Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A block of mass  is placed over a frictionless bar as shown in the figure. Initially, the

A block of mass 1 \mathrm{kg} is placed over a frictionless bar as shown in the figure. Initially, the block was at rest, and spring was in its natural length. If a horizontal bar starts rotating about a vertical axis about its one end with constant angular velocity of 10 \mathrm\ {rad} / \mathrm{s}. Then the required spring constant to keep the block in equilibrium:

Option: 1

10\ N/m


Option: 2

100\ N/m


Option: 3

5\ N/m


Option: 4

50\ N/m


Answers (1)

best_answer

In case of equilibrium,

F_{\text {net }}=0 \\ m a=k x \\ m\omega^{2}x=kx \\ k=m\omega^2\\ k=(1)(10)^2\\ k=10\ N/m

 

Posted by

rishi.raj

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

Maharashtra NEET UG Counselling 2024: Medical college sued over MBBS fees
anam-khan

Karnataka NEET UG 2024 final seat allotment results out at kea.kar.nic.in after 2 pm today
anam-khan

NEET UG 2024: MCC withdraws two seats from MBBS, BDS, BSc Nursing special stray vacancy round

Latest Question