$A$ null point is found at $200\,cm$ in a potentiometer when the 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 $..............\,\Omega$.

If the resistivity of a potentiometer wire is $\rho$ and the area of cross-section is $A$,then what will be the potential gradient along the wire,given that a current $I$ flows through it?

Two cells of e.m.f. $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When a potentiometer is connected between points $A$ and $B$, the balancing length of the potentiometer wire is $412 \text{ cm}$. When the same potentiometer is connected between points $A$ and $C$, the balancing length is $103 \text{ cm}$. The ratio $E_1 : E_2$ is:

To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

In a potentiometer arrangement,a cell of $emf$ $1.25\,V$ gives a balance point at $35.0\,cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\,cm,$ the $emf$ of the second cell is ............... $V$.

Two cells having unknown e.m.f.s $E_{1}$ and $E_{2}$ $(E_{1} > E_{2})$ are connected in a potentiometer circuit so as to assist each other. The null point is obtained at $490 \ cm$ from the higher potential end. When cell $E_{2}$ is connected so as to oppose cell $E_{1}$,the null point is obtained at $90 \ cm$ from the same end. The ratio of the e.m.f.s of the two cells $(\frac{E_{1}}{E_{2}})$ is: