In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)

Potentiometer measures the potential difference more accurately than a voltmeter because

$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

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

$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $3 \text{ V}$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)