$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

In a potentiometer experiment, the balancing length with a cell is $560 \, cm$. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. Find the internal resistance of the cell.

$A$ $10 \, m$ long wire of resistance $20 \, \Omega$ is connected in series with a battery of e.m.f. $3 \, V$ (negligible internal resistance) and a resistance of $10 \, \Omega$. The potential gradient along the wire is (in $ \, V/m$)

In the determination of the internal resistance of a cell with a potentiometer,the error in the measurement of the balancing length is $\pm 1 \text{ mm}$. When the cell alone is connected in the circuit,the balancing length is obtained at $60 \text{ cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$,the balancing length is obtained at $50 \text{ cm}$. The error in the determination of the internal resistance of the cell is (in $\%$)

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 circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be