A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is

A. $${\text{T}} = \frac{{2\omega }}{{{\pi ^2}}}$$ B. $${\text{T}} = \frac{{2\pi }}{\omega }$$ C. $${\text{T}} = \frac{2}{\omega }$$ D. $${\text{T}} = \frac{\pi }{{2\omega }}$$

A. 0.1 rad/sec B. 1 rad/sec C. 10 rad/sec D. 100 rad/sec

A. Always directed away from the centre, the point of reference B. Proportional to the square of the distance from the point of reference C. Proportional to the distance from the point of reference and directed towards it D. Inversely proportion to the distance from the point of reference

A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$ B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$ C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$ D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$ E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$