Engineering Mechanics The centre of gravity a T-section 100 mm × 150 mm × 50 mm from its bottom is 125 mm 87.5 mm 50 mm 75 mm 125 mm 87.5 mm 50 mm 75 mm ANSWER DOWNLOAD EXAMIANS APP
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 (in watts) T.ω/75 (in kilowatts) T.ω/4500 (in kilowatts) T.ω/60 (in watts) T.ω (in watts) T.ω/75 (in kilowatts) T.ω/4500 (in kilowatts) T.ω/60 (in watts) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A ladder is resting on a smooth ground and leaning against a rough vertical wall. The force of friction will act Downward at its upper end Towards the wall at its upper end Upward at its upper end Away from the wall at its upper end Downward at its upper end Towards the wall at its upper end Upward at its upper end Away from the wall at its upper end ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics Whenever a force acts on a body and the body undergoes a displacement, then Power is being transmitted None of these Work is said to be done Body has kinetic energy of translation Power is being transmitted None of these Work is said to be done Body has kinetic energy of translation ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The moment of inertia of a sphere of mass 'm' and radius 'r', about an axis tangential to it, is 2mr²/3 2mr²/5 7mr²/3 7mr²/5 2mr²/3 2mr²/5 7mr²/3 7mr²/5 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If ‘P’ is the force acting on the body, ‘m’ is the mass of the body and ‘a’ is the acceleration of the body, then according to Newton's second law of motion, P + m.a = 0 P × m.a = 0 P/m.a = 0 P - m.a = 0 P + m.a = 0 P × m.a = 0 P/m.a = 0 P - m.a = 0 ANSWER DOWNLOAD EXAMIANS APP
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