Machine Design In order to obtain bolt of uniform strength Tighten die bolt properly Drill an axial hole through head up to threaded portion so that shank area is equal to root area of thread Increase its length Increase shank diameter Tighten die bolt properly Drill an axial hole through head up to threaded portion so that shank area is equal to root area of thread Increase its length Increase shank diameter ANSWER DOWNLOAD EXAMIANS APP
Machine Design An open coiled helical compression spring 'A' of mean diameter 50 mm is subjected to an axial load ‘W’. Another spring 'B' of mean diameter 25 mm is similar to spring 'A' in all respects. The deflection of spring 'B' will be __________ as compared to spring 'A'. One-half One-eighth Double One-fourth One-half One-eighth Double One-fourth ANSWER DOWNLOAD EXAMIANS APP
Machine Design According to Indian standards, total numbers of fundamental deviations are 25 20 10 15 25 20 10 15 ANSWER DOWNLOAD EXAMIANS APP
Machine Design The rolling contact bearings are known as Thick lubricated bearings Plastic bearings Thin lubricated bearings Antifriction bearings Thick lubricated bearings Plastic bearings Thin lubricated bearings Antifriction bearings 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
Machine Design If T₁ and T₂ are the tensions on the tight and slack sides of the belt respectively, and Tc is the centrifugal tension, then initial tension in the belt is equal to (T₁ + T₂ + Tc)/2 (T₁ - T₂ + Tc)/2 T₁ - T₂ + Tc T₁ + T₂ + Tc (T₁ + T₂ + Tc)/2 (T₁ - T₂ + Tc)/2 T₁ - T₂ + Tc T₁ + T₂ + Tc ANSWER DOWNLOAD EXAMIANS APP
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