A ring of radius $0.5\, m$ and mass $10 \,kg$ is rotating about its diameter with an angular velocity of $20 \,rad/s.$  Its kinetic energy is .......... $J$

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively as shown is the fig. The velocities of two discs at the points $A$ and $B$ of the inclined planes are related as

A disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

A disc of radius $1\,m$ and mass $4\,kg$ rolls on a horizontal plane without slipping in such a way that its centre of mass moves with a speed of $10\,cm/\sec .$ Its rotational kinetic energy is

A uniform disk of mass $m$ and radius $R$ rolls without slipping down an incline plane of length $l$ and inclination $\theta$. Initially the disk was at rest at the top of the incline plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to :-

A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement($s$) is/are correct, when the rod makes an angle $60^{\circ}$ with vertical ? [ $g$ is the acceleration due to gravity]

$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3 g }{4}$

$(2)$ The angular acceleration of the rod will be $\frac{2 g }{ L }$

$(3)$ The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$

$(4)$ The normal reaction force from the floor on the rod will be $\frac{ Mg }{16}$