In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell 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 along with $E_1$ such that the two cells oppose each other, and the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is

In a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

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

The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.