(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 \[{{I}_{0}}\]. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then
[JEE Main Online Paper (Held On 11-Jan-2019 Morning]
A) \[I=\frac{15}{16}{{I}_{0}}\]
B) \[I=\frac{9}{16}{{I}_{0}}\]
C) \[I=\frac{3}{4}{{I}_{0}}\]
D) \[I=\frac{{{I}_{0}}}{4}\]
Correct Answer: A
Solution :
Moment of inertia at triangular lamina, \[{{I}_{0}}=KM{{a}^{2}}\]where K= constant of proportionality Now moment of inertia of small lamina, \[I'=K\frac{M}{4}{{\left( \frac{a}{2} \right)}^{2}}=\frac{KM{{a}^{2}}}{16}\] \[I'=\frac{{{I}_{0}}}{16}\] So, moment of inertia of remaining part, \[I-I'=\frac{15{{I}_{0}}}{16}\]
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