Question

A standard cell of $1.08V$ is balanced by the potential difference across $90cm$ of a $1$ meter, long wire supplied by a cell of emf $2V$ through a series resistor of resistance $2\Omega$ . If the internal resistance of cell in the primary circuit is zero then the resistance per unit length (in $\Omega m^{-1}$ ) of potentiometer wire is

Answer 1

Answer: 3

$E=X.l(X=\text{Potential}\text{gradient})$
$E=\left(\frac{E_o}{L}\times \frac{R_w}{R_w+r+R_h}\right)\times l\left(\because r=0,R_h=2\Omega \right)$
$\therefore 1.08=\left(\frac{2}{1m}\times \frac{R_w}{R_w+2}\right)\times 0.90m$
$\frac{1.08}{2\times 0.90}=\frac{R_w}{R_w+2}$
Hence : $0.6=\frac{R_w}{R_w+2}$
$0.6R_w+1.2=R_w\Rightarrow 1.2=0.4R_w\Rightarrow 3=R_w$
$\therefore$ Resistance per unit length is $3\Omega m^{-1}$