$A$ thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$. Its angular momentum is $L_A$ and $L_B$ computed about points $A$ and $B$, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is :

$A$ particle having mass $4 \ kg$ is moving with velocity $(4 \hat{i} + 2 \hat{j}) \ m/s$. Find the angular momentum of the particle about the origin when it is at position $(1, 1, 0) \ m$.

$A$ particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the $XOY$ plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is .......

The position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$, where $b$ is a constant. At an instant, $\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be:

$A$ particle is moving along a straight line parallel to $x-$ axis with constant velocity. Its angular momentum about the origin

$A$ particle of mass $0.5\, kg$ is rotating in a circular path of radius $2\, m$ and centripetal force on it is $9\, N$. Its angular momentum (in $J\cdot s$) is: