Engineering Mechanics The unit of angular acceleration is m/s2 rad/s2 m/s N-m m/s2 rad/s2 m/s N-m ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A semicircular disc rests on a horizontal surface with its top flat surface horizontal and circular portion touching down. The coefficient of friction between semi circular disc and horizontal surface is µ. This disc is to be pulled by a horizontal force applied at one edge and it always remains horizontal. When the disc is about to start moving, its top horizontal force will None of these Slant down towards direction of pull Slant up towards direction of pull Remain horizontal None of these Slant down towards direction of pull Slant up towards direction of pull Remain horizontal ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The equivalent length of a simple pendulum which gives the same frequency as compound pendulum is (kG² + h²)/h² h/(kG² + h²) h²/(kG² + h²) (kG² + h²)/h (kG² + h²)/h² h/(kG² + h²) h²/(kG² + h²) (kG² + h²)/h ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The resolved part of the resultant of two forces inclined at an angle 'θ' in a given direction is equal to The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The algebraic sum of the resolved parts of the forces in the given direction The difference of the forces multiplied by the cosine of θ The sum of the resolved parts of the forces in the given direction The sum of the forces multiplied by the sine of θ The algebraic sum of the resolved parts of the forces in the given direction The difference of the forces multiplied by the cosine of θ ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A trolley wire weighs 1.2 kg per meter length. The ends of the wire are attached to two poles 20 meters apart. If the horizontal tension is 1500 kg find the dip in the middle of the span 5.0 cm 3.0 cm 4.0 cm 2.5 cm 5.0 cm 3.0 cm 4.0 cm 2.5 cm ANSWER DOWNLOAD EXAMIANS APP
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 D/(d₁ + d₂) 2D/(d₁ + d₂) D/(d₁ - d₂) 2D/(d₁ - d₂) D/(d₁ + d₂) 2D/(d₁ + d₂) D/(d₁ - d₂) 2D/(d₁ - d₂) ANSWER DOWNLOAD EXAMIANS APP
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