Applied Mechanics and Graphic Statics ‘μ’ is coefficient of friction. A wheeled vehicle travelling on a circular level track will slip and overturn simultaneously if the ratio of its wheel distance to the height of its centroid, is ½μ 3μ μ 2μ ½μ 3μ μ 2μ ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The shape of a suspended cable for a uniformly distributed load over it is Cubic parabola Circular Parabolic Catenary Cubic parabola Circular Parabolic Catenary ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One end of an elastic string of natural length ‘l’ 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 simple harmonic motion A damped oscillatory motion A rectilinear motion with constant speed None of these A simple harmonic motion A damped oscillatory motion A rectilinear motion with constant speed ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A cube on a smooth horizontal surface Cannot be in unstable equilibrium Can be in any of these states Cannot be in neutral equilibrium Cannot be in stable equilibrium Cannot be in unstable equilibrium Can be in any of these states Cannot be in neutral equilibrium Cannot be in stable equilibrium ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A train weighing 196 tonnes experiences a frictional resistance of 5(11/22) per tonne. The speed of the train at the top of a down gradient 1 in 78.4 is 36 km/hour. The speed of the train after running 1 km down the slope, is 5 √10 m/sec 10 √5 m/sec 5 √3 m/sec 3 √5 m/sec 5 √10 m/sec 10 √5 m/sec 5 √3 m/sec 3 √5 m/sec 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 = v₁ - v₂/u₂ - u₁ e = v₁ - v₂/u₂ + u₁ e = v₂ - v₁/u₁ - u₂ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ - u₁ e = v₁ - v₂/u₂ + u₁ e = v₂ - v₁/u₁ - u₂ e = u₂ - u₁/v₁ - v₂ ANSWER DOWNLOAD EXAMIANS APP
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