$A$ battery of $6\,V$ is connected to the terminals of a $3\,m$ long wire of uniform thickness and resistance of the order of $100\,\Omega$. The difference of potential between two points separated by $50\,cm$ on the wire will be .......... $V$.

$A$ wire of length $10 \ m$ and resistance $30 \ \Omega$ is connected to a battery of $emf$ $2.5 \ V$ and internal resistance $5 \ \Omega$ through an external resistance $R$. If the potential gradient along the wire is $50 \ \mu V/mm$,then $R = $ ................. $\Omega$.

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is:

In a potentiometer experiment,the balancing length for a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes half the initial balancing length. The internal resistance of the cell is: (in $Omega$)

$A$ cell can be balanced against $110 \ cm$ and $100 \ cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \ \Omega$. Its internal resistance is

When a cell of $E.M.F.$ $E_1$ is connected to a potentiometer wire,the balancing length is $l_1$. Another cell of $E.M.F.$ $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is