Physics 1 Marks Question

Yes, it is an example of pure rotation.
The axis of rotation passes through the pivot of the pendulum, perpendicular to the plane containing the pendulum. 

The body with the smallest moment of inertia will roll down taking the smallest time. Here, the solid sphere has the lowest moment of inertia among all the other bodies. So, it will roll down taking the least time.

Yes, such motion is possible if the translation takes place along the axis of rotation. This type of motion is called screw motion. 

Ice caps near the poles concentrate the mass of water at the poles through which axis of rotation of the Earth passes. If the ice melts, water will spread across the globe due to hydrostatic equilibrium and tend to move to the equatorial areas of the Earth due to centrifugal force of rotation. Mass, now being distributed more along the equator, will increase MI of the Earth and this in turn will decrease the angular velocity of the Earth. Decrease in angular velocity will increase the duration of day-night.

When the balance is kept at an angle, there is a net extra torque given to one of its arm. When the extra torque is removed, the balance becomes torque free and sum of all the torque acting on it is zero.
But balance kept at an angle has got a greater potential energy compared to the balance kept horizontal. The potential energy acquired is due to the initial torque applied on it. This displaces the balance by an angle. As soon as the body is set free to rotate, the body tends to have the lowest potential energy. Thus, potential energy starts converting in to kinetic energy, but on the other side, kinetic energy converts into potential energy when the other arm of the balance is raised. This energy transformation oscillates the balance. But in this process, friction with the air and fulcrum dissipates energy converting into heat. Finally, the balance loses the energy and becomes horizontal, or attains equilibrium.

The centre of mass (CM) of a rectangular block lies in the middle of the block. When the block is projected less than half of its length (CM being over the table), no net force acts on it.
Thus, no net torque acts upon the body.But if the block is projected more than half of its length outside the table (CM being outside the table), gravitational force acts along the CM of the block. This force produces a moment along the edge of the table. This rotates the block, and as a result, it falls down.

No, angular momentum is dependent on the position vector of the particle, angle between the radius vector and the linear velocity of the particle. So, there may be finite angular momentum along any different point even if it is zero at a particular point. If angular momentum is zero along O but finite along O.

Some points on the equator of the sphere has got vertical velocity with respect to the direction of motion of the sphere.

It will require more force to set the bar into rotation by clamping at A and then clamping at B.

Explanation:
Since the rod has mass density increasing towards B, the Center of Mass (CM) of the rod is near B. If the rod is clamped along A, the distance of CM of the rod from the pivot will be greater when the rod is clamped along B.
Greater distance of CM from the Center of rotation increases the moment of inertia of the rod and hence more torque will be necessary to rotate the bar about A. Greater torque implies greater force will be necessary to rotate it.
F_A = Force required to rotated the rod clamped at A.
R_A = Distance of CM from pivot A.
M = Mass of the rod.
F_B = Force required to rotate the rod clamped at B.
R_B = Distance of CM from pivot B.
We have R_A > R_B.
We have to find the torque required to rotate rod clamped at A to produce angular acceleration a.
T_A = MR_A²a = R_AF_A
⇒ F_A = MR_Aa
We have to find torque required to rotate rod clamped at B to produce angular acceleration a.
T_B = MR_B²a = R_BF_B
⇒ F_B = MR_Ba
On comparing, since R_A > R_B, we get:
F_A > F_B