In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

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:

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?

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ and the balancing length is $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is:

The figure shows a simple potentiometer circuit for measuring a small $e.m.f.$ produced by a thermocouple. The potentiometer wire $PQ$ has a resistance of $5 \,\Omega$ and the driver cell has an $e.m.f.$ of $2 \, V$. If a balance point is obtained at $0.600 \, m$ along $PQ$ when measuring an $e.m.f.$ of $6.00 \, mV$,what is the value of resistance $R$ in $\Omega$?

In the experiment to determine the internal resistance $(r)$ of a cell $(E_1)$ using a potentiometer,the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of the internal resistance $(r)$ of the cell is: