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## 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

Answer: c [Reason:] Work to be done = wQHg w = 1 Kg/litre Q = 50 litres per min H = 20 + 5 + 5 = 30m Work output/min = 1 x 50 x 30 x 9.81 Nm/min Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60 = 0.302 kW.

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

Answer: a [Reason:] Work per revolution = Tϴ = 15 x 2п = 94.24 Nm per revolution P = Work/sec = 94.24 x 1.5 = 141.36 W.

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

Answer: b [Reason:] Power = Tω = Frω = Fv F = T1 – T2 = 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.

Answer: d [Reason:] I = mk2 = 2000 kgm2 T = Iα α = T/I = 1200/2000 = 0.6 rad/s2 ω = ω0 + 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

Answer: a [Reason:] I = 2/5 mr2 = 49 kgm2 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

Answer: a [Reason:] Force required to punch one hole = area shared x ultimate shear strength = пdt x Ss Ss = 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

Answer: a [Reason:] PZ= 65, ZX = 90, tilt after 6 hours = 90 – ZY =? In triangle PZX, Cos PX = Cos Z Sin PZ Cos PX = Cos 40 Sin 65 Cos PX = 0.694272 PX = PY = 46.031 Again, in triangle PZX, Cos P = – Cot PZ Cot PX Cos P = – Cot 65 Cot 46.031 P or angle ZPX = 116.732 Angle XPY = 6 hours = 90. Thus, angle ZPY = 26.732 In triangle PZY, Cos ZY = Cos P Sin PZ Sin PY + Cos PZ Cos PY Cos ZY = Cos 26.732 Sin 65 Sin 46.031 + Cos 65 Cos 46.031 ZY = 28.839 and tilt = 61.16 up.

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

Answer: b [Reason:] Steering means is to turn the vehicle to either left or right.

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

Answer: a [Reason:] When the rotor rotates in the anticlockwise direction, when viewed from the stern and the ship is steering to the left, then the effect of reactive gyroscopic couple will be to lower the bow and raise the stern. When the ship is steering to the right under similar conditions, then the effect of reactive gyroscopic couple will be to raise the bow and lower the stern.

## 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

Answer: c [Reason:] Power of Governor: The work done by the governor on the sleeve to its equilibrium position for the fractional change in speed of governor is known as power of governor. It is actually a work done. Power = Main force × Sleeve movement.

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

Answer: d [Reason:] The governor is said to be Isochronous if the equilibrium speed is constant for all radii of rotation of balls.

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

Answer: b [Reason:] Isochronism in governor means constant equilibrium speed for all the radii of rotation.

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

Answer: c [Reason:] Mass of flyballs(m) = 2.5 kg and N speed of governor = 75 r.p.m. So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s

Height of the governor, h = g/ω2 = 9.81/2.5п2 = 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.

Answer: a [Reason:] Mass of flyballs(m) = 2.5 kg and N speed of governor = 75 r.p.m. So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s

Height of the governor, h = g/ω2 = 9.81/2.5п2 = 0.159m = 159 mm

Speed of the governor when balls rise by 20 mm The height of the governor h reduces to h1 = 159 – 20 = 139 mm = 0.139 m Hence, the angular velocity ω1 corresponding to height h1, ω12 = g/h1= 9.81/0.139 = 70.5 ω1 = 8.4 rad/s

Speed in r.p.m., N1 = 60ω1/2п = 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.

Answer: c [Reason:] Mass of flyballs(m) = 2.5 kg and N speed of governor = 75 r.p.m. So the angular velocity of the governor ω = 2пN/60 = 2п75/60 = 2.5п rad/s Height of the governor, h = g/ω2 = 9.81/2.5п2 = 0.159m = 159 mm

Speed of the governor when balls fall by 20 mm The height of the governor h increases to h2 = 159 + 20 = 179 mm = 0.179 m Hence, the angular velocity ω2 corresponding to height h2, ω22 = g/h2= 9.81/0.179 = 54.7 ω2 = 7.4 rad/s Speed in r.p.m., N2 = 60ω2/2п = 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

Answer: c [Reason:] Work to be done = wQHg w = 1 Kg/litre Q = 50 litres per min H = 20 + 5 + 5 = 30m Work output/min = 1 x 50 x 30 x 9.81 Nm/min Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60 = 0.302 kW.

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

Answer: a [Reason:] Work per revolution = Tϴ = 15 x 2п = 94.24 Nm per revolution P = Work/sec = 94.24 x 1.5 = 141.36 W.

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

Answer: b [Reason:] Power = Tω = Frω = Fv F = T1 – T2 = 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.

Answer: d [Reason:] I = mk2 = 2000 kgm2 T = Iα α = T/I = 1200/2000 = 0.6 rad/s2 ω = ω0 + 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

Answer: a [Reason:] I = 2/5 mr2 = 49 kgm2 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

Answer: a [Reason:] Force required to punch one hole = area shared x ultimate shear strength = пdt x Ss Ss = 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
b) steam lap
c) exhaust lap
d) none of the mentioned

Answer: b [Reason:] Steam lap is the distance by which the outer edge of the D-slide valve overlaps the steam port.

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

Answer: b [Reason:] In actual practise, the displacement of the D-slide valve is greater than the steam lap by a distance known as lead of the valve.

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

Answer: a [Reason:] In actual practise, the displacement of the D-slide valve is greater than the steam lap by a distance known as lead of the valve.

4. The D-slide valve is also known as
c) piston slide valve
d) none of the mentioned

Answer: b [Reason:] Since the steam is admitted from outside the steam chest, therefore, D-slide valve is also known as outside admission valve.

5. The piston slide valve is an inside admission valve.
a) True
b) False

Answer: a [Reason:] Since the steam enters from the inside of the two pistons, therefore, the piston valve is also known as inside admission valve.

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

Answer: a [Reason:] Steam lap or exhaust lap is the distance by which the outer edge of the D-slide valve overlaps the steam port.

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

Answer: d [Reason:] All the statements are correct.

8. The distance moved by the valve from one end to the other end is called valve travel.
a) True
b) False

Answer: a [Reason:] Valve travel is the distance moved by the valve from one end to the other end.

9. The throw of the eccentric is equal of the valve travel.
a) True
b) False

Answer: b [Reason:] The throw of the eccentric is equal to half of the valve travel.

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

Answer: b [Reason:] 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 reduce length of effective stroke of piston. In a steam engine, one method of obtaining the earlier cut-off with a simple slide valve may be obtained by increasing the angle of advance of the eccentric but reducing the throw of the eccentric and keeping the steam lap and exhaust lap constant. The method will cause withdrawing or throttling of steam.

## 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°

Answer: c [Reason:] The primary unbalanced force is maximum, when θ = 0° or 180°. Thus, the primary force is maximum twice in one revolution of the crank. The maximum primary unbalanced force is given by FP(max) = m⋅ω2⋅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

Answer: b [Reason:] To balance the reciprocating masses partially is known as partial balancing.

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°

Answer: b [Reason:] Due to the partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke.

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

Answer: b [Reason:] The secondary unbalanced force is maximum, when θ = 0°, 90°,180° and 360°. Thus, the secondary force is maximum four times in one revolution of the crank. Keeping large distance between connecting rod and crank, minimises the effect of secondary forces.

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 mrω2a
b) 1 – c / √2mrω2a
c) √2(1 – c)mrω2a
d) none of the mentioned

Answer: b [Reason:] The swaying couple is maximum or minimum when (cosθ+sinθ) is maximum or minimum. For (cosθ+sinθ) to be maximum or minimum ∴ tanθ =1 or θ = 45° or 225° Maximum value of the swaying couple = 1 – c / √2mrω2a

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°

Answer: d [Reason:] The swaying couple is maximum or minimum when (cosθ+sinθ) is maximum or minimum. For (cosθ+sinθ) to be maximum or minimum ∴ tanθ =1 or θ = 45° or 225°

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°

Answer: d [Reason:] The tractive force is maximum or minimum when (cos θ – sin θ ) is maximum or minimum. For (cos θ – sin θ ) to be maximum or minimum, ∴ tanθ = −1 or θ =135° or 315°

8. The swaying couple is due to the
a) primary unbalanced force
b) secondary unbalanced force
c) two cylinders of locomotive
d) partial balancing

Answer: a [Reason:] The unbalanced forces along the line of stroke for the two cylinders constitute a couple. This couple has swaying effect about a vertical axis, and tends to sway the engine alternately in clockwise and anticlockwise directions. Hence the couple is known as swaying couple.

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

Answer: c [Reason:] The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. The effect of hammer blow is to cause the variation in pressure between the wheel and the rail.

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)