Engineering Mechanics The velocity ratio of a differential wheel and axle with 'D' as the diameter of effort wheel and d1 and d2 as the diameters of larger and smaller axles respectively, is 2D/(d₁ + d₂) 2D/(d₁ - d₂) D/(d₁ + d₂) D/(d₁ - d₂) 2D/(d₁ + d₂) 2D/(d₁ - d₂) D/(d₁ + d₂) D/(d₁ - d₂) 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
Engineering Mechanics For a self locking machine, the efficiency must be Greater than 50% Less than 50% 1 Equal to 50% Greater than 50% Less than 50% 1 Equal to 50% ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The horizontal range of a projectile (R) is given by R = u² sin2α/g R = u² cosα/g R = u² cos2α/g R = u² sinα/g R = u² sin2α/g R = u² cosα/g R = u² cos2α/g R = u² sinα/g ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The total momentum of a system of masses (i. e. moving bodies) in any one direction remains constant, unless acted upon by an external force in that direction. This statement is called Newton's first law of motion Principle of conservation of energy Newton's second law of motion Principle of conservation of momentum Newton's first law of motion Principle of conservation of energy Newton's second law of motion Principle of conservation of momentum ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The centre of gravity of a uniform lamina lies at The bottom surface All of these The centre of heavy portion The midpoint of its axis The bottom surface All of these The centre of heavy portion The midpoint of its axis ANSWER DOWNLOAD EXAMIANS APP
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