In a potentiometer experiment,when key $K_1$ is closed,the balance length is $100 \, cm$. What will be the balancing length when key $K_2$ is closed? ................ $cm$

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.

In a potentiometer, the null point is received at the $7^{th}$ wire. If we now want to shift the null point to the $9^{th}$ wire, what should we do?

Balancing point of a potentiometer shifts from a length of $60 \ cm$ to $40 \ cm$ by shunting the cell with a $4 \ \Omega$ resistance. What is the internal resistance of the cell (in $\Omega$)?