Q63 / 90 · JEE Main 2019 (10 Apr Shift 2) · Physics

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M8 and is converted into uniform disc of radius  2R . The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio  I1/I2 is given by:

  • Correct answer
    A
    140
  • B
    185
  • C
    65
  • D
    285

Solution

Moment of inertia of disc about the axis passing its centre and perpendicular to its surface, I1=7M82R22=7MR24.
Radius of small sphere r is related as
M43πR3=M843πr3
r=R2
Moment of inertia of sphere about its axis passing through its centre∴I2=25M8R22=MR280
I1I2=74×801=140

Hence, the correct answer is option: 'A'.

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