The length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

$A$ potentiometer wire of length $100\, cm$ has a resistance of $10\, \Omega$. It is connected in series with a resistance $R$ and an accumulator of emf $2\, V$ and of negligible internal resistance. $A$ source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance $R$?

$A$ potentiometer is used to measure the potential difference between $A$ and $B$,and the null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. Find the ratio $\frac{E_{2}}{E_{1}}$,given that $E_{1} > E_{2}$.

In a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.

$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.