Question

The diameter of a wire of length 100 cm is measured with the help of a screw gauge. The main scale reading is 1 mm and circular scale is reading is 25. Pitch of the screw gauge is 1 mm and the total number of divisions on the circular scale is 100. This wire is used in an experiment of determination of Young's modulus of a wire by Searle's method. The following data is available: elongation in the wire I = 0.125 cm under the tension of 50 N, least count for measuring normal length of wire is 0.4 cm and for elongation in the wire is 0.005 cm. The maximum error(in %) in the calculating value of Young's modulus (Y), assuming that the force is measured very accurately, is

• A. 6.0
• B. 5.0
• C. 4.0
• D. 3.0

Answer 1

Answer: (A)

6.0

Answer 2

The formula for Young's modulus is $Y = \frac{FL}{Al}$, where $A = \frac{\pi d^2}{4}$. Thus, $Y = \frac{4FL}{\pi d^2 l}$. The maximum percentage error in Y, assuming force F is accurate, is given by $\frac{\Delta Y}{Y} \times 100\% = \left(\frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}\right) \times 100\%$. The diameter $d$ is calculated from screw gauge readings: $d = 0.125$ cm with an error $\Delta d = 0.001$ cm. The relative error is $\frac{\Delta d}{d} = \frac{0.001}{0.125} = 0.008$. The length $L = 100$ cm with error $\Delta L = 0.4$ cm, so $\frac{\Delta L}{L} = \frac{0.4}{100} = 0.004$. The elongation $l = 0.125$ cm with error $\Delta l = 0.005$ cm, so $\frac{\Delta l}{l} = \frac{0.005}{0.125} = 0.04$. Substituting these values: $\frac{\Delta Y}{Y} \times 100\% = (0.004 + 2 \times 0.008 + 0.04) \times 100\% = (0.004 + 0.016 + 0.04) \times 100\% = 0.060 \times 100\% = 6.0\%$.