Question

$A$ student performs an experiment to determine the Young's modulus of a wire, exactly 2 $m$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8mm$ with an uncertainty of $\pm 0.05$ mm at a load of exactly $1.0kg$. The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm 0.01$ mm. Take $g=9.8m/s^2$ (exact). The Young's modulus obtained from the reading is

• A. $(2.0\pm 0.3)\times {10}^{11}N/m^2$
• B. $(2.0\pm 0.2)\times {10}^{11}N/m^2$
• C. $(2.0\pm 0.1)\times {10}^{11}N/m^2$
• D. $(2.0\pm 0.05)\times {10}^{11}N/m^2$

Answer 1

The correct option is B $(2.0\pm 0.2)\times {10}^{11}N/m^2$

Young's modulus $=\frac{FL}{Al}$

$Y=\frac{mg\times 2}{\pi \times (\frac{d}{2}{)}^2\times 0.8\times {10}^{-3}}=1.96\times {10}^{11}N/m^2$

$\frac{\Delta Y}{Y}=\frac{2\Delta d}{d}+\frac{\Delta \left(\Delta l\right)}{\Delta l}$

$=\frac{2\left(0.01\right)}{0.4}+\frac{0.05}{0.8}$

$=0.1125$

$\Delta Y=.1125Y=0.2205\times {10}^{11}N/m^2$

$Y=(2\pm 0.2)\times {10}^{11}N/m^2$