Question

A wire $AB$ (of length $1m,$ area of cross section $\pi m^2$ ) is used in potentiometer experiment to calculate emf and internal resistance $\left(r\right)$ of battery. The emf and internal resistance of driving battery are $15V$ and $3\Omega$ respectively. The resistivity of wire $AB$ varies as $\rho ={\rho }_{0}x,$ where $x$ is distance from $A$ in meters and ${\rho }_0=24\pi \Omega .$ The distance of null point from $A$ is obtained at $\sqrt{\frac{2}{3}}m$ when switch ${}^{\prime }{S}^{\prime }$ is open. Find the value of $E.$

Answer 1

Answer: 8

$R_{AB}=\underset{0}{\overset{1}{\int }}\frac{{\rho }_{0}XdX}{A}$
$=\underset{0}{\overset{1}{\int }}\frac{24\pi XdX}{A}=12\Omega$
$R_{AG}=\underset{0}{\overset{\sqrt{\frac{2}{3}}}{\int }}\frac{24\pi XdX}{\pi }=\frac{24}{2}{\left[X^2\right]}_0^{\sqrt{\frac{2}{3}}}$
$=12\times \frac{2}{3}\Omega$
$i_{in}$ wire $=\frac{15}{3+12}=1A$
$V_{AG}=1\times 8=8V$