$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length '$x$' cm. If the length of the potentiometer wire is increased by a few cm without changing the cell or the driving source, the balancing length will:

The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because:

$A$ battery of internal resistance $1 \, \Omega$ and $emf$ $3 \, V$ sends a current through $1 \, m$ of uniform wire of resistance $5 \, \Omega$. The poles of a cell of $emf$ $1.4 \, V$ are connected to two points on the wire such that no current passes through this cell. The potential gradient of the wire is: