$A$ pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force $F$ starting from rest. As the time increases,

$A$ solid sphere of mass $M$,radius $R$ and having moment of inertia about an axis passing through the centre of mass as $I$,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 $I$. Then,the radius of the disc will be

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

$A$ thin uniform straight rod of mass $2 \, kg$ and length $1 \, m$ is free to rotate about its upper end when at rest. It receives an impulsive blow of $10 \, Ns$ at its lowest point,normal to its length as shown in the figure. The kinetic energy of the rod just after impact is ........ $J$.

$A$ particle of mass $M=0.2 \ kg$ is initially at rest in the $xy$-plane at a point $(x=-l, y=-h)$,where $l=10 \ m$ and $h=1 \ m$. The particle is accelerated at time $t=0$ with a constant acceleration $a=10 \ m/s^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin,in $SI$ units,are represented by $\vec{L}$ and $\vec{\tau}$,respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x, y$ and $z$-directions,respectively. If $\hat{k}=\hat{i} \times \hat{j}$,then which of the following statement$(s)$ is(are) correct?
$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t=2 \ s$.
$(B)$ $\vec{\tau}=2 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(C)$ $\vec{L}=4 \hat{k}$ when the particle passes through the point $(x=l, y=-h)$.
$(D)$ $\vec{\tau}=\hat{k}$ when the particle passes through the point $(x=0, y=-h)$.

The graph between $\log_e L$ and $\log_e P$ will be (where $L$ is angular momentum and $P$ is linear momentum):