1. Given

• Mass of spherical shell: $m = 1\,\text{kg}$
• Radius: $R$
• Angular speed while rolling: $\omega$
• It is rolling without slipping on a horizontal plane.

We need the magnitude of angular momentum about origin $O$.

2. Angular momentum about a point

For a rolling body, angular momentum about point $O$ is the sum of:

$\vec L_O = \vec L_{\text{orbital}} + \vec L_{\text{spin}}$

where

• $\vec L_{\text{orbital}} = \vec r_{CM} \times m\vec v_{CM}$
• $\vec L_{\text{spin}} = I_{CM}\,\vec\omega$

3. Orbital angular momentum

Since the shell rolls without slipping,

$v_{CM} = R\omega$

From the figure (implied standard rolling geometry), the perpendicular distance from origin $O$ to the line of motion of the center of mass is $R$.

Hence,

$L_{\text{orbital}} = m v_{CM} R = 1 \cdot (R\omega) \cdot R = R^2\omega$

4. Spin angular momentum about center of mass

For a spherical shell,

$I_{CM} = \frac{2}{3}mR^2$

Since $m=1$,

$I_{CM} = \frac{2}{3}R^2$

Thus,

$L_{\text{spin}} = I_{CM}\omega = \frac{2}{3}R^2\omega$

5. Direction check

For rolling as shown, the orbital and spin angular momenta about $O$ are in the same direction, so magnitudes add:

$L_O = R^2\omega + \frac{2}{3}R^2\omega$

$L_O = \frac{5}{3}R^2\omega$

This is given as

$L_O = \frac{a}{3}R^2\omega$

So,

$\frac{a}{3} = \frac{5}{3} \Rightarrow a=5$

6. Option check

• A: $2$ ❌
• B: $3$ ❌
• C: $5$ ✅
• D: $4$ ❌

So the correct option is C.