Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

$A$ plank with a uniform sphere placed on it rests on a smooth horizontal plane. The plank is pulled to the right by a constant force $F$. If the sphere does not slip over the plank,which of the following statements is incorrect?

$A$ homogeneous disc of mass $2 \ kg$ and radius $15 \ cm$ is rotating about its axis (which is fixed) with an angular velocity $4 \ rad/s$. The linear momentum of the disc is

The torque acting on a body about a point is equal to $\overrightarrow{A} \times \overrightarrow{L}$,where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum about that point. This implies that:

$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?

$A$ rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. $A$ particle of mass $m$ travelling along the surface hits at one end of the rod with velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of $x$ will be ..... .