Theory of Machine For fluctuating loads, well suited bearing is Roller bearing Needle roller bearing Ball bearing Thrust bearing Roller bearing Needle roller bearing Ball bearing Thrust bearing ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine Klein's construction can be used when Crank has uniform angular acceleration Crank has a uniform angular velocity Crank has uniform angular velocity and angular acceleration Crank has non-uniform velocity Crank has uniform angular acceleration Crank has a uniform angular velocity Crank has uniform angular velocity and angular acceleration Crank has non-uniform velocity ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine When a point moves along a straight line, its acceleration will have Radial and tangential components both Radial component only Coriolis component only Tangential component only Radial and tangential components both Radial component only Coriolis component only Tangential component only ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine In considering friction of a V-thread, the virtual coefficient of friction (μ₁) is given by μ₁ = μ/cos β μ₁ = μ cos β μ₁ = μ sin β μ₁ = μ/sin β μ₁ = μ/cos β μ₁ = μ cos β μ₁ = μ sin β μ₁ = μ/sin β ANSWER DOWNLOAD EXAMIANS APP
Theory of Machine A rigid body, under the action of external forces, can be replaced by two masses placed at a fixed distance apart. The two masses form an equivalent dynamical system, if The center of gravity of the two masses coincides with that of the body The sum of the two masses is equal to the total mass of body The sum of mass moment of inertia of the masses about their center of gravity is equal to the mass moment of inertia of the body All of these The center of gravity of the two masses coincides with that of the body The sum of the two masses is equal to the total mass of body The sum of mass moment of inertia of the masses about their center of gravity is equal to the mass moment of inertia of the body All of these 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) - 3j - h n = 3(l - 1) - 3j - h n = 2(l - 1) - 2j - h n = 3(l - 1) - 2j - h n = 2(l - 1) - 3j - h n = 3(l - 1) - 3j - h n = 2(l - 1) - 2j - h ANSWER DOWNLOAD EXAMIANS APP
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