Error Contribution Analysis in Searles Youngs Modulus Experiment

To find the contributions of the errors in measuring $d$ and $l$ to the maximum probable fractional error in the determination of Young's modulus, we begin with the given formula: $Y = \frac{4MLg}{\pi l d^2}$

Taking the natural logarithm on both sides: $\ln Y = \ln\left(\frac{4g}{\pi}\right) + \ln M + \ln L - \ln l - 2\ln d$

Differentiating to express the fractional error in $Y$: $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + \frac{\Delta l}{l} + 2\frac{\Delta d}{d}$

Both the diameter $d$ and the elongation $l$ are measured using instruments with: $\text{Pitch} = 0.5\text{ mm}$ $\text{Number of divisions on the circular scale} = 100$

The least count (absolute uncertainty) for both instruments is: $\Delta d = \Delta l = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}$

Now, we calculate the individual fractional error contributions to $\frac{\Delta Y}{Y}$:

1. Contribution due to the error in measuring elongation $l$: $\left(\frac{\Delta Y}{Y}\right)_l = \frac{\Delta l}{l} = \frac{0.005\text{ mm}}{0.25\text{ mm}} = \frac{1}{50} = 0.02$

2. Contribution due to the error in measuring diameter $d$: $\left(\frac{\Delta Y}{Y}\right)_d = 2\frac{\Delta d}{d} = 2 \times \frac{0.005\text{ mm}}{0.5\text{ mm}} = 2 \times \frac{1}{100} = 0.02$

Comparing the two contributions: $\left(\frac{\Delta Y}{Y}\right)_l = \left(\frac{\Delta Y}{Y}\right)_d$

Thus, the fractional contributions to the maximum probable error in the determination of $Y$ arising from errors in the measurement of $d$ and $l$ are identical.

Therefore, the correct option is (A).