Engineering Mechanics The moment of inertia of a solid sphere of mass ‘m’ and radius ‘r’ is mr² 2mr²/3 mr²/2 2mr²/5 mr² 2mr²/3 mr²/2 2mr²/5 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resolved part of the resultant of two forces inclined at an angle 'θ' in a given direction is equal to The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The difference of the forces multiplied by the cosine of θ The algebraic sum of the resolved parts of the forces in the given direction The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The difference of the forces multiplied by the cosine of θ The algebraic sum of the resolved parts of the forces in the given direction ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics One joule is equal to 10 N-m 100 N-m 0.1 N-m 1 N-m 10 N-m 100 N-m 0.1 N-m 1 N-m ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics One kg force is equal to 9.8 N 8.9 N 7.8 N 12 N 9.8 N 8.9 N 7.8 N 12 N ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The impact between two lead spheres is approximately equal to an __________ impact. Solid Inelastic Elastic None of these Solid Inelastic Elastic None of these 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 = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 IP = IG / Ah2 IP = Ah2 / IG IP = IG + Ah2 IP = IG - Ah2 ANSWER DOWNLOAD EXAMIANS APP
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