# Multiple choice question for engineering

## Set 1

1. A pantograph consists of

a) 4 links

b) 6 links

c) 8 links

d) 10 links

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2. Which of these mechanisms gives an approximately straight line?

a) hart

b) watt

c) peaucellier

d) tchebicheff

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3. Which of these mechanism has six links?

a) tchebicheff

b) hart

c) peaucellier

d) watt

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4. Which of these mechanisms use two identical mechanisms?

a) hart

b) watt

c) peaucellier

d) none of the mentioned

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5. The Davis steering gear is not used because

a) it has turning pairs

b) it is costly

c) it has sliding pair

d) it does not fulfil the condition of correct gearing

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6. The Davis steering gear fulfils the condition of correct steering at

a) two positions

b) three positions

c) all positions

d) one positions

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7. The Ackermann steering gear fulfils the condition of correct steering at

a) no position

b) one position

c) three positions

d) all positions

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8. A Hooke’s joint is used to join two shafts which are

a) aligned

b) intersecting

c) parallel

d) none of the mentioned

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9. The maximum velocity of the driven shaft of a Hooke’s joint is

a) ω_{1} cosα

b) ω_{1}/cosα

c) ω_{1} sinα

d) ω_{1}/sinα

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_{1(max)}= ωcosα/ 1 – sin

^{2}α = ωcosα/cos

^{2}α = ω/cosα.

10. The maximum velocity of the driven shaft of a Hooke’s joint is at ϴ equal to

a) 0^{0} and 180^{0}

b) 90^{0} and 270^{0}

c) 90^{0} and 180^{0}

d) 180^{0} and 270^{0}

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^{0},180

^{0}and 360

^{0}.

## Set 2

1. The direction of Corioli’s component of acceleration is the direction

a) of relative velocity vector for the two coincident points rotated by 90^{0} in the direction of the angular velocity of the rotation of the link

b) along the centripetal acceleration

c) along tangential acceleration

d) along perpendicular to angular velocity

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2. In a shaper mechanism, the Corioli’s component of acceleration will

a) not exist

b) exist

c) depend on position of crank

d) none of the mentioned

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3. The magnitude of tangential acceleration is equal to

a) velocity^{2} x crank radius

b) velocity^{2}/ crank radius

c) (velocity/ crank radius)^{2}

d) velocity x crank radius^{2}

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^{2}/ crank radius. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

4. Tangential acceleration direction is

a) along the angular velocity

b) opposite to angular velocity

c) perpendicular to angular velocity

d) all of the mentioned

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5. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is

a) Vω

b) 2Vω

c) Vω/2

d) 2V/ω

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^{2}/ crank radius. The magnitude of the Corioli’s component of acceleration of a slider moving at velocity V on a link rotating at angular speed ω is 2Vω.

6. In a rotary engine the angular velocity of the cylinder center line is 25 rad/sec and the relative velocity of a point on the cylinder center line w.r.t. cylinder is 10 m/sec. Corioli’s acceleration will be

a) 500m/sec^{2}

b) 250m/sec^{2}

c) 1000m/sec^{2}

d) 2000m/sec^{2}

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^{2}.

7. Corioli’s component is encountered in

a) quick return mechanism of shaper

b) four bar chain mechanism

c) slider crank mechanism

d) all of the mentioned

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8. Klein’s construction gives a graphical construction for

a) slider-crank mechanism

b) velocity polygon

c) acceleration polygon

d) none of the mentioned

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9. The velocity of a slider with reference to a fixed point about which a bar is rotating and slider sliding on the bar will be

a) parallel to bar

b) perpendicular to bar

c) both of the mentioned

d) none of the mentioned

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10. Klien’s construction can be used to determine acceleration of various parts when the crank is at

a) inner dead center

b) outer dead center

c) right angles to the link of the stroke

d) all of the mentioned

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11. The number of dead centers in a crank driven slider crank mechanism are

a) 0

b) 2

c) 4

d) 6

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12. Corioli’s component acts

a) perpendicular to sliding surfaces

b) along sliding surfaces

c) both of the mentioned

d) all of the mentioned

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13. The sense of Coriol’s component is such that it

a) leads the sliding velocity vector by 90^{0}

b) lags the sliding velocity vector by 90^{0}

c) is along the sliding velocity vector by 90^{0}

d) leads the sliding velocity vector by 180^{0}

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14. Klien’s construction can be used when

a) crank has a uniform angular velocity

b) crank has non-uniform velocity

c) crank has uniform angular acceleration

d) crank has uniform angular velocity and angular acceleration

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15. Klein’s construction is useful to determine

a) velocity of various parts

b) acceleration of various parts

c) displacement of various parts

d) angular acceleration of various parts

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## Set 3

1. The total number of instantaneous centres for a mechanism consisting of n links are

a) n/2

b) n

c) n-1

d) n(n-1)/2

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2. According to Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on

a) straight line

b) parabolic curve

c) triangle

d) rectangle

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3. Which of the following property of the instantaneous center is correct?

a) A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.

b) The two rigid links have no linear velocity relative to each other at the instantaneous centre.

c) The velocity of the instantaneous centre relative to any third link is same whether the instantaneous centre is regarded as a point on the first link or on the second rigid link.

d) all of the mentioned

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4. The magnitude of velocities of the points on a rigid link is

a) directly proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.

b) directly proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.

c) inversely proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.

d) inversely proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.

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5. In a mechanism, the fixed instantaneous centres are those centres which

a) remain in the same place for all configurations of the mechanism

b) vary with the configuration of the mechanism

c) moves as the mechanism moves, but joints are of permanent nature

d) none of the mentioned

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6. The instantaneous centres, which moves as the mechanism moves but joints are of permanent nature, are called permanent instantaneous centres.

a) True

b) False

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7. The instantaneous centres which vary with the configuration of mechanism, are called

a) permanent instantaneous centres

b) fixed instantaneous centres

c) neither fixed nor permanent instantaneous centres

d) none of the mentioned

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8. When two links are connected by a pin joint, their instantaneous centre lies

a) on their point of contact

b) at the centre of curvature

c) at the centre of circle

d) at the pin joint

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9. The two links are said to have a pure rolling contact, when their instantaneous centre __________ on their point of contact.

a) lies

b) does not lie

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10. When a slider moves on a fixed link having ____________ their instantaneous center lies at infinity.

a) straight surface

b) curved surface

c) oval surface

d) none of the mentioned

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11. When a slider moves on a fixed link having curved surface, their instantaneous centre lies

a) on their point of contact

b) at the centre of curvature

c) at the centre of circle

d) at the pin joint

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12. A slider moving on a fixed link having constant radius of curvature will have its instantaneous centre at the center of the circle.

a) True

b) False

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13. The instantaneous center of a rigid thin disc rolling on a plane rigid surface is located at

a) the centre of the disc

b) the point of contact

c) an infinite distance on the plane surface

d) the point on the circumference situated vertically opposite to the contact point

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## Set 4

1. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the acceleration.

a) 0.3 m/s^{2}

b) 0.4 m/s^{2}

c) 0.5 m/s^{2}

d) 0.6 m/s^{2}

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^{2}= u

^{2}+ 2as or, 20

^{2}= 0 + 2a x 500 = 1000a or, a = 20

^{2}/1000 = 0.4 m/s

^{2}

2. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. Calculate the time taken to attain the speed.

a) 50 s

b) 60 s

c) 70 s

d) 80 s

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^{2}= u

^{2}+ 2as or, 20

^{2}= 0 + 2a x 500 = 1000a or, a = 20

^{2}/1000 = 0.4 m/s

^{2}

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

3. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. If a further acceleration raises the speed to 90 km. p.h. in 10 seconds, find this acceleration and the further distance moved.

a) 0.3 m/s^{2}

b) 0.4 m/s^{2}

c) 0.5 m/s^{2}

d) 0.6 m/s^{2}

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^{2}= u

^{2}+ 2as or, 20

^{2}= 0 + 2a x 500 = 1000a or, a = 20

^{2}/1000 = 0.4 m/s

^{2}

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds.
Given : Initial velocity, u = 72 km.p.h. = 20 m/s ;
Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s
Let a = Acceleration of the car.
We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m^{2}

4. A car starts from rest and accelerates uniformly to a speed of 72 km. p.h. over a distance of 500 m. A further acceleration raises the speed to 90 km. p.h. in 10 seconds.The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds. Find the distance travelled during braking.

a) 200 m

b) 300 m

c) 225 m

d) 335 m

### View Answer

^{2}= u

^{2}+ 2as or, 20

^{2}= 0 + 2a x 500 = 1000a or, a = 20

^{2}/1000 = 0.4 m/s

^{2}

Let t = Time taken by the car to attain the speed. We know that v = u + a.t ∴ 20 = 0 + 0.4 × t or t = 20/0.4 = 50 s

Now consider the motion of the car from 72 km.p.h. to 90 km.p.h. in 10 seconds. Given : Initial velocity, u = 72 km.p.h. = 20 m/s ; Final velocity, v = 96 km.p.h. = 25 m/s ; t = 10 s Let a = Acceleration of the car.

We know that v = u + a.t
25 = 20 + a × 10 or a = (25 – 20)/10 = 0.5 m^{2}

We know that distance moved by the car,
s = ut + 1/2 at^{2} = 20 x 10 + 1/2 0.5(10)^{2} = 225 m

5. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?

a) 10.475 rad/s^{2}

b) 11.475 rad/s^{2}

c) 12.475 rad/s^{2}

d) 13.475 rad/s^{2}

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_{0}= 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/s

^{2}. We know that ω = ω

_{0}+ α.t or 209.5 = 0 + α × 20 ∴ α = 209.5 / 20 = 10.475 rad/s

^{2}

6. 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) 333.4

b) 444.4

c) 555.4

d) 666.4

a) 10.475 rad/s^{2}

b) 11.475 rad/s^{2}

c) 12.475 rad/s^{2}

d) 13.475 rad/s^{2}

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_{0}= 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/s

^{2}. We know that ω = ω

_{0}+ α.t or 209.5 = 0 + α × 20 ∴ α = 209.5 / 20 = 10.475 rad/s

^{2}

We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s),
θ = (ω_{0} + 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

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?

a) 288.6 m/s

b) 388.6 m/s

c) 488.6 m/s

d) 188.6 m/s

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_{0}= 1200 r.p.m. or ω

_{0}= 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,
v_{0} = r . ω_{0} = 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 the end of the interval ?

a) 235.5 m/s

b) 335.5 m/s

c) 435.5 m/s

d) 535.5 m/s

### View Answer

_{0}= 1200 r.p.m. or ω

_{0}= 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,
v_{0} = r . ω_{0} = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v_{5} = r . ω = 1.5 × 157 = 235.5 m/s

9. 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 normal component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

a) 2.7 m/s^{2}

b) 3.7 m/s^{2}

c) 4.7 m/s^{2}

d) 5.7 m/s^{2}

### View Answer

_{0}= 1200 r.p.m. or ω

_{0}= 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,
v_{0} = r . ω_{0} = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v_{5} = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration.
We know that ω = ω_{0}+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s^{2}

Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s^{2}

10. 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 tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

a) 18287 m/s^{2}

b) 18387 m/s^{2}

c) 18487 m/s^{2}

d) 18587 m/s^{2}

### View Answer

_{0}= 1200 r.p.m. or ω

_{0}= 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

_{0} = r . ω_{0} = 1.5 × 125.7 = 188.6 m/s

Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v_{5} = r . ω = 1.5 × 157 = 235.5 m/s

Let α = Constant angular acceleration.
We know that ω = ω_{0}+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s^{2}

Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
∴ Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s^{2}

Radial acceleration = ω^{2} . r = (157)^{2} 0.75 = 18 487 m/s^{2}

## Set 5

1. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate mechanical advantage.

a) 6

b) 7

c) 8

d) 9

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2. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate velocity ratio.

a) 6

b) 7

c) 8

d) 9

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3. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate efficiency at this load.

a) 44.44%

b) 55.55%

c) 66.66%

d) 77.77%

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4. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate effect of friction.

a) 10 N

b) 20 N

c) 30 N

d) 40 N

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_{P}= P – W/V.R. = 60 – 360/9 = 20 N.

5. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine law of the machine.

a) P = 1/10W +30

b) P = 1/20W +30

c) P = 1/30W +30

d) P = 1/40W +30

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Subtracting (i) and (ii), we get 1 = 10 m or, m = 1/10 Putting this value in equation (i), we get 50 = 200 x 1/10 + C C = 30

Hence, the machine follows the laws P = 1/10W +30.

6. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 200 N.

a) 10 %

b) 15 %

c) 20 %

d) 25 %

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7. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 300 N.

a) 10 %

b) 15 %

c) 20 %

d) 25 %

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8. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.

a) 30 N

b) 35 N

c) 40 N

d) 45 N

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_{P}= P – W/ V.R. = 50 – 200/20 = 40 N.

9. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.

a) 30 N

b) 35 N

c) 40 N

d) 45 N

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_{P}= P – W/ V.R. = 60 – 300/20 = 45 N.

10. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine maximum efficiency which can be expected from this machine.

a) 30 %

b) 40 %

c) 50 %

d) 60 %