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 0.4mm with an uncertainty of $\pm \ 0.01\ mm.$ Take $g = 9.8\ {ms}^{- 2}$ (exact). The Young’s modulus obtained from the reading is
Explanation
Young’s modulus $Y = \frac{FL}{Al} = \frac{4FL}{\pi d^{2}l}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{(4)(1.0 \times 9.8)(2)}{\pi\left( 0.4 \times 10^{- 3} \right)^{2}\left( 0.8 \times 10^{- 3} \right)}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.0 \times 10^{11}\ {Nm}^{- 2}$ Further, $\frac{\mathrm{\Delta}Y}{Y} = 2\left( \frac{\mathrm{\Delta}d}{d} \right) + \left( \frac{\mathrm{\Delta}l}{l} \right)$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}Y = \left\{ 2\left( \frac{\mathrm{\Delta}d}{d} \right) + \left( \frac{\mathrm{\Delta}l}{l} \right) \right\}\ Y$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\{ 2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8} \right\} \times 2.0 \times 10^{11}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.2 \times 10^{11}\ {Nm}^{- 2}$ Or $\ \ \ \ \ \ \ \ \ \ \ (Y + \mathrm{\Delta}Y) = (2 + 0.2) \times 10^{11}\ {Nm}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.