Engineering Mechanics The time of flight (t) of a projectile on an upward inclined plane is(where u = Velocity of projection, α = Angle of projection, and β = Inclination of the plane with the horizontal.) t = 2u sin (α + β)/g cos β t = g cos β/2u sin (α - β) t = g cos β/2u sin (α + β) t = 2u sin (α - β)/g cos β t = 2u sin (α + β)/g cos β t = g cos β/2u sin (α - β) t = g cos β/2u sin (α + β) t = 2u sin (α - β)/g cos β ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Mass moment of inertia of a uniform thin rod of mass M and length (l) about its mid-point and perpendicular to its length is (2/3) Ml² (1/12) Ml² (3/4) Ml² (1/3) Ml² (2/3) Ml² (1/12) Ml² (3/4) Ml² (1/3) Ml² ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics For any system of coplanar forces, the condition of equilibrium is that the Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of the horizontal components of all the forces should be zero All of these Algebraic sum of moments of all the forces about any point should be zero Algebraic sum of the vertical components of all the forces should be zero Algebraic sum of the horizontal components of all the forces should be zero All of these Algebraic sum of moments of all the forces about any point should be zero ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity ratio of a differential wheel and axle with 'D' as the diameter of effort wheel and d1 and d2 as the diameters of larger and smaller axles respectively, is D/(d₁ + d₂) D/(d₁ - d₂) 2D/(d₁ + d₂) 2D/(d₁ - d₂) D/(d₁ + d₂) D/(d₁ - d₂) 2D/(d₁ + d₂) 2D/(d₁ - d₂) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The rate of doing work is known as Power None of these Kinetic energy Potential energy Power None of these Kinetic energy Potential energy 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
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