Four spheres,each of mass $M$ and diameter $2a$,are placed at the corners of a square of side $b$. The moment of inertia of this system about an axis along one of the sides of the square is:

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes:

Two thin circular discs of mass $m$ and $4m$,having radii of $a$ and $2a$,respectively,are rigidly fixed by a massless,rigid rod of length $l=\sqrt{24}a$ through their centers. This assembly is laid on a firm and flat surface,and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point $O$ is $\vec{L}$ (see the figure). Which of the following statement$(s)$ is(are) true?
$(A)$ The center of mass of the assembly rotates about the $z$-axis with an angular speed of $\omega/5$
$(B)$ The magnitude of angular momentum of center of mass of the assembly about the point $O$ is $81ma^2\omega$
$(C)$ The magnitude of angular momentum of the assembly about its center of mass is $17ma^2\omega/2$
$(D)$ The magnitude of the $z$-component of $\vec{L}$ is $55ma^2\omega$

On a solid sphere lying on a horizontal surface,a force $F$ is applied at a height of $R/2$ from the centre of mass. The initial acceleration of a point at the top of the sphere is (there is no slipping at any point).

$A$ hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

$A$ uniform sphere of radius $R$ is placed on a rough horizontal surface and given a linear velocity $v_0$ and angular velocity $\omega_0$ as shown. The sphere comes to rest after moving some distance to the right. It follows that: