In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

In the experiment to determine the internal resistance $(r)$ of a cell $(E_1)$ using a potentiometer,the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of the internal resistance $(r)$ of the cell is:

$A$ potentiometer wire of length $4 \text{ m}$ and resistance $5 \Omega$ is connected in series with a resistance of $992 \Omega$ and a cell of e.m.f. $4 \text{ V}$ with internal resistance $3 \Omega$. The length of $0.75 \text{ m}$ on the potentiometer wire balances the e.m.f. of: (in $\text{ mV}$)

In the given potentiometer circuit,the balancing length for points $B$ and $C$ is $40 \, cm$. What is the balancing length for points $C$ and $D$ in $cm$?

The length of a potentiometer wire is $l$. $A$ cell of $emf$ $E$ is balanced at a length $l/3$ from the positive end of the wire. If the length of the wire is increased by $l/2$,at what distance will the same cell give a balance point?

$A$ potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ammeter $A$ in the circuit reads $1.0\, A$ when the two-way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when the two-way key $K_3$ is plugged in between $2$ and $1$,while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\frac{R_1}{R_2}$ is found to be