ABC is a plane lamina of the shape of an equilateral triangle. D , E are mid points of AB,AC and G is the centroid of the lamina . Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is `I_0` If part ADE is removed, the moment of inertia of a remaining part about the same axis is `(NI_0)/16` where N is an integer . Value of N is

The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 0.4 kg m^2 . Calculate its moment of inertia about tangent parallel to the plane

The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is 0.4 kg m^2 . Calculate its moment of inertia about its diameter

In the figure shown ABCD is a lamina of mass 12 kg . The moment of inertia of the given body about an axis passing through D, and perpendicular to the plane of the body is

In the figure shown ABCD is a lamina of mass 12 kg . The moment of inertia of the given body about an axis passing through D, and perpendicular to the plane of the body is

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 to 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 :

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 to 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 :