The Law of Polygon of Forces states that
A. If a polygon representing the forces acting at point in a body is closed, the forces are in equilibrium
B. If forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order, then the resultant of the forces will be represented in magnitude and direction by t
C. If forces acting on a point can be represented of a polygon taken in order, their sides of a polygon taken in order, their resultant will be represented in magnitude and direction by the closing side
D. If forces acting on a point can be represented in magnitude and direction by the sides of a polygon in order, the forces are in equilibrium
Answer: Option C
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. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
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