Engineering Mechanics Moment of inertia of a circular section about an axis perpendicular to the section is πd3/32 πd3/16 πd4/32 πd4/64 πd3/32 πd3/16 πd4/32 πd4/64 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity ratio of a simple wheel and axle with D and d as the diameters of effort wheel and load axle, is D / d D × d D + d D - d D / d D × d D + d D - d ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The law of motion involved in the recoil of gun is Newton's first law of motion Newton's second law of motion None of these Newton's third law of motion Newton's first law of motion Newton's second law of motion None of these Newton's third law of motion ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The centre of percussion is below the centre of gravity of the body and is at a distance equal to h × kG h/kG h2/kG kG2/h h × kG h/kG h2/kG kG2/h ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics In order to completely specify angular displacement by a vector, it must fix All of these Direction of the axis of rotation Sense of angular displacement Magnitude of angular displacement All of these Direction of the axis of rotation Sense of angular displacement Magnitude of angular displacement 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 = (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 y = x. tanα + (gx²/2u² cos²α) ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS