Theory of Machine Cylindrical cams can be classified as Reciprocating Circular None of these Tangent Reciprocating Circular None of these Tangent ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The frictional torque transmitted in a truncated conical pivot bearing, considering uniform wear, is (2/3). μ W cosec α [(r₁³ - r₂³)/(r₁² - r₂²)] (1/2). μ W cosec α [(r₁³ - r₂³)/(r₁² - r₂²)] (2/3).μ W cosec α (r₁ + r₂) (1/2). μ W cosec α (r₁ + r₂) (2/3). μ W cosec α [(r₁³ - r₂³)/(r₁² - r₂²)] (1/2). μ W cosec α [(r₁³ - r₂³)/(r₁² - r₂²)] (2/3).μ W cosec α (r₁ + r₂) (1/2). μ W cosec α (r₁ + r₂) ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The primary unbalanced force due to inertia of reciprocating parts in a reciprocating engine is given by (where m = Mass of reciprocating parts, ω = Angular speed of crank, r = Radius of crank, θ = Angle of inclination of crank with the line of stroke, and n = Ratio of the length of connecting rod to radius of crank) m.ω².r (sin 2θ/n) m.ω².r cosθ m.ω².r sinθ m.ω².r (cos 2θ/n) m.ω².r (sin 2θ/n) m.ω².r cosθ m.ω².r sinθ m.ω².r (cos 2θ/n) ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Whitworth quick return mechanism is obtained by inversion of Five link mechanism Slider crank mechanism Roller cam mechanism Kinematic chain Five link mechanism Slider crank mechanism Roller cam mechanism Kinematic chain ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The Coriolis component of acceleration acts Along the sliding surface Perpendicular to the sliding surface Parallel to the sliding surface At 45° to the sliding surface Along the sliding surface Perpendicular to the sliding surface Parallel to the sliding surface At 45° to the sliding surface ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Two heavy rotating masses are connected by shafts of lengths l₁, l₂ and l₃ and the corresponding diameters are d₁, d₂ and d₃. This system is reduced to a torsionally equivalent system having uniform diameter d = d₁ of the shaft. The equivalent length of the shaft is l₁ + l₂ + l₃ l = l₁ + l₂.(d₁/d₂)³ + l₂.(d₁/d₃)³ (l₁ + l₂ + l₃)/3 l = l₁ + l₂.(d₁/d₂)⁴ + l₃.(d₁/d₃)⁴ l₁ + l₂ + l₃ l = l₁ + l₂.(d₁/d₂)³ + l₂.(d₁/d₃)³ (l₁ + l₂ + l₃)/3 l = l₁ + l₂.(d₁/d₂)⁴ + l₃.(d₁/d₃)⁴ ANSWER DOWNLOAD EXAMIANS APP