Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively $360 \ cm$ and $420 \ cm$. If the emf of $A$ is $2.4 \ V$,the emf of the second cell $B$ is (in $V$)

The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

$A$ battery of emf $10 \, V$ is connected to a uniform wire $AB$ of $1 \, m$ length and having a resistance of $10 \, \Omega$ in series with a $10 \, \Omega$ resistor as shown in the figure. Two cells of emf $2 \, V$ each, having internal resistance $2 \, \Omega$ each, are connected in parallel as shown in the figure. If the galvanometer shows null deflection at point $J$ on the wire, then the distance of point $J$ from the point $B$ is. (in $ \, cm$)

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

$A$ potentiometer wire has a length of $100 \ cm$ and is connected to a cell of $emf \ E$. It is used to measure the $emf \ E_0$ of a battery with an internal resistance of $0.5 \ \Omega$. If the balance point is obtained at a distance of $\ell = 30 \ cm$ from the positive terminal,the $emf \ E_0$ of the battery is:

Let $A$ be the cross-sectional area and $\rho$ be the specific resistance (resistivity) of a potentiometer wire. If $I$ is the current passing through the wire,then the potential gradient along the length of the wire is