As shown in figure, a mass $m$ = $500\  g$ hangs from the rim of a wheel of radius $r$ = $20\  cm$. When released from rest, the mass falls $2.0\  m$ in $8\  sec$. Then moment of inertia of the wheel is.......... $kg-m^2$. $(g = 10\  m/s^2)$

Two like parallel forces $20 \,N$ and $30 \,N$ act at the ends $A$ and $B$ of a rod $1.5 \,m$ long. The resultant of the forces will act at the point ........

A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$, the initial angular acceleration of the rod will be 

A uniform rod of length $'l'$ is pivoted at one of its ends on a vertical shaft of negligible radius When the shaft rotates at angular speed $\omega$ the rod makes an angle $\theta$ with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper ) $\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the CM. The value of $\theta$ is then such that:

A uniform rod of length $L$ and weight $W$ is suspended horizontally by two vertical ropes as shown. The first rope is attached to the left end of the rod while the second rope is attached a distance $L /4$ from the right end. The tension in the second rope is 

A solid cylinder of mass $m$ is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is:

[The coefficient of static friction, $\mu_{ s },$ is $\left.0.4\right]$