Physics System of Particles and Rotational Motion

A massless equilateral triangle EFG of side ' $a$ ' (As shown in figure) has three particles of mass $m$ situated at its vertices. The moment of inertia of the system about the line EX perpendicular to $EG$ in the plane of EFG is $\frac{N}{20}m{a}^2$ where $N$  is an integer. The value of $\mathrm{N}$  is                     

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)   $\frac{m{l}^2}{12}{\omega }^2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \mathrm{c}\mathrm{o}\mathrm{s}?\theta$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces   ${\mathrm{F}}_{\mathrm{H}}$ and   ${\mathrm{F}}_{\mathrm{V}}$ about the $\mathrm{C}\mathrm{M}$   . The value of $\theta$ is then such that:

Which of the following are correct expressions for torque acting on a body?

A. τ = r × L
B. τ = d/dt (r × p)
C. τ = r × dp/dt
D. τ = Iα
E. τ = r × F
(r = position vector; p = linear momentum; L = angular momentum; α = angular acceleration; I = moment of inertia; F = force; t = time)
Choose the correct answer from the options given below:

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre

Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is I = M((R²/4) + (L²/12)). If such a cylinder is to be made for a given mass of a material, the ratio L/R for it to have minimum possible I is

A square shaped hole of side l = a/2 is carved out at a distance d = a/2 from the centre O of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is -a/X, value of X (to the nearest integer) is :

Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass 'm' and has another weight of mass 2m hung at a distance of 75 cm from A. The tension in the string at A is: