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QJEEM17Ph1a37: JEE Main 2017 Physics - Class 11 JEEM17Ph1a37 Solution: A circular hole of radius is made in a thin uniform disc having …

A circular hole of radius R4{R \over 4} is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

JEE Main 2017 (Online) 9th April Morning Slot Physics - Rotational Motion Question 181 English

Multiple Choice Options

  • A 219 MR2256{{219\,M{R^2}} \over {256}}
  • B 237 MR2512{{237\,M{R^2}} \over {512}}
  • C 19 MR2512{{19\,M{R^2}} \over {512}}
  • D 197 MR2256{{197\,M{R^2}} \over {256}}

Step-by-Step Verified Solution

Mass of removed disc = M16{M \over {16}} Radius of removed disc = R4{R \over 4}

Moment of inertia of removed disc about it's own axis (O')

= 12{1 \over 2} ×\times M16{M \over {16}} ×\times (R4)2{\left( {{R \over 4}} \right)^2} = MR2512{{M{R^2}} \over {512}}

Moment of inertia of removed disc about O,

= Icm + mx2

= MR2512{{M{R^2}} \over {512}} + M16(3R4)2{M \over {16}}{\left( {{{3R} \over 4}} \right)^2}

= 19MR2512{{19M{R^2}} \over {512}}

Moment of inertia of complete disc about point 'O',

I = MR22{{M{R^2}} \over 2}

∴   \therefore\,\,\, Moment of Inertia of remaining disc

= MR22{{M{R^2}} \over 2} −- 19MR2512{{19M{R^2}} \over {512}}

= 237MR2512{{237M{R^2}} \over {512}}

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