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NEET-XII-Physics

exam-1 year:2016

with Solutions -
  • #1
    From a disc of radius R and mass M, a circular hole
    of diameter R, whose rim passes through the centre
    is cut. What is the moment of inertia of the
    remaining part of the disc about a perpendicular
    axis, passing through the centre ?
    (1) 15 ``MR^2``/32
    (2) 13 ``MR^2``/32
    (3) 11 ``MR^2``/32
    (4) 9 ``MR^2``/32
    digAnsr:   2
    Ans : 2
    Sol.
    R
    R
    M
    ITotal disc =
    2MR
    2
    MRemoved =
    M
    4
    (Mass  area)
    IRemoved (about same Perpendicular axis)
    =
    2M (R /2)
    4 2
    +
     
     
     
    2
    M R
    4 2
    =
    23MR
    32
    IRemaing disc = ITotal - IRemoved
    =
    2MR
    2
    -
    3
    32
    MR2 =
    213 MR
    32