Applied Mechanics and Graphic Statics The units of moment of inertia of an area are kg/m2 m3 kg/m m4 kg/m2 m3 kg/m m4 ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics When a body of mass M1 is hanging freely and another of mass M2 lying on a smooth inclined plane(α) are connected by a light index tensile string passing over a smooth pulley, the acceleration of the body of mass M1, will be given by g(M1 + M2 sin α)/(M1 + M2) m/sec g(M2 × M1 sin α)/(M2 - M1) m/sec² g(M2 + M1 sin α)/(M1 + M2) m/sec² g(M1 - M2 sin α)/(M1 + M2) m/sec² g(M1 + M2 sin α)/(M1 + M2) m/sec g(M2 × M1 sin α)/(M2 - M1) m/sec² g(M2 + M1 sin α)/(M1 + M2) m/sec² g(M1 - M2 sin α)/(M1 + M2) m/sec² ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If ‘v’ and ‘ω’ are linear and angular velocities, the centripetal acceleration of a moving body along the circular path of radius ‘r’, will be r/v² v²/r r/ω² ω²/r r/v² v²/r r/ω² ω²/r ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Two circular discs of same weight and thickness are made from metals having different densities. Which disc will have the larger rotational inertia about its central axis? Disc with smaller density None of these Both discs will have same rotational inertia Disc with larger density Disc with smaller density None of these Both discs will have same rotational inertia Disc with larger density ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If the angle of projection is double the angle of inclination (α) of the plane on which particle is projected, the ratio of times of flight up the inclined plane and down the inclined plane, will be ½ cos α ½ sin α ½ tan α 2 cos α ½ cos α ½ sin α ½ tan α 2 cos α ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics ‘u₁’ and ‘u₂’ are the velocities of approach of two moving bodies in the same direction and their corresponding velocities of separation are ‘v₁’ and ‘v₂’. As per Newton's law of collision of elastic bodies, the coefficient of restitution (e) is given by e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ - u₁ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ + u₁ e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ - u₁ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ + u₁ ANSWER DOWNLOAD EXAMIANS APP
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