A particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then

$A$ 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 $z$ -axis in $kg-m^2/s$ is

A particle of mass $2\, kg$ is moving such that at time $t$, its position, in meter, is given by $\overrightarrow r \left( t \right) = 5\hat i - 2{t^2}\hat j$ . The angular momentum of the particle at $t\, = 2\, s$ about the origin in $kg\, m^{-2}\, s^{-1}$ is

Obtain the relation between angular momentum of a particle and torque acting on it. 

A binary star consists of two stars $\mathrm{A}$ (mass $2.2 \mathrm{M}_5$ ) and $\mathrm{B}$ (mass $11 \mathrm{M}_5$ ), where $\mathrm{M}_5$ is the mass of the sun. They are separated by distance $\mathrm{d}$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star $\mathrm{A}$ to the angular momentum of star $\mathrm{B}$ about the centre of mass is

A particle of mass $'{m}'$ is moving in time $'t'$ on a trajectory given by

$\overrightarrow{{r}}=10 \alpha {t}^{2}\, \hat{{i}}+5 \beta({t}-5)\, \hat{{j}}$

Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for ${t}=0$ at time ${t}=$ .....$seconds.$