Q. Find the moment of inertia of a plate cut in shape of a right angled triangle of mass `M` side AC=BC=`a` about an axis perpendicular to the plane of the plate and passing through the mid point of side AB:-

A) `(Ma^2)/18`

B) `(Ma^2)/6`

C) `(Ma^2)/3`

D) `(2Ma^2)/3`

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A plate of mass M is in the shape of an isosceles right triangle with the equal sides having length a.

The moment of inertia of the plate about a point at the mid point of the hypotenuse has to be determined.

The center of mass of the triangle lies at the centroid. Draw a line between the points C and O. The length of this line is `sqrt(a^2 - (2a^2)/4)` = `sqrt(a^2/2)` = `a/sqrt 2` . The centroid of the triangle lies on this line at a distance `(a/sqrt 2)/3 = a/(3*sqrt 2)` from the point O. The moment of inertia of the triangle about the point O is `I = M*r^2` , where r is the distance of the center of mass from O. The moment of inertia is `M*(a/(3*sqrt 2))^2` = `(M*a^2)/18`

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Sorry, this response is not correct. The correct moment of inertia is `(M*a^2)/12` but that does not seem to be one of the options.

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