$A$ 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 $\text{ V}$)

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

$A$ galvanometer can be used as a voltmeter by connecting a

$A$ potentiometer having a potential gradient of $2\, mV/cm$ is used to measure the potential difference across a resistance of $10\, \Omega$. If a length of $50\, cm$ of the potentiometer wire is required to obtain the null point,the current passing through the $10\, \Omega$ resistor is (in $mA$):

$A$ potentiometer wire of length $4 \, m$ and resistance $5 \, \Omega$ is connected in series with a resistance of $992 \, \Omega$ and a cell of e.m.f. $4 \, V$ with internal resistance $3 \, \Omega$. The length of $0.75 \, m$ on the potentiometer wire balances the e.m.f. of (in $ \, mV$)

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?