Theory of Machine The natural frequency of free transverse vibrations due to uniformly distributed load acting over a simply supported shaft is (where δS = Static deflection of simply supported shaft due to uniformly distributed load) 0.5615/ √δS 0.6253/ √δS 0.571/ √δS 0.4985/ √δS 0.5615/ √δS 0.6253/ √δS 0.571/ √δS 0.4985/ √δS ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine A thin circular disc is rolling with a uniform linear speed, along a straight path on a plane surface. Which of the following statement is correct in this regard? All points of the disc have the same velocity The centre of the disc has zero acceleration The centre of the disc has centrifugal acceleration The point on the disc making contact with the plane surface has zero acceleration All points of the disc have the same velocity The centre of the disc has zero acceleration The centre of the disc has centrifugal acceleration The point on the disc making contact with the plane surface has zero acceleration ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The instantaneous center of a rigid thin disc rolling on a plane rigid surface is located at The center of the disc An infinite distance on the plane surface The point on the circumference situated vertically opposite to the contact point The point of contact The center of the disc An infinite distance on the plane surface The point on the circumference situated vertically opposite to the contact point The point of contact ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine When the crank is at the inner dead centre, in a reciprocating steam engine, then the velocity of the piston will be Minimum Zero None of these Maximum Minimum Zero None of these Maximum ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine A cam mechanism imparts following motion All of these Reciprocating Oscillating Rotating All of these Reciprocating Oscillating Rotating ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine In a disc clutch, if there are n₁ number of discs on the driving shaft and n₂ number of discs on the driven shaft, then the number of pairs of contact surfaces will be n₁ + n₂ n₁ + n₂ + 1 n₁ + n₂ - 1 n₁ + n₂ - 2 n₁ + n₂ n₁ + n₂ + 1 n₁ + n₂ - 1 n₁ + n₂ - 2 ANSWER DOWNLOAD EXAMIANS APP
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