Engineering Mechanics Moment of inertia of a circular section about an axis perpendicular to the section is πd4/32 πd4/64 πd3/32 πd3/16 πd4/32 πd4/64 πd3/32 πd3/16 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) IP = Ah2 / IG IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 IP = Ah2 / IG IP = IG - Ah2 IP = IG / Ah2 IP = IG + Ah2 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The rate of change of momentum is directly proportional to the impressed force, and takes place in the same direction in which the force acts. This statement is known as Newton's first law of motion Newton's second law of motion Newton's third law of motion None of these Newton's first law of motion Newton's second law of motion Newton's third law of motion None of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The horizontal range of a projectile (R) is given by R = u² sin2α/g R = u² sinα/g R = u² cos2α/g R = u² cosα/g R = u² sin2α/g R = u² sinα/g R = u² cos2α/g R = u² cosα/g ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity ratio in case of an inclined plane inclined at angle 'θ' to the horizontal and weight being pulled up the inclined plane by vertical effort is cosθ cosecθ tanθ sinθ cosθ cosecθ tanθ sinθ 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
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