$A$ wire of length $10 \ m$ and resistance $30 \ \Omega$ is connected to a battery of $emf$ $2.5 \ V$ and internal resistance $5 \ \Omega$ through an external resistance $R$. If the potential gradient along the wire is $50 \ \mu V/mm$,then $R = $ ................. $\Omega$.

In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

Two students $X$ and $Y$ perform a potentiometer experiment separately and the null point was obtained as shown in the diagram. During the experiment:
$(i)$ $X$ increases the value of $R$ (resistance).
(ii) $Y$ decreases the value of $S$ (resistance) without any other change.
The position of the null point obtained by students $X$ and $Y$ respectively:

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 length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.