$A$ 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 $\Omega$)

$A$ 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

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

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

In a potentiometer experiment, the null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased without changing the cell or the driving source, the balancing length will: