Applied Mechanics and Graphic Statics The centre of gravity of a triangle is at the point where three Medians of the triangle meet Bisectors of the angle of the triangle meet None of these Perpendicular bisectors of the sides of the triangle meet Medians of the triangle meet Bisectors of the angle of the triangle meet None of these Perpendicular bisectors of the sides of the triangle meet ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The shape of a suspended cable for a uniformly distributed load over it is Cubic parabola Circular Catenary Parabolic Cubic parabola Circular Catenary Parabolic ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Periodic time of body moving with simple harmonic motion, is Inversely proportional to its angular velocity Directly proportional to its angular velocity Inversely proportional to the square of its angular velocity Directly proportional to the square of its angular velocity Inversely proportional to its angular velocity Directly proportional to its angular velocity Inversely proportional to the square of its angular velocity Directly proportional to the square of its angular velocity ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 2 kg -m2 and its rate of rotation 2 revolutions/see. The angular momentum of the body in kg-m2/sec is 4 8TC 8 6 7i 4 8TC 8 6 7i ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A trolley wire weighs 1 kg per metre length. The ends of the wire are attached to two poles 20 m apart. If the horizontal tension is 1000 kg, the central dip of the cable is 2 cm 3 cm 4 cm 5 cm 2 cm 3 cm 4 cm 5 cm ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics When a body of mass M1 is hanging freely and another of mass M2 lying on a smooth inclined plane(α) are connected by a light index tensile string passing over a smooth pulley, the acceleration of the body of mass M1, will be given by g(M2 × M1 sin α)/(M2 - M1) m/sec² g(M1 - M2 sin α)/(M1 + M2) m/sec² g(M2 + M1 sin α)/(M1 + M2) m/sec² g(M1 + M2 sin α)/(M1 + M2) m/sec g(M2 × M1 sin α)/(M2 - M1) m/sec² g(M1 - M2 sin α)/(M1 + M2) m/sec² g(M2 + M1 sin α)/(M1 + M2) m/sec² g(M1 + M2 sin α)/(M1 + M2) m/sec ANSWER DOWNLOAD EXAMIANS APP
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