Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A bead of mass slides along a straight frictionless rigid wire rotating in a horizontal plane with a

A bead of mass \mathrm{m} slides along a straight frictionless rigid wire rotating in a horizontal plane with a constant angular speed then the axes of rotation is perpendicular to the wire and \mathrm{p} asses through one end of the wire. If r is the distance of the mass from the axis of rotation and \mathrm{v} is its speed then the magnitude of coriolis force is -
 

Option: 1

\mathrm{\frac{m v^2}{r}}
 


Option: 2

\mathrm{\frac{2 m v^2}{r}}
 


Option: 3

\mathrm{mv\omega}
 


Option: 4

\mathrm{2mv\omega}


Answers (1)

best_answer

Use, Coriolis force \mathrm{ =2 m \overrightarrow{\omega} \times \overrightarrow{v} }

                            \mathrm{ =2 m \omega v \sin \theta}

Since the axes of rotation is perpendicular to the wire.

                \mathrm{\theta =90^{\circ} }

\therefore  coriolis force \mathrm{=2 \mathrm{~m} \omega v }

Since \mathrm{ \sin 90^{\circ} =1 }

Hence option 4 is correct.


 

Posted by

mansi

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

NEET UG 2026 correction facility opens at neet.nta.nic.in; exam on May 3
anam-khan

NEET 2026 LIVE: NTA application correction window closes today; admit card, exam dates, pattern
anam-khan

NEET 2026 registration deadline extended to March 11 for MBBS, BDS; exam on May 3

Latest Question