Question

To measure the diameter of a wire, a screw gauge is used. In a complete radiation, spridle of the screw gauge advances by $\frac{1}{2}$ mm and its circular scale has 50 deviations. The main scale a graduated to $\frac{1}{2}$mm. If the wire is put between the jaws, 4 main scale divisions are clearly visible and 1 dimensions of circular scale co-inside with the reference line. The resistance of the wire is measured to be $(10\Omega \pm 1$. Length of the wire is measured to be 10 cm using a scale of least circuit 1 mm. Maximum permissible error in resistivity measurement is:

• A. 1.5%
• B. 2%
• C. 2.9%
• D. 3%

Answer 1

The correct option is C 2.9%

$\begin{array}{cc}\text{pitch}& =\frac{1}{2}mm=0.5mm\\ \text{division on circular scale}& =50\\ \text{Least count}L\cdot C& =\frac{\text{pitch}}{n}\\ & =\frac{0.5}{50}\\ & =0.01mm\end{array}$

$\text{value of}1\text{main scale division}=0.5mm$

$\text{Resistance}R=\rho \frac{l}{A}$

$\begin{array}{c}\text{taking log both sides}\\ \ln R=\ln \rho +\ln \ell -\ln A\end{array}$

differentiating buth sides

$\frac{dR}{12}=\frac{d\rho }{\rho }+\frac{dl}{l}\frac{dA}{A}$

for error equation, error is always added,

'

$\frac{dR}{R}=\frac{d\rho }{\rho }+\frac{dl}{l}+\frac{dA}{A}-\left(i\right)$

$\text{Reading of screw gauge}=\frac{M\cdot S\cdot R}{}+\text{C.S}R$

$\begin{array}{c}=4\times 0.5mm+1\times 0.01\\ =2.01mm\end{array}$

$\begin{array}{cc}A& =\pi r^2\\ dA& =2\pi rdr\end{array}$

$\frac{dA}{A}=\frac{2\pi rdr}{\pi r^2}$$\frac{2dr}{r}=\frac{2d\left(2r\right)}{2r}$

$\begin{array}{c}\frac{d\rho }{\rho }=\frac{dR}{R}+\frac{dl}{l}+\frac{dA}{A}\\ \frac{dp}{p}=\frac{dR}{R}+\frac{dl}{l}+2\frac{d\left(2r\right)}{2r}\end{array}$

$\frac{dp}{\rho }=\frac{0.1}{10}+\frac{0.1}{10}+2\times \frac{0.01}{2.01}$$=0.29$