JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 17)
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 :
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$${7 \over 8}$$ Io
$${15 \over 16}$$ Io
$${{3\,{{\rm I}_o}} \over 4}$$
$${{31\,{{\rm I}_o}} \over 32}$$
Explanation
Let, side of triangle ABC = $$\ell $$
According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,
IO = $${1 \over {12}}$$ m$$\ell $$2
In, triangle DEF,
DE = DF = EF = $${1 \over 2}$$ AB = $${1 \over 2}$$ $$\ell $$
$$\therefore\,\,\,$$ moment of inertia of triangle DEF,
IDEF = $${1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}$$
= $${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$$
= $${{{I_O}} \over {16}}$$
$$\therefore\,\,\,$$ Moment of inertia of the remaining part,
Iremain = IO $$-$$ $${{{{\rm I}_O}} \over {16}}$$ = $${{15\,{{\rm I}_O}} \over {16}}$$
According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,
IO = $${1 \over {12}}$$ m$$\ell $$2
In, triangle DEF,
DE = DF = EF = $${1 \over 2}$$ AB = $${1 \over 2}$$ $$\ell $$
$$\therefore\,\,\,$$ moment of inertia of triangle DEF,
IDEF = $${1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}$$
= $${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$$
= $${{{I_O}} \over {16}}$$
$$\therefore\,\,\,$$ Moment of inertia of the remaining part,
Iremain = IO $$-$$ $${{{{\rm I}_O}} \over {16}}$$ = $${{15\,{{\rm I}_O}} \over {16}}$$
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