Engineering Mechanics The centre of percussion is below the centre of gravity of the body and is at a distance equal to kG2/h h/kG h × kG h2/kG kG2/h h/kG h × kG h2/kG ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A force while acting on a body may Give rise to the internal stresses in it Balance the forces, already acting on it All of these Change its motion Give rise to the internal stresses in it Balance the forces, already acting on it All of these Change its motion ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resultant of the two forces ‘P’ and ‘Q’ is ‘R’. If ‘Q’ is doubled, the new resultant is perpendicular to ‘P’. Then P = Q Q = 2R Q = R None of the listed here P = Q Q = 2R Q = R None of the listed here ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If a suspended body is struck at the center of percussion, then the pressure on die axis passing through the point of suspension will be Infinity Zero Minimum Maximum Infinity Zero Minimum Maximum 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
Engineering Mechanics If the resultant of two equal forces has the same magnitude as either of the forces, then the angle between the two forces is 90° 30° 120° 60° 90° 30° 120° 60° ANSWER DOWNLOAD EXAMIANS APP
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