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Multiple choice question for engineering

Set 1

1. A spur gear of 40 teeth is machined in a gear hobbing machine using a double start hob cutter. The speed ratio between the hob and the blank is
a) 1:20
b) 1:40
c) 40:1
d) 20:1

View Answer

Answer: d

2. A helical gear has the active face width equal to b pitch p and helix angle α. What should be the minimum value of b in order that contact is maintained across the entire active face of the gear?
a) p cosα
b) p secα
c) p tanα
d) p cotα

View Answer

Answer: d [Reason:] b > p/tanα b > p cotα.

3. Match the type of gears with their most appropriate description.
Type of gear Description

P. Helical 1. Axes non parallel and intersecting
Q. Spiral 2. Axes parallel and teeth are inclined to the axis
R. Hypoid 3. Axes parallel and teeth are parallel to the axis
S. Rack and pinion 4. Axes are perpendicular and intersecting,and teeth are inclined to the axis
5. Axes are perpendicular and used for large speed reduction
6. Axes parallel and one of the gears has infinite radius
a) P-2, Q- 4, R- 1, S- 6
b) P-1, Q- 4, R- 5, S- 6
c) P-2, Q- 6, R- 4, S- 2
d) P-6, Q- 3, R- 1, S- 5

View Answer

Answer: a [Reason:] Helical – Axes parallel and teeth are inclined to the axis Spiral – Axes are perpendicular and intersecting,and teeth are inclined to the axis Hypoid – Axes non parallel and intersecting Rack and pinion – Axes parallel and one of the gears has infinite radius.

4. Tooth interference in an external in volute spur gear pair can be reduced by
a) Decreasing center distance between gear pair
b) Decreasing module
c) Decreasing pressure angle
d) Increasing number of gear teeth

View Answer

Answer: d [Reason:] There are several ways to avoid interfering: i. Increase number of gear teeth ii. Modified involutes iii. Modified addendum iv. Increased centre distance .

5. Interference in a pair of gears is avoided, if the addendum circles of both the gears intersect common tangent to the base circles within the points of tangency.
a) True
b) False

View Answer

Answer: a

6. Twenty degree full depth involute profiled 19-tooth pinion and 37-tooth gear are in mesh. If the module is 5 mm, the centre distance between the gear pair will be
a) 140 mm
b) 150 mm
c) 280 mm
d) 300 mm

View Answer

Answer: a [Reason:] Centre distance = D1 + D2/2 = mT1 + mT2/2 = 5(19 + 37)/2 = 140 mm.

7. If the drive efficiency is 80%, then torque required on the input shaft to create 1000 N output thrust is
a) 20 Nm
b) 25 Nm
c) 32 Nm
d) 50 Nm

View Answer

Answer: b [Reason:] Module m = 2, D/T = 2 ∴ D = 80 × 2 = 160 mm 2F = 1000, or F = 500 N Let T1 be the torque applied by motor. TT2 be the torque applied by gear. ∴ Power transmission = 80%

Now, T1ω12 x ω2/0.8 T1 = 25 N-m.

8. If the pressure angle of the rack is 20°, then force acting along the line of action between the rack and the gear teeth is
a) 250 N
b) 342 N
c) 532 N
d) 600 N

View Answer

Answer: c [Reason:] Pcos φ = F ∴ Force acting along the line of action, P = F/cos φ = 500/cos20° = 532 N.

9. Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 1200 rpm and transmits a torque of 20 Nm. The torque transmitted by the gear is
a) 6.6 Nm
b) 20 Nm
c) 40 Nm
d) 60Nm

View Answer

Answer: d [Reason:] We know, N1/N2 = T1/T2 N1 = speed of pinion N2 = speed of gear wheel T1 = number of teeth of gear T2 = number of teeth of pinion

1200/N2 = 120/40 N2 = 400 r.p.m. Since power transmitted by both gear will be equal T1ω12 x ω2

torque transmitted by gear, = 60 N-m.

10. The velocity ratio between pinion and gear in a gear drive is 2.3, the module of teeth is 2.0 mm and sum of number of teeth on pinion and gear is 99. What is the centre distance between pinion and the gear?
a) 49.5 mm
b) 99 mm
c) 148.5 mm
d) 198 mm

View Answer

Answer: b [Reason:] Centre distance = D1 + D2/2 = mT1 + mT2/2 = m/2(T1 + T2) = 2/2 x 99 = 99 mm.

Set 2

1. In a crank and slotted lever quick-return motion, the distance between the fixed centres is 150 mm and the length of the driving crank is 75mm. the ratio of the time taken on the cutting and return strokes is
a) 1.5
b) 2.0
c) 2.2
d) 2.93

View Answer

Answer: b [Reason:] OA = 150mm AB = 75 mm sin (90 – α/2) = AB/OA cosα/2 = 1/2

therefore, α = 1200 β = 3600 – 1200 = 2400 Quick return ratio = β/α = 240/120 = 2.

2. A helical coil spring of stiffness k is cut to two equal halves and then these are connected in parallel to support a vibrating mass m. The angular frequency of vibration, ωn is
a) √k/m
b) √2k/m
c) √4k/m
d) √k/4m

View Answer

Answer: c [Reason:] stiffness ∞ 1/length keq = 2k + 2k = 4k ω = √4k/m.

3. Consider the following statements:
In a slider-crank mechanism, the slider is at its dead centre position when the
(i) slider velocity is zero
(ii) slider velocity is maximum
(iii) slider acceleration is zero
(iv) slider acceleration is maximum
Which of the above statements are correct?
a) 1 and 4
b) 1 and 3
c) 2 and 3
d) 2 and 4

View Answer

Answer: a [Reason:] At dead centre velocity is zero, because instantaneous acceleration is maximum.

4. Which one of the following mechanisms is an inversion of double slider-crank chain?
a) Elliptic trammels
b) Beam engine
c) Oscillating cylinder engine
d) Coupling rod of a locomotive

View Answer

Answer: a [Reason:] Elliptic trammels are inversion of double slider crank chain.

5. The number of instantaneous centres of rotation for a 10-link kinematic chain is
a) 36
b) 90
c) 120
d) 45

View Answer

Answer: d [Reason:] Instantaneous centre = n(n – 1)/2 = 45.

6. A slider moves with uniform velocity on a revolving link of length r with angular velocity ω. The Coriolis acceleration component of a point on the slider relative to a coincident point on the link is equal to
a) ω parallel to the link
b) 2ω perpendicular to the link
c) ω perpendicular to the link
d) 2ω parallel to the link

View Answer

Answer: b [Reason:] coriolis components of acceleration for slider is perpendicular to the link.

7. In a crank and slotted lever type quick return mechanism, the link moves with an angular velocity of 20 rad/s, while the slider moves with the linear velocity of 1.5 m/s. The magnitude and direction of Coriolis component of acceleration with respect to angular velocity are
a) 30 m/s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity
b) 30 m/s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity
c) 60 m/s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity
d) 60 m/s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity.

View Answer

Answer: c [Reason:] Magnitude of Coriolis acceleration = 2Vω = 2 x 1.5 x 20 = 60 m/s2 Rotate the link in direction of rotation and see the direction of linear velocity that will be direction of Coriolis acceleration.

8. Which of the following are associated with Ackerman steering mechanism used in automobiles?
1. Has both sliding and turning pairs
2. Less friction and hence long life
3. Mechanically correct in all positions
4. Mathematically not accurate except in three positions.
5. Has only turning pairs 6. Controls movement of two front wheels
a) 2, 4, 5 and 6
b) 1, 2, 3 and 6
c) 2, 3, 5 and 6
d) 1, 2, 3 and 5

View Answer

Answer: a [Reason:] Ackerman steering mechanism has no sliding mechanism & not suitable for all the positions.

9. The crankshaft of reciprocating engine having a 20 cm crank and 100 cm connecting rod rotates at 210 r.p.m. When the crank angle is 45o, the velocity of piston is nearly
a) 1.8m/s
b) 1.9m/s
c) 3.5m/s
d) 19m/s

View Answer

Answer: c [Reason:] Vp = ωr(sinϴ + sin2ϴ/2n) ω = 2пN/60 = 21.98rad/sec r = 6.2m n = l/r = 100/20 = 5

therefore, Vp = 21.980 x 0.2(sin45o + sin90o/2 x 5) = 3.5m/sec.

Set 3

1. The instantaneous centers of a slider moving in a curved surface lies at
a) infinity
b) their point of contact
c) the center of curvature
d) the pin point

View Answer

Answer: c [Reason:] The instantaneous centers of a slider moving in a linear guide lies at infinity. The instantaneous centers of a slider moving in a curved surface lies at the center of curvature.

2. The fixed instantaneous center of mechanism
a) varies with the configuration
b) remains at the same place for all configurations
c) all of the mentioned
d) none of the mentioned

View Answer

Answer: b [Reason:] The fixed instantaneous center of mechanism remains at the same place for all configurations. The instantaneous centres which vary with the configuration of the mechanism, are called neither fixed nor permanent instantaneous centres.

3. The instantaneous center of rotation of a circular disc rolling on a straight path is
a) at the center of the disc
b) at their point of contact
c) at the center of gravity of the disc
d) at infinity

View Answer

Answer: b [Reason:] The space centrode of a circular disc rolling on a straight path is a straight line.

4. The locus of instantaneous center of a moving body relative to a fixed body is known as the
a) space centrode
b) body centrode
c) moving centrode
d) none of the mentioned

View Answer

Answer: a [Reason:] The locus of the instantaneous centre in space during a definite motion of the body is called the space centrode and the locus of the instantaneous centre relative to the body itself is called the body centrode.

5. The space centrode of a circular disc rolling on a straight path is
a) circle
b) parabola
c) a straight line
d) none of the mentioned

View Answer

Answer: c [Reason:] The instantaneous center of rotation of a circular disc rolling on a straight path is at their point of contact.

6. The component of the acceleration directed towards the center of rotation of a revolving body is known as ____________ component.
a) tangential
b) centripetal
c) coriolis
d) none of the mentioned

View Answer

Answer: b [Reason:] The centripetal or radial component, which is perpendicular to the velocity of the particle at the given instant. The tangential component, which is parallel to the velocity of the particle at the given instant.

7. At an instant the link AB of length r has an angular velocity ω and an angular acceleration α. What is the total acceleration of AB?
a) [(ω2r)2 + αr)2]1/2
b) [(ωr)2 + αr)2]1/2
c) [(ω2r)2 + αr)]1/2
d) [(ω2r)2 + αr)2]1/2

View Answer

Answer: a

8. At an instant, if the angular velocity of a link is clockwise then the angular acceleration will be
a) clockwise
b) counterclockwise
c) in any direction
d) none of the mentioned

View Answer

Answer: c

9. Angular acceleration of a link AB is given by
a) centripetal acceleration/length
b) tangential acceleration/length
c) total acceleration/length
d) none of the mentioned

View Answer

Answer: b

10. A slider moves with uniform velocity v on a revolving link of length r with angular velocity of ω. The Coriolis acceleration component of a point on the slider relative to a coincident point on the link is equal to
a) 2rω parallel to the link
b) 2ωv perpendicular to the link
c) 2rω perpendicular to the link
d) 2ωv parallel to the link

View Answer

Answer: b [Reason:] The coriolis component is always perpendicular to the link and is given by 2ωv.

Set 4

1. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?
a) 10.475 rad/s2
b) 12 rad/s2
c) 14 rad/s2
d) 15 rad/s2

View Answer

Answer: a [Reason:] Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s Angular acceleration Let α = Angular acceleration in rad/s2 We know that ω = ω0 + α.t or 209.5 = 0 + α × 20 or α = 209.5 / 20 = 10.475 rad/s2

2. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
a) 400
b) 300
c) 333.4
d) 200

View Answer

Answer: c [Reason:] Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s), θ = (ω0 + ω )t/2 = (0 + 209.5)20/2 = 2095 rad Since the angular distance moved by the wheel during one revolution is 2π radians, therefore number of revolutions made by the wheel, n = θ /2π = 2095/2π = 333.4

3. The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned

View Answer

Answer: a [Reason:] The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.

4. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned

View Answer

Answer: b [Reason:] The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.

5. When a particle moves along a straight path, then the radius of curvature is
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned

View Answer

Answer: c [Reason:] When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero.

6. When a particle moves with a uniform velocity, then dv/dt will be
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned

View Answer

Answer: b [Reason:] When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration.

7. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s

View Answer

Answer: a [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s Linear velocity at the beginning We know that linear velocity at the beginning, v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s

8. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at end of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s

View Answer

Answer: b [Reason:] Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s Linear velocity at the end of 5 seconds We also know that linear velocity after 5 seconds, v5 = r . ω = 1.5 × 157 = 235.5 m/s

Set 5

1. Match list I with list II

List I List II
A. Law of correct steering 1. f = 3(n – 1) – 2j
B. Displacement relation of Hook’e joint 2. x = R[ (1 – cosϴ) + sin2ϴ/2n].
C. Relation between kinematic pairs and links 3. cotɸ – cotɸ = c/b
D. Displacement equation of reciprocating engine piston 4. tanϴ = tanɸ cosα

a) A-1,B-4,C-3,D-2
b) A-1,B-2,C-3,D-4
c) A-3,B-4,C-1,D-2
d) A-3,B-2,C-1,D-4

View Answer

Answer: c [Reason:] Law of correct steering – cotɸ – cotɸ = c/b Displacement relation of Hook’e joint – tanϴ = tanɸ cosα Relation between kinematic pairs and links – f = 3(n – 1) – 2j Displacement equation of reciprocating engine piston – x = R[ (1 – cosϴ) + sin2ϴ/2n].

2. Consider the following statements regarding motions in machines:
1. Tangential acceleration is a function of angular velocity and the radial acceleration is a function of angular acceleration.
2. The resultant acceleration of a point A with respect to a point B on a rotating link is perpendicular to AB.
3. The direction of the relative velocity of a point A with respect to a point B on a rotating link is perpendicular.

Which of these statements is/are correct?
a) 1 alone
b) 2 and 3
c) 1 and 2
d) 3 alone

View Answer

Answer: d [Reason:] Only statement 1 is correct.

3.The speed of driving shaft of a Hooke’s Joint of angle 19.50 is 500 r.p.m. The maximum speed of the driven shaft is nearly
a) 168 r.p.m.
b) 444 r.p.m.
c) 471 r.p.m.
d) 531 r.p.m.

View Answer

Answer: d [Reason:] Nmax = N/cosα = 500/cos19.50 = 531 r.p.m.

4. In a slider crank mechanism. the maximum acceleration of slider is obtained when the crank is
a) at the inner dead centre position
b) at the outer dead centre position
c) exactly midway position between the two dead centres
d) slightly in advance of the midway position between the two dead centres

View Answer

Answer: b

5. In a shaper machine, the mechanism for tool feed is
a) Geneva mechanism
b) Whitworth mechanism
c) Ratchet and Pawl mechanism
d) Ward Leonard system

View Answer

Answer: c [Reason:] The crank shaper, in which the tool carrier is driven forward and backward by an oscillating arm operated by a crankpin in the main driving gear, or “bull wheel,” and in which the feed is transmitted to the worktable by ratchet-and-pawl mechanism, is so commonly used as to be termed standard.

6. The instantaneous centre of rotation of a rigid thin disc rolling without slip on a plane rigid surface is located at
a) the centre of the disc
b) an infinite distance perpendicular to the plane surface
c) the point of contact
d) the point on the circumference situated vertically opposite to the contact point

View Answer

Answer: a [Reason:] The instantaneous centre of rotation of a rigid thin disc without slip is located at the centre of the disc.

7. Match list I with list II

List I List II
A. Sliding pair 1. A road roller rolling over the ground
B. Revolute pair 2. Crank shaft in a journal bearing in an engine
C. Rolling pair 3. Ball and socket joint
D. Spherical pair 4. Piston and cylinder
5. Nut and screw
a) A-5,B-2,C-4,D-3
b) A-4,B-3,C-1,D-2
c) A-5,B-3,C-4,D-2
d) A-4,B-2,C-1,D-3

View Answer

Answer: d [Reason:] Sliding pair – Piston and cylinder Revolute pair – Crank shaft in a journal bearing in an engine Rolling pair – A road roller rolling over the ground Spherical pair – Ball and socket joint.

8. Slider crank chain is an inversion of the four bar mechanism.
a) True
b) False

View Answer

Answer: a [Reason:] Slider crank chain often finds applications in most of the reciprocating machinery.

9. f = 3 (n – 1) – 2j. In the Gruebler’s equation for planer mechanisms given, j is the
a) Number of mobile links
b) Number of links
c) Number of lower pairs
d) Length of the longest link

View Answer

Answer: c

10. Which of the following are examples of forced closed kinematic pairs?
1. Cam and roller mechanism
2. Door closing mechanism
3. Slider crank mechanism
4. Automotive clutch operating mechanism
Select the correct answer using the codes given below:
a) 1,2 and 4
b) 1 and 3
c) 2,3 and 4
d) 1,2,3 and 4

View Answer

Answer: c [Reason:] Except statement 1 all are examples of forced closed kinematic pairs.

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