$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $Omega$)

$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?

$A$ 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:

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$,then using the same cell,the null point is obtained at: