$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)

The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be

$A$ resistance of $4\,\Omega$ and a wire of length $5\,m$ and resistance $5\,\Omega$ are joined in series and connected to a cell of $e.m.f.$ $10\,V$ and internal resistance $1\,\Omega$. $A$ parallel combination of two identical cells is balanced across $300\,cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$.

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

$A$ potentiometer balances at $44 \ cm$ when a cell of internal resistance $1 \ \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \ cm$,the resistance to be connected in parallel to the cell is: (in $\Omega$)

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 $E_1$.