Engineering Mechanics The moment of inertia of a thin disc of mass ‘m’ and radius ‘r’, about an axis through its center of gravity and perpendicular to the plane of the disc is mr²/8 mr²/6 mr²/2 mr²/4 mr²/8 mr²/6 mr²/2 mr²/4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A semicircular disc rests on a horizontal surface with its top flat surface horizontal and circular portion touching down. The coefficient of friction between semi circular disc and horizontal surface is µ. This disc is to be pulled by a horizontal force applied at one edge and it always remains horizontal. When the disc is about to start moving, its top horizontal force will Remain horizontal Slant down towards direction of pull None of these Slant up towards direction of pull Remain horizontal Slant down towards direction of pull None of these Slant up towards direction of pull 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 centre of gravity of a right circular solid cone is at a distance of _________ from its base, measured along the vertical axis.(where h = Height of a right circular solid cone.) h/2 h/6 h/4 h/3 h/2 h/6 h/4 h/3 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be Towards the wall at its upper end Away from the wall at its upper end Upwards at its upper end Downwards at its upper end Towards the wall at its upper end Away from the wall at its upper end Upwards at its upper end Downwards at its upper end ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics In a screw jack, the effort required to lift the load is given by (where W = Load lifted, α = Helix angle, and φ = Angle of friction.) P = W cos (α + φ) P = W tan (α - φ) P = W tan (φ - α) P = W tan (α + φ) P = W cos (α + φ) P = W tan (α - φ) P = W tan (φ - α) P = W tan (α + φ) ANSWER DOWNLOAD EXAMIANS APP
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