On a mass ‘m’ describing a circular path of radius ‘r’, the centrifugal force
A. Acts tangentially to the circular path
B. Acts towards the centre of rotation
C. Acts away from the centre of rotation
D. Is $$\frac{{{\text{m}}{{\text{w}}^{2{\text{r}}}}}}{g}{\text{kfg}}$$
Answer: Option C
Solution(By Examveda Team)
The centrifugal force on a mass 'm' describing a circular path of radius 'r' acts away from the centre of rotation. It is given by the formula:F = (m * w^2 * r) / g
Where:
F
is the centrifugal forcem
is the massw
is the angular velocityr
is the radius of the circular pathg
is the acceleration due to gravitykfg
is not relevant to the context and seems to be a typo or irrelevant notation in the provided options.Therefore, the correct option is:
Option C: Acts away from the centre of rotation
Join The Discussion
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. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
Option C is correct.