When a uniform solid sphere and a disc of the same mass and of the same radius rolls down a rough inclined plane from rest to the same distance, then the ratio of the time taken by them is
$15 : 14$
$15^2 : 14^2$
$\sqrt {14} :\sqrt {15} $
$14 : 15$
Let $F $ be the force acting on a particle having position vector $\vec r$ and $\vec T$ be the torque of this force about the origin. Then ..
The given figure shows a disc of mass $M$ and radius $R$ lying in the $x-y$ plane with its centre on $x$ axis at a distance a from the origin. then the moment of inertia of the disc about the $x-$ axis is
$A$ particle of mass $m$ is projected with a velocity $u$ making an angle $45^o$ with the horizontal. The magnitude of the torque due to weight of the projectile, when the particle is at its maximum height, about the point of projectile
An object slides down a smooth incline and reaches the bottom with velocity $v$. If same mass is in the form of a ring and it rolls down an inclined plane of same height and angle of inclination, then its velocity at the bottom of inclined plane will be ............
The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through the rod from a point at a distance of $L/3$ from one of its ends perpendicular to the rod is