$A$ particle of mass $2 \ kg$ located at the position $(\hat{i} + \hat{j}) \ m$ has a velocity $2(\hat{i} - \hat{j} + \hat{k}) \ m/s$. Its angular momentum about the $z$-axis in $kg \cdot m^2/s$ is:

$A$ particle of mass $m$ is moving along a line $y = x + a$ with a constant velocity $v$. The angular momentum of the particle about the origin is

$A$ particle moves along a circular path with decreasing speed. Hence,

$A$ particle starts from the point $(0, 8) \, m$ and moves with a uniform velocity of $\vec{v} = 3 \hat{i} \, m/s$. What is the angular momentum of the particle about the origin after $5 \, s$ (mass of the particle is $1 \, kg$)?

$A$ thin uniform rod of length $L$ and mass $m$ is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$, with the rod remaining aligned with the string. In such a case, the value of $n$ is:

$A$ 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 .......