$A$ potentiometer wire,$10 \, m$ long,has a resistance of $40 \, \Omega$. It is connected in series with a resistance box and a $2 \, V$ storage cell. If the potential gradient along the wire is $0.1 \, mV/cm$,the resistance unplugged in the box is .............. $\Omega$.

$A$ potentiometer is used to measure the potential difference between $A$ and $B$. 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 $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

In a potentiometer arrangement,a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.

In a potentiometer circuit,a cell of $2\,V$ $e.m.f.$ and $5\,\Omega$ internal resistance is connected to a uniform wire of length $1000\,cm$ and resistance $15\,\Omega$. The potential gradient of the wire is:

$A$ potentiometer wire has length $4\, m$ and resistance $8\, \Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$,so as to get a potential gradient of $1\, mV$ per $cm$ on the wire is ............. $\Omega$.

$A$ circuit for the measurement of resistance by a potentiometer is shown. The galvanometer is first connected at point $A$ and zero deflection is observed at a length $PJ = 10 \ cm$. In the second case,it is connected at point $C$ and zero deflection is observed at a length of $30 \ cm$ from $P$. Then the unknown resistance $X$ is: