$A$ wire of length $100\, cm$ is connected to a cell of $emf$ $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential drop of $1\, mV/cm$ is ............... $\Omega$.

$A$ 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, the null point is received at the $7^{th}$ wire. If we now want to shift the null point to the $9^{th}$ wire, what should we do?

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to:

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

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: