Engineering Mechanics The power developed by a body acted upon by a torque 'T' Newton meter (N - m) and revolving at ω radian/s is given by T.ω/75 (in kilowatts) T.ω (in watts) T.ω/60 (in watts) T.ω/4500 (in kilowatts) T.ω/75 (in kilowatts) T.ω (in watts) T.ω/60 (in watts) T.ω/4500 (in kilowatts) 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 h2/kG h × kG h/kG kG2/h h2/kG h × kG h/kG kG2/h 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 vertex and parallel to the base, is __________ than that passing through its C.G. and parallel to the base. Nine times Six times Two times Four times Nine times Six times Two times Four times ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The rate of doing work is known as Power Potential energy None of these Kinetic energy Power Potential energy None of these Kinetic energy 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 (1/12) Ml² (1/3) Ml² (2/3) Ml² (3/4) Ml² (1/12) Ml² (1/3) Ml² (2/3) Ml² (3/4) Ml² 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
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