Engineering Mechanics In order to determine the effects of a force, acting on a body, we must know Magnitude of the force Nature of the force i.e. whether the force is push or pull Line of action of the force All of these Magnitude of the force Nature of the force i.e. whether the force is push or pull Line of action of the force All of these ANSWER DOWNLOAD EXAMIANS APP
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 mr²/6 mr²/4 mr²/2 mr²/8 mr²/6 mr²/4 mr²/2 mr²/8 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Non-coplanar concurrent forces are those forces which Do not meet at one point and their lines of action do not lie on the same plane Meet at one point and their lines of action also lie on the same plane Do not meet at one point, but their lines of action lie on the same plane Meet at one point, but their lines of action do not lie on the same plane Do not meet at one point and their lines of action do not lie on the same plane Meet at one point and their lines of action also lie on the same plane Do not meet at one point, but their lines of action lie on the same plane Meet at one point, but their lines of action do not lie on the same plane ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The time of flight (t) of a projectile on a horizontal plane is given by t = 2u. cosα/g t = 2u/g.sinα t = 2u. tanα/g t = 2u. sinα/g t = 2u. cosα/g t = 2u/g.sinα t = 2u. tanα/g t = 2u. sinα/g ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The moment of inertia of a sphere of mass 'm' and radius 'r', about an axis tangential to it, is 7mr²/3 7mr²/5 2mr²/3 2mr²/5 7mr²/3 7mr²/5 2mr²/3 2mr²/5 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The Cartesian equation of trajectory is (where u = Velocity of projection, α = Angle of projection, and x, y = Co-ordinates of any point on the trajectory after t seconds.) y = x. tanα - (gx²/2u² cos²α) y = (gx²/2u² cos²α) - x. tanα y = x. tanα + (gx²/2u² cos²α) y = (gx²/2u² cos²α) + x. tanα y = x. tanα - (gx²/2u² cos²α) y = (gx²/2u² cos²α) - x. tanα y = x. tanα + (gx²/2u² cos²α) y = (gx²/2u² cos²α) + x. tanα ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS