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## 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

Answer: c [Reason:] These two types of movements are possible in carriage. Carriage provides various movements for the cutting tool.

2. Carriage is the part of lathe which slides over the bed ways.
a) true
b) false

Answer: a [Reason:] This is the function of carriage. It can be locked on the bed at any desired position by tightening the carriage lock-screw.

3. The compound rest may be clamped at_____
a) 0 degree
b) 90 degree
c) 180 degree
d) any degree between 0 and 360

Answer: d [Reason:] Compound rest may be clamped at any required position between 0 and 360 degree. It can be tightened by T-bolts at that position.

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

Answer: a [Reason:] Longitudinal feed is always done parallel to the axis of the work as per the classification of carriage movements.

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

Answer: b [Reason:] Only top slide assists in feeding the tool to the work. We can give manual operation in the top slide.

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

Answer: b [Reason:] Angular feed is done with the help of the compound slide movement positioned at an angle to the axis of the work.

7. In how many parts, carriage can be divided?
a) 2
b) 3
c) 4
d) 5

Answer: d [Reason:] Carriage can be divided in saddle, cross-slide, compound rest swivel and top slide, tool post and apron.

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

Answer: a [Reason:] Saddle is H shaped casting and has v guide grooves and flat grooves. Saddle is mounted on the spindle.

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

Answer: c [Reason:] In saddle, grooves are machined at the bottom corresponding to the lathe bed ways. The surface of the bed in contact with the sliding units of the lathe are known as guideways or slideways or bedways.

10. The cross slide functions_____ to the lathe axis.
a) parallel
b) perpendicular
c) anti-parallel
d) none of the mentioned

Answer: b [Reason:] The cross slide functions perpendicular to the lathe axis. The bottom of the cross-slide has got a dovetail groove machined, which corresponds to the external dovetail machined on the saddle.

## 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

Answer: a [Reason:] Table top drilling machine can be categorized as general purpose drilling machine as it is not used for any specific application.

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

Answer: a [Reason:] Manual feed is given in this machine. We don’t deal with very heavy jobs in table top drilling machine so manual feed can perform job sufficiently.

3. Power produced by table top drilling machine is comparatively low.
a) true
b) false

Answer: a [Reason:] This is true. Power produced is comparatively low because manual feed is given to this machine, not powered feed.

4. In table top drilling machine, feed can be operated by_____
a) gear mechanism
b) lever
c) spindle
d) none of the mentioned

Answer: b [Reason:] The feed can be operated by lever in this machine. This lever can be operated manually to provide feed.

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

Answer: b [Reason:] It can produce diameter up to 10 mm. As it is operated by manual feed, its power production is low so it is not used for producing large diameters.

6. Table top drilling machine can be mounted on the table.
a) true
b) false

Answer: a [Reason:] It can be mounted or clamped on the table to accommodate with job.

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

Answer: a [Reason:] Table tap drilling machine is comparatively small in terms of height. It is hardly 1 to 3 feet long in height.

8. In table top drilling machine, belt-pulley system operates______
a) lever
b) spindle
c) both lever and spindle
d) none of the mentioned

Answer: b [Reason:] Spindle is operated by this belt-pulley system. Lever can be operated manually for feed.

9. Drill is mounted on lever in table top drilling machine.
a) true
b) false

Answer: b [Reason:] This is false. Drill is mounted on spindle in table top drilling machine.

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

Answer: b [Reason:] All mentioned are true accept the sentence which says that it is usede for mass production. It cannot be used for mass production. It is very small and used only for small job work.

## Machine Design MCQ Set 3

1. If the reflux is total, minimum stages are used.
a) True
b) False

Answer: a [Reason:] For total reflux no distillate formation; all the condensed vapour are sent back to fractionator.

2. At Total reflux
a) Residue is nil
b) Distillate is nil
c) Reflux is nil
d) All of the mentioned

Answer: b [Reason:] Total reflux there condensed vapours are sent back to column itself.

3. Reflux ratio is infinity at
a) Total reflux
b) Minimum reflux
c) Part reflux
d) None of the mentioned

Answer: a [Reason:] Reflux ratio becomes infinity at distillate is equals to zero. Since reflux ratio = Reflux/distillate At distillate = 0; Reflux ratio= infinity.

4. Estimate the condenser requirement for total reflux.
a) Small condenser
b) Medium condenser
c) Large condenser
d) Partial condenser

Answer: c [Reason:] Since if the large condenser are used, vapour gets condensed and becomes reflux where distillate is zero.

5. For minimum reflux ratio the number of trays becomes infinite.
a) True
b) False

Answer: a [Reason:] For minimum reflux ratio, all the condensed vapour becomes distillate; no reflux. So it requires infinite number of trays.

6. For a minimum reflux ratio ________ condenser or reboiler loads are used.
a) Less
b) More
c) Infinity
d) Partial

Answer: a [Reason:] Since for minimum reflux ratio no reflux is produced, it’s enough to use minimum load condenser or reboiler.

7. Find the representation X.

a) Partial condenser
b) Total condenser
c) Reboiler
d) None of the mentioned

Answer: b [Reason:] All the condensed vapours are refluxed (no distillate).

8. For finding the exact number of trays the reflux ratio should be between minimum and infinity.
a) True
b) False

Answer: a [Reason:] The minimum trays cannot be determined; also infinite trays are impossible; so the trays should be minimum and infinity.

9. The reflux ratio increases by increasing the reflux and decreasing the distillate.
a) True
b) False

Answer: a [Reason:] Reflux ratio= reflux/distillate.

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

Answer: c [Reason:] Since there is no reflux, all the condensed vapours are distillate. Then it represents minimum reflux ratio.

## 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

Answer: d [Reason:] All mentioned are the methods of indexing. Generally direct indexing is used but when the indexing required for numbers is beyond the the range of direct indexing, other methods are used.

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

Answer: a [Reason:] Direct indexing method is also known as rapid indexing method because it is more fast as compare to others.

3. Simple indexing method is also known as plain indexing method.
a) true
b) false

Answer: a [Reason:] Simple indexing method is also known as plain indexing method, which is used for the simple task.

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

Answer: b [Reason:] Direct indexing is used when a large number of identical pieces are indexed by very small divisions. It is generally not used when divisions are large

5. In direct indexing, the operation may be performed in_______
c) both plain and universal dividing head
d) none of the mentioned

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

Answer: a [Reason:] As per the working of direct indexing, the worm and worm wheel are first disengaged. This is done in a manner similar to that used in the back gear of a lathe by turning a handle which operates an eccentric bushing.

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

Answer: a [Reason:] The required number of divisions on the work is obtained by means of the rapid index plate, which is generally fitted to the front end of the spindle nose.

8. In direct indexing, the plate has ______ equally spaced holes.
a) 12
b) 24
c) 48
d) none of the mentioned

Answer: b [Reason:] In direct indexing, the plate has twenty four equally spaced holes into any one of which a spring loaded pin is pushed to lock the spindle with the frame.

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

Answer: b [Reason:] It is the formula for direct indexing. Various types of indexing methods are there like compound indexing, angular indexing and so on.

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

Answer: b [Reason:] From the formula 24/n, wher n=number of divisions required. So, by dividing 24 by 4, we can easily get 6 as an answer.

## 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?

Answer: c [Reason:] Given: d = 300 mm or r = 150 mm = 0.15 m ; m = 5 kg ; l = 600 mm = 0.6 m ; N = 300 r.p.m. or ω = 2π × 300/60 = 31.42 rad/s We know that the mass moment of inertia of the disc, about an axis through its centre of gravity and perpendicular to the plane of disc,

I = m.r2/2 = 5(0.15)2/2 = 0.056 kg-m2 and couple due to mass of disc, C = m.g.l = 5 × 9.81 × 0.6 = 29.43 N-m Let ωP = Speed of precession We know that couple (C), 29.43 = I.ω.ωP = 0.056 × 31.42 × ωP = 1.76 ωP ωP = 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

Answer: a [Reason:] Given : R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ; N = 2400 r.p.m. or ω = 2π × 2400/60 = 251 rad/s We know that mass moment of inertia of the engine and the propeller, I = mk2 = 36 kg-m2 and angular velocity of precession, ωP = v/R = 55.6/50 = 1.11 rad/s We know that gyroscopic couple acting on the aircraft, C = I. ω. ωP = 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

Answer: b [Reason:] Given: m = 8 t = 8000 kg ; k = 0.6 m ; N = 1800 r.p.m. or ω = 2π × 1800/60 = 188.5 rad/s ; v = 100 km/h = 27.8 m/s ; R = 75 m We know that mass moment of inertia of the rotor, I = mk2 = 2880 kg-m2 and angular velocity of precession, ωP = v / R = 27.8 / 75 = 0.37 rad/s We know that gyroscopic couple, C = I.ω.ωP = 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

Answer: d [Reason:] Given: N = 1500 r.p.m. or ω = 2π × 1500/60 = 157.1 rad/s; m = 750 kg; ωP = 1 rad/s; k = 250 mm = 0.25 m We know that mass moment of inertia of the rotor, I = mk2 = 46.875 kg-m2 Gyroscopic couple transmitted to the hull (i.e. body of the sea vessel), C = I.ω.ωP = 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

Answer: b [Reason:] Given : m = 3500 kg ; k = 0.45 m; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s When the ship is steering to the left Given: R =100 m ; v = km/h = 10 m/s We know that mass moment of inertia of the rotor, I = mk2 = 708.75 kg-m2 and angular velocity of precession, ωp = v/R = 10/100 = 0.1 rad/s Gyroscopic couple, C = I.ω.ωp = 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

Answer: a [Reason:] Given: tp = 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, ω1 = 2π / tp = 2π / 40 = 0.157 rad/s We know that maximum angular velocity of precession, ωp = φ.ω1 = 0.105 × 0.157 = 0.0165 rad/s Gyroscopic couple, C = I.ω.ωp = 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

Answer: c [Reason:] Given : m = 20 t = 20 000 kg ; k = 0.6 m ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad ; tp = 30 s We know that mass moment of inertia of the rotor, I = m.k2 = 20 000 (0.6)2 = 7200 kg-m2 and angular velocity of the simple harmonic motion, ω1 = 2π / tp = 2π/30 = 0.21 rad/s Maximum angular velocity of precession, ωPmax = φ.ω1 = 0.105 × 0.21 = 0.022 rad/s We know that maximum gyroscopic couple, Cmax = I.ω.ωPmax = 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.

Answer: d [Reason:] We know that maximum angular acceleration during pitching = φ(ω1)2 = 0.105 (0.21)2 = 0.0046 rad/s2

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