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QJEEM17Pheea0: JEE Main 2017 Physics - Class 11 JEEM17Pheea0 Solution: Moment of inertia of an equilateral triangular lamina ABC,…

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Rotational Motion Question 185 English

Multiple Choice Options

  • A 78{7 \over 8} Io
  • B 1516{15 \over 16} Io
  • C 3 Io4{{3\,{{\rm I}_o}} \over 4}
  • D 31 Io32{{31\,{{\rm I}_o}} \over 32}

Step-by-Step Verified Solution

Let, side of triangle ABC = ℓ\ell

According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,

IO = 112{1 \over {12}} m ℓ\ell 2

In, triangle DEF,

DE = DF = EF = 12{1 \over 2} AB = 12{1 \over 2} ℓ\ell

∴   \therefore\,\,\, moment of inertia of triangle DEF,

IDEF = 112×m4×(ℓ2)2{1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}

= 112×mℓ216{1 \over {12}} \times {{m{\ell ^2}} \over {16}}

= IO16{{{I_O}} \over {16}}

∴   \therefore\,\,\, Moment of inertia of the remaining part,

Iremain = IO −- IO16{{{{\rm I}_O}} \over {16}} = 15 IO16{{15\,{{\rm I}_O}} \over {16}}

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