A rigid body of moment of inertial `I` is projected with velocity `V` making an angle of `45^(@)` with horizontal. The magnitude of angular momentum of the projectile about the point of projection when the body is its maximum height is given by `(IV^(3))/(2sqrt(2)gR^(2))` where `R` is the radius of the rigid body. the ridid body is

Options

A. sphere
B. spherical shell
C. disc
D. none of these

Answer

C

Explanation

At highest point, velocity `=V cos 45^(@)=V//sqrt(2)`
`L=mV/(sqrt(2)).rbot,` Here `r_(bot)=h=(V^(2)sin^(2)45^(@))/(2g)=(V^(2))/(4g)`
`L=(mV^(3))/(4sqrt(2)g)=(IV^(3))/(2sqrt(2)gR^(2))` (given) , So, `m/2=I/(R^(2))`
`I=1/2mR^(2)` it is a disc.