$A$ potentiometer wire of length $1 \ m$ is connected in series with $495 \ \Omega$ resistance and $2 \ V$ battery. If $0.2 \ mV/cm$ is the potential gradient,then the resistance of the potentiometer wire is: (in $Omega$)

If the length of a potentiometer wire is increased,then the length of the previously obtained balance point will

In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.

$A$ potentiometer wire has length $L$. For a given cell of emf $E$,the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of the potentiometer wire is increased by $50 \%$,then for the same cell,the balance point is obtained at length

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)

$A$ 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: