In a potentiometer experiment,the balancing length for a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes half the initial balancing length. The internal resistance of the cell is: (in $\Omega$)

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.

$A$ potentiometer is used to measure the potential difference between $A$ and $B$,and the null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. Find the ratio $\frac{E_{2}}{E_{1}}$,given that $E_{1} > E_{2}$.

What is the advantage of using thick metallic strips to join wires in a potentiometer?

$A$ 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 $\Omega$)

In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell is