Q. 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 $\pm 0.01mm$. Take $g = 9.8\ m/s^{2}$ (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) $(2.0 \pm 0.05) \times 10^{11}N/m^{2}$
Explanation
(b) $Y = \frac{FL}{Al} = \frac{4FL}{\pi l^{2}l};F = mg$
Where L = length of the wire
$l =$ elongation of the wire
d = diameter of the wire
substituting the values, we get $Y = 2 \times 10^{11}N/m^{2}$
$\Rightarrow \frac{\mathrm{\Delta}Y}{Y} = 2\frac{\mathrm{\Delta}d}{d} + \frac{\mathrm{\Delta}l}{l} = 2\left( \frac{0.01}{0.4} \right) + \frac{0.05}{0.8} = \frac{9}{80}$
$\Rightarrow \mathrm{\Delta}Y = \frac{9}{80} \times Y = \frac{9}{80} \times 2 \times 10^{11} = 0.2 \times 10^{11}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.