In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

$A$ $6\,V$ battery of negligible internal resistance is connected across a uniform wire $AB$ of length $100\,cm$. The positive terminal of another battery of $emf$ $4\,V$ and internal resistance $1\,\Omega$ is joined to the point $A$ as shown in the figure. Take the potential at $B$ to be zero. At which point $D$ of the wire $AB$,measured from $A$,is the potential equal to the potential at $C$? ...................... $cm$ (approximately)

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:

$A$ potentiometer is used to measure the potential difference between $A$ and $B$,and the null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. Find the ratio $\frac{E_{2}}{E_{1}}$,given that $E_{1} > E_{2}$.

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

In the figure,the potentiometer wire of length $l = 100 \, cm$ and resistance $R = 9 \, \Omega$ is joined to a cell of emf $E_1 = 10 \, V$ and internal resistance $r_1 = 1 \, \Omega$. Another cell of emf $E_2 = 5 \, V$ and internal resistance $r_2 = 2 \, \Omega$ is connected as shown. The galvanometer $G$ will show no deflection when the length $AC$ is ............... $cm$.