This set of Machine Dynamics MCQs focuses on “Wilson-Hartnell and Pickering Governor”.

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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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/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

Explanation: 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/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

Explanation: 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/ω

^{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

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: Power = Tω

= Frω = Fv

F = T

_{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

Explanation: I = mk

^{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

Explanation: I = 2/5 mr

^{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

Explanation: Force required to punch one hole = area shared x ultimate shear strength = пdt x S

_{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.

**Sanfoundry Global Education & Learning Series – Machine Dynamics.**

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