# Multiple choice question for engineering

## Set 1

1. The relation between the number of links (l) and the number of binary joints ( j) for a kinematic chain having constrained motion is given by j = 3/2 I -2 If the left hand side of this equation is greater than right hand side, then the chain is

a) locked chain

b) completely constrained chain

c) successfully constrained chain

d) incompletely constrained chain

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2. In a kinematic chain, a quaternary joint is equivalent to

a) one binary joint

b) two binary joints

c) three binary joints

d) four binary joints

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3. If n links are connected at the same joint, the joint is equivalent to

a) (n – 1) binary joints

b) (n – 2) binary joints

c) (2n – 1) binary joints

d) none of the mentioned

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4. In a 4 – bar linkage, if the lengths of shortest, longest and the other two links are denoted by s, l, p and q, then it would result in Grashof’s linkage provided that

a) l + p < s + q

b) l + s < p + q

c) l + p = s + q

d) none of the mentioned

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5. A kinematic chain is known as a mechanism when

a) none of the links is fixed

b) one of the links is fixed

c) two of the links are fixed

d) all of the links are fixed

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6. The Grubler’s criterion for determining the degrees of freedom (n) of a mechanism having plane motion is

a) n = (l – 1) – j

b) n = 2 (l – 1) – 2j

c) n = 3 (l – 1) – 2j

d) n = 4 (l – 1) – 3j

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7. The mechanism forms a structure, when the number of degrees of freedom (n) is equal to

a) 0

b) 1

c) 2

d) – 1

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8. In a four bar chain or quadric cycle chain

a) each of the four pairs is a turning pair

b) one is a turning pair and three are sliding pairs

c) three are turning pairs and one is sliding pair

d) each of the four pairs is a sliding pair.

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9. Which of the following is an inversion of single slider crank chain ?

a) Beam engine

b) Watt’s indicator mechanism

c) Elliptical trammels

d) Whitworth quick return motion mechanism

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10. Which of the following is an inversion of double slider crank chain ?

a) Coupling rod of a locomotive

b) Pendulum pump

c) Elliptical trammels

d) Oscillating cylinder engine

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## Set 2

1. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate kinetic energy of rotation of the wheels and axles at a speed of 9 km/h.

a) 7670 N-m

b) 8670 N-m

c) 9670 N-m

d) 6670 N-m

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_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,
I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,
ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

2. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate total kinetic energy of road roller.

a) 25170 N-m

b) 35170 N-m

c) 45170 N-m

d) 55170 N-m

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_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,
I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,
ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,
E_{3} = 1/2 mv^{2} = 1/2 x 1200 (2.5)^{2} = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,
E_{4} = Kinetic energy of translation + Kinetic energy of rotation
= E_{3} + E = 37 500 + 7670 = 45 170 N-m

3. A road roller has a total mass of 12 tonnes. The front roller has a mass of 2 tonnes, a radius of gyration of 0.4 m and a diameter of 1.2 m. The rear axle, together with its wheels, has a mass of 2.5 tonnes, a radius of gyration of 0.6 m and a diameter of 1.5 m. Calculate braking force required to bring the roller to rest from 9 km/h in 6 m on the level.

a) 5528.3 N

b) 6528.3 N

c) 7528.3 N

d) 8528.3 N

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_{1}= 2 t = 2000 kg ; k

_{1}= 0.4 m ; d

_{1}= 1.2 m or r

_{1}= 0.6 m ; m

_{2}= 2.5 t = 2500 kg ; k

_{2}= 0.6 m ; d

_{2}= 1.5 m or r

_{2}= 0.75 m ; v = 9 km/h = 2.5 m/s; s = 6 m

Kinetic energy of rotation of the wheels and axles

We know that mass moment of inertia of the front roller,
I_{1} = m_{1}(k_{1})^{2} = 2000 (0.4)^{2} = 320 kg-m^{2}

and mass moment of inertia of the rear axle together with its wheels,
I_{2} = m_{2} (k_{2})^{2} = 2500 (0.6)^{2} = 900 kg -m^{2}

Angular speed of the front roller,
ω_{1} = v/r_{1} = 2.5/0.6 = 4.16 rad/s
and angular speed of rear wheels,

ω_{2} = v/r_{2} = 2.5/0.75 = 3.3 rad/s

We know that kinetic energy of rotation of the front roller,

E_{1} =1/2 I_{1} (ω_{1})^{2} = 1/2 × 320(4.16)^{2} 2770 N-m

and kinetic energy of rotation of the rear axle together with its wheels,

E_{2} =1/2 I_{2} (ω_{2})^{2} = 1/2 × 900(3.3)^{2} 4900 N-m

∴ Total kinetic energy of rotation of the wheels,
E = E_{1} + E_{2} = 2770 + 4900 = 7670 N-m

We know that the kinetic energy of motion (i.e. kinetic energy of translation) of the road roller,
E_{3} = 1/2 mv^{2} = 1/2 x 1200 (2.5)^{2} = 37500 N-m

This energy includes the kinetic energy of translation of the wheels also, because the total mass (m) has been considered.

∴ Total kinetic energy of road roller,
E_{4} = Kinetic energy of translation + Kinetic energy of rotation
= E_{3} + E = 37 500 + 7670 = 45 170 N-m

Let F = Braking force required to bring the roller to rest, in newtons. We know that the distance travelled by the road roller, s = 6 m … (Given) ∴ Work done by the braking force = F × s = 6 F N-m This work done must be equal to the total kinetic energy of road roller to bring the roller to rest, i.e. 6 F = 45 170 or F = 45 170/6 = 7528.3 N

4. A haulage rope winds on a drum of radius 500 mm, the free end being attached to a truck. The truck has a mass of 500 kg and is initially at rest. The drum is equivalent to a mass of 1250 kg with radius of gyration 450 mm. The rim speed of the drum is 0.75 m/s before the rope tightens. By considering the change in linear momentum of the truck and in the angular momentum of the drum, find the speed of the truck when the motion becomes steady.

a) 0.502 m/s

b) 0.602 m/s

c) 0.702 m/s

d) 0.802 m/s

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_{1}= 500 kg ; m

_{2}= 1250 kg ; k = 450 mm = 0.45 m ; u = 0.75 m/s

We know that mass moment of inertia of drum,

I_{2} = m_{2}.k^{2} = 1250 (0.45)^{2} = 253 kg-m^{2}

Let v = Speed of the truck in m/s, and F = Impulse in rope in N-s.

We know that the impulse is equal to the change of linear momentum of the truck. Therefore
F = m_{1}.v = 500 v N-s
and moment of impulse = Change in angular momentum of drum
i.e. F x r = I_{1} (ω_{2} − ω_{1}) = I_{2}(u – v/r)
500v x 0.5 = 253(0.75 – v/0.5)
or, 250v = 380 − 506v

∴ 250 v + 506 v = 380 or v = 380/756 = 0.502 m/s

5. An electric motor drives a machine through a speed reducing gear of ratio 9:1. The motor armature, with its shaft and gear wheel, has moment of inertia 0.6 kg-m^{2}. The rotating part of the driven machine has moment of inertia 45 kg-m^{2}. The driven machine has resisting torque of 100 N-m and the efficiency of reduction gear is 95%. Find the power which the motor must develop to drive the machine at a uniform speed of 160 r.p.m.

a) 1764 W

b) 2764 W

c) 3764 W

d) 4764 W

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_{A}= 0.6 kg-m

^{2}; I

_{B}= 45 kg-m

^{2}; T

_{B}= 100 N-m; η = 95% = 0.95; N = 160 r.p.m. ; N

_{1}= 0 ; N

_{2}= 60 r.p.m. T

_{A}= 30 N-m

We know that the power which the motor must develop,

P = 2πN T_{B}/60× η
= 2π × 160 × 100/60 x 0.95
= 1764 W

6. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Angular acceleration of the flywheel.

a) 0.6 rad/s^{2}

b) 0.8 rad/s^{2}

c) 0.10 rad/s^{2}

d) none of the mentioned

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^{2}= 2500×12 = 2500 kg-m

^{2}

We also know that torque ( T ),
1500 = I .α = 2500 × α
or α = 1500 / 2500 = 0.6 rad/s^{2}

7. The flywheel of a steam engine has a radius of gyration of 1 m and mass 2500 kg. The starting torque of the steam engine is 1500 N-m and may be assumed constant. Determine : Kinetic energy of the flywheel after 10 seconds from the start.

a) 50 kJ

b) 60 kJ

c) 45 kJ

d) none of the mentioned

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^{2}= 2500×12 = 2500 kg-m

^{2}

We also know that torque ( T ),
1500 = I .α = 2500 × α
or α = 1500 / 2500 = 0.6 rad/s^{2}

Kinetic energy of the flywheel after 10 seconds from start
First of all, let us find the angular speed of the flywheel (ω^{2}) after t = 10 seconds from the start (i.e. ω^{1} = 0 ).
We know that ω^{2} = ω^{1} + α.t = 0 + 0.6 × 10 = 6 rad/s

∴ Kinetic energy of the flywheel,
E = 1/2 I(ω^{2})^{2} = 1/2 x 2500 x 6^{2} = 45 000J
= 45 kJ

8. Which of the following objects have momentum?

a) An electron is orbiting the nucleus of an atom.

b) A UPS truck is stopped in front of the school building.

c) The high school building rests in the middle of town.

d) None of the mentioned

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9. A truck driving along a highway road has a large quantity of momentum. If it moves at the same speed but has twice as much mass, its momentum is ________________

a) zero

b) quadrupled

c) doubled

d) unchanged

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10. A ball is dropped from the same height upon various flat surfaces. For the same collision time, impulses are smaller when the most bouncing take place.

a) True

b) False

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## Set 3

1. The mathematical form of the bevel tooth profile which closely resembles a spherical involute but is fundamentally different is

a) crown

b) back cone

c) octoid

d) none of the mentioned

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2. The angle formed between an element of the pitch cone and the bevel gear axis is

a) shaft angle

b) pitch angle

c) spiral angle

d) none of the mentioned

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3. The angle between the tooth trace and an element of the pitch cone, corresponding to helix angle in helical gears is

a) shaft angle

b) pitch angle

c) spiral angle

d) none of the mentioned

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4. The diameter and plane of rotation surface or shaft centre which is used for locating the gear blank during fabrication of the gear teeth is known as

a) crown

b) back cone

c) generating mounting surface

d) none of the mentioned

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5. The sharp corner orming the outside diameter is

a) crown

b) back cone

c) octoid

d) none of the mentioned

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6. The angle between elements of the face cone and pitch cone is

a) addendum angle

b) pitch angle

c) spiral angle

d) none of the mentioned

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7. The angle between mating bevel gear axes, also the sum of the two pitch angles is

a) shaft angle

b) pitch angle

c) spiral angle

d) none of the mentioned

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8. The length of teeth along the cone distance is

a) crown

b) face width

c) octoid

d) none of the mentioned

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9. The angle of a cone whose elements are tangent to a sphere containing a trace of the pitch circle is

a) crown

b) back cone

c) octoid

d) none of the mentioned

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10. The angle between elements of the root cone and pitch cone is

a) addendum angle

b) dedendum angle

c) spiral angle

d) none of the mentioned

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## Set 4

1. In its simplest form, a cam mechanism consists of following number of links

a) 1

b) 2

c) 3

d) 4

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2. Which of the following mechanisms produces mathematically an exact straight line motion?

a) Grasshopper mechanism

b) Watt mechanism

c) Peaucellier’s mechanism

d) Tchabichiff mechanism

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3. In a mechanism, usually one link is fixed. If the fixed link is changed in a kinematic chain, then relative motion of other links

a) will remain same

b) will change

c) will not occur

d) none of the mentioned

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4. A kinematic chain requires at least

a) 2 links and 3 turning pairs

b) 3 links and 4 turning pairs

c) 4 links and 4 turning pairs

d) 5 links and 4 turning pairs

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5. In a drag link quick return mechanism, the shortest link is always fixed. The sum of the shortest and longest link is

a) equal to sum of other two

b) greater than sum of other two

c) less than sum of other two

d) none of the mentioned

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6. The following is the inversion of slider crank chain mechanism

a) whitworth quick return mechanism

b) hand pump

c) oscillating cylinder engine

d) all of the mentioned

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7. Kinematic pairs are those which have

a) two elements held together mechanically

b) two elements having relative motion

c) two elements having Coroli’s component

d) none of the mentioned

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8. According to criterion of constraint by A.W. Klein

a) J + 1/2 H = 3/2L – 2

b) H + 1/2J = 2/3L – 2

c) J + 1/2H = 3/2L – 1

d) J + 3/2H = 1/2L – 2

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9. A quarternary joint is equivalent to

a) one binary joint

b) two binary joints

c) three binary joints

d) four binary joints

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10. A typewriter mechanism has 7 number of binary joints, six links and none of higher pairs. The mechanism is

a) kinematically sound

b) not sound

c) soundness would depend upon which link is kept fixed

d) none of the mentioned

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Hence, it is Kinematically sound.

## Set 5

1. In a coupling rod of a locomotive, each of the four pairs is a _____________ pair.

a) sliding

b) turning

c) rolling

d) screw

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2. Whitworth quick return motion mechanism consists of three turning pairs and one sliding pair.

a) True

b) False

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3. In a single slider crank chain

a) each of the four pairs is a turning pair

b) one is a turning pair and three are sliding pairs

c) two are turning pairs and two are sliding pairs

d) three are turning pairs and one is a sliding pair

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4. The mechanism consisting of three turning pairs and one sliding pair, is called a

a) single slider crank chain

b) whitworth quick return motion mechanism

c) crank and slotted lever quick return motion mechanism

d) all of the mentioned

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5. Which of the following is an inversion of a single slider crank chain?

a) Pendulum pump

b) Oscillating cylinder engine

c) Rotary internal combustion engine

d) All of the mentioned

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6. A point on a connecting link of a double slider crank mechanism traces a

a) straight line path

b) hyperbolic path

c) parabolic path

d) elliptical path

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7. The whitworth quick return motion mechanism is formed in a slider crank chain when the

a) coupler link is fixed

b) longest link is a fixed link

c) slider is a fixed link

d) smallest link is a fixed link

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8. Scotch yoke mechanism is used to generate

a) sine functions

b) square roots

c) logarithms

d) inversions

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9. Which of the following is an inversion of a double slider crank chain?

a) Oldham’s coupling

b) Elliptical trammel

c) Scotch yoke mechanism

d) All of the mentioned

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10. Whitworth quick return motion mechanism is an inversion of a double slider crank chain.

a) True

b) False

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11. A rotary internal combustion engine has __________ cylinders.

a) four

b) five

c) six

d) seven