$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is

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

For a cell,the balancing length is $0.60 \, m$. For another cell having an electromotive force $(emf)$ $0.1 \, V$ less than the first one,the balancing length is $0.55 \, m$. What are the $emf$ values of the two cells?

$A$ cell in a secondary circuit gives a null deflection for $2.5 \ m$ length of a potentiometer wire having a total length of $10 \ m$. If the length of the potentiometer wire is increased by $1 \ m$ without changing the cell in the primary circuit,the new position of the null point is: (in $m$)

In a potentiometer arrangement,a cell of $emf$ $1.25\,V$ gives a balance point at $35.0\,cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\,cm,$ the $emf$ of the second cell is ............... $V$.

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?