The moment of inertia of a sphere of mass $M$ and radius $R$ is $I$. Which graph represents the variation of $I$ with respect to $R$,keeping $M$ constant?

$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

$A$ uniform rod $AB$ of length $L$ and mass $M$ is lying on a smooth table. $A$ small particle of mass $m$ strikes the rod with a velocity $v_0$ at point $C$ at a distance $x$ from the centre $O$. The particle comes to rest after the collision. The value of $x$ such that point $A$ of the rod remains stationary just after the collision is:

$A$ man spinning in free space changes the shape of his body,e.g.,by spreading his arms or curling up. By doing this,he can change his

$A$ block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially,the right edge of the block is at $x=0$,in a coordinate system fixed to the table. $A$ point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block,its position is $x$ and the velocity is $v$. At that instant,which of the following options is/are correct?
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$.
$[B]$ The position of the point mass is: $x=-\sqrt{2} \frac{mR}{M+m}$.
$[C]$ The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
$[D]$ The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.

$Assertion$ : If polar ice melts, days will be shorter.
$Reason$ : Moment of inertia decreases and thus angular velocity increases.