$A$ 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 $\,V$)

$A$ 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:

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

$A$ potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ammeter $A$ in the circuit reads $1.0\, A$ when the two-way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when the two-way key $K_3$ is plugged in between $2$ and $1$,while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\frac{R_1}{R_2}$ is found to be

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

What is the advantage of using thick metallic strips to join wires in a potentiometer?