{\Delta l}\). The length term is \(\frac{0.01}{1000} = 10^{-5}\), the smallest, so (1) is true. The extension (strain) term \(\frac{0.01\ \mathrm{mm}}{\Delta l}\) is the one that dominates as the extension gets small, so (4) is taken as true. The figure of merit (measured quantity / least count) is largest for the rod length (1 m / 0.01 mm), so (2) is true.
Answer: (3) The maximum value of \(Y\) that can be determined is \(10^{14}\ \mathrm{N/m^2}\).
\(Y = \frac{50\pi\times10^3\times 1}{\pi\times25\times10^{-6}\,\Delta l} = \frac{2\times10^{9}}{\Delta l}\)
The smallest measurable extension is the least count, \(\Delta l = 0.01\) mm \(= 10^{-5}\) m, so \(Y_{max} = 2\times10^{14}\ \mathrm{N/m^2}\), not \(10^{14}\ \mathrm{N/m^2}\). Hence statement (3) is false.
Check the others: \(\frac{\Delta Y}{Y} = \frac{\Delta l_{rod}}{l} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta l)}{\Delta l}\). The length term is \(\frac{0.01}{1000} = 10^{-5}\), the smallest, so (1) is true. The extension (strain) term \(\frac{0.01\ \mathrm{mm}}{\Delta l}\) is the one that dominates as the extension gets small, so (4) is taken as true. The figure of merit (measured quantity / least count) is largest for the rod length (1 m / 0.01 mm), so (2) is true.
Answer: (3) The maximum value of \(Y\) that can be determined is \(10^{14}\ \mathrm{N/m^2}\).
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