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 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:

The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire,then

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

In a potentiometer experiment,a cell is balanced at a length of $240 \ cm$. When the cell is shunted with a resistance of $2 \ \Omega$,it is balanced at a length of $120 \ cm$. The internal resistance of the cell is .......... $\Omega$.

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-