Category: Machine Kinematics Objective Questions

250+ TOP MCQs on Grubler’s Criterion for Plane Mechanisms and Answers

Machine Kinematics Multiple Choice Questions on ” Grubler’s Criterion for Plane Mechanisms”.

1. The approximate straight line mechanism is a
a) 4 bar linkage
b) 6 bar linkage
c) 8 bar linkage
d) 3 bar linkage
Answer: a
Clarification: The straight line mechanism has 4 bar linkage.

2. Open pairs are those which have
a) point or line contact between the two elements when in motion
b) surface contact between the two elements when in motion
c) elements of pairs not held together mechanically
d) two elements that permit relative motion
Answer: c
Clarification: Two elemets which are not held together form open pairs.
Two elements held together mechanically form a closed pair.

3. Peaucellier mechanism has
a) 8 links
b) 6 links
c) 4 links
d) 5 links
Answer: a
Clarification: Peaucellier mechanism has 8 links.
Hart mechanism has 6 links.

4. Hart mechanism has
a) 8 links
b) 6 links
c) 4 links
d) 5 links
Answer: b
Clarification: Hart mechanism has 6 links.
Peaucellier mechanism has 8 links.

5. A chain comprises of 5 links having 5 joints. Is it kinematic chain?
a) Yes
b) No
c) It is a marginal case
d) None of the mentioned
Answer: b
Clarification: Kinematic chain is one which satisfies
L = 2/3 (J + 2),
which is not satisfied in this case.

6. In the following equation [ L = 2/3(J + 2)] to determine whether or not the given chain in kinematic, higher pair is treated equivalent to
a) two lower pairs and an additional links
b) two higher pairs and two additional links
c) one lower pairs and two additional links
d) none of the mentioned
Answer: a
Clarification: In the above equation, each higher pair is taken equivalent to two lower pairs and an additional link.
If one of the links of the constrained chain is fixed, it results into mechanism.

7. The main disadvantage of the sliding pair is that it is
a) bulky
b) difficult to manufacture
c) wears rapidly
d) both a and c
Answer: d
Clarification: The sliding pair wears rapidly due to friction and is bulky also.

8. For a kinematic chain to be considered as mechanism
a) two links should be fixed
b) one link should be fixed
c) none of the links should be fixed
d) there is no such criterion
Answer: b
Clarification: If one of the links of the constrained chain is fixed, it results into mechanism.

9. An eccentric sheave pivoted at one point rotates and transmits oscillatory motion to a link whose one end is pivoted and other end is connected to it. This mechanism has
a) 2 links
b) 3 links
c) 4 links
d) 5 links
Answer: c
Clarification: Eccentric sheave is equivalent to 2 links, 1 link is due to oscillatory link, and one is fixed link.

10. Whitworth quick return mechanism is obtained by inversion of
a) slider crank mechanism
b) kinematic chain
c) five link mechanism
d) roller cam mechanism
Answer: a
Clarification: None.

250+ TOP MCQs on Velocities in a Slider Crank Mechanism and Motion of a Link and Answers

Machine Kinematics Multiple Choice Questions on “Velocities in a Slider Crank Mechanism and Motion of a Link”.

1. The lengths of the links of a 4- bar linkage with revolute pairs only are p,q,r and s units. Given that p < q < r < s and s+p < q+r which of these links should be the fixed one, for obtaining a ‘double crank’ mechanism?
a) ink of length p
b) link of length q
c) link of length r
d) link of length s
Answer: a
Clarification: For Double crank mechanism Shortest link is fixed.
Here shortest link is ‘P’.

2. For a four-bar linkage in toggle position, the value of mechanical advantage is
a) 0.0
b) 0.5
c) 1.0
d) ∞
Answer: d
Clarification: At Toggle position output velocity is zero
And hence, mechanical advantage = input velocity/output velocity = ∞.

3. In a slider-crank mechanism, the crank is rotating with an angular velocity of 20 rad/s in counterclockwise direction. At the instant when the crank is perpendicular to the direction of the piston movement, velocity of the piston is 2 m/s. Radius of the crank is
a) 100 cm
b) 10 cm
c) 1 cm
d) 0.1 cm
Answer: b
Clarification: Vp = ωr(sinϴ + sin2ϴ/2n)
In this case ϴ = 900
Vp = ωr
r = Vp/ω = 2/20 = 0.1 m or 10 cm.

4. In a single link robotic arm the end-effector slides upward along the link with a velocity of 0.5 m/s while the link rotates about revolute joint with an angular speed of 1 rad/sec. When the end-effector is at a distance of 1 m from the joint, the acceleration experienced by the end-effector will be
a) 1 m/s2
b) 1.41 m/s2
c) 1.71 m/2
d) 2 m/2
Answer: a
Clarification: a = 2ωV = 2 x 1 x 0.5 = 1 m/s2.

5. For the same crank length and uniform angular velocity of the crank in an offset slider crank mechanism, if the connecting rod length is increased by 1.5 times, the velocity of piston will
a) remain unchanged
b) increase 1.5 times
c) decrease by 1.5 times
d) increase by 1.5√2 times
Answer: c
Clarification: V1 = ωr(sinϴ + sin2ϴ/2n)
V2 = ωr(sinϴ + sin2ϴ/3n)
from these two equation, V21
V2 will decrease but correct quantification can not be done with available data.
Among the available options, best answer is (c).

6. It is planned to construct a four-bar mechanism ABCD with length AB= 60mm, BC = 100mm, CD = 70 mm and fixed link AD = 200 mm. If at least one link is required to have a complete rotation, this mechanism is
a) of crank-rocker type
b) of double-crank type
c) of double rocker type
d) impossible to construct
Answer: c
Clarification: S + L = 60 + 200 = 260 mm
P + Q = 100 + 70 = 170 mm
From grashoff equality when S + L > P + Q
So always double rocker.

7. The number of links in a planer mechanism with revolute joints having 10 instantaneous centres is
a) 3
b) 4
c) 5
d) 6
Answer: c
Clarification: n(n – 1)/2 = 10
n(n – 1) = 20
n = 5.

8. A weston differential pulley block consists of a lower block and upper block. The upper block has two cogged grooves, one of which has a radius of 150 mm and the other a radius of 125 mm. If the efficiency of the machine is 50% calculate the effort required to raise a load of 1.5 kN.
a) 250 N
b) 300 N
c) 350 N
d) 400 N
Answer: a
Clarification: We know that in case of a Weston differential pulley block,
V.R. = 2D/D – d = 2 x 300/300 -250 = 12
Using the relation, Efficiency = M.A./V.R. x 100
or, 50 = M.A./12 x 100
M.A. = 6

Again, M.A. = W/P
6 = 1.5 x 1000/ P
P = 250 N.

9. Following are the specifications of a single purchase crab:
Diameter of load drum, d = 200 mm
Length of lever, l = 1.2 m
No. of teeth on pinion, T1 = 10
No. of teeth on spur wheel, T2 = 100
Find the velocity ratio of this machine.
a) 100
b) 110
c) 120
d) 130
Answer: c
Clarification: V.R. = 2l/d x T2/T1
= 2 x 120/20 x 100/10 = 120.

10. On a machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the law of the machine.
a) P = 1/100W + 60
b) P = 1/100W + 70
c) P = 1/100W + 80
d) P = 1/100W + 90
Answer: c
Clarification: Let the law of machine be P = mW + C
where P = effort applied, W = load lifted and m and C being constants.
when P = 100 N W = 3000 N
when P = 160 N W = 9000 N
Putting these values in the law of machine.
100 = 3000m + C …………(i)
160 = 9000m + C …………(ii)

Subtracting (i) and (ii), we get
60 = 6000 m
or, m = 1/100
Putting this value in equation (i), we get
100 = 3000 x 1/100 + C
C = 70

Hence, the machine follows the laws
P = 1/100W +70.

11. Following are the specifications of a single purchase crab:
Diameter of load drum, d = 200 mm
Length of lever, l = 1.2 m
No. of teeth on pinion, T1 = 10
No. of teeth on spur wheel, T2 = 100
On this machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the efficiency at 3000N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %
Answer: d
Clarification: V.R. = 2l/d x T2/T1
= 2 x 120/20 x 100/10 = 120
M.A. = W/P = 3000/100 = 30

Efficiency = M.A./V.R. = 30/120 = 0.25 = 25%.

12. Following are the specifications of a single purchase crab:
Diameter of load drum, d = 200 mm
Length of lever, l = 1.2 m
No. of teeth on pinion, T1 = 10
No. of teeth on spur wheel, T2 = 100
On this machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the efficiency at 9000N.
a) 30 %
b) 40 %
c) 46.8 %
d) 56.8 %
Answer: d
Clarification: V.R. = 2l/d x T2/T1
= 2 x 120/20 x 100/10 = 120
M.A. = W/P = 9000/160 = 900/16

Efficiency = M.A./V.R. = 900/16 x 120 = 0.468 = 46.8%.

250+ TOP MCQs on Coefficient of Friction and Answers

Machine Kinematics Multiple Choice Questions on “Coefficient of Friction”.

1. A block whose mass is 650 gm is fastened to aspring constant K equals 65 N/m whose other end is fixed. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a smooth surface, and released from rest at t = 0. The maximum speed ‘S’ of the oscillating block is
a) 11 cm /sec
b) 11 m/sec
c) 11 mm/sec
d) 1.1 m/sec
Answer: d
Clarification: Maximum speed of the spring will be at x = 0, when total spring energy will be converted into kinematic energy of mass.
1/2 kx2 = 1/2 mv2
1/2 x 65 x(0.11)2 = 1/2 x (0.650)V2
V = 1.1 m/sec.

2. Which of the following statements regarding laws governing the friction between dry surfaces are correct?
a) The friction force is directly proportional to the normal force.
b) The friction force is dependent on the materials of the contact surfaces.
c) The friction force is independent of the area of contact.
d) all of the mentioned
Answer: d
Clarification: Following are the laws of solid friction :
1. The force of friction is directly proportional to the normal load between the surfaces.
2. The force of friction is independent of the area of the contact surface for a given normal load.
3. The force of friction depends upon the material of which the contact surfaces are made.
4. The force of friction is independent of the velocity of sliding of one body relative to the other body.

3. If two bodies one light and other heavy have equal kinetic energies, which one has a greater momentum
a) heavy body
b) light body
c) both have equal momentum
d) it depends on the actual velocities
Answer: a
Clarification: 1/2m1V12 = 1/2m2V22
v2/V1 = √m1/m2
For momentum ratio, M1/M2 = √m1/m2.

4. A heavy block of mass m is slowly placed on a conveyer belt moving with speed v. If coefficient of friction between block and the belt is μ, the block will slide on the belt through distance
a) v/μg
b) v2/√μg
c) (v/μg)2
d) v2/2μg
Answer: d
Clarification: Retardation due to friction force = μg
V2 = 2.μg s
s = v2/2μg.

5. A car moving with uniform acceleration cover 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. The acceleration of the car is
a) 7 m/s2
b) 50 m/s2
c) 25 m/s2
d) 10 m/s2
Answer: d
Clarification: s = ut + 1/2at2
at t = 5 sec, s = 450
450 = u(5) + 1/2a(25)
at t = 10 sec, s = 450 + 700 = 1150
1150 = u(10) = 1/2a(100)
a = 10 m/sec2.

6. A particle starts with a velocity 2 m/sec and moves on a straight-line track with retardation 0.1 m/s2. The time at which the particle is 15 m from the startin g point would be
a) 10 s
b) 20 s
c) 50 s
d) 40 s
Answer: a
Clarification: u = 2m/sec a = 0.1 m/sec
s = ut – 1/2at2
15 = 2t – 1/2(0.1)t2
t = 30, t = 10.

7. Two particles with masses in the ratio 1 : 4 are moving with equal kinetic energies. The magnitude of their linear momentums will conform to the ratio
a) 1 : 8
b) 1 : 2
c) √2 : 1
d) √2
Answer: b
Clarification: 1/2m1V12 = 1/2m2V22
m1/m2 = (V2/V1)2 = 1/4
V2/V1 = 1/2
L1/L2 = m1V1/m2V2 = 1/2.

8. A stone is projected horizontally from a cliff at 10 m/sec and lands on the ground below at 20 m from the base of the cliff. Find the height of the cliff.
a) 18 m
b) 20 m
c) 22 m
d) 24 m
Answer: b
Clarification: h = 1/2at2
h = 1/2 x 10 x 4 = 20m.

9. A car moving with speed u can be stopped in minimum distance x when brakes are applied. If the speed becomes n times, the minimum distance over which the car can be stopped would take the value
a) x/n
b) nx
c) x/n2
d) n2x
Answer: d
Clarification: x = u2/2g
x’ = (nu)2/2g
x’ = n2x.

10. Ratio of the radii of planes P1 and P2 is k and ratio of the accelerations due to gravity on them is s. Ratio of escape velocities from them will be
a) ks
b) √ks
c) √k/s
d) √s/k
Answer: d
Clarification: v1/v2 = √g1/g2 √R2/R1
v1/v2 = √s/√k = √s/k.

250+ TOP MCQs on Toothed Gearing – 2 and Answers

Machine Kinematics Questions and Answers for Entrance exams on “Toothed Gearing – 2”.

1. The condition of correct gearing is
a) pitch line velocities of teeth be same
b) radius of curvature of two profiles be same
c) common normal to the pitch surface cuts the line of centres at a fixed point
d) none of the mentioned
Answer: c
Clarification: The fundamental condition of correct gearing is the common normal at the point of contact between a pair of teeth must always pass through the pitch point.

2. Law of gearing is satisfied if
a) two surfaces slide smoothly
b) common normal at the point of contact passes through the pitch point on the line joining the centres of rotation
c) number of teeth = P.C.D. / module
d) addendum is greater than dedendum
Answer: b
Clarification: Law of gearing says that the common normal at the point of contact between a pair of teeth must always pass through the pitch point.

3. Involute profile is preferred to cyloidal because
a) the profile is easy to cut
b) only one curve is required to cut
c) the rack has straight line profile and hence can be cut accurately
d) none of the mentioned
Answer: b
Clarification: The face and flank of involute teeth are generated by a single curve where as in cycloidal
gears, double curves (i.e. epi-cycloid and hypo-cycloid) are required for the face and flank respectively.
Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the basic rack has straight teeth and the same can be cut with simple tools.

4. The contact ratio for gears is
a) zero
b) less than one
c) greater than one
d) none of the mentioned
Answer: c
Clarification: The theoretical minimum value for the contact ratio is one, that is there must always be at least one pair of teeth in contact for continuous action.

5. The maximum length of arc of contact for two mating gears, in order to avoid interference, is
a) (r + R) sin φ
b) (r + R) cos φ
c) (r + R) tan φ
d) none of the mentioned
Answer: c
Clarification: Interference may only be prevented, if the addendum circles of the two mating gears cut the
common tangent to the base circles between the points of tangency.
maximum length of arc of contact = (r + R) tan φ
where r = Pitch circle radius of pinion,
R = Pitch circle radius of driver, and
φ = Pressure angle.

6. When the addenda on pinion and wheel is such that the path of approach and path of recess are half of their maximum possible values, then the length of the path of contact is given by
a) (r + R) sin φ/2
b) (r + R) cos φ/2
c) (r + R) tan φ/2
d) (r + R) cot φ/2
Answer: a
Clarification: In case the addenda on pinion and wheel is such that the path of approach and path of recess are half of their maximum possible values, then
Path of approach, KP = 1/2 MP.

7. Interference can be avoided in involute gears with 20° pressure angle by
a) cutting involute correctly
b) using as small number of teeth as possible
c) using more than 20 teeth
d) using more than 8 teeth
Answer: c
Clarification: None.

8. The ratio of face width to transverse pitch of a helical gear with α as the helix angle is normally
a) more than 1.15/tan α
b) more than 1.05/tan α
c) more than 1/tan α
d) none of the mentioned
Answer: a
Clarification: None.

9. The maximum efficiency for spiral gears is
a) sin (θ + φ ) + 1/ cos (θ − φ ) + 1
b) cos (θ − φ) + 1/sin (θ + φ ) + 1
c) cos (θ + φ ) + 1/ cos (θ − φ ) + 1
d) cos (θ − φ) + 1/cos (θ + φ ) + 1
Answer: c
Clarification: ηmax = cos (θ + φ ) + 1/ cos (θ − φ ) + 1
where θ = Shaft angle, and φ = Friction angle.

10. For a speed ratio of 100, smallest gear box is obtained by using
a) a pair of spur gears
b) a pair of helical and a pair of spur gear compounded
c) a pair of bevel and a pair of spur gear compounded
d) a pair of helical and a pair of worm gear compounded
Answer: d
Clarification: None.

To practice all areas of Machine Kinematics for Entrance exams,

250+ TOP MCQs on Standard Spur Gears-1 and Answers

Machine Kinematics Multiple Choice Questions on “Standard Spur Gears – 1”.

1. Lewis equation for design of involute gear tooth predicts the dynamic load capacity of a cantilever beam of uniform strength.
a) True
b) False
Answer: b
Clarification: Lewis equation for design of involute gear tooth predicts the static load capacity of a cantilever beam of uniform strength.

2. For a pair of gears in mesh, pressure angle and module must be different to satisfy the condition of interchangeability and correct gearing.
a) True
b) False
Answer: b
Clarification: For a pair of gears in mesh, pressure angle and module must be same to satisfy the condition of interchangeability and correct gearing.

3. In skew bevel gears, the axes are
a) non-parallel and non-intersecting, and teeth are curved
b) non-parallel and non-intersecting, and teeth are straight
c) intersecting, and teeth are curved and oblique
d) intersecting, and teeth are curved and can be ground
Answer: a
Clarification: Skew bevel gears imply non-parallel, non-intersecting and curved them.

4. In case the number of teeth on two bevel gears in mesh is 30 and 60 respectively, then the pitch cone angle of the gear will be
a) tan-12
b) п/2 + tan-12
c) п/2 – tan-10.5
d) tan-10.5
Answer: d
Clarification: tanɸ = NP/NG
NP = No. of teeth of pinion
NG = No. of teeth of gear
ɸ = Pitch cone angle
ɸ = tan-1(30/60)
ɸ = tan-10.5.

5. Consider the following statements:
The axes of spiral bevel gear are non-parallel and intersecting.
1. The most common pressure angle for spiral bevel gear is 20o.
2. The most common spiral angle for spiral bevel gear is 35o.
3. Spiral bevel gears are generally interchangeable.
4. Spirals are noisy and recommended for low speeds of 10 m/s.

Which of the above statements are correct?
a) 1 and 4
b) 1 and 2
c) 2 and 3
d) 3 and 4
Answer: a
Clarification: Commonly used pressure angle is 20o, spiral gear operation is noisy hence recommended for low speed operation.

6. Consider the following statements:
In case of helical gears, teeth are cut at an angle to the axis of rotation of the gears.
1. Helix angle introduces another ratio called axial contact ratio.
2. Transverse contact ratio is equal to axial contact ratio in helical gears.
3. Large transverse contact ratio does not allow multiple teeth to share the load.
4. Large axial contact ratio will cause larger axial force component.

Which of the above statements are correct?
a) 1 and 2
b) 2 and 3
c) 1 and 4
d) 3 and 4
Answer: c
Clarification: Transverse contact ratio may not be equal to axial contact ratio. Larger contact ratio means sharing of multiple teeth.

7. A worm gear set is designed to have pressure angle of 300 which is equal to the helix angle. The efficiency of the worm gear set at an interface friction of 0.05 is
a) 87.9%
b) 77.9%
c) 67.9%
d) 57.9%
Answer: a
Clarification: ȵ = tanλ(cosɸ – μtanλ)/cosɸtanλ + μ
tanλ = -μ + √1 + μ2
tanλ = -(0.05) + √1 + (0.05)2
tanλ = 0.9512

ȵ = 89.1%

So most suitable option is ‘a’.

8. The use of straight or curved external gear teeth in mesh with internal teeth in gear and spline couplings is specifically employed accommodate
a) torsional misalignment
b) parallel misalignment
c) angular misalignment
d) substantial axial movements between shafts
Answer: c
Clarification: Straight or curved external gear teeth are used to correct the angular misalignment.

9. A planetary gear train is gear train having
a) a relative motion of axes and the axis of at least one of the gears also moves relative to the frame
b) no relative motion of axes and no relative motion of axes with respect to the frame
c) no relative motion of axes and the axis of at least one of the gears also moves relative to the frame
d) a relative motion of axes and none of the axes of gears has relative motion with the frame
Answer: a
Clarification: Planetary gear train ensures the relative motion of at least one axis w.r.t. the frame.

10. The load on a gear tooth is 50 kN. If the gear is transmitting a torque of 6000 Nm, the diameter of the gear is approximately
a) 0.5 m
b) 0.75 m
c) 1 m
d) 0.25 m
Answer: d
Clarification: T = w cosɸ x d/2
6000 = 50000 x 0.94 x d/2
d = 0.25 m.