Theory of Machine The frictional torque transmitted in a truncated conical pivot bearing, considering uniform pressure, is (1/2). μ W cosecα [(r₁³ - r₂³)/(r₁² - r₂²)] (1/2). μ W cosecα (r₁ + r₂) (2/3). μ W cosecα [(r₁³ - r₂³)/(r₁² - r₂²)] (2/3). μ W cosecα (r₁ + r₂) (1/2). μ W cosecα [(r₁³ - r₂³)/(r₁² - r₂²)] (1/2). μ W cosecα (r₁ + r₂) (2/3). μ W cosecα [(r₁³ - r₂³)/(r₁² - r₂²)] (2/3). μ W cosecα (r₁ + r₂) ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Which of the following statements regarding laws governing the friction between dry surfaces are correct? All of these The friction force is directly proportional to the normal force. The friction force is dependent on the materials of the contact surfaces. The friction force is independent of me area of contact. All of these The friction force is directly proportional to the normal force. The friction force is dependent on the materials of the contact surfaces. The friction force is independent of me area of contact. ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The velocity of the rubbing surface __________ with the distance from the axis of the bearing. Increases Decreases Remain same None of these Increases Decreases Remain same None of these ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The cam and follower is an example of Rolling pair Sliding pair Lower pair Higher pair Rolling pair Sliding pair Lower pair Higher pair ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine To obviate axial thrust, following gear drive is used Double helical gears having identical teeth Single helical gear in which one of the teeth of helix angle a is more Mutter gears Double helical gears having opposite teeth Double helical gears having identical teeth Single helical gear in which one of the teeth of helix angle a is more Mutter gears Double helical gears having opposite teeth ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine The Kutzbach criterion for determining the number of degrees of freedom (n) is (where l = number of links, j = number of joints and h = number of higher pairs) n = 3(l - 1) - 2j - h n = 2(l - 1) - 2j - h n = 3(l - 1) - 3j - h n = 2(l - 1) - 3j - h n = 3(l - 1) - 2j - h n = 2(l - 1) - 2j - h n = 3(l - 1) - 3j - h n = 2(l - 1) - 3j - h ANSWER DOWNLOAD EXAMIANS APP
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