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If 'a' the length of side of the cube, the distance between the body-centred atom and one corner atom in the cube will be:[A] $\\dfrac{4}{\\sqrt{3}}a$ [B] $\\dfrac{\\sqrt{3}}{4}a$ [C] $\\dfrac{\\sqrt{3}}{2}a$[D] $\\dfrac{2}{\\sqrt{3}}a$
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AB = diagonal (axis) moment of inertia of all the edges of cube
there is a cubic frame on each corner a pont massm are kept tell about momet of inertia if axis is pass through body diagonal of cubeeach side of cube 1u06l1vv -Physics -
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