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:

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:

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

The length of a potentiometer wire is $1200 \; cm$ and it carries a current of $60 \; mA$. For a cell of $emf \; 5 \; V$ and internal resistance of $20 \; \Omega$,the null point on it is found to be at $1000 \; cm$. The resistance of the whole wire is .............. $\Omega$.

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is: