## Machine Design MCQ Set 1

1. Carriage provides various movements for the cutting tool______

a) manually

b) by power feed

c) manually and by power feed both

d) none of the mentioned

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2. Carriage is the part of lathe which slides over the bed ways.

a) true

b) false

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3. The compound rest may be clamped at_____

a) 0 degree

b) 90 degree

c) 180 degree

d) any degree between 0 and 360

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4. Which type of feed is always done parallel to the axis of work?

a) longitudinal feed

b) cross feed

c) angular feed

d) none of the mentioned

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5. Which of the following assists in feeding the tool to the work?

a) swivel base

b) top slide

c) swivel base and top slide both

d) none of the mentioned

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6. Which type of feed is done with the help of compound slide movement?

a) longitudinal feed

b) angular feed

c) cross feed

d) none of the mentioned

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7. In how many parts, carriage can be divided?

a) 2

b) 3

c) 4

d) 5

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8. Saddle is_____ shaped casting and has_____ guide grooves and flat grooves.

a) H and V

b) V and H

c) H and H

d) V and V

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9. In saddle, grooves are machined at the ______ corresponding to the lathe bed ways.

a) top face

b) middle face

c) bottom face

d) none of the mentioned

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10. The cross slide functions_____ to the lathe axis.

a) parallel

b) perpendicular

c) anti-parallel

d) none of the mentioned

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## Machine Design MCQ Set 2

1. Table top drilling machine can be categorized as_______

a) general purpose drilling machine

b) specific purpose drilling machine

c) can’t say anything

d) none of the mentioned

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2. Which type of feed can be given in table top drilling machine?

a) manual

b) power

c) both manual and power feed

d) none of the mentioned

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3. Power produced by table top drilling machine is comparatively low.

a) true

b) false

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4. In table top drilling machine, feed can be operated by_____

a) gear mechanism

b) lever

c) spindle

d) none of the mentioned

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5. Generally, table top drilling machine can produce diameters up to______

a) 1 mm

b) 10 mm

c) 100 mm

d) none of the mentioned

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6. Table top drilling machine can be mounted on the table.

a) true

b) false

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7. Table top drilling machine is generally______ in terms of height.

a) small

b) big

c) can’t say anything

d) none of the mentioned

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8. In table top drilling machine, belt-pulley system operates______

a) lever

b) spindle

c) both lever and spindle

d) none of the mentioned

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9. Drill is mounted on lever in table top drilling machine.

a) true

b) false

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10. Which of the following is not true for table top drill machine?

a) it is used in only small jobs

b) it is used for mass production

c) odd shape jobs are not machined

d) none of the mentioned

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## Machine Design MCQ Set 3

1. If the reflux is total, minimum stages are used.

a) True

b) False

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2. At Total reflux

a) Residue is nil

b) Distillate is nil

c) Reflux is nil

d) All of the mentioned

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3. Reflux ratio is infinity at

a) Total reflux

b) Minimum reflux

c) Part reflux

d) None of the mentioned

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4. Estimate the condenser requirement for total reflux.

a) Small condenser

b) Medium condenser

c) Large condenser

d) Partial condenser

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5. For minimum reflux ratio the number of trays becomes infinite.

a) True

b) False

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6. For a minimum reflux ratio ________ condenser or reboiler loads are used.

a) Less

b) More

c) Infinity

d) Partial

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7. Find the representation X.

a) Partial condenser

b) Total condenser

c) Reboiler

d) None of the mentioned

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8. For finding the exact number of trays the reflux ratio should be between minimum and infinity.

a) True

b) False

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9. The reflux ratio increases by increasing the reflux and decreasing the distillate.

a) True

b) False

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10. Find the co-relation of the representation.

a) Maximum reflux ratio

b) Optimum reflux ratio

c) Minimum reflux ratio

d) None of the mentioned

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## Machine Design MCQ Set 4

1. Which of the following is not the method of indexing?

a) compound indexing

b) differential indexing

c) angular indexing

d) none of the mentioned

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2. Which of the following indexing method is also known as rapid indexing?

a) direct indexing method

b) angular indexing method

c) compound indexing method

d) none of the mentioned

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3. Simple indexing method is also known as plain indexing method.

a) true

b) false

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4. Direct indexing is used when a large number of identical pieces are indexed by very large divisions.

a) true

b) false

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5. In direct indexing, the operation may be performed in_______

a) plain dividing head

b) universal dividing head

c) both plain and universal dividing head

d) none of the mentioned

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6. When using a universal dividing head, the worm and worm wheel are first disengaged.

a) true

b) false

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7. The rapid index plate is generally fitted to the ______ end of the spindle nose.

a) front

b) rear

c) can’t say anything

d) none of the mentioned

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8. In direct indexing, the plate has ______ equally spaced holes.

a) 12

b) 24

c) 48

d) none of the mentioned

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9. If n=number of divisions required, then find the number of holes to be moved in direct indexing.

a) 12/n

b) 24/n

c) 12*n

d) 24*n

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10. If number of divisions required=4, then find the number of holes to be moved in direct indexing.

a) 12

b) 6

c) 4

d) none of the mentioned

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## Machine Design MCQ Set 5

1. A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end of an arm of length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what speed will it precess about the vertical axis?

a) 14.7 rad/s

b) 15.7 rad/s

c) 16.7 rad/s

d) 17.7 rad/s

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I = m.r^{2}/2 = 5(0.15)^{2}/2 = 0.056 kg-m^{2} and couple due to mass of disc, C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m
Let ω_{P} = Speed of precession
We know that couple (C),
29.43 = I.ω.ω_{P} = 0.056 × 31.42 × ω_{P} = 1.76 ω_{P} ω_{P} = 29.43/1.76 = 16.7 rad/s

2. An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at 200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft.

a) 10.046 kN-m

b) 11.046 kN-m

c) 12.046 kN-m

d) 13.046 kN-m

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^{2}= 36 kg-m

^{2}and angular velocity of precession, ω

_{P}= v/R = 55.6/50 = 1.11 rad/s We know that gyroscopic couple acting on the aircraft, C = I. ω. ω

_{P}= 36 × 251.4 × 1.11 = 100 46 N-m = 10.046 kN-m

3. The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at 1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius.

a) 100.866 kN-m

b) 200.866 kN-m

c) 300.866 kN-m

d) 400.866 kN-m

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^{2}= 2880 kg-m

^{2}and angular velocity of precession, ω

_{P}= v / R = 27.8 / 75 = 0.37 rad/s We know that gyroscopic couple, C = I.ω.ω

_{P}= 2880 × 188.5 × 0.37 = 200 866 N-m = 200.866 kN-m

4. The heavy turbine rotor of a sea vessel rotates at 1500 r.p.m. clockwise looking from the stern, its mass being 750 kg. The vessel pitches with an angular velocity of 1 rad/s. Determine the gyroscopic couple transmitted to the hull when bow is rising, if the radius of gyration for the rotor is 250 mm.

a) 4.364 kN-m

b) 5.364 kN-m

c) 6.364 kN-m

d) 7.364 kN-m

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_{P}= 1 rad/s; k = 250 mm = 0.25 m We know that mass moment of inertia of the rotor, I = mk

^{2}= 46.875 kg-m

^{2}Gyroscopic couple transmitted to the hull (i.e. body of the sea vessel), C = I.ω.ω

_{P}= 46.875 × 157.1 × 1= 7364 N-m = 7.364 kN-m

5. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.

a) 11.27 kN-m

b) 22.27 kN-m

c) 33.27 kN-m

d) 44.27 kN-m

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^{2}= 708.75 kg-m

^{2}and angular velocity of precession, ω

_{p}= v/R = 10/100 = 0.1 rad/s Gyroscopic couple, C = I.ω.ω

_{p}= 708.75 × 314.2 × 0.1 = 22 270 N-m = 22.27 kN-m

6. The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple upon the ship when the ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity. The period of pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching is 12 degrees.

a) 3.675 kN-m

b) 4.675 kN-m

c) 5.675 kN-m

d) 6.675 kN-m

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_{p}= 40 s Since the total angular displacement between the two extreme positions of pitching is 12° (i.e. 2φ = 12°), therefore amplitude of swing, φ = 12 / 2 = 6° = 6 × π/180 = 0.105 rad and angular velocity of the simple harmonic motion, ω

_{1}= 2π / t

_{p}= 2π / 40 = 0.157 rad/s We know that maximum angular velocity of precession, ω

_{p}= φ.ω

_{1}= 0.105 × 0.157 = 0.0165 rad/s Gyroscopic couple, C = I.ω.ω

_{p}= 708.75 × 314.2 × 0.0165 = 3675 N-m = 3.675 kN-m

7. The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum gyroscopic couple.

a) 11.185 kN-m

b) 22.185 kN-m

c) 33.185 kN-m

d) 44.185 kN-m

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_{p}= 30 s We know that mass moment of inertia of the rotor, I = m.k

^{2}= 20 000 (0.6)

^{2}= 7200 kg-m

^{2}and angular velocity of the simple harmonic motion, ω

_{1}= 2π / t

_{p}= 2π/30 = 0.21 rad/s Maximum angular velocity of precession, ω

_{Pmax}= φ.ω

_{1}= 0.105 × 0.21 = 0.022 rad/s We know that maximum gyroscopic couple, C

_{max}= I.ω.ω

_{Pmax}= 7200 × 209.5 × 0.022 = 33 185 N-m = 33.185 kN-m

8. The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum angular acceleration of the ship during pitching.

a) 0.0016 rad/s^{2}

b) 0.0026 rad/s^{2}

c) 0.0036 rad/s^{2}

d) 0.0046 rad/s^{2}

### View Answer

_{1})

^{2}= 0.105 (0.21)

^{2}= 0.0046 rad/s

^{2}

9. A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effect when the ship sails at a speed of 30 km/h and steers to the left in a curve having 60 m radius.

a) 38.5 kN-m

b) 48.5 kN-m

c) 58.5 kN-m

d) 68.5 kN-m

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_{p}= v/R = 8.33/60 = 0.14 rad/s and mass moment of inertia of the rotor, I = m.k

^{2}= 5000(0.5)2 = 1250 kg-m

^{2}Gyroscopic couple, C = I.ω.ω

_{p}= 1250 × 220 × 0.14 = 38 500 N-m = 38.5 kN-m

10. A ship propelled by a turbine rotor which has a mass of 5 tonnes and a speed of 2100 r.p.m. The rotor has a radius of gyration of 0.5 m and rotates in a clockwise direction when viewed from the stern. Find the gyroscopic effect when the ship pitches 6 degree above and 6 degree below the horizontal position. The bow is descending with its maximum velocity. The motion due to pitching is simple harmonic and the periodic time is 20 seconds.

a) 6075 N-m

b) 7075 N-m

c) 8075 N-m

d) 9075 N-m

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_{1}= 2π / t

_{p}= 2π / 20 = 0.3142 rad/s and maximum angular velocity of precession, ω

_{Pmax}= φ.ω

_{1}= 0.105 × 0.3142 = 0.033 rad/s Maximum gyroscopic couple, C

_{max}= I.ω.ω

_{Pmax}= 1250 × 220 × 0.033 = 9075 N-m