$A$ potentiometer wire is $4 \,m$ long and a potential difference of $3 \,V$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \,cm$ of the potentiometer wire is: (in $V$)

Balancing point of a potentiometer shifts from a length of $60 \ cm$ to $40 \ cm$ by shunting the cell with a $4 \ \Omega$ resistance. What is the internal resistance of the cell (in $\Omega$)?

$A$ potentiometer wire has length $10\, m$ and resistance $20\,\Omega$. $A$ $2.5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega$ series resistance. The potential gradient on the wire will be

In the arrangement shown in the figure,when the switch $S_2$ is open,the galvanometer shows no deflection for $l = L/2$. When the switch $S_2$ is closed,the galvanometer shows no deflection for $l = 5L/12$. The internal resistance $(r)$ of the $6\, V$ cell and the $emf$ $E$ of the other battery are respectively:

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$,then using the same cell,the null point is obtained at:

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should: