Question

A standard cell of $1.08 \, V$ is balanced by the potential difference across $90 \, cm$ 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

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