In the potentiometer circuit as shown in the figure,the balance length $l = 60 \ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5 \ \Omega$,the balance length $l' = 50 \ cm$. The internal resistance of the cell $C'$ is : .............. $\Omega$

In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)

$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $Omega$)

$A$ potentiometer is connected between $A$ and $B$ and the balance point is obtained at $203.6 \, cm$. When the end of the potentiometer connected to $B$ is shifted to $C$,then the balance point is obtained at $24.6 \, cm$. If now the potentiometer be connected between $B$ and $C$,the balance point will be at ................. $cm$.