A Seconds pendulum executes
A. 0.5 beat per second
B. 1.0 beat per second
C. 2.0 beats per second
D. 2.5 beats per second
Answer: Option B
Solution (By Examveda Team)
The correct answer is Option B: 1.0 beat per secondHere's why:
A seconds pendulum is defined as a pendulum that takes exactly one second to swing in one direction.
This means it takes one second to complete half of its full swing (from one extreme point to the other).
A "beat" refers to one such swing.
Since it takes 1 second for one beat, it means it has 1.0 beat per second.
This Question Belongs to Civil Engineering >> Applied Mechanics And Graphic Statics
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Comments (1)
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. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
Correct option is B please correct it in your system