When the speed of a turbine is 6000 rpm, it takes _____ seconds for the turbine to complete one revolution. a)0.1 sec b)0.0001sec c)10sec d)0.001sec

When the speed of a turbine is 6000 rpm, it takes _____ seconds for the turbine to complete one
revolution.
a)0.1 sec
b)0.0001sec
c)10sec
d)0.001sec
by

1 Answer

Answer :

c)10sec
by

Related questions

Two mail trains Pand Q of 300 m and 600m, run at speed of 130 km/hr and 160 km/hr respectively, in the direction opposite direction to each other. Find the time required to cross each other after the moment they met? A) 11.17 sec B) 12sec C) 9.10sec D) 21sec

Description : Two mail trains Pand Q of 300 m and 600m, run at speed of 130 km/hr and 160 km/hr respectively, in the direction opposite direction to each other. Find the time required to cross each other after the moment they met? A) 11.17 sec B) 12sec C) 9.10sec D) 21sec

Last Answer :  ANSWER:A Explanation:  Length of train P = 300 m  Length of train Q = 600 m  Speed of train p = 130 kmph  Convert km/hr into m/s,  Speed of train p =(130 *5/18)m/s  = 650/18 m/s  = 36.11 m/ ... train B)  = (300+600)m = 900 m We know, time =distance /speed  = 900/ 80.55  = 11.17seconds

A goods train 110m long is running with the speed of 30 km/hr. in what time will it pass a boy who is running at 3.5 km/hr in the direction opposite to that in which the train is going. A) 9 sec B) 10sec C) 12sec D) 13sec

Description : A goods train 110m long is running with the speed of 30 km/hr. in what time will it pass a boy who is running at 3.5 km/hr in the direction opposite to that in which the train is going. A) 9 sec B) 10sec C) 12sec D) 13sec

Last Answer : ANSWER : C Explanation:  Length of the train = 110 m  Speed of the train = 30 km/hr  Speed of the boy = 3.5 km/hr  Speed of the train relative to man = (30 + 3.5) km/hr  = 33.5 km/hr  Convert km/ ... cover 110 m at 9.30 m/s  = 110 / 9.30  = 11.8 seconds  = 12 sec(approx.).

Rotor shaft of a large electric motor supported between short bearings at both ends shows a deflection of 1.8 mm in middle of motor. Assuming rotor to be perfectly balanced and supported at knife edges at both ends, likely critical speed (rpm) of shaft isA 350 B 4430 C 705 D 2810

Description : Rotor shaft of a large electric motor supported between short bearings at both ends shows a deflection of 1.8 mm in middle of motor. Assuming rotor to be perfectly balanced and supported at knife edges at both ends, likely critical speed (rpm) of shaft isA 350 B 4430 C 705 D 2810

Last Answer : C 705

A flywheel starting from rest and accelerating uniformly performs 25 revolutions in 5 sec. Its angular velocity in rpm after 10 seconds would be a.50 rpm b.20 rpm c.25 rpm d.60 rpm e.45 rpm

Description : A flywheel starting from rest and accelerating uniformly performs 25 revolutions in 5 sec. Its angular velocity in rpm after 10 seconds would be a.50 rpm b.20 rpm c.25 rpm d.60 rpm e.45 rpm

Last Answer : b. 20 rpm

Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. A 25.62 rad/sec B 20.78 rad/sec C 14.4 rad/sec D 15.33 rad/sec

Description : Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. A 25.62 rad/sec B 20.78 rad/sec C 14.4 rad/sec D 15.33 rad/sec

Last Answer : A 25.62 rad/sec

Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. A 21 rad/sec B 25.62 rad/sec C 20.22 Hz D 3.15 Hz

Description : Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. A 21 rad/sec B 25.62 rad/sec C 20.22 Hz D 3.15 Hz

Last Answer : B 25.62 rad/sec

Determine the viscous damping coefficient if damper offers resistance 0.05N at constant velocity 0.04m/sec A 0.8N-sec/m B 1.5N-sec/m C 2.5N-sec/m D 1.25N-sec/m

Description : Determine the viscous damping coefficient if damper offers resistance 0.05N at constant velocity 0.04m/sec A 0.8N-sec/m B 1.5N-sec/m C 2.5N-sec/m D 1.25N-sec/m

Last Answer : D 1.25N-sec/m

A 1 kg mass is suspended by a spring having a stiffness of 0.4 N/mm. Determine the natural frequency. A 20 rad/sec B 30 rad/sec C 20 Hz D 30 Hz

Description : A 1 kg mass is suspended by a spring having a stiffness of 0.4 N/mm. Determine the natural frequency. A 20 rad/sec B 30 rad/sec C 20 Hz D 30 Hz

Last Answer : B 30 rad/sec

In above numerical what will be the new time period if the spring constant is decreased by 50%. A 0.5 sec B 0.353 sec C 0.125 sec D 0.533 sec

Description : In above numerical what will be the new time period if the spring constant is decreased by 50%. A 0.5 sec B 0.353 sec C 0.125 sec D 0.533 sec

Last Answer : B 0.353 sec

A spring mass system has time period of oscillation of 0.25 sec. What will be the natural frequency of the system? A 1 Hz B 2 rad sec C 4 rad/sec D 4 Hz

Description : A spring mass system has time period of oscillation of 0.25 sec. What will be the natural frequency of the system? A 1 Hz B 2 rad sec C 4 rad/sec D 4 Hz

Last Answer : D 4 Hz

Calculate damped natural frequency, if a spring mass damper system is subjected to periodic disturbing force of 30 N. Damping coefficient is equal to 0.76 times of critical damping coefficient and undamped natural frequency is 5 rad/sec A. 3.99 rad/sec B. 2.13 rad/sec C. 4.12 rad/sec D. 3.24 rad/sec

Description : Calculate damped natural frequency, if a spring mass damper system is subjected to periodic disturbing force of 30 N. Damping coefficient is equal to 0.76 times of critical damping coefficient and undamped natural frequency is 5 rad/sec A. 3.99 rad/sec B. 2.13 rad/sec C. 4.12 rad/sec D. 3.24 rad/sec

Last Answer : D. 3.24 rad/sec

Calculate natural frequency of damped vibration, if damping factor is 0.52 and A natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. ( A )25.62 rad/sec ( B )20.78 rad/sec ( C )14.4 rad/sec ( D )15.33 rad/sec

Description : Calculate natural frequency of damped vibration, if damping factor is 0.52 and A natural frequency of the system is 30 rad/sec which consists of machine supported on springs and dashpots. ( A )25.62 rad/sec ( B )20.78 rad/sec ( C )14.4 rad/sec ( D )15.33 rad/sec

Last Answer : ( A )25.62 rad/sec

Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequ the system is 30 rad/sec which consists of machine supported on springs and dashpots. a. 25.62 rad/secb. 20.78 rad/sec c. 14.4 rad/sec d. 15.33 rad/sec

Description : Calculate natural frequency of damped vibration, if damping factor is 0.52 and natural frequ the system is 30 rad/sec which consists of machine supported on springs and dashpots. a. 25.62 rad/secb. 20.78 rad/sec c. 14.4 rad/sec d. 15.33 rad/sec

Last Answer : a. 25.62 rad/sec

find the value of logarithmic decrement of a vibratory system if its natural frequency is 10rad/sec, its mass is 10kg and its damping constant is 100N.s/m a) 36.27 b)362.7 c)0.3627 d)3.627

Description : find the value of logarithmic decrement of a vibratory system if its natural frequency is 10rad/sec, its mass is 10kg and its damping constant is 100N.s/m a) 36.27 b)362.7 c)0.3627 d)3.627

Last Answer : d)3.627

A car weighing 1000kg deflects its springs by 0.4cm under its load. Determine the natural frequency of 2 car in vertical direction take g=10N/m a) 25 rad/sec b)50 rad/sec c) 2 rad/sec d)none

Description : A car weighing 1000kg deflects its springs by 0.4cm under its load. Determine the natural frequency of 2 car in vertical direction take g=10N/m a) 25 rad/sec b)50 rad/sec c) 2 rad/sec d)none

Last Answer : b)50 rad/sec

Two Pelton turbines A and B have the same specific speed and are working under the same head. Turbine A produces 400 kW at 1000 rpm. If turbine B produces 100kW, then its rpm is (a) 4000 (b) 2000 (c) 1500 (d) 3000

Description : Two Pelton turbines A and B have the same specific speed and are working under the same head. Turbine A produces 400 kW at 1000 rpm. If turbine B produces 100kW, then its rpm is (a) 4000 (b) 2000 (c) 1500 (d) 3000

Last Answer : (b) 2000

What is the safe maximum tripping speed of a turbine operating at 2500 rpm?

Description : What is the safe maximum tripping speed of a turbine operating at 2500 rpm?

Last Answer : The rule is to trip at 10 percent overspeed. Therefore, 2500 x 1.10 = 2750 rpm.

A hydraulic turbine having rated speed of 250 rpm is connected to a synchronous generator. In order to

Description : A hydraulic turbine having rated speed of 250 rpm is connected to a synchronous generator. In order to produce power at 50 Hz, the number of poles required in the generator are (A) 6 (B) 12 (C) 16 (D) 24

Last Answer : A hydraulic turbine having rated speed of 250 rpm is connected to a synchronous generator. In order to produce power at 50 Hz, the number of poles required in the generator are 24

Whenthecrankshaft of 4-stroke running at 1500 rpm thecamshaftwillrun at A. 3000 rpm B. 6000 rpm C. 1500 rpm D. 750 rpm

Description : Whenthecrankshaft of 4-stroke running at 1500 rpm thecamshaftwillrun at A. 3000 rpm B. 6000 rpm C. 1500 rpm D. 750 rpm

Last Answer : ANSWER : D

In above numerical what will be the frequency corresponding to the peak amplitude A 14.18rad/sec B 24.13rad/sec C 20.22rad/sec D 22.32rad/sec

Description : In above numerical what will be the frequency corresponding to the peak amplitude A 14.18rad/sec B 24.13rad/sec C 20.22rad/sec D 22.32rad/sec

Last Answer : A 14.18rad/sec

Calculate the value of critical damping coefficient if a vibrating system has mass of 4kg and stiffness of 100N/m A 20 N-sec/m B 40 N-sec/m C 60 N-sec/m D 80 N-sec/m

Description : Calculate the value of critical damping coefficient if a vibrating system has mass of 4kg and stiffness of 100N/m A 20 N-sec/m B 40 N-sec/m C 60 N-sec/m D 80 N-sec/m

Last Answer : B 40 N-sec/m

The unit of the viscous damping coefficient is A N-m/sec B m/N-sec C N-sec-m D N-sec/m

Description : The unit of the viscous damping coefficient is A N-m/sec B m/N-sec C N-sec-m D N-sec/m

Last Answer : D N-sec/m

Determine natural frequency of a system, which has equivalent spring stiffness of 43200 N/m and mass of 12 kg. A 40.22 rad/sec B 40 Hz C 60 Hz D 60 rad/sec

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 43200 N/m and mass of 12 kg. A 40.22 rad/sec B 40 Hz C 60 Hz D 60 rad/sec

Last Answer : D 60 rad/sec

Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 –3 m 3. Natural frequency of the system = 60 rad/sec A. 100.5 x 10 3 N-s/m B. 80 x 10 3 N-s/m C. 42 x 10 3 N-s/m D. None of the above

Description : Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 -3 m 3. Natural frequency of the system = 60 rad/sec ... /m B. 80 x 10 3 N-s/m C. 42 x 10 3 N-s/m D. None of the above

Last Answer : C. 42 x 10 3 N-s/m

In a 2-mass 3 spring vibrating system the two masses each are of 9.8 kg coupling spring is having a stiffness of 3430 N/m whereas the other two springs have each a stiffness of 8820 N/m. The two natural frequencies in rad /sec are A) 10 & 20 B) 20 & 30 C) 30 & 40D) 40 & 50

Description : In a 2-mass 3 spring vibrating system the two masses each are of 9.8 kg coupling spring is having a stiffness of 3430 N/m whereas the other two springs have each a stiffness of 8820 N/m. The two natural frequencies in rad /sec are A) 10 & 20 B) 20 & 30 C) 30 & 40D) 40 & 50

Last Answer : C) 30 & 40

A rotary system has a damping coefficient of 40 N-m-sec/rad. The damping torque at a velocity of 2 rad/s, will be A) 20 N-m B) 40 N-m C) 80 N-m D) 100 N-m

Description : A rotary system has a damping coefficient of 40 N-m-sec/rad. The damping torque at a velocity of 2 rad/s, will be A) 20 N-m B) 40 N-m C) 80 N-m D) 100 N-m

Last Answer : C) 80 N-m

The units of viscous damping coefficient is A) N-m/sec B) m/N-sec C) N-sec/m D) N-m-sec

Description : The units of viscous damping coefficient is A) N-m/sec B) m/N-sec C) N-sec/m D) N-m-sec

Last Answer : C) N-sec/m

In spring mass experiment, the natural frequency of 10 kg mass was found to be 12 rad/sec. the stiffness of the spring is A. 800 N/m B. 1200 N/m C. 1440 N/m D. 2000 N/m

Description : In spring mass experiment, the natural frequency of 10 kg mass was found to be 12 rad/sec. the stiffness of the spring is A. 800 N/m B. 1200 N/m C. 1440 N/m D. 2000 N/m

Last Answer : C. 1440 N/m

The unit of natural frequency is A. Rad/sec B. Hz C. Both D. No unit

Description : The unit of natural frequency is A. Rad/sec B. Hz C. Both D. No unit

Last Answer : C. Both

Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 –3 m 3. Natural frequency of the system = 60 rad/sec a. 100.5 x 10 3 N-s/m b. 80 x 10 3 N-s/m c. 42 x 10 3 N-s/m d. None of the above

Description : Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 -3 m 3. Natural frequency of the system = 60 rad/sec ... /m b. 80 x 10 3 N-s/m c. 42 x 10 3 N-s/m d. None of the above

Last Answer : c. 42 x 10 3 N-s/m

A spring-mass system has a natural frequency of 10 rad/sec. When the spring constant is reduced by 800 N/m, the frequency is altered by 45 percent. Find the mass and spring constant of the original system. a)11.47kg and 1147.95N/m b)8.95kg and 895.25N/m c) 7.265kg and 726.5N/m d)None

Description : A spring-mass system has a natural frequency of 10 rad/sec. When the spring constant is reduced by 800 N/m, the frequency is altered by 45 percent. Find the mass and spring constant of the original system. a)11.47kg and 1147.95N/m b)8.95kg and 895.25N/m c) 7.265kg and 726.5N/m d)None

Last Answer : a)11.47kg and 1147.95N/m

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as * 1 point (A) simple pendulum (B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as * 1 point (A) simple pendulum (B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Last Answer : (C) compound pendulum

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as A. simple pendulum B. torsional pendulum C. compound pendulum D. second’s pendulum

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as A. simple pendulum B. torsional pendulum C. compound pendulum D. second’s pendulum

Last Answer : C. compound pendulum

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as (A) simple pendulum B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as (A) simple pendulum B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Last Answer : (C) compound pendulum

Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further, consider that a normal instruction takes one micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available? (A) 60 sec (B) 30 sec (C) 45 sec (D) 10 sec

Description : Suppose that the number of instructions executed between page faults is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also ... memory were available? (A) 60 sec (B) 30 sec (C) 45 sec (D) 10 sec

Last Answer : Answer: C Explanation: T = Ninstr x 1µs + 15,000 x 2,000 µs = 60s Ninstr x 1µs = 60,000,000 µs - 30,000,000 µs = 30,000,000 µs Ninstr = 30,000,000 The number of instruction ... doesn't mean that the program runs twice as fast as on the first system. Here, the performance increase is of 25%.

A flywheel weighs 400 kg and has a radius of gyration of 30 cm about its axis of rotation. It is mounted on a shaft subjected to a moment of 40 kgm. If the flywheel starts from restits speed after 3 seconds wil be a.10.9 rad/sec b.19.0 rad/sec c.51.2 rad/sec d.3.33 rad/sec e.16.35 rad/sec

Description : A flywheel weighs 400 kg and has a radius of gyration of 30 cm about its axis of rotation. It is mounted on a shaft subjected to a moment of 40 kgm. If the flywheel starts from restits speed after 3 seconds wil be a.10.9 rad/sec b.19.0 rad/sec c.51.2 rad/sec d.3.33 rad/sec e.16.35 rad/sec

Last Answer : d. 3.33 rad/sec

A 3-phase 50HZ SCIM takes a power input of 30 KW at its full load speed of 1440 rpm. Total stator losses are 1 KW. The slip and rotor ohmic losses at full load are a) 0.02, 600 W b) 0.04, 580 W c) 0.04, 1160 W d) 0.04, 1200 W

Description : A 3-phase 50HZ SCIM takes a power input of 30 KW at its full load speed of 1440 rpm. Total stator losses are 1 KW. The slip and rotor ohmic losses at full load are a) 0.02, 600 W b) 0.04, 580 W c) 0.04, 1160 W d) 0.04, 1200 W

Last Answer : Ans: 0.04, 1160 W

The reciprocal of the interval of time by a vibrating body to complete a cycle is called A Period B Frequency C Resonance D None of the mentioned

Description : The reciprocal of the interval of time by a vibrating body to complete a cycle is called A Period B Frequency C Resonance D None of the mentioned

Last Answer : B Frequency

Time taken to complete one cycle is known as A Resonance B Frequency C Period D Damping

Description : Time taken to complete one cycle is known as A Resonance B Frequency C Period D Damping

Last Answer : C Period

Time required to complete one cycle is known as A. Idle time B. Periodic time C. Both D. None

Description : Time required to complete one cycle is known as A. Idle time B. Periodic time C. Both D. None

Last Answer : B. Periodic time

Reduction in vibration amplitude after one complete cycle of single degree free vibration with dry friction damping is_____, if where F"= frictional force between mass and surface and k =stiffness of the system. a)4F/k a b) 2f/K C) 3F/k D)8F/k

Description : Reduction in vibration amplitude after one complete cycle of single degree free vibration with dry friction damping is_____, if where F"= frictional force between mass and surface and k =stiffness of the system. a)4F/k a b) 2f/K C) 3F/k D)8F/k

Last Answer : a)4F/k

In a 2280 m race Dinesh beats Aarav by 360 m or 6 seconds. In another race on the same track at the same speeds. Aarav and karthick start at one end while Dinesh starts at the opposite end. How many metres would Aarav have covered ,by the time Dinesh meets karthick given that Dinesh speed is 16 m/sec more than that of karthick a) 1140 m b) 2280 m c) 2460 m d) 1180 m

Description : In a 2280 m race Dinesh beats Aarav by 360 m or 6 seconds. In another race on the same track at the same speeds. Aarav and karthick start at one end while Dinesh starts at the opposite end. How many metres would Aarav ... is 16 m/sec more than that of karthick a) 1140 m b) 2280 m c) 2460 m d) 1180 m

Last Answer : B Aarav 's speed =360/6 =60 m/s Time taken by Aarav to cover 2280 m =2280/60= 38 seconds Dinesh 's speed = 2280/60=38 m/s karthick 's speed = 22m/s Time taken by Dinesh to meet karthick in ... direction: Distance covered by Aarav: =2280/(38+22) =2280/60 =38 seconds =(38*60)m =2280 m

If a train crosses a tree in 24 seconds while travelling at a speed of 90km/hr, then in how much time will the train cross a telegraph post of length 1200m at same speed? A) 26 sec B) 48 sec C) 72 sec D) 38 sec

Description :  If a train crosses a tree in 24 seconds while travelling at a speed of 90km/hr, then in how much time will the train cross a telegraph post of length 1200m at same speed? A) 26 sec B) 48 sec C) 72 sec D) 38 sec 

Last Answer : ANSWER:C Explanation: Speed of the train = 90 km/hr Converting the speed into m/sec 90 km/hr = (90* 5/18)m/s = 25m/s length of the train = distance travelled we know distance = speed * ...  Time = distance / speed  = ( 1800/25) sec  = 72 seconds  Hence the required answer is 72 seconds

A dc shunt generator has a speed of 800 rpm when delivering 20 A to the load at the terminal voltage of 220V. If the same machine is run as a motor it takes a line current of 20A from 220V supply. The speed of the machine as a motor will be (A) 800 rpm. (B) more than 800 rpm. (C) less than 800 rpm. (D) both higher or lower than 800 rpm.

Description : A dc shunt generator has a speed of 800 rpm when delivering 20 A to the load at the terminal voltage of 220V. If the same machine is run as a motor it takes a line current of 20A from 220V supply. The speed of ... . (B) more than 800 rpm. (C) less than 800 rpm. (D) both higher or lower than 800 rpm.

Last Answer : (C) less than 800 rpm.

If the static deflection is 1.665×10 -3 m, calculate the critical speed of the shaft in rps. Centre of disc at 0.25m away from centre of axis of shaft. A. 8.64 B. 9.64 C. 10.64 D. 12.2

Description : If the static deflection is 1.665×10 -3 m, calculate the critical speed of the shaft in rps. Centre of disc at 0.25m away from centre of axis of shaft. A. 8.64 B. 9.64 C. 10.64 D. 12.2

Last Answer : D. 12.2

In measuring critical speed of shaft experiment, it was found that the frequency ratio is 0.707 when the eccentricity is 0.05 m. what will be the displacement of the system. A. 0.05 m B. 0.005 m C. 0.5 m D. Infinite

Description : In measuring critical speed of shaft experiment, it was found that the frequency ratio is 0.707 when the eccentricity is 0.05 m. what will be the displacement of the system. A. 0.05 m B. 0.005 m C. 0.5 m D. Infinite

Last Answer : A. 0.05 m

Eddy current damping is an example of _____ A Coulomb damping B Hysteresis damping C Viscous damping D Dry friction damping

Description : Eddy current damping is an example of _____ A Coulomb damping B Hysteresis damping C Viscous damping D Dry friction damping

Last Answer : C Viscous damping

Eddy current damping is an example of _____ A) Coulomb damping B) Hysteresis damping C) Viscous damping D) Dry friction damping

Description : Eddy current damping is an example of _____ A) Coulomb damping B) Hysteresis damping C) Viscous damping D) Dry friction damping

Last Answer : C) Viscous damping