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