## Machine Design MCQ Set 1

1. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required to drive the pump.

a) 0.102 kW

b) 0.202 kW

c) 0.302 kW

d) 0.402 kW

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2. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?

a) 141.36 W

b) 241.36 W

c) 341.36 W

d) 441.36 W

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3. Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

a) 46.548 kW

b) 56.548 kW

c) 66.548 kW

d) 76.548 kW

### View Answer

_{1}– T

_{2}= 2200 – 1000 = 1200N v = пDN/60 = п x 600 x 1500/1000 x 60 = 47.12 m/s P = 1200 x 47.12/1000 = 56.548 kW.

4. The flywheel of an engine has a mass of 200 kg and radius of gyration equal to 1 m. The average torque on the flywheel is 1200 Nm. Find the angular acceleration of flywheel and the angular speed after 10 seconds starting from rest.

a) 3 rad/s

b) 4 rad/s

c) 5 rad/s

d) 6 rad/s

### View Answer

^{2}= 2000 kgm

^{2}T = Iα α = T/I = 1200/2000 = 0.6 rad/s

^{2}ω = ω

_{0}+ αt = 0 + 0.6 x 10 = 6 rad/s.

5. Calculate the moment of inertia and radius of gyration of a solid sphere of mass 10 kg and diameter 6.5m about its centroidal axis.

a) 2.055 m

b) 3.055 m

c) 4.055 m

d) 5.055 m

### View Answer

^{2}= 49 kgm

^{2}k = 0.6325 x 3.25 = 2.055 m.

6. Calculate the work done per minute by a punch tool making 20 working strokes per min when a 30 mm diametre hole is punched in 5 mm thick plate with ultimate shear strength og 450 Mpa in each stroke.

a) 10.69 kNm

b) 20.69 kNm

c) 30.69 kNm

d) 40.69 kNm

### View Answer

_{s}S

_{s}= п x 30 x 5 x 450/1000 = 212.05 kN Work done/min = Average force x Thickness of plate x No. of holes/min = 212.05/2 x 5/1000 x 20 = 10.69 kNm.

7. In latitude 25.0 S, SA (spin axis) of a FG (free gyro) is in position S40E and horizontal. Find the tilt after 6 hours.

a) 61.16 up

b) 61.16 down

c) 51.15 up

d) none of the mentioned

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8. The steering of a ship means

a) movement of a complete ship up and down in vertical plane about transverse axis

b) turning of a complete ship in a curve towards right r left, while it moves forward

c) rolling of a complete ship sideways

d) none of the mentioned

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9. When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be

a) to move the ship towards starboard

b) to move the ship towards port side

c) to raise the bow and lower the stern

d) to raise the stern and lower the bow

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## Machine Design MCQ Set 2

1. The power of a governor is the work done at

a) the governor balls for change of speed

b) the sleeve for zero change of speed

c) the sleeve for a given rate of change of change

d) each governor ball for given percentage change of speed

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2. In a governor, if the equilibrium speed is constant for all radii of rotation of balls, the governor is said to be

a) Stable

b) unstable

c) inertial

d) isochronous

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3. A governor is said to be isochronous when the equilibrium speed is

a) variable for different radii of rotation of governor balls

b) constant for all radii of rotation of the balls within the working range

c) constant for particular radii of rotation of governor balls

d) constant for only one radius of rotation of governor balls

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4. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the height of the governor, when it is running at 75 r.p.m.

a) 139 mm

b) 149 mm

c) 159 mm

d) 169 mm

### View Answer

Height of the governor, h = g/ω^{2} = 9.81/2.5п^{2} = 0.159m = 159 mm.

5. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the sped of the governor, when the balls rise by 20 mm.

a) 80.2 r.p.m.

b) 90.2 r.p.m.

c) 100.2 r.p.m.

d) 110.2 r.p.m.

### View Answer

Height of the governor, h = g/ω^{2} = 9.81/2.5п^{2} = 0.159m = 159 mm

Speed of the governor when balls rise by 20 mm
The height of the governor h reduces to h_{1} = 159 – 20 = 139 mm = 0.139 m
Hence, the angular velocity ω_{1} corresponding to height h_{1},
ω_{1}^{2} = g/h_{1}= 9.81/0.139 = 70.5
ω_{1} = 8.4 rad/s

Speed in r.p.m., N_{1} = 60ω_{1}/2п = 60 x 8.4/2п = 80.2 r.p.m.

6. A centrifugal Watt governor is fitted with two balls, each of mass 2.5 kg. Find the sped of the governor, when the balls fall by 20 mm.

a) 60 r.p.m.

b) 70 r.p.m.

c) 70.2 r.p.m.

d) 80 r.p.m.

### View Answer

^{2}= 9.81/2.5п

^{2}= 0.159m = 159 mm

Speed of the governor when balls fall by 20 mm
The height of the governor h increases to h_{2} = 159 + 20 = 179 mm = 0.179 m
Hence, the angular velocity ω_{2} corresponding to height h_{2},
ω_{2}^{2} = g/h_{2}= 9.81/0.179 = 54.7
ω_{2} = 7.4 rad/s
Speed in r.p.m., N_{2} = 60ω_{2}/2п = 60 x 7.4/2п = 70.7 r.p.m.

7. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required todrive the pump.

a) 0.102 kW

b) 0.202 kW

c) 0.302 kW

d) 0.402 kW

### View Answer

8. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?

a) 141.36 W

b) 241.36 W

c) 341.36 W

d) 441.36 W

### View Answer

9. Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.

a) 46.548 kW

b) 56.548 kW

c) 66.548 kW

d) 76.548 kW

### View Answer

_{1}– T

_{2}= 2200 – 1000 = 1200N v = пDN/60 = п x 600 x 1500/1000 x 60 = 47.12 m/s P = 1200 x 47.12/1000 = 56.548 kW.

10. The flywheel of an engine has a mass of 200 kg and radius of gyration equal to 1 m. The average torque on the flywheel is 1200 Nm. Find the angular acceleration of flywheel and the angular speed after 10 seconds starting from rest.

a) 3 rad/s

b) 4 rad/s

c) 5 rad/s

d) 6 rad/s

### View Answer

^{2}= 2000 kgm

^{2}T = Iα α = T/I = 1200/2000 = 0.6 rad/s

^{2}ω = ω

_{0}+ αt = 0 + 0.6 x 10 = 6 rad/s.

11. Calculate the moment of inertia and radius of gyration of a solid sphere of mass 10 kg and diameter 6.5m about its centroidal axis.

a) 2.055 m

b) 3.055 m

c) 4.055 m

d) 5.055 m

### View Answer

^{2}= 49 kgm

^{2}k = 0.6325 x 3.25 = 2.055 m.

12. Calculate the work done per minute by a punch tool making 20 working strokes per min when a 30 mm diametre hole is punched in 5 mm thick plate with ultimate shear strength og 450 Mpa in each stroke.

a) 10.69 kNm

b) 20.69 kNm

c) 30.69 kNm

d) 40.69 kNm

### View Answer

_{s}S

_{s}= п x 30 x 5 x 450/1000 = 212.05 kN Work done/min = Average force x Thickness of plate x No. of holes/min = 212.05/2 x 5/1000 x 20 = 10.69 kNm.

## Machine Design MCQ Set 3

1. In a steam engine, the distance by which the outer edge of the D-slide valve overlaps the steam port is called

a) lead

b) steam lap

c) exhaust lap

d) none of the mentioned

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2. In a steam engine, the distance by which the inner edge of the D-slide valve overlaps the steam port is called lead.

a) True

b) False

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3. The displacement of the D-slide valve is ______________ the steam lap by a distance known as lead of the valve.

a) greater than

b) less than

c) equal to

d) none of the mentioned

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4. The D-slide valve is also known as

a) inside admission valve

b) outside admission valve

c) piston slide valve

d) none of the mentioned

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5. The piston slide valve is an inside admission valve.

a) True

b) False

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6. The distance by which the _____________ of the D-slide valve overlaps the steam port is called exhaust lap.

a) inner edge

b) outer edge

c) centre edge

d) none of the mentioned

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7. Which of the following statement is correct?

a) The power absorbed in operating the piston valve is less than D-slide valve.

b) The wear of the piston valve is less than the wear of the D-slide valve.

c) The D-slide valve is also called outside admission valve.

d) all of the mentioned

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8. The distance moved by the valve from one end to the other end is called valve travel.

a) True

b) False

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9. The throw of the eccentric is equal of the valve travel.

a) True

b) False

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10. In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve is by increasing the angle of advance of the eccentric while the throw of the eccentric, steam lap and exhaust lap are kept constant. The method will

a) cause withdrawing or throttling of steam

b) reduce length of effective stroke of piston

c) reduce maximum opening opening of port to steam

d) all of the mentioned

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## Machine Design MCQ Set 4

1. The primary unbalanced force is maximum when the angle of inclination of the crank with the line of stroke is

a) 0°

b) 90°

c) 180°

d) 360°

### View Answer

_{P(max)}= m⋅ω

^{2}⋅r

2. The partial balancing means

a) balancing partially the revolving masses

b) balancing partially the reciprocating masses

c) best balancing of engines

d) all of the mentioned

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3. In order to facilitate the starting of locomotive in any position, the cranks of a locomotive, with two cylinders, are placed at ______________ to each other.

a) 45°

b) 90°

c) 120°

d) 180°

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4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to

a) minimise the effect of primary forces

b) minimise the effect of secondary forces

c) have perfect balancing

d) start the locomotive quickly

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5. If c be the fraction of the reciprocating parts of mass m to be balanced per cyclinder of a steam locomotive with crank radius r, angular speed ω, distance between centre lines of two cylinders a, then the magnitude of the maximum swaying couple is given by

a) 1 – c / 2 mrω^{2}a

b) 1 – c / √2mrω^{2}a

c) √2(1 – c)mrω^{2}a

d) none of the mentioned

### View Answer

^{2}a

6. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to

a) 45° and 135°

b) 90° and 135°

c) 135° and 225°

d) 45° and 225°

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7. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to

a) 90° and 225°

b) 135° and 180°

c) 180° and 225°

d) 135° and 315°

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8. The swaying couple is due to the

a) primary unbalanced force

b) secondary unbalanced force

c) two cylinders of locomotive

d) partial balancing

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9. In a locomotive, the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke, is known as

a) tractive force

b) swaying couple

c) hammer blow

d) none of the mentioned

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10. The effect of hammer blow in a locomotive can be reduced by

a) decreasing the speed

b) using two or three pairs of wheels coupled together

c) balancing whole of the reciprocating parts

d) both (a) and (b)