Engineering Mechanics Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is bh3/4 bh3/8 bh3/12 bh3/36 bh3/4 bh3/8 bh3/12 bh3/36 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Moment of inertia of a triangular section of base (b) and height (h) about an axis passing through its C.G. and parallel to the base, is bh3/8 bh3/36 bh3/12 bh3/4 bh3/8 bh3/36 bh3/12 bh3/4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The point, through which the whole weight of the body acts, irrespective of its position, is known as Center of mass Moment of inertia Center of percussion Center of gravity Center of mass Moment of inertia Center of percussion Center of gravity ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A lead ball with a certain velocity is made to strike a wall, it falls down, but rubber ball of same mass and with same velocity strikes the same wall, it rebounds. Select the correct reason from the following: The change in momentum suffered by rubber ball is more than the lead ball None of these Both the balls undergo an equal change in momentum The change in momentum suffered by rubber ball is less than the lead ball The change in momentum suffered by rubber ball is more than the lead ball None of these Both the balls undergo an equal change in momentum The change in momentum suffered by rubber ball is less than the lead ball ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics On the ladder resting on the ground and leaning against a smooth vertical wall, the force of friction will be Perpendicular to the wall at its upper end Downwards at its upper end Zero at its upper end Upwards at its upper end Perpendicular to the wall at its upper end Downwards at its upper end Zero at its upper end Upwards at its upper end 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
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