Question

The diameter of a wire of length $100cm$ is measured with the help of a screw gauge. The main scale reading is $1mm$ and circular scale reading is $25$. Pitch of the screw gauge is $1mm$ and the total number of divisions on the circular scale is $100$. The wire is used in an experiment for determination of Young’s modulus of the wire by Searle’s method. The following data are available: elongation in the wire $\Delta l=0.125cm$ under the tension of $50N$, least count for measuring normal length of wire is $0.01cm$ and for elongation in the wire is $0.001cm$. The maximum error in calculating the value of Young’s modulus $\left(Y\right)$, assuming that the force is measured very accurately, is $\frac{(8\times n)}{10\%}$ , where n is very nearly an integer. Find the value of $n$.

Answer 1

Given, main scale reading $=1mm$,
circular scale reading $=25$,
Pitch of the screw gauge $=1mm$,
The total number of divisions on the circular scale $=100$.
$LC$ of screw gauge $=\frac{1}{100}=0.01mm$
Total diameter reading,

$d=$ Main scale reading $+LC\times$ (Circular scale reading )

$d=1mm+0.01\times 25=1.25mm$

Young’s modulus is given by
$Y=\frac{F}{A}\frac{l}{\Delta l}$

$\therefore \frac{\Delta Y}{Y}\times 100=(\frac{\Delta A}{A}+\frac{\Delta l}{l}+\frac{\Delta \left(\Delta l\right)}{\Delta l})\times 100$

$(\frac{2\Delta d}{d}+\frac{\Delta l}{l}+\frac{\Delta \left(\Delta l\right)}{\Delta l})\times 100$

$(\frac{2\times 0.01}{1.25}+\frac{0.01}{100}+\frac{0.001}{0.125})\times 100$

$\frac{\Delta Y}{Y}\times 100=\frac{30125}{12500}\%\dots \left(ii\right)$

Given,

$\frac{\Delta Y}{Y}\times 100=\frac{8}{10}\times n\%\dots \left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$, we get

$n=3.0125\approx 3$

Final answer:$n=3$