Question

In an experiment of determine the Young's modulus of wire of a length exactly $1\, m$, the extension in the length of the wire is measured as $0.4\, mm$ with an uncertainty of $\pm 0.02\, mm$ when a load of $1 \,kg$ is applied. The diameter of the wire is measured as $0.4 \,mm$ with an uncertainty of $\pm 0.01 \,mm$. The error in the measurement of Young's modulus $(\Delta Y )$ is found to be $X \times 10^{10} Nm ^{-2}$ . The value of $x$ is _______
[Take $g =10\, m / s ^2$ ]

Answer 1

$ L =1 m $
$ \Delta L =0.4 \times 10^{-3} m $
$ m =1 \,kg $
$ d =0.4 \times 10^{-3} m $
$ \frac{ F }{ A }= Y \frac{\Delta L }{ L }$
$ Y =\frac{ FL }{ A \Delta L }=\frac{( mg ) \cdot(1)}{\left(\frac{\pi d ^2}{4}\right) 0.4 \times 10^{-3}} $
$\Rightarrow \frac{10 \times 4}{\pi\left(0.4 \times 10^{-3}\right)^2 \times 0.4 \times 10^{-3}}$
$ Y =\frac{40}{\pi\left(0.4 \times 10^{-3}\right)^3} $
$ Y =\frac{40 \times 7}{22 \times 64 \times 10^{-3} \times 10^{-9}} $
$ Y =0.199 \times 10^{-12} N / m ^2$
$ \frac{\Delta Y }{ Y }=\frac{\Delta F }{ F }+\frac{\Delta L }{ L }+\frac{\Delta A }{ A }+\frac{\Delta(\Delta L )}{(\Delta L )} $
$ =\frac{0.02}{0.4}+2 \frac{\Delta d }{ d }=\frac{0.2}{4}+2 \times \frac{0.01}{0.4}$
$ =\frac{0.1}{2}+\frac{0.1}{2}=0.1$
$ \Rightarrow \Delta Y =0.1 \times Y $
$=0.199 \times 10^{11}=1.99 \times 10^{10}$