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 are available: elongation in the wire $I=0.125$ cm under the tension of 50 N, least count for measuring normal length of wire is 0.01 cm and for elongation in the wire is 0.001 cm. The maximum error in the calculating value of Young's modulus (Y), assuming that the force is measured very accurately, is $\frac{8n}{10}\%$. Find the value of n. (roundoff value of n to nearest integer).

Answer 1

3

Answer 2

The formula for Young's modulus (Y) is given by $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} = \frac{FL}{Al}$, where F is the applied force, L is the original length of the wire, A is the cross-sectional area, and l is the elongation.

The cross-sectional area is $A = \frac{\pi d^2}{4}$, where d is the diameter of the wire. So, $Y = \frac{4FL}{\pi d^2 l}$.

To find the maximum percentage error in Y, we use the formula for propagation of errors. Assuming the force F is measured very accurately ($\Delta F = 0$) and $\pi$ is a constant with no error: $\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$ Since $\frac{\Delta F}{F} = 0$, we have: $\frac{\Delta Y}{Y} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$

We are given the following data:

1. Length of the wire, $L = 100$ cm. The least count for measuring the normal length is 0.01 cm. So, the error in length is $\Delta L = 0.01$ cm. The relative error in length is $\frac{\Delta L}{L} = \frac{0.01}{100} = 0.0001$.

2. Diameter of the wire, d, is measured with a screw gauge. Pitch of the screw gauge = 1 mm. Total number of divisions on the circular scale = 100. Least count of the screw gauge = $\frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} = 0.001 \text{ cm}$. Main scale reading (MSR) = 1 mm = 0.1 cm. Circular scale reading (CSR) = 25. Diameter $d = \text{MSR} + \text{CSR} \times \text{LC} = 0.1 \text{ cm} + 25 \times 0.001 \text{ cm} = 0.1 \text{ cm} + 0.025 \text{ cm} = 0.125 \text{ cm}$. The error in diameter measurement is equal to the least count of the screw gauge. So, $\Delta d = 0.001$ cm. The relative error in diameter is $\frac{\Delta d}{d} = \frac{0.001}{0.125} = \frac{1}{125} = 0.008$.

3. Elongation in the wire, $l = 0.125$ cm. The least count for measuring the elongation is 0.001 cm. So, the error in elongation is $\Delta l = 0.001$ cm. The relative error in elongation is $\frac{\Delta l}{l} = \frac{0.001}{0.125} = \frac{1}{125} = 0.008$.

Now, calculate the maximum relative error in Y: $\frac{\Delta Y}{Y} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$ $\frac{\Delta Y}{Y} = 0.0001 + 2(0.008) + 0.008$ $\frac{\Delta Y}{Y} = 0.0001 + 0.016 + 0.008$ $\frac{\Delta Y}{Y} = 0.0241$

The maximum percentage error in Y is $\left(\frac{\Delta Y}{Y}\right) \times 100\%$. Percentage error $= 0.0241 \times 100\% = 2.41\%$.

The problem states that the maximum error in the calculating value of Young's modulus is $\frac{8n}{10}\%$. So, $2.41\% = \frac{8n}{10}\%$. $2.41 = \frac{8n}{10}$ $24.1 = 8n$ $n = \frac{24.1}{8} = 3.0125$.

Rounding off the value of n to the nearest integer, we get $n = 3$.