Rotational Motion

JEE Main level, single correct, medium

81 of 280

Question 81 of 280

An equilateral triangle ABC\mathrm{ABC} is cut from a thin solid sheet of wood. (See figure) D,E\mathrm{D}, \mathrm{E} and F\mathrm{F} are the mid-points of its sides as shown and G\mathrm{G} is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G\mathrm{G} and perpendicular to the plane of the triangle is I0\mathrm{I}_{0}. If the smaller triangle DEF\mathrm{DEF} is removed from ABC\mathrm{ABC}, the moment of inertia of the remaining figure about the same axis is II. Then

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Answer: (A) I=1516I0\mathrm{I}=\frac{15}{16} \mathrm{I}_{0}

Let mass of the larger triangle =M=\mathrm{M} Side of larger triangle =ℓ=\ell Moment of inertia of larger triangle =ma2=\mathrm{ma}^{2}

Mass of smaller triangle =M4=\frac{\mathrm{M}}{4}

Length of smaller triangle =ℓ2=\frac{\ell}{2}

Moment of inertia of removed triangle =M4(a2)2=\frac{\mathrm{M}}{4}\left(\frac{\mathrm{a}}{2}\right)^{2}

∴Iremoved Ioriginal =M4M⋅(a2)2(a)2\therefore \frac{\mathrm{I}_{\text {removed }}}{\mathrm{I}_{\text {original }}}=\frac{\frac{\mathrm{M}}{4}}{\mathrm{M}} \cdot \frac{\left(\frac{\mathrm{a}}{2}\right)^{2}}{(\mathrm{a})^{2}}

Iremoved =I016I_{\text {removed }}=\frac{I_{0}}{16}

So,I=I0−I016=15I016\mathrm{So}, \mathrm{I}=\mathrm{I}_{0}-\frac{\mathrm{I}_{0}}{16}=\frac{15 \mathrm{I}_{0}}{16}

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