Engineering Mechanics A force while acting on a body may Balance the forces, already acting on it All of these Give rise to the internal stresses in it Change its motion Balance the forces, already acting on it All of these Give rise to the internal stresses in it Change its motion ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The centre of gravity a T-section 100 mm × 150 mm × 50 mm from its bottom is 87.5 mm 75 mm 50 mm 125 mm 87.5 mm 75 mm 50 mm 125 mm ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The loss of kinetic energy during inelastic impact, is given by (where m1 = Mass of the first body,m2 = Mass of the second body, and u1 and u2 = Velocities of the first and second bodies respectively.) [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The forces, which meet at one point, but their lines of action do not lie in a plane, are called Coplanar non-concurrent forces Intersecting forces Non-coplanar non-concurrent forces Non-coplanar concurrent forces Coplanar non-concurrent forces Intersecting forces Non-coplanar non-concurrent forces Non-coplanar concurrent forces 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 = (gx²/2u² cos²α) + x. tanα y = x. tanα - (gx²/2u² cos²α) y = x. tanα + (gx²/2u² cos²α) y = (gx²/2u² cos²α) - x. tanα y = (gx²/2u² cos²α) + x. tanα y = x. tanα - (gx²/2u² cos²α) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The unit of work in S.I. units is erg kg-m Newton joule erg kg-m Newton joule ANSWER DOWNLOAD EXAMIANS APP