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 ............... $V$.

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 ........... $\Omega$.

In a potentiometer arrangement,a cell gives a balancing point at $75 \, cm$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3:2$,the difference in the balancing length of the potentiometer wire in above two cases will be ......... $cm$.

$A$ potentiometer circuit shown in the figure is set up to measure the $e.m.f.$ of a cell $E$. As the point $P$ moves from $X$ to $Y$, the galvanometer $G$ shows deflection always in one direction, but the deflection decreases continuously until $Y$ is reached. In order to obtain a balance point between $X$ and $Y$, it is necessary to

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to:

$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be: