Two concentric spherical conducting shells of radii $R$ and $2R$ carry charges $Q$ and $2Q$ respectively, change in electric potential on the outer shell when both are connected by a conducting wire \[(k = \dfrac{1}{{4\pi \varepsilon _0}})\] is:
$\left( A \right)$ $Zero$
$\left( B \right)$ $\dfrac{{3kQ}}{{2r}}$
$\left( C \right)$ $\dfrac{{kQ}}{R}$
$\left( D \right)$ $\dfrac{{2kQ}}{R}$

Hint: Conductors contain mobile electric carriers. In metallic conductors, these charge carriers are electrons. Inside the conductor, the electrostatic field is zero.
Since the conducting shell is spherical by calculating the potential of the shell we get the answer.

Complete step by step answer:
The electrical potential, the amount of work necessary to transfer a unit charge against an electrical field from a reference point to a particular point
Here we take the spherical shell with the radius r.
Now calculating the potential of the charged spherical shell with radius r is as follows,
\[V = \dfrac{1}{{4\pi \varepsilon _0}}\dfrac{q}{r}\]
Since, \[(k = \dfrac{1}{{4\pi \varepsilon _0}})\]then,
\[V = k\dfrac{q}{r}\]
Since the two spherical shells carry charges Q and 2Q then if we add both the potentials we get the following equation \[{V_1}\].
\[\Rightarrow {V_1} = k\dfrac{Q}{{2R}} + k\dfrac{{2Q}}{{2R}}\]
\[\Rightarrow {V_1} = k\dfrac{{3Q}}{{2R}}\]
Now charge remains constant before and after connecting them, thus the final charge \[ = \]\[3Q\].
\[\Rightarrow {V_2} = k\dfrac{{3Q}}{{2R}}\]
Hence, by subtracting both the values we get the change in the potential and since the charge remains the same we get the answer zero.
$\Rightarrow \Delta V = {V_2} - {V_1} = 0$.

Thus the correct option will be $\left( A \right)$, thus making the other three options invalid.

Note: The electric potential or voltage is the difference in the potential energy per unit charge between two locations in an electric field.
In other words, it can be defined as the difference between the two charged bodies' electric potential.
The electron will accelerate toward a higher potential due to its negative charge.
The change in potential energy is due to the charge times the potential difference.
The change in potential energy equals the gain in the kinetic energy, which can also be used to find the speed.