Engineering Mechanics The moment of inertia of a thin rod of mass 'm' and length 'l', about an axis through its centre of gravity and perpendicular to its length is ml2/ 6 ml2/8 ml2/12 ml2/4 ml2/ 6 ml2/8 ml2/12 ml2/4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Moment of inertia of a rectangular section having width (b) and depth (d) about an axis passing through its C.G. and parallel to the depth (d), is db³/36 bd³/36 bd³/12 db³/12 db³/36 bd³/36 bd³/12 db³/12 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The acceleration of a particle moving with simple harmonic motion, at any instant is given by ω2/y ω2.y ω.y ω3.y ω2/y ω2.y ω.y ω3.y ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity of a body on reaching the ground from a height h, is 2g.√h √(gh) √(2gh) 2.√(gh) 2g.√h √(gh) √(2gh) 2.√(gh) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to the law of moments, if a number of coplanar forces acting on a particle are in equilibrium, then Their algebraic sum is zero Their lines of action are at equal distances The algebraic sum of their moments about any point is equal to the moment of their resultant force about the same point The algebraic sum of their moments about any point in their plane is zero Their algebraic sum is zero Their lines of action are at equal distances The algebraic sum of their moments about any point is equal to the moment of their resultant force about the same point The algebraic sum of their moments about any point in their plane is zero ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 IP = Ah2 / IG IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 IP = Ah2 / IG ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS