Applied Mechanics and Graphic Statics A disc of mass 4 kg, radius 0.5 m and moment of inertia 3 kg.m² rolls on a horizontal surface so that its center moves with speed 5 m/see. Kinetic energy of the disc is 200 J 150 J 400 J 50 J 200 J 150 J 400 J 50 J ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics When a body slides down an inclined surface, the acceleration (f) of the body, is given by f = g sin θ f = g tan θ f = g f = g cos θ f = g sin θ f = g tan θ f = g f = g cos θ ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One end of a light string 4 m in length is fixed to a point on a smooth wall and the other end fastened to a point on the surface of a smooth sphere of diameter 2.25 m and of weight 100 kg. The reaction between the sphere and the wall of the arrangement made is 108.5 kg 110 kg 105.5 kg 102.5 kg 108.5 kg 110 kg 105.5 kg 102.5 kg ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A Second's pendulum gains 2 minutes a day. To make it to keep correct time its length Is not changed but weight of the bob is decreased Must be increased Must be decreased Is not changed but weight of the bob is increased Is not changed but weight of the bob is decreased Must be increased Must be decreased Is not changed but weight of the bob is increased ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One end of an elastic string of natural length / and modulus X is kept fixed while to the other end is attached a particle of mass m which is hanging freely under gravity. The particle is pulled down vertically through a distance x, held at rest and then released. The motion is None of these a rectilinear motion with constant speed a simple harmonic motion a damped oscillatory motion None of these a rectilinear motion with constant speed a simple harmonic motion a damped oscillatory motion 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 = 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 e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ + u₁ ANSWER DOWNLOAD EXAMIANS APP
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