Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me please, Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M(see figure) The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M(see figure) The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

 

  • Option 1)

     

    \frac{152}{15}MR^{2}

  • Option 2)

  • Option 3)

    \frac{209}{15}MR^{2}

  • Option 4)

    \frac{17}{15}MR^{2}

Answers (1)

best_answer

 

Paraller Axis Theorem -

I_{b\: b'}=I_{a\: a'}+Mh^{2}

- wherein

b\: b' is axis parallel to a\: a' & a\: a' an axis passing through centre of mass.

 

 

 

Moment of inertia for solid sphere -

I=\frac{2}{5} MR^{2}

 

 

- wherein

About a diameter.

I_{ball} = \frac{2}{5}MR^{2}+3(2R)^{2}

          =\frac{22}{5}MR^{2}

2 balls are there

\therefore \frac{44}{5}MR^{2} = I_{balls}

I_{rod} = \frac{M\left ( 2R \right )^{2}}{12}

Isystem = Iballs + Irod

           =\frac{137}{15}MR^{2}

 

 


Option 1)

 

\frac{152}{15}MR^{2}

Option 2)

Option 3)

\frac{209}{15}MR^{2}

Option 4)

\frac{17}{15}MR^{2}

Posted by

admin

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

JEE Main 2025 session 2 in less than a month; how to crack ‘difficult, lengthy’ mathematics
anam-khan

Top engineering deemed universities in India for BTech admission with NIRF ranking
anam-khan

JEE Main 2025 session 2 application correction deadline today; edit fields at jeemain.nta.nic.in

Latest Question