Theory of Machine The maximum efficiency of spiral gears is (where θ = Shaft angle, and φ = Friction angle) cos (θ - φ) + 1/ cos (θ + φ) + 1 cos (θ + φ) + 1/ cos (θ - φ) + 1 sin (θ + φ) + 1/ cos (θ - φ) + 1 cos (θ - φ) + 1/ sin (θ + φ) + 1 cos (θ - φ) + 1/ cos (θ + φ) + 1 cos (θ + φ) + 1/ cos (θ - φ) + 1 sin (θ + φ) + 1/ cos (θ - φ) + 1 cos (θ - φ) + 1/ sin (θ + φ) + 1 ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Which of the following is an example of a higher pair? Belt and rope drive All of the listed here Ball and roller bearing Toothed gearing Belt and rope drive All of the listed here Ball and roller bearing Toothed gearing ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Module of a gear is 2D/T D/T T/D 2T/D 2D/T D/T T/D 2T/D ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine In a cam drive with uniform velocity follower, the sharp corners of the displacement diagram are rounded off at the beginning and at the end of each stroke. This is done In order to have acceleration in beginning and retardation at the end of stroke within the finite limits Because of difficulty in manufacturing cam profile Because the uniform velocity motion is a partial parabolic motion Because of loose contact of follower with cam surface In order to have acceleration in beginning and retardation at the end of stroke within the finite limits Because of difficulty in manufacturing cam profile Because the uniform velocity motion is a partial parabolic motion Because of loose contact of follower with cam surface ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine A body of weight W is required to move up the rough inclined plane whose angle of inclination with the horizontal is α. The effort applied parallel to the plane is given by (where μ = tan φ = Coefficient of friction between the plane and the body) P = W tan (α + φ) P = W tan α P = W (sin α + μ cos α) P = W (cos α + μ sin α) P = W tan (α + φ) P = W tan α P = W (sin α + μ cos α) P = W (cos α + μ sin α) ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The essential condition of placing the two masses, so that the system becomes dynamically equivalent, is (where l₁ and l₂ = Distance of two masses from the centre of gravity of the body, and kG = Radius of gyration of the body) l₂ = kG l₁l₂ = kG l₁l₂ = kG² l₁ = kG l₂ = kG l₁l₂ = kG l₁l₂ = kG² l₁ = kG ANSWER DOWNLOAD EXAMIANS APP
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