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QJEEM19Pha89b: JEE Main 2019 Physics - Class 11 JEEM19Pha89b Solution: An equilateral triangle ABC is cut from a thin solid sheet of…

An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. then :

JEE Main 2019 (Online) 11th January Morning Slot Physics - Rotational Motion Question 169 English

Multiple Choice Options

  • A I=I04{\rm I} = {{{{\rm I}_0}} \over 4}
  • B I=1516I0{\rm I} = {{15} \over {16}}{{\rm I}_0}
  • C I=916I0{\rm I} = {9 \over {16}}{{\rm I}_0}
  • D I=34I0{\rm I} = {3 \over 4}{{\rm I}_0}

Step-by-Step Verified Solution

Suppose M is mass and a is side of larger triangle, then M4{M \over 4} and a2{a \over 2} will be mass and side length of smaller triangle.

IremovedIoriginal=M4(a2)2M(a)2{{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}

Iremoved=I016{{\rm I}_{removed}} = {{{{\rm I}_0}} \over {16}}

So, I{\rm I} = I{\rm I} 0 −- I016{{{{\rm I}_0}} \over {16}} = 15I016{{15{{\rm I}_0}} \over {16}}

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