$A$ potentiometer has a potential gradient of $2 \, mV/cm$. It is used to measure the potential difference across a $10 \, \Omega$ resistor. If the length of the potentiometer wire required to obtain the null point is $50 \, cm$,the current flowing through the $10 \, \Omega$ resistor is ............. $mA$.

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?

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 ............... $cm$.

In an experiment to find the emf of a cell using a potentiometer, the length of the null point for a cell of emf $1.5 \text{ V}$ is found to be $60 \text{ cm}$. If this cell is replaced by another cell of emf $E$, the length of the null point increases by $40 \text{ cm}$. The value of $E$ is $x/10 \text{ V}$. The value of $x$ is