Angular momentum of a planet of mass m orbiting around sun is J, areal velocity of its radius vector will be

The correct option is J/2m.

Here is the detailed step-by-step derivation to find the areal velocity of the planet:

1. Understanding Areal Velocity
Areal velocity is the rate at which area is swept out by the radius vector of a planet orbiting the sun. Let the planet of mass $m$ be at a position vector $r$ relative to the sun. In an infinitesimally small time interval $dt$, the planet moves by a small displacement vector $dr$.

The area $dA$ of the triangular region swept out by the radius vector during this time interval is given by the cross product of the vectors:
$dA=\frac{1}{2}|r\times dr|$

To find the rate of area swept per unit time (areal velocity), we divide both sides by $dt$:
$\frac{dA}{dt}=\frac{1}{2}|r\times \frac{dr}{dt}|$

Since the rate of change of displacement is velocity ($v=\frac{dr}{dt}$), we get:
$\frac{dA}{dt}=\frac{1}{2}|r\times v|$

2. Expressing in Terms of Angular Momentum
The angular momentum $J$ of the planet orbiting the sun is defined as:
$J=r\times p$

where $p=mv$ is the linear momentum of the planet. Substituting this in, we obtain:
$J=r\times \left(mv\right)=m(r\times v)$

Taking the magnitude on both sides gives:
$J=m|r\times v|$

Rearranging the equation to solve for the magnitude of the cross product:
$|r\times v|=\frac{J}{m}$

3. Final Substitution
Now, substitute this relation back into our expression for areal velocity:
$\frac{dA}{dt}=\frac{1}{2}\left(\frac{J}{m}\right)=\frac{J}{2m}$

Thus, the areal velocity of the planet's radius vector is $\frac{J}{2m}$.