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

$A$ null point is found at $200\,cm$ in a potentiometer when the cell in the secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting,the null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.

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:

In a potentiometer arrangement,a cell of $emf$ $1.25\; V$ gives a balance point at $35.0\; cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\; cm ,$ what is the $emf$ of the second cell in $V$?

What is the advantage of using thick metallic strips to join wires in a potentiometer?

The length of a wire of a potentiometer is $100 \, cm$,and the $emf$ of its standard cell is $E \, volt$. It is employed to measure the $emf$ of a battery whose internal resistance is $0.5 \, \Omega$. If the balance point is obtained at $l = 30 \, cm$ from the positive end,the $emf$ of the battery is (where $i$ is the current in the potentiometer wire).