In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)

$A$ 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 $cm$)

$A$ potentiometer wire of length $100 \, cm$ has a resistance of $10 \, \Omega$. It is connected in series with a resistance $R$ and a cell of $emf$ $2 \, V$ and of negligible internal resistance. $A$ source of $emf$ $10 \, mV$ is balanced against a length of $40 \, cm$ of the potentiometer wire. What is the value of external resistance $R$?

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

$A$ potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. $A$ constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is ... $\Omega$.

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.