JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 7)
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 :
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$${\rm I} = {{{{\rm I}_0}} \over 4}$$
$${\rm I} = {{15} \over {16}}{{\rm I}_0}$$
$${\rm I} = {9 \over {16}}{{\rm I}_0}$$
$${\rm I} = {3 \over 4}{{\rm I}_0}$$
توضیح
Suppose M is mass and a is side of larger triangle, then $${M \over 4}$$ and $${a \over 2}$$ will be mass and side length of smaller triangle.
$${{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}$$
$${{\rm I}_{removed}} = {{{{\rm I}_0}} \over {16}}$$
So, $${\rm I}$$ = $${\rm I}$$ 0 $$-$$ $${{{{\rm I}_0}} \over {16}}$$ = $${{15{{\rm I}_0}} \over {16}}$$
$${{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}$$
$${{\rm I}_{removed}} = {{{{\rm I}_0}} \over {16}}$$
So, $${\rm I}$$ = $${\rm I}$$ 0 $$-$$ $${{{{\rm I}_0}} \over {16}}$$ = $${{15{{\rm I}_0}} \over {16}}$$