The $e.m.f.$ of a standard cell balances across $150 \ cm$ length of a potentiometer wire. When a resistance of $2 \ \Omega$ is connected as a shunt across the cell,the balance point is obtained at $100 \ cm$. The internal resistance of the cell is .............. $\Omega$.

In a potentiometer circuit, there is a cell of $e.m.f.$ $2\, V$, a resistance of $5\, \Omega$ and a wire of uniform thickness of length $1000\, cm$ and resistance $15\, \Omega$. The potential gradient in the wire is:

$A$ potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ammeter $A$ in the circuit reads $1.0\, A$ when the two-way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when the two-way key $K_3$ is plugged in between $2$ and $1$,while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\frac{R_1}{R_2}$ is found to be

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.

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 ................. $\Omega$.

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: