Engineering Mechanics Moment of inertia of a circular section about its diameter (d) is πd3/32 πd4/32 πd4/64 πd3/16 πd3/32 πd4/32 πd4/64 πd3/16 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 = Ah2 / IG IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 IP = Ah2 / IG IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics In order to determine the effects of a force, acting on a body, we must know Magnitude of the force All of these Nature of the force i.e. whether the force is push or pull Line of action of the force Magnitude of the force All of these Nature of the force i.e. whether the force is push or pull Line of action of the force ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The forces which do not meet at one point and their lines of action do not lie on the same plane are known as None of these Coplanar non-concurrent forces Non-coplanar concurrent forces Coplanar concurrent forces None of these Coplanar non-concurrent forces Non-coplanar concurrent forces Coplanar concurrent forces ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The C.G. of a solid hemisphere lies on the central radius 3r At distance — from the plane base 3r At distance — from the plane base 3r At distance — from the plane base At distance — from the plane base 3r At distance — from the plane base 3r At distance — from the plane base 3r At distance — from the plane base At distance — from the plane base 3r ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is bh3/12 bh3/8 bh3/36 bh3/4 bh3/12 bh3/8 bh3/36 bh3/4 ANSWER DOWNLOAD EXAMIANS APP
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