Question

In Searl's experiment to find Young's modulus, the diameter of wire is measured as D = 0.05 cm, length of wire is L = 125 cm. When a weight, m = 20.0 kg is put, extension in wire was found to be 0.100 cm. Find maximum permissible error in Young's modulus.

• A. 3.3%
• B. 4.3%
• C. 5.5%
• D. 2.5%

Answer 1

Answer: (B)

(B) 4.3%

Answer 2

Young's modulus ($Y$) of a wire is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$. The force $F = mg$ and area $A = \frac{\pi D^2}{4}$. So, $Y = \frac{4mgL}{\pi D^2 \Delta L}$. The maximum permissible error in $Y$ is given by the formula for propagation of errors: $\frac{\Delta Y}{Y} \times 100\% = \left( \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta (\Delta L)}{\Delta L} \right) \times 100\%$ Assuming the error in diameter measurement is the dominant factor and considering the given options, the maximum permissible error is approximately 4.3%. This implies a specific precision for the diameter measurement that leads to this result.