$A$ 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 .............. $\Omega$.

The length of a potentiometer wire is $2.5 \ m$ and its resistance is $8 \ \Omega$. $A$ cell of negligible internal resistance and emf of $2.5 \ V$ is connected in series with a resistance of $242 \ \Omega$ in the primary circuit. The potential difference between two points separated by a distance of $20 \ cm$ on the potentiometer wire is (in $mV$)

In a potentiometer arrangement,$E_1$ is the cell establishing the current in the primary circuit,$E_2$ is the cell to be measured,$AB$ is the potentiometer wire,and $G$ is a galvanometer. Which of the following are the essential conditions for a balance point to be obtained?

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

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 $60 \; cm$. If the internal resistance of the cell is $\frac{N}{10} \; \Omega$,where $N$ is an integer,then the value of $N$ is:

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?