$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

$A$ potentiometer wire has a length of $4 \ m$ and a resistance of $10 \ \Omega$. It is connected to a cell of $2 \ V$ emf. The potential gradient (potential difference per unit length) of the wire is: (in $V/m$)

The sliding contact of a potentiometer is in the middle of the potentiometer wire having a total resistance $R_p = 1 \Omega$,as shown in the figure. An external resistance of $R_e = 2 \Omega$ is connected via the sliding contact. Find the total current drawn from the $0.9 \text{ V}$ battery. (in $\text{ A}$)

The $e.m.f.$ of a standard cell balances across $150 \ cm$ length of a potentiometer wire. When a resistance of $2 \ \Omega$ is connected as a shunt across the cell,the balance point is obtained at $100 \ cm$. The internal resistance of the cell is .............. $\Omega$.

In a potentiometer circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is: