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JEE Main 20204 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2020 · 4 September · Morning Shift · Physics · Rotational Motion

ABC is a plane lamina of the shape of an equilateral triangle. D, E...

ABCABC is a plane lamina of the shape of an equilateral triangle. D,ED, E are mid-points of AB,ACAB, AC and GG is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through GG and perpendicular to the plane ABCABC is I0{I}_{0}. If part ADEADE is removed, the moment of inertia of the remaining part about the same axis is NI016\dfrac{N{I}_{0}}{16} where NN is an integer. Value of NN is:

ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of

Correct answer

11

Step-by-step solution

ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of

Let mass of lamina =m= m, and length of side =l= l, then moment of inertila of lamina about an axis passing through GG and perpendicular to the plane.

I0∝ml2{I}_{0} ∝ m{l}^{2}

I0=kml2{I}_{0} = km{l}^{2}

ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of

Let moment of inertila of DEF=I1DEF = {I}_{1} about GG

then, I1=I016{I}_{1} = \dfrac{{I}_{0}}{16}

then,IADE=IBDF=IEFC=I2{I}_{ADE} = {I}_{BDF} = {I}_{EFC} = {I}_{2}

3I2+I1=I03{I}_{2} + {I}_{1} = {I}_{0}

3I2+I016=I03{I}_{2} + \dfrac{{I}_{0}}{16} = {I}_{0}

⇒⇒ I2=5I016{I}_{2} = \dfrac{5{I}_{0}}{16}

So, Moment of inertila of DECB=2I2+I12(5I016)+(I016)DECB = 2{I}_{2} + {I}_{1}2\left(\dfrac{5{I}_{0}}{16}\right) + \left(\dfrac{{I}_{0}}{16}\right)

=11I016= \dfrac{11{I}_{0}}{16}

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