A particle of mass `m=3kg` moves along a straight line `4y-3x=2` where `x` and `y` are in metre, with constant velocity `v=5ms^(-1)` the magnitude of angular momentum about the origin is

Options

A. `12kgm^(2)s^(-1)`
B. `6.0kgm^(2)s^(-1)`
C. `4.5mgm^(2)s^(-1)`
D. `8.0kgm^(2)s^(-1)`

Answer

B

Explanation

`y=(3)/(4)x+0.5`
slope `=tantheta=(3)/(4)`
`thereforetheta=37^(@)`
`r_(bot)=0.5sin53^(@)=0.4m`
`L=mvr_(bot)=(3)(5)(0.4)`
`=6kgm^(2)s^(-1)`