Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is........$times$

Two spherical conductors $A$ and $B$ of radius $a$ and $b (b > a)$ are placed in air concentrically $B$ is given charge $+ Q$ coulomb and $A$ is grounded. The equivalent capacitance of these is

The radius of a metallic sphere if its capacitance is $1/9\,F$, is

Capacitance of an isolated conducting sphere of radius $R_{1}$ becomes $n$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $\left(\frac{ R _{2}}{ R _{1}}\right)$ is:

Two identical charged spherical drops each of capacitance $C$ merge to form a single drop. The resultant capacitance is

A particle of charge $Q$ and mass $M$ moves in a circular path of radius $R$ in a uniform magnetic field of magnitude $B$. The same particle now moves with the same speed in a circular path of same radius $R$ in the space between the cylindrical electrodes of the cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $ 3R/2.$ Then the potential difference between the capacitor electrodes must be