Machine Design In a flange coupling, the keys are staggered at __________ along the circumference of the shafts in order to divide the weakening effect caused by key ways. 90° 180° 135° 160° 90° 180° 135° 160° ANSWER DOWNLOAD EXAMIANS APP
Machine Design The belt slip occurs due to Driving pulley too small Loose belt Any one of the above Heavy load Driving pulley too small Loose belt Any one of the above Heavy load ANSWER DOWNLOAD EXAMIANS APP
Machine Design For same pulley diameters, center distance, belt speed and belt and pulley materials, Power transmission does not depend upon open and crossed types of constructions Crossed belt drive transmits more power than open belt drive Open belt drive transmits more power than crossed belt drive Open and crossed belt drives transmit same power Power transmission does not depend upon open and crossed types of constructions Crossed belt drive transmits more power than open belt drive Open belt drive transmits more power than crossed belt drive Open and crossed belt drives transmit same power ANSWER DOWNLOAD EXAMIANS APP
Machine Design When the shaft rotates in anticlockwise direction at slow speed in a bearing, it will Move towards right of the bearing making no metal to metal contact Move towards left of the bearing making metal to metal contact Move towards right of the bearing making the metal to metal contact Have contact at the bottom most of the bearing Move towards right of the bearing making no metal to metal contact Move towards left of the bearing making metal to metal contact Move towards right of the bearing making the metal to metal contact Have contact at the bottom most of the bearing ANSWER DOWNLOAD EXAMIANS APP
Machine Design In a partial journal bearing, the angle of contact of the bearing with the journal is 120° 360° 180° 270° 120° 360° 180° 270° ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ratio of belt tensions (p1/p2) considering centrifugal force in flat belt is given byWhere m = mass of belt per meter (kg/m)v = belt velocity (m/s)f = coefficient of frictiona = angle of wrap (radians) (P1 - mv²)/ (P2 - mv²) = e–ᶠα (P1 - mv²)/ (P2 - mv²) = eᶠα P1 / P2 = e–ᶠα P1 / P2 = eᶠα (P1 - mv²)/ (P2 - mv²) = e–ᶠα (P1 - mv²)/ (P2 - mv²) = eᶠα P1 / P2 = e–ᶠα P1 / P2 = eᶠα ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS