In a potentiometer, when the cell in the secondary circuit is shunted with a $4 \ \Omega$ resistance, the balance is obtained at a length of $120 \ cm$ of the wire. Now, when the same cell is shunted with a $12 \ \Omega$ resistance, the balance point shifts to a length of $180 \ cm$. The internal resistance of the cell is . . . . . . $\Omega$.

$A$ 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 $cm$)

$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

In a potentiometer of $10$ wires,the balance point is obtained on the $6^{\text{th}}$ wire. To shift the balance point to the $8^{\text{th}}$ wire,we should:

$A$ potentiometer wire of length $1\,m$ and resistance $10\,\Omega$ is connected in series with a cell of $emf$ $2\,V$ with internal resistance $1\,\Omega$ and a resistance box including a resistance $R$. If the potential difference between the ends of the wire is $1\,mV$,the value of $R$ is ............. $\Omega$.

In the experiment of calibration of a voltmeter,a standard cell of $e.m.f. = 1.1 \text{ V}$ is balanced against $440 \text{ cm}$ of a potentiometer wire. The potential difference across a resistance is found to balance against $220 \text{ cm}$ of the wire. The corresponding reading of the voltmeter is $0.5 \text{ V}$. The error in the reading of the voltmeter will be ................. $V$.