A planet of mass m is moving in an elliptical path about the sun. Its maximum and minimum distances from the sun are r₁ and r₂ respectively. If Mₛ is the mass of sun then the angular momentum of this planet about the center of sun will be

The correct option is:
$m\sqrt{\frac{2G{M}_s{r}_1{r}_2}{r_1+r_2}}$

To find the angular momentum of the planet, we use the principles of conservation of angular momentum and conservation of mechanical energy. Let the mass of the planet be $m$ and the mass of the sun be $M_s$.

At the points of maximum and minimum distances (aphelion and perihelion), the velocity vectors of the planet are perpendicular to the position vectors relative to the sun.
Let $v_1$ be the speed of the planet when it is at the maximum distance $r_1$, and $v_2$ be the speed at the minimum distance $r_2$.

By conservation of angular momentum $L$ about the center of the sun:
$L=m{v}_1{r}_1=m{v}_2{r}_2$
From this relation, we can express $v_2$ in terms of $v_1$:
$v_2=v_1\frac{r_1}{r_2}$

Now, by the conservation of total mechanical energy of the planet-sun system:
$E_1=E_2$
$\frac{1}{2}m{v}_1^2-\frac{G{M}_{s}m}{r_1}=\frac{1}{2}m{v}_2^2-\frac{G{M}_{s}m}{r_2}$

We can divide both sides by the mass of the planet $m$:
$\frac{1}{2}{v}_1^2-\frac{G{M}_s}{r_1}=\frac{1}{2}{v}_2^2-\frac{G{M}_s}{r_2}$

Substitute $v_2=v_1\frac{r_1}{r_2}$ into the energy equation:
$\frac{G{M}_s}{r_2}-\frac{G{M}_s}{r_1}=\frac{1}{2}{\left(v_1\frac{r_1}{r_2}\right)}^2-\frac{1}{2}{v}_1^2$
$G{M}_s\left(\frac{1}{r_2}-\frac{1}{r_1}\right)=\frac{1}{2}{v}_1^2\left(\frac{r_1^2}{r_2^2},-,1\right)$
$G{M}_s\left(\frac{r_1-r_2}{r_1{r}_2}\right)=\frac{1}{2}{v}_1^2\left(\frac{r_1^2-r_2^2}{r_2^2}\right)$

Since $r_1^2-r_2^2=(r_1-r_2)(r_1+r_2)$, we can simplify the equation by canceling $(r_1-r_2)$ from both sides (since $r_1\ne r_2$):
$\frac{G{M}_s}{r_1{r}_2}=\frac{1}{2}{v}_1^2\frac{r_1+r_2}{r_2^2}$

Solving for $v_1^2$:
$v_1^2=\frac{2G{M}_s{r}_2^2}{r_1{r}_2(r_1+r_2)}=\frac{2G{M}_s{r}_2}{r_1(r_1+r_2)}$
Taking the square root to find $v_1$:
$v_1=\sqrt{\frac{2G{M}_s{r}_2}{r_1(r_1+r_2)}}$

Finally, substitute this expression for $v_1$ back into the formula for angular momentum $L$:
$L=m{v}_1{r}_1$
$L=m{r}_1\sqrt{\frac{2G{M}_s{r}_2}{r_1(r_1+r_2)}}$
$L=m\sqrt{\frac{2G{M}_s{r}_1^2{r}_2}{r_1(r_1+r_2)}}$
$L=m\sqrt{\frac{2G{M}_s{r}_1{r}_2}{r_1+r_2}}$