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

The material of the wire of a potentiometer is

$A$ potentiometer wire of length $10 \,m$ and resistance $20 \,\Omega$ is connected in series with a $25 \,V$ battery and an external resistance $30 \,\Omega$. $A$ cell of emf $E$ in the secondary circuit is balanced by a $250 \,cm$ long potentiometer wire. The value of $E$ (in volt) is $\frac{x}{10}$. The value of $x$ is.......

$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

$AB$ is a potentiometer wire as shown in the figure. If the value of $R$ is increased,in which direction will the balance point $J$ shift?