$A$ 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 $E_{1} > E_{2}$.

$A$ potentiometer wire is $4 \,m$ long and a potential difference of $3 \,V$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \,cm$ of the potentiometer wire is: (in $\,V$)

In the experiment to determine the internal resistance $(r)$ of a cell $(E_1)$ using a potentiometer,the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of the internal resistance $(r)$ of the cell is:

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.

$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)

As shown in the figure,a potentiometer wire of resistance $20\,\Omega$ and length $300\,cm$ is connected with a resistance box ($R$.$B$.) and a standard cell of emf $4\,V$. For a resistance '$R$' of the resistance box introduced into the circuit,the null point for a cell of $20\,mV$ is found to be $60\,cm$. The value of '$R$' is $.....\Omega$