$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

In an experiment of a potentiometer for measuring the internal resistance of a primary cell,a balancing length $\ell$ is obtained on the potentiometer wire when the cell is in an open circuit. Now,the cell is short-circuited by a resistance $R$. If $R$ is equal to the internal resistance of the cell,the balancing length on the potentiometer wire will be:

For comparing the $e.m.f.$'s of two cells with a potentiometer,a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful?

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

The potential gradient along the length of a uniform wire is $10\,V/m$. $B$ and $C$ are two points at $30\,cm$ and $60\,cm$ on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be ............. $V$.