$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a cell in the secondary circuit is shunted by $5 \ \Omega$. When a resistance of $15 \ \Omega$ is used for shunting,the null point moves to $300 \ cm$. The internal resistance of the cell is: (in $\Omega$)

Potentiometer measures the potential difference more accurately than a voltmeter because

The arrangement as shown in the figure is called:

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 $E_1$.

$A$ 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 $\Omega$)

In the figure,the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. The cell $C$'s $emf$ is $\frac{\varepsilon}{2}$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is