If the radius of the earth is 600 km the height of a mountain above sea level at the top of which a beat seconds pendulum at sea level, looses 27 seconds a day, 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. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$