Q6 · Errors

Problem

Young’s modulus Y = (4MgL)/($\pi$d²λ) from Searle’s method. Compare error contributions from d and λ measurements (same screw gauge: pitch 0.5 mm, 100 divisions).

Options

Option	Value
(A)	Errors from d and λ are the same
(B)	Error from d is twice that from λ
(C)	Error from λ is twice that from d
(D)	Error from d is four times that from λ

1. Propagation

ΔY/Y = Δλ/λ + 2(Δd/d). Both gauges give same fractional least count.

$$\frac{\Delta Y}{Y} = \frac{\Delta\lambda}{\lambda} + 2\frac{\Delta d}{d}$$

2. Conclusion

With identical instruments, both contributions are equal.

Hints

• Y ∝ λ/d² — use fractional error propagation.

Answer

(A)