McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Mechanical Engineering .
| 101. |
A 3-phase induction motor with its rotor blocked behaves similar to a |
| A. | transformer under short circuit of secondary terminals |
| B. | transformer under open circuit of secondary |
| C. | synchronous motor under slip test |
| D. | synchronous motor under open circuit |
| Answer» A. transformer under short circuit of secondary terminals | |
| Explanation: it is analogous to transformer under the shorted terminals of the secondary. | |
| 102. |
No load current in induction motor is 10- 20% of full load current and the no load current of transformer is 2-6%. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the air gap in induction motor is more prominent than transformer and so the flux requirement will also be more. | |
| 103. |
no load estimated current for the machine? |
| A. | 12 a |
| B. | 20 a |
| C. | 30 a |
| D. | 5 a |
| Answer» A. 12 a | |
| Explanation: no load current in induction motor is 10-20% of full load current. | |
| 104. |
The no load current of the transformer is very less due to |
| A. | mutual flux having low reluctance iron core |
| B. | mutual flux having high reluctance iron core |
| C. | leakage flux having low reluctance iron core |
| D. | leakage flux having high reluctance iron core |
| Answer» A. mutual flux having low reluctance iron core | |
| Explanation: it is due to the low reluctance path, the flux requirement is low in the transformer. | |
| 105. |
The no load current of the induction motor is high due to |
| A. | long and high reluctance path between stator and rotor |
| B. | mutual flux having moderate reluctance path between stator and rotor |
| C. | leakage flux having low reluctance iron core |
| D. | leakage flux having high reluctance iron core |
| Answer» A. long and high reluctance path between stator and rotor | |
| Explanation: it is due to the higher | |
| 106. |
At no load induction motor has possible power factor as |
| A. | 0.2 |
| B. | 0.5 |
| C. | 0.65 |
| D. | 0 |
| Answer» A. 0.2 | |
| Explanation: at no load the lower power factor is low and lagging in nature. | |
| 107. |
Mechanically air gaps in induction motor are kept very low to avoid |
| A. | lower power factor |
| B. | lagging nature |
| C. | magnetizing current |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: air gap is kept lower to avoid the low value power factor, lagging behaviour and to reduce the magnetizing current. | |
| 108. |
The low no load power factor |
| A. | reduces full load operating pf |
| B. | increases full load operating pf |
| C. | reduces full load excitation voltage |
| D. | increases full load excitation voltage |
| Answer» A. reduces full load operating pf | |
| Explanation: the low no-load power factor reduces full load operating pf. | |
| 109. |
An induction motor when started on load, it does not accelerate up to full speed but runs at 1/7th of the rated speed. The motor is said to be |
| A. | locking |
| B. | plumming |
| C. | crawling |
| D. | cogging |
| Answer» C. crawling | |
| Explanation: running stably at low speed around 1/7th of normal motor speed is known as crawling. | |
| 110. |
The great advantage of the double squirrel-cage induction motor over single cage rotor is that its |
| A. | efficiency is higher |
| B. | power factor is higher |
| C. | slip is larger |
| D. | starting current is lower |
| Answer» D. starting current is lower | |
| Explanation: the outer cage has higher resistance and low reactance while vice versa for the inner cage. so at starting the current is confined to outer cage with the reduction in starting current improving the torque. | |
| 111. |
The rotor of 3-phase slip ring induction motor is fed from a 3-phase supply with its stator winding short circuited having rotor rotating clockwise at a speed of Nr, then the |
| A. | speed of air gap field w.r.t. stator is ns-nr anticlockwise |
| B. | speed of air gap field w.r.t. stator is ns-nr clockwise |
| C. | speed of airgap field w.r.t rotor is ns clockwise |
| D. | speed of airgap field w.r.t. stator is ns-nr clockwise |
| Answer» A. speed of air gap field w.r.t. stator is ns-nr anticlockwise | |
| Explanation: the air gap field of stator = - (ns-nr). | |
| 112. |
(II) If a large 3–phase squirrel-cage induction motor with no load is started at full voltage, it will be damaged. |
| A. | i is true, ii is false |
| B. | i is true and ii is also true |
| C. | i is false, ii is true |
| D. | i and ii are false |
| Answer» A. i is true, ii is false | |
| Explanation: large current of short duration are not harmful to induction motor but it may cause voltage drop in the power supply. | |
| 113. |
The PWM control of DC motor varies |
| A. | linearly with speed |
| B. | inversely with speed |
| C. | parabolically with speed |
| D. | exponentially with speed |
| Answer» A. linearly with speed | |
| Explanation: in pwm technique duty ratio is a linear function with respect to speed. | |
| 114. |
Ward-Leonard system of system of speed control is not recommended for |
| A. | constant speed operation |
| B. | wide speed |
| C. | frequent-motor reversed |
| D. | very slow speed |
| Answer» A. constant speed operation | |
| Explanation: ward-leonard system of system of speed control is not recommended for constant speed operation. | |
| 115. |
Mark the wrong option. Which of the following cause and effect behaviour in speed control is correct when field resistance is increased. |
| A. | decrease in flux |
| B. | increase in armature current |
| C. | increase in emf |
| D. | decrease in speed |
| Answer» C. increase in emf | |
| Explanation: when field resistance is increased, emf actually decreases, not increases due to fall in flux. | |
| 116. |
resistance will |
| A. | decrease the speed of motor |
| B. | increase the speed of motor |
| C. | not have signicant effect on speed |
| D. | no effect |
| Answer» A. decrease the speed of motor | |
| Explanation: at low speed increase in armature current caused by decrease in emf, will not be enough to compensate for decrease in flux in induced torque equation. | |
| 117. |
by |
| A. | armature voltage control |
| B. | field resistance control |
| C. | any of the methods |
| D. | none of the mentioned |
| Answer» A. armature voltage control | |
| Explanation: for the small dc motors it is difficult to vary the speed using their field control. | |
| 118. |
must be ensured that |
| A. | it is used on separately excited machine |
| B. | it is used on shunt machine |
| C. | it is used on series machine |
| D. | any of the mentioned machine |
| Answer» A. it is used on separately excited machine | |
| Explanation: it must be separately excited mahcine to implement the armature voltage control. | |
| 119. |
works well for speed |
| A. | below, above |
| B. | above, below |
| C. | base speed. |
| Answer» A. below, above | |
| Explanation: after the base speed is attained, the power consumed becomes constant as the speed remains constant in a shunt machine. | |
| 120. |
The torque limit of speed for a shunt motor |
| A. | remains constant till base speed |
| B. | remains constant after base speed |
| C. | varies linearly after base speed |
| D. | varies inversely till base speed |
| Answer» A. remains constant till base speed | |
| Explanation: the torque limit of the shunt motor remains constant till the base speed. | |
| 121. |
A laboratory group was working with a set of 3-hp shunt DC motor. But there was a mistake that it was fused with a 0.3A fuse instead 3A fuse. Then it was started |
| A. | a flash occurred instantly |
| B. | it ran for 3s and fuse was blown |
| C. | it ran normal |
| D. | none of the mentioned |
| Answer» B. it ran for 3s and fuse was blown | |
| Explanation: due to the wrong usage of the fuse wire, it could not sustain the current and in few seconds it was blown off. | |
| 122. |
avoided by using |
| A. | a turn or two of cumulative compounding |
| B. | compensating winding |
| C. | stablizer winding |
| D. | all of the mentioned |
| Answer» B. compensating winding | |
| Explanation: the runaway in large machine is avoided by using compensating winding. | |
| 123. |
fast when under-loaded? |
| A. | due to weak field |
| B. | due to high line voltage |
| C. | due to brush-shifted to neutral |
| D. | all of the mentioned |
| Answer» A. due to weak field | |
| Explanation: all of the reasons cause the dc motor to increase in speed when under loaded. | |
| 124. |
Concentrated winding differ from distributed winding with the concern of |
| A. | identical magnetic axis |
| B. | two magnetic axis |
| C. | no magnetic axis |
| D. | physical spacing |
| Answer» A. identical magnetic axis | |
| Explanation: concentrated winding have all turns on one magnetic axis. | |
| 125. |
We can place closed windings in |
| A. | ac commutator machine |
| B. | stepper motor |
| C. | ac machine |
| D. | dc machine |
| Answer» A. ac commutator machine | |
| Explanation: closed windings are placed only in commutator as it does not need any reconnections later. | |
| 126. |
Which of the following machines can be used to place open slot winding? |
| A. | ac machine |
| B. | ac commutator machine, ac machine |
| C. | dc machines |
| D. | all of the mentioned |
| Answer» A. ac machine | |
| Explanation: only ac machines are used for open type of winding inside the machine. | |
| 127. |
The simplex lap winding has the range of winding pitch of |
| A. | (-2,2) |
| B. | (-1,1) |
| C. | more than 2 |
| D. | less than 1 |
| Answer» A. (-2,2) | |
| Explanation: for a progressive winding, forward pitch – backward pitch = winding pitch | |
| 128. |
A 200V dc machine has 4 poles and 40 coils, having simplex lap winding. The number of commutator segments which required in the given machine will be? |
| A. | 40 |
| B. | 20 |
| C. | 80 |
| D. | 26 |
| Answer» A. 40 | |
| Explanation: it is a simplex winding and so one segment of commutator will be needed for each coil. | |
| 129. |
Commutator segments in a DC shunt machine is equal to the number of |
| A. | coil sides |
| B. | turns |
| C. | coils |
| D. | slots |
| Answer» A. coil sides | |
| Explanation: commutator segments are placed as per the coil sides in the machine. | |
| 130. |
We can employ dummy coils in a DC machine to |
| A. | compensate reactance voltage |
| B. | reduce armature reaction |
| C. | provide mechanical balance to the armature |
| D. | improve the waveforms generated inside the commutator |
| Answer» A. compensate reactance voltage | |
| Explanation: dummy coils are placed inside so that to provide mechanical support to dc machine and they do not take part in the emf development. | |
| 131. |
In AC machines we should prefer double layer winding over single layer windings because |
| A. | it requires identical coils |
| B. | it is economical to use |
| C. | it offers lower leakage reactance |
| D. | all of the mentioned |
| Answer» A. it requires identical coils | |
| Explanation: double layer winding enables more usage of the space and lesser air gaps so the leakage flux also reduces and efficiency improves. | |
| 132. |
For an electrical machine with C number of coils and P poles, the distance between the coils connected by an equalizer ring is? |
| A. | c/p |
| B. | c/2 |
| C. | 2c/p |
| D. | c/2p |
| Answer» C. 2c/p | |
| Explanation: backward pitch for an equalizer ring = number of coils/pole pairs = c/p/2 = 2c/p. | |
| 133. |
With a P-pole DC machine with armature current Ia, the current per brush arm for a lap connected windings is? |
| A. | ia/p |
| B. | ia/2p |
| C. | 2ia/p |
| D. | ia/4p |
| Answer» C. 2ia/p | |
| Explanation: current per brush (lap winding) | |
| 134. |
unused component in the machine, then we must |
| A. | s/p/2 should not be an integer |
| B. | s/p/2 should be integer |
| C. | s/p should be integer |
| D. | s/p should not be an integer |
| Answer» A. s/p/2 should not be an integer | |
| Explanation: if slots per pole pair is not integer than all the winding will be completed without adding any dummy coil. | |
| 135. |
While doing regular checks on the dc machine with the lap connected winding, it is reported to have ammeter fluctuations, this can be due to |
| A. | different air gaps under poles |
| B. | variable reluctances in the core |
| C. | irregular design deformations |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: in a lap winding configuration, if the machine as air gap variations due to the design aspects, it will give circulating current in the windings. | |
| 136. |
with the remaining brushes, voltage and power that could be obtained from the machine are |
| A. | e, p |
| B. | e, 2p/3 |
| C. | e/p, 2p/3 |
| D. | e, p/3 |
| Answer» A. e, p | |
| Explanation: taking kirchoff’s law into study here, the voltage across the winding will remain same. | |
| 137. |
Ward Leonard method is |
| A. | armature control method |
| B. | field control method |
| C. | combination of armature control method and field control method |
| D. | totally different from armature and field control method |
| Answer» C. combination of armature control method and field control method | |
| Explanation: ward leonard method is the combination of armature control method and field control method, which can also be called as voltage control method. this is the most efficient method of speed control over wide range. | |
| 138. |
Reducing the armature voltage will give us |
| A. | variable torque speed control |
| B. | constant torque speed control |
| C. | variable and constant both can be achieved |
| D. | cannot comment on torque |
| Answer» B. constant torque speed control | |
| Explanation: as seen from speed torque characteristics, reducing armature voltage | |
| 139. |
Reducing the field current will give us |
| A. | constant torque and variable power speed control |
| B. | constant torque speed control |
| C. | variable power speed control |
| D. | constant power speed control |
| Answer» B. constant torque speed control | |
| Explanation: as seen from speed torque characteristics, reducing field current will increase the speed of the motor above base value but power will remain same. thus, reducing armature voltage will give constant power speed control, with variable torque. | |
| 140. |
Speed-power characteristic for Ward Leonard speed control method |
| A. | will start from origin |
| B. | will start from some positive value on power axis |
| C. | will start from some positive value on speed axis |
| D. | depends on other parameters |
| Answer» A. will start from origin | |
| Explanation: speed power characteristic of dc motor is plotted when, ward leonard speed control method is employed. for speed equal to zero, which is less than base speed, we get constant torque but variable power operation. thus, power will start increasing from origin. | |
| 141. |
Efficiency of Ward Leonard method is |
| A. | higher than rheostatic control method but lower than shunted field control method |
| B. | lower than rheostatic control method |
| C. | higher than rheostatic control method and shunted field control method |
| D. | depends on load |
| Answer» C. higher than rheostatic control method and shunted field control method | |
| Explanation: unlike all other methods, external resistance is not added in the circuit of control system. thus, efficiency of ward leonard control method is always highest at various different speeds. | |
| 142. |
Ward Leonard method is an ideal choice for motor which undergoes frequent starting, stopping, speed reversal. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: absence of external resistance increases efficiency. also, when the generator emf becomes less than the back emf of the motor, electrical power flows back from motor to generator, is converted to | |
| 143. |
Starting gear used in Ward Leonard method |
| A. | is of small size |
| B. | is of large size |
| C. | size depends on application |
| D. | is absent |
| Answer» D. is absent | |
| Explanation: no special starting gear is required in ward leonard method of speed control. as the induced voltage by generator is gradually raised from zero, the motor starts up smoothly. speed reversal is smoothly carried out. | |
| 144. |
To get the speed of DC motor below the normal speed without wastage of electrical energy we use |
| A. | ward leonard control |
| B. | rheostatic control |
| C. | any of the ward leonard or rheostatic method can be used |
| D. | not possible |
| Answer» A. ward leonard control | |
| Explanation: ward leonard method of speed control is most efficient method of speed control in all aspects. we can get constant torque operation and constant power operations as well, with this method. | |
| 145. |
can give uniform speed variation |
| A. | in both directions |
| B. | in one direction |
| C. | below normal speed only |
| D. | above normal speed only. |
| Answer» A. in both directions | |
| Explanation: speed control by ward leonard method, gives uniform speed variation in both the directions and above and below of normal speed as well. speed reversal is carried out smoothly by this control method. | |
| 146. |
control method is |
| A. | high initial cost |
| B. | high maintenance cost |
| C. | low efficiency at high loads |
| D. | high cost, high maintenance and low efficiency |
| Answer» D. high cost, high maintenance and low efficiency | |
| Explanation: ward leonard speed control method requires large number of building blocks like generators, motors. thus, installing cost and maintenance cost of the whole unit is very high. apart from cost, it gives low efficiency at very high loads. | |
| 147. |
TOPIC 5.1 SPEED CONTROL OF THREE PHASE INDUCTION MOTOR |
| A. | slip frequency |
| B. | supply frequency |
| C. | the frequency corresponding to rotor speed |
| D. | zero |
| Answer» A. slip frequency | |
| Explanation: the relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced | |
| 148. |
An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is |
| A. | 2 |
| B. | 4 |
| C. | 3 |
| D. | 1 |
| Answer» A. 2 | |
| Explanation: given a number of poles = 8. supply frequency is 50 hz. rotor speed is 720 rpm. ns = 120×f÷p=120×50÷8 = 750 rpm. s=ns-nr÷ns = 750 – 720÷750 = .04. | |
| 149. |
Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm. |
| A. | 1.56 kgm2 |
| B. | 1.47 kgm2 |
| C. | 1.38 kgm2 |
| D. | 1.48 kgm2 |
| Answer» A. 1.56 kgm2 | |
| Explanation: the moment of inertia of the thin spherical shell can be calculated using the formula i=mr2×.66. the mass of the thin spherical shell and diameter is given. i= | |
| 150. |
Calculate the phase angle of the sinusoidal waveform z(t)=78sin(456πt+2π÷78). |
| A. | π÷39 |
| B. | 2π÷5 |
| C. | π÷74 |
| D. | 2π÷4 |
| Answer» A. π÷39 | |
| Explanation: sinusoidal waveform is generally expressed in the form of v=vmsin(ωt+α) where vm represents peak value, ω represents angular frequency, α represents a phase difference. | |
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