Young's Modulus By Searle's Method

NCERT Class 11, Chapter 9 defines Young's modulus as $Y=\frac{FL}{A\Delta L}$, which rearranged gives the extension $\Delta L=\frac{FL}{AY}$ for an applied force $F$. This shows the extension is directly proportional to the original length $L$ and inversely proportional to the cross-sectional area $A$. Hence a long wire (large $L$) and a thin wire (small $A$) together maximise the extension for a given load, making it large enough to be measured accurately against the small least count of the micrometer in Searle's apparatus. The option favouring a short thick wire is wrong because both a smaller length and a larger area reduce extension, giving a value too tiny to read precisely. The option claiming dimensions have no effect is wrong because the extension formula depends explicitly on both length and area. The option about breaking safely is wrong because wire choice aims at measurability, not deliberate failure. Searle's apparatus uses two identical wires side by side, one loaded as the experimental wire and one carrying a reference, so that thermal expansion and support sag affect both equally and cancel out, isolating the true elastic extension. The extension is read with a spherometer or micrometer whose least count is around 0.01 mm, so a larger extension keeps the measurement well above this resolution limit. A consistency check confirms that increasing $L$ and decreasing $A$ both enlarge $\Delta L$, which is exactly what accurate measurement requires while keeping the wire within its elastic limit.