To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $Omega$)

$A$ potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. $A$ cell $C$ having $emf$ $\varepsilon/2$ and internal resistance $3r$ is connected as shown in the figure. The length $AJ$ at which the galvanometer shows no deflection is

$A$ cell in the 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 $m$)

$A$ potentiometer wire of length $100 \, cm$ has a resistance of $10 \, \Omega$. It is connected in series with a resistance $R$ and a cell 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 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$)

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-