Millikan's Oil Drop Method

Cathode rays moving with same velocity $v$ describe an approximate circular path of radius $r \mathrm{metre}$ in an electric field of strength $x \mathrm{volt} {\mathrm{metre}}^{-1}$. If the speed of the cathode rays is doubled to $2v$, the value of electric field (in $\mathrm{volt} {\mathrm{metre}}^{-1}$) needed, so that the rays describe the same approximate circular path.

In Millikan's oil drop experiment, an oil drop is observed to move vertically upward. The upward motion of the drop is due to,

In Millikan oil-drop experiment a drop of charge $Q$ and radius $r$ is kept constant between two plates of potential difference of $800 \mathrm{V}$. Then charge on another drop of radius $2r$ which is kept constant with a potential difference of $3200 \mathrm{V}$ is

An oil drop carrying a charge $q$ has a mass $m \mathrm{kg}\text{.}$ It is falling freely in air with terminal speed $v$. The electric field required to make the drop move upwards with the same speed is

Doubly ionised helium atom and hydrogen ions are accelerated, from rest, through the same potential difference. The ratio of final velocities of helium and hydrogen is

An ionisation chamber, with parallel conducting plates as anode and cathodes have singly charged positive $5\times {10}^7$ ions per ${\mathrm{cm}}^3\text{.}$ The electrons, are moving towards the anode with velocity $0.4 \mathrm{m} {\mathrm{s}}^{-1}\text{.}$ The current density from anode to cathode is $4 \mathrm{\mu A} {\mathrm{m}}^{-2}\text{.}$ The velocity of positive ions moving towards cathode is

Which of these is non-divisible

It is possible to understand nuclear fission on the basis of the _____.

In Millikan's oil drop experiment, the charge on an oil drop is calculated to be $6.35\times {10}^{-19} \mathrm{C}$. The number of excess electrons on the drop is