Theory of Machine In order to give a complete secondary balance of a multi-cylinder inline engine, Both (A) and (B) None of these The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero The algebraic sum of the secondary forces must be equal to zero Both (A) and (B) None of these The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero The algebraic sum of the secondary forces must be equal to zero ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The Whitworth quick return motion mechanism is formed in a slider crank chain when the Longest link is a fixed link Slider is a fixed link Coupler link is fixed Smallest link is a fixed link Longest link is a fixed link Slider is a fixed link Coupler link is fixed Smallest link is a fixed link ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Module of a gear is D/T 2T/D 2D/T T/D D/T 2T/D 2D/T T/D ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine A body having moment of inertia o:m2 is rotating at 210 RPM and r with another body at rest having I 40 kg m2. The resultant speed after ing will be 90 RPM data are insufficient 80 RPM 100 RPM 90 RPM data are insufficient 80 RPM 100 RPM ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine In order to give the primary balance of the reciprocating parts of a multi-cylinder inline engines The algebraic sum of the primary forces must be equal to zero Both (A) and (B) None of these The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero The algebraic sum of the primary forces must be equal to zero Both (A) and (B) None of these The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero 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 l = l₁ + l₂.(d₁/d₂)⁴ + l₃.(d₁/d₃)⁴ l = l₁ + l₂.(d₁/d₂)³ + l₂.(d₁/d₃)³ l₁ + l₂ + l₃ (l₁ + l₂ + l₃)/3 l = l₁ + l₂.(d₁/d₂)⁴ + l₃.(d₁/d₃)⁴ l = l₁ + l₂.(d₁/d₂)³ + l₂.(d₁/d₃)³ l₁ + l₂ + l₃ (l₁ + l₂ + l₃)/3 ANSWER DOWNLOAD EXAMIANS APP
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