The potential energy of a particle falling through a straight shaft drilled through the earth (assumed homogenous and spherical) is proportional to (where r is the distance of’the particle from centre of the earth)

In case of S.H.M. the period of oscillation (T), is given by

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

B. $${\text{T}} = \frac{{2\pi }}{\omega }$$

C. $${\text{T}} = \frac{2}{\omega }$$

D. $${\text{T}} = \frac{\pi }{{2\omega }}$$

The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be

A. 0.1 rad/sec

B. 1 rad/sec

C. 10 rad/sec

D. 100 rad/sec

A body is said to move with Simple Harmonic Motion if its acceleration, is

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

The resultant of two forces P and Q acting at an angle $$\theta $$, is

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. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$