Machine Design Belt slip may occur due to All of these Heavy load Loose belt Driving pulley too small All of these Heavy load Loose belt Driving pulley too small ANSWER DOWNLOAD EXAMIANS APP
Machine Design A shaft is subjected to a maximum bending stress of 80 N/mm² and maximum shearing stress equal to 30 N/mm² at a particular section. If the yield point in tension of the material is 280 N/mm² and the maximum shear stress theory of failure is used, then the factor of safety obtained will be 3.5 2.8 3 2.5 3.5 2.8 3 2.5 ANSWER DOWNLOAD EXAMIANS APP
Machine Design For a shoe brake, the equivalent coefficient of friction is equal to (where μ = Actual coefficient of friction, and θ = Semi-block angle) 2μ sinθ/(θ + sinθ) 4μ sinθ/(2θ + sin 2θ) μ sinθ/(2θ + sin 2θ) 4μ sinθ/(θ + sinθ) 2μ sinθ/(θ + sinθ) 4μ sinθ/(2θ + sin 2θ) μ sinθ/(2θ + sin 2θ) 4μ sinθ/(θ + sinθ) ANSWER DOWNLOAD EXAMIANS APP
Machine Design A coupling used to connect two perfectly aligned shafts, is Compression coupling Flange coupling Muff coupling All of these Compression coupling Flange coupling Muff coupling All of these ANSWER DOWNLOAD EXAMIANS APP
Machine Design In case of pressure vessels having closed ends, the fluid pressure induces Shear stress None of these Longitudinal stress Circumferential stress Shear stress None of these Longitudinal stress Circumferential stress ANSWER DOWNLOAD EXAMIANS APP
Machine Design Two closely coiled helical springs with stiffness k₁ and k₂ respectively are connected in series. The stiffness of an equivalent spring is given by (k₁ + k₂)/ (k₁ k₂) (k₁ - k₂)/ (k₁ k₂) (k₁ - k₂)/ (k₁ + k₂) (k₁ k₂)/ (k₁ + k₂) (k₁ + k₂)/ (k₁ k₂) (k₁ - k₂)/ (k₁ k₂) (k₁ - k₂)/ (k₁ + k₂) (k₁ k₂)/ (k₁ + k₂) ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS