Question

A student performs an experiment to determine the Young's modulus of a wire, exactly $2m$ along. 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 $0.05mm$ at a load of exactly $1.0kg$. The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $0.01mm$. 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 $(2.0\pm 0.2)\times {10}^{11}N{m}^{-2}$

$Y=4\frac{\mathrm{Mg}L}{x{d}^{2}d}$

$\text{Young's modulus Y, length L,extension in length (l), diameler of wire i d}$

$L=2m$ $\text{exactly}$

$0=0.8\pm 0.05m$

$m=1.0lcg$ $\text{exactly}$

$d=0.4+0.01mm$

$g=9.8m/s^2$$\text{exactly}$

$Fromeqn1$

$Y=\frac{4\times 1\times 9.8\times 2}{\pi \times {(0.4\times {10}^{-3})}^2\times 0.8}=2.0\times 10N/m^2$

$\text{taking log in equation}1\log Y=\log \frac{Y}{\pi }+\log M+\log g+\log L-2\log d-\log e$

$\text{differentiating}$

$\frac{dY}{Y}=-2\frac{d\left(d\right)}{d}-\frac{dl}{l}$

$\text{error always added}$

$\begin{array}{cc}\frac{dY}{Y}& =\frac{2d\left(d\right)}{d}+\frac{d\left(l\right)}{d}\\ dY& =Y(\frac{2d\left(d\right)}{d}+\frac{d\left(l\right)}{l})\\ & =2\times {10}^{11}(2\times \frac{0.01}{0.4}+\frac{0.05}{0.8})\\ & =2\times {10}^{11}\left(0.1125\right)\\ & =0.225\times {10}^{11}\\ & \end{array}$