To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

In the given figure,there is a circuit of a potentiometer of length $AB = 10 \, m$. The resistance per unit length is $0.1 \, \Omega/cm$. $A$ battery of $6 \, V$ and an internal resistance of $20 \, \Omega$ is connected across $AB$. The maximum value of emf that can be measured by this potentiometer is (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:

$A$ potentiometer wire of length $100 \ cm$ and resistance $3 \ \Omega$ is connected in series with a resistance of $8 \ \Omega$ and an accumulator of $4 \ V$ whose internal resistance is $1 \ \Omega$. $A$ cell of e.m.f. $E$ is balanced by $50 \ cm$ length of the wire. The e.m.f. of the cell is: (in $V$)

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

The figure shows a simple potentiometer circuit for measuring a small $e.m.f.$ produced by a thermocouple. The potentiometer wire $PQ$ has a resistance of $5 \,\Omega$ and the driver cell has an $e.m.f.$ of $2 \, V$. If a balance point is obtained at $0.600 \, m$ along $PQ$ when measuring an $e.m.f.$ of $6.00 \, mV$,what is the value of resistance $R$ in $\Omega$?