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 $0.8\, mm$ with an uncertainty of $\pm 0.05 \, mm $ at a load of exactly $1.0\, kg$. The student also measures the diameter of the wire to be $0.4\, mm$ with an uncertainty of $\pm 0.01 \, mm .$ Take $g = 9.8 \, m/s^2 $ (exact). The Young's modulus obtained from the reading is close to

• A. $(2.0 \pm; 0.3)10^{11}N/m^2$
• 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

Answer: (B)

$Y= \frac {FL}{Al}=\frac {4FL}{\pi d^2l} $
$=\frac {(4)(1.0 \times 9.8)(2)}{\pi (0.4 \times 10^{-3})^2(0.8 \times 10^{-3})}$
$ =2.0 \times 10^{11} \, N/m^2 $
Further $\frac {\Delta;Y}{Y}=2 \left(\frac {\Delta;d}{d}\right)+ \left(\frac {\Delta;l}{l} \right) $
$\therefore \Delta ; Y=\left\{2\left(\frac{\Delta ; d}{d}\right)+\left(\frac{\Delta ; l}{l}\right)\right\} Y$
$ =\left\{2 \times \frac {0.01}{0.4}+ \frac {0.05}{0.8}\right\}\times 2.0 \times 10^{11}$
$ =0.225 \times 10^{11} \, N/m^2 $
$ =0.2\times 10^{11} \, N/m^2$ (By rounding off )
or $(Y+ \Delta;Y)=(2+0.2) \times 10^{11} \, N/m^2 $