Engineering Mechanics The centre of gravity a T-section 100 mm × 150 mm × 50 mm from its bottom is 75 mm 50 mm 87.5 mm 125 mm 75 mm 50 mm 87.5 mm 125 mm 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 The moment of inertia of a square of side (a) about an axis through its center of gravity is a4/12 a4/4 a4/8 a4/36 a4/12 a4/4 a4/8 a4/36 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics For any system of coplanar forces, the condition of equilibrium is that the All of these Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of the horizontal components of all the forces should be zero Algebraic sum of moments of all the forces about any point should be zero All of these Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of the horizontal components of all the forces should be zero Algebraic sum of moments of all the forces about any point should be zero ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resultant of two equal forces ‘P’ making an angle ‘θ’, is given by 2P tanθ/2 2P cosθ/2 2P cotθ/2 2P sinθ/2 2P tanθ/2 2P cosθ/2 2P cotθ/2 2P sinθ/2 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The centre of percussion of the homogeneous rod of length ‘L’ suspended at the top will be 2L/3 L/3 3L/4 L/2 2L/3 L/3 3L/4 L/2 ANSWER DOWNLOAD EXAMIANS APP
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