Kinematics Question 727
Question: A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of the particle in $ m/s^{2} $ it’s [MNR 1991; UPSEAT 2000; DPMT 1999; RPET 2003; Pb. PET 2004]
Options:
A) $ {{\pi }^{2}} $
B) $ 8{{\pi }^{2}} $
C) $ 4{{\pi }^{2}} $
D) $ 2{{\pi }^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ n=2 $ , $ \omega =2\pi \times 2=4\pi rad/s^{2} $ So acceleration $ ={{\omega }^{2}}r $
$ ={{(4\pi )}^{2}}\times \frac{25}{100}m/s^{2}=4{{\pi }^{2}} $
