McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Mechanical Engineering .
| 101. |
Isolation factor is twice the transmissibility ratio. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 102. |
The height of a Watt’s governor is |
| A. | directly proportional to speed |
| B. | directly proportional to (speed) |
| C. | inversely proportional to speed |
| D. | inversely proportional to (speed) |
| Answer» D. inversely proportional to (speed) | |
| 103. |
A Watt’s governor can work satisfactorily at speeds from |
| A. | 60 to 80 r.p.m |
| B. | 80 to 100 r.p.m |
| C. | 100 to 200 r.p.m |
| D. | 200 to 300 r.p.m |
| Answer» A. 60 to 80 r.p.m | |
| 104. |
The ratio of height of Porter governor to the height of Watt’s governor is |
| A. | m/m + m |
| B. | m/ m + m |
| C. | m + m/m |
| D. | m + m/m |
| Answer» C. m + m/m | |
| 105. |
When the sleeve of a porter governor moves upwards, the governor speed |
| A. | increases |
| B. | decreases |
| C. | remains unaffected |
| D. | first increases and then decreases |
| Answer» A. increases | |
| 106. |
When the sleeve of a Porter governor moves downwards, the governor speed |
| A. | increases |
| B. | decreases |
| C. | remains unaffected |
| D. | first increases and then decreases |
| Answer» B. decreases | |
| 107. |
In a Porter governor, the balls are attached to the extension of lower links. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 108. |
A Hartnell governor is a |
| A. | dead weight governor |
| B. | pendulum type governor |
| C. | spring loaded governor |
| D. | inertia governor |
| Answer» C. spring loaded governor | |
| 109. |
A Watt’s governor is a spring loaded governor. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 110. |
Which of the following is a pendulum type governor? |
| A. | watt’s governor |
| B. | porter governor |
| C. | hartnell governor |
| D. | none of the mentioned |
| Answer» A. watt’s governor | |
| 111. |
The axis of precession is to the plane in which the axis of spin is going to rotate. |
| A. | parallel |
| B. | perpendicular |
| C. | spiral |
| D. | none of the mentioned |
| Answer» B. perpendicular | |
| 112. |
A disc is a spinning with an angular velocity ω rad/s about the axis of spin. The couple applied to the disc causing precession will be |
| A. | 1/2 iω2 |
| B. | iω2 |
| C. | 1/2 iωωp |
| D. | iωωp |
| Answer» D. iωωp | |
| 113. |
The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the left. The effect of gyroscopic couple on the aeroplane will be |
| A. | to dip the nose and tail |
| B. | to raise the nose and tail |
| C. | to raise the nose and dip of the tail |
| D. | to dip the nose and raise the tail |
| Answer» C. to raise the nose and dip of the tail | |
| 114. |
The engine of an aeroplane rotates in clockwise direction when seen from the tail end and the aeroplane takes a turn to the right. The effect of gyroscopic couple on the aeroplane will be to dip the nose and raise the tail. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 115. |
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 or left, while it moves forward |
| C. | rolling of a complete ship side-ways |
| D. | none of the mentioned |
| Answer» B. turning of a complete ship in a curve towards right or left, while it moves forward | |
| 116. |
The rolling of a complete ship side-ways is known as pitching of a ship. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 117. |
The rotor of a ship rotates in clockwise direction when viewed from stern and the ship takes a right turn. The effect of gyroscopic couple acting on it will be to raise the stern and lower the bow. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 118. |
The pitching of a ship is assumed to take place with simple harmonic motion. |
| A. | true |
| B. | false |
| Answer» A. true | |
| 119. |
When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be |
| A. | to move the ship towards star-board |
| 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. to move the ship towards star-board | |
| 120. |
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. 0.302 kw | |
| 121. |
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. 56.548 kw | |
| 122. |
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 |
| Answer» D. 6 rad/s | |
| 123. |
6 x 10 = 6 rad/s. |
| A. | 2.055 m |
| B. | 3.055 m |
| C. | 4.055 m |
| D. | 5.055 m |
| Answer» A. 2.055 m | |
| 124. |
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. 10.69 knm | |
| 125. |
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. 61.16 up | |
| 126. |
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. turning of a complete ship in a curve towards right r left, while it moves forward | |
| 127. |
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. to move the ship towards starboard | |
| 128. |
24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m |
| A. | 1.35 m |
| B. | 1.42 m |
| C. | 1.48 m |
| D. | 1.50 m |
| Answer» B. 1.42 m | |
| 129. |
24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m |
| A. | 52.7 |
| B. | 49.5 |
| C. | 59.5 |
| D. | 56.5 |
| Answer» C. 59.5 | |
| 130. |
24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m |
| A. | 16.782 |
| B. | 17.824 |
| C. | 15.142 |
| D. | 17.161 |
| Answer» B. 17.824 | |
| 131. |
24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m |
| A. | 19000 n |
| B. | 19064 n |
| C. | 19032 n |
| D. | 20064 n |
| Answer» B. 19064 n | |
| 132. |
66. |
| A. | 4/3 |
| B. | 3/2 c) 3/4 |
| C. | d) 1 |
| Answer» B. 3/2 c) 3/4 | |
| 133. |
The starting torque of the steam engine is 1500 N-m and may be assumed constant, using this data find the angular acceleration of the flywheel in rad/s2. |
| A. | 0.4 |
| B. | 0.6 |
| C. | 0.3 |
| D. | 1.2 |
| Answer» B. 0.6 | |
| 134. |
1m then find the other balancing mass situated at a distance of 0.2m. |
| A. | 80 |
| B. | 40 |
| C. | 20 |
| D. | 10 |
| Answer» D. 10 | |
| 135. |
1m. One of the balancing mass is 30 Kg with RoR of 10cm. |
| A. | 70 |
| B. | 35 |
| C. | 20 |
| D. | 10 |
| Answer» B. 35 | |
| 136. |
2m and r=0.3m, l=1m, find l2. |
| A. | 0.26m |
| B. | 0.52m |
| C. | 1.04m |
| D. | 0.13m |
| Answer» A. 0.26m | |
| 137. |
2m and r=0.3m, l=1m, find l2. |
| A. | 0.26m |
| B. | 0.52m |
| C. | 1.04m |
| D. | 0.13m |
| Answer» D. 0.13m | |
| 138. |
3m, l=1m, find l2 = 0.5m, find r1 in m. |
| A. | 1.5 |
| B. | 0.75 |
| C. | 3 |
| D. | 6 |
| Answer» B. 0.75 | |
| 139. |
2m, 0.15m, 0.25m and 0.3m. Angles between consecutive masses = 45, 75 and 135 degrees. |
| A. | 116 |
| B. | 58 |
| C. | 232 |
| D. | 140 |
| Answer» A. 116 | |
| 140. |
2m, 0.15m, 0.25m and 0.3m. Angles between consecutive masses = 45, 75 and 135 degrees. |
| A. | 201.48 |
| B. | 200.32 |
| C. | 210.34 |
| D. | 202.88 |
| Answer» A. 201.48 | |
| 141. |
The system is automatically statically balanced. |
| A. | 1, 2, 3 and 4 |
| B. | 1, 2, and 3 only |
| C. | 2, 3 and 4 only |
| D. | 1, 3 and 4 only |
| Answer» D. 1, 3 and 4 only | |
| 142. |
6m c = 2/3 |
| A. | 6978 |
| B. | 7574 |
| C. | 6568 |
| D. | 7374 |
| Answer» A. 6978 | |
| 143. |
6m c = 2/3 |
| A. | 13956 |
| B. | 17574 |
| C. | 16568 |
| D. | 17374 |
| Answer» A. 13956 | |
| 144. |
6m c = 2/3 |
| A. | -6978 |
| B. | -7574 |
| C. | -6568 |
| D. | -7374 |
| Answer» A. -6978 | |
| 145. |
5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons? |
| A. | 736 |
| B. | 836 |
| C. | 936 |
| D. | 636 |
| Answer» B. 836 | |
| 146. |
5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons? |
| A. | 7172 |
| B. | 1672 |
| C. | 1122 |
| D. | 1272 |
| Answer» B. 1672 | |
| 147. |
5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in kN? |
| A. | 2.238 |
| B. | 2.508 |
| C. | 2.754 |
| D. | 2.908 |
| Answer» B. 2.508 | |
| 148. |
6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor (d ) and damped natural frequency (fn), respectively, are |
| A. | 0.471 and 1.19 hz |
| B. | 0.471 and 7.48 hz |
| C. | 0.666 and 1.35 hz |
| D. | 0.666 and 8.50 hz |
| Answer» A. 0.471 and 1.19 hz | |
| 149. |
1 mm. |
| A. | 0.1 |
| B. | 0.2 |
| C. | 0.5 |
| D. | 0.6 |
| Answer» D. 0.6 | |
| 150. |
Calculate the natural longitudinal frequency in Hz. |
| A. | 575 |
| B. | 625 |
| C. | 525 |
| D. | 550 |
| Answer» A. 575 | |
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