The length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

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:

The length of a potentiometer wire is $1200 \; cm$ and it carries a current of $60 \; mA$. For a cell of $emf \; 5 \; V$ and internal resistance of $20 \; \Omega$,the null point on it is found to be at $1000 \; cm$. The resistance of the whole wire is .............. $\Omega$.

In a potentiometer, the null point is received at the $7^{th}$ wire. If we now want to shift the null point to the $9^{th}$ wire, what should we do?