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?

With a potentiometer,null points were obtained at $140 \, cm$ and $180 \, cm$ with cells of $emf$ $1.1 \, V$ and one unknown $X \, V$. The unknown $emf$ is .............. $V$.

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

In the given arrangement,$E_1 = 5 \, V$ and $E_2 = 7 \, V$. The balancing length is $6 \, m$. If the terminals of $E_2$ are reversed,then the new balancing length will be:

For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

In a potentiometer experiment,a wire of length $10 \ m$ and resistance $5 \ \Omega$ is connected to a cell of emf $2.2 \ V$. If the potential difference between two points separated by a distance of $660 \ cm$ on the potentiometer wire is $1.1 \ V$,then the internal resistance of the cell is: (in $\Omega$)