Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A particle moves in a circle of radius 5 cm with constant speed and time period 0.2s. The acceleration of the particle i

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2\pis. The acceleration of the particle is:

Option: 1

5 m/s2


Option: 2

15 m/s2


Option: 3

25 m/s2


Option: 4

36 m/s2


Answers (1)

best_answer

 

 v = R \omega = R\frac{2\pi }{T} = \frac{2\pi R}{T}

v= \frac{2\pi \times 0.05}{0.2\pi } = 0.5\ ms^{-1}

\therefore a = \frac{v^{2}}{R} = \frac{\left ( 0.5\right )^{2}}{5\times 10^{-2}} = 5\ ms^{-2}

Posted by

himanshu.meshram

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Bihar NEET PG 2024 stray vacancy round registration starts tomorrow; schedule out
anam-khan

NEET UG must for foreign MBBS graduates to obtain eligibility certificate; NMC regulations explained
anam-khan

Top private medical colleges in India accepting NEET UG score for admission; NIRF rankings

Latest Question