In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)

In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.

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 $\text{ A}$)

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to: