Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • I have a doubt, kindly clarify. - Oscillations and Waves - JEE Main-6

A thin uniform rod of length l is pivoted at its upper end. It is free to swing in vertical plane. Its time period for oscillation of small amplitude is 

  • Option 1)

    2\pi\sqrt{\frac{l}{g}}

  • Option 2)

    2\pi\sqrt{\frac{2l}{3g}}

  • Option 3)

    2\pi\sqrt{\frac{3l}{2g}}

  • Option 4)

    2\pi\sqrt{\frac{l}{3g}}

 

Answers (1)

best_answer

T = 2\pi\sqrt{\frac{I}{Mg\frac{l}{2}}} = 2\pi\sqrt{\frac{m\frac{l^{2}}{3}}{mg\frac{l}{2}}} \Rightarrow T = 2\pi\sqrt{\frac{2l}{3g} }

 

Time period of physical pendulum -

T= 2\pi \sqrt{\frac{I}{mgl}}

- wherein

I= moment of inertia about point of hinge

l= Seperation between point of suspension & centre of mass

 

 

 


Option 1)

2\pi\sqrt{\frac{l}{g}}

This is incorrect.

Option 2)

2\pi\sqrt{\frac{2l}{3g}}

This is correct.

Option 3)

2\pi\sqrt{\frac{3l}{2g}}

This is incorrect.

Option 4)

2\pi\sqrt{\frac{l}{3g}}

This is incorrect.

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question