$A$ thin uniform rod of mass $M$ and length $L$ has its moment of inertia $I_1$ about its perpendicular bisector. The rod is bent in the form of a semicircular arc. Now its moment of inertia through the centre of the semicircular arc and perpendicular to its plane is $I_2$. The ratio of $I_1 : I_2$ will be

$A$ 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 $\frac{3g}{4}$
$(2)$ The angular acceleration of the rod will be $\frac{3\sqrt{3}g}{4L}$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$

Fill in the blanks:
$(1)$ In rotational motion,the role played by ............ is analogous to the role played by mass in linear motion.
$(2)$ For a rigid body in rotational motion,if a particle at a distance of $10 \ cm$ from the fixed axis of rotation has an angular velocity of $10 \ rad/s$,then the linear velocity of a particle at a distance of $5 \ cm$ from the axis of rotation is ............
$(3)$ The $SI$ unit $J \cdot s^{-2}$ is the unit of the physical quantity ............
$(4)$ The condition for a body to roll without slipping down an inclined plane with friction is ............

$A$ 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 is moving in a uniform circular motion with angular momentum $L$. If the frequency of motion is doubled and its kinetic energy is halved,the new angular momentum will be ...

$A$ uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(CM)$ at a distance $L$ from the wall. The disk rolls without slipping with velocity $\vec{V}_0 = V_0 \hat{i}$. The coefficient of friction is $\mu$.
$1.$ The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is
$(A) -kx$ $(B) -2kx$ $(C) -\frac{2kx}{3}$ $(D) -\frac{4kx}{3}$
$2.$ The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to
$(A) \sqrt{\frac{k}{M}}$ $(B) \sqrt{\frac{2k}{M}}$ $(C) \sqrt{\frac{2k}{3M}}$ $(D) \sqrt{\frac{4k}{3M}}$
$3.$ The maximum value of $V_0$ for which the disk will roll without slipping is
$(A) \mu g \sqrt{\frac{M}{k}}$ $(B) \mu g \sqrt{\frac{M}{2k}}$ $(C) \mu g \sqrt{\frac{3M}{k}}$ $(D) \mu g \sqrt{\frac{5M}{2k}}$