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 an experiment with a potentiometer,$V_B = 10 \, V$ and the variable resistance is adjusted to $R = 50 \, \Omega$ (figure). $A$ student wanting to measure the voltage $E_1$ of a battery (approx. $8 \, V$) finds no null point. He then diminishes $R$ to $10 \, \Omega$ and is able to locate the null point on the last $(4^{th})$ segment of the potentiometer wire. Find the resistance of the potentiometer wire and the potential drop per unit length.

The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \text{ m}$ and its area of cross-section is $8 \times 10^{-6} \text{ m}^2$. If $0.2 \text{ A}$ current is flowing through the wire, the potential gradient of the wire is:

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.

$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)