Applied Mechanics and Graphic Statics The velocity ratio of the differential wheel and axle is R/r1 - r2 2R/r1 2R/r1 + r2 3R/r1 - r2 R/r1 - r2 2R/r1 2R/r1 + r2 3R/r1 - r2 ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A satellite goes on moving along its orbit round the earth due to Centripetal force Centrifugal force Gravitational force None of these Centripetal force Centrifugal force Gravitational force None of these ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is 40 kg 100 kg 10 kg 60 kg 40 kg 100 kg 10 kg 60 kg ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The resultant of two forces ‘P’ and ‘Q’ acting at an angle ‘θ’, is P² + Q² + 2PQ tan θ √(P² + Q² + 2PQ cos θ) P² + Q² + 2P sin θ P² + Q² + 2PQ cos θ P² + Q² + 2PQ tan θ √(P² + Q² + 2PQ cos θ) P² + Q² + 2P sin θ P² + Q² + 2PQ cos θ ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If two bodies of masses M1 and M2(M1 > M2) are connected by alight inextensible string passing over a smooth pulley, the tension in the string, will be given by T = g(M2 - M1)/(M1 + M2) T = g(M1 - M2)/(M1 + M2) T = g(M2 + M1)/(M2 - M1) T = g(M1 + M2)/(M1 × M2) T = g(M2 - M1)/(M1 + M2) T = g(M1 - M2)/(M1 + M2) T = g(M2 + M1)/(M2 - M1) T = g(M1 + M2)/(M1 × M2) ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A stone is whirled in a vertical circle, the tension in the string, is maximum When the string is horizontal When the stone is at the lowest position At all the positions When the stone is at the highest position When the string is horizontal When the stone is at the lowest position At all the positions When the stone is at the highest position ANSWER DOWNLOAD EXAMIANS APP
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