$A$ null point is obtained at $200 \ cm$ on a potentiometer wire when a 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: (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

$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $16 \, \Omega$. The driving cell has an e.m.f. of $5 \, V$ and an internal resistance of $4 \, \Omega$. When two cells of e.m.f.s $1.3 \, V$ and $1.1 \, V$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively:

$A$ 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:

In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)

The balancing length for a cell is $560 \; cm$ in a potentiometer experiment. When an external resistance of $10 \; \Omega$ is connected in parallel to the cell,the balancing length changes by $60 \; cm$. If the internal resistance of the cell is $\frac{N}{10} \; \Omega$,where $N$ is an integer,then the value of $N$ is: