MCQOPTIONS
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MCQOPTIONS
This section includes 2114 Mcqs, each offering curated multiple-choice questions to sharpen your Unit Processes knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
The plot between carter’s coefficient and the relative flux density is the carter’s fringe curve. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 252. |
What is the formula of the flux density in the gap at a distance ‘x’ from the centre of the pole? |
| A. | flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap |
| B. | flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole / length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap |
| C. | flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole / maximum flux density in air gap |
| D. | flux density in the gap at a distance ‘x’ from the centre of the pole = 1/ length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap |
| Answer» C. flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole / maximum flux density in air gap | |
| 253. |
How is the flux distributed in the field form? |
| A. | to reduce the high voltage |
| B. | to reduce the high current |
| C. | to reduce the harmonics |
| D. | to keep the total reluctance low |
| Answer» E. | |
| 254. |
Rate of change of concentration per unit length in a semiconductor is called as. |
| A. | Concentration change |
| B. | Concentration mixture |
| C. | Concentration gradient |
| D. | Concentration variant |
| Answer» D. Concentration variant | |
| 255. |
When there is an open circuit what will be the net hole current. |
| A. | 5A |
| B. | 0.05A |
| C. | 0.5A |
| D. | 0A |
| Answer» E. | |
| 256. |
The drift velocity is 5V and the applied electric field intensity 20v/m what will be the mobility of charge carriers. |
| A. | 100 m2/ (vs) |
| B. | 4 m2/ (vs) |
| C. | 15 m2/ (vs) |
| D. | 0.25 m2/ (vs) |
| Answer» E. | |
| 257. |
Which of the following is reverse biased?a) |
| A. | A) |
| B. | B) |
| C. | C) |
| D. | D) |
| Answer» D. D) | |
| 258. |
Which of the factors doesn’t change the diode current. |
| A. | Temperature |
| B. | External voltage applied to the diode |
| C. | Boltzmann‘s constant |
| D. | Resistance |
| Answer» E. | |
| 259. |
What is the thickness of ‘space charge region’ or ‘transition region’ in P-N junction diode? |
| A. | 1 micron |
| B. | 5 micron |
| C. | 10 micron |
| D. | 2.876 micron |
| Answer» B. 5 micron | |
| 260. |
In nominal-T method of solution of medium transmission line, resistance and inductive reactance of the line is divided into two half’s. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 261. |
In a nominal-π method the resistance of line is assumed to be __________ |
| A. | Distributed uniformly from sending end to receiving end |
| B. | Divided into two halves |
| C. | Lumped at the generation end |
| D. | Lumped at middle of the line |
| Answer» E. | |
| 262. |
In nominal-π method of solution of medium transmission line, capacitance is assumed to be: |
| A. | Distributed uniformly from sending end to receiving end |
| B. | Divided into two halves |
| C. | Lumped at the generation end |
| D. | Lumped at middle of generation and load end |
| Answer» C. Lumped at the generation end | |
| 263. |
In medium transmission lines capacitive current is ________________ |
| A. | Less than short transmission lines |
| B. | Equal to short transmission lines |
| C. | More than short transmission lines |
| D. | More than long transmission lines |
| Answer» D. More than long transmission lines | |
| 264. |
With what energy must the radiation be given to image a bone of thickness 5 cm which has covering of skin of thickness of 2 cm on the both sides and the emerging intensity of the X – Ray is 200MeV. (impedance for bone = b for skin = s ) |
| A. | 2000e9b) 200e(4s + 5 |
| B. | 200e(4s + 5b) |
| C. | 20/e |
| D. | 2e |
| Answer» C. 20/e | |
| 265. |
In a hypothetical radioactive material, the total number of active photons are 20000 and the decay constant is found out to be 4.916 X 10-17 per second. How much of the material will be left in a 100 years? (1 year = 365 days. Leap year is not assumed in the calculations) |
| A. | 1589.99999 |
| B. | 19,999.9999 |
| C. | 19.999999 |
| D. | 123.99999 |
| Answer» C. 19.999999 | |
| 266. |
The two known units of radioactivity and the relation between the two are _______ |
| A. | Curie and Becquerel 1 Ci = 3.7 x 1010 Bq |
| B. | Curie and Becquerel 1 Bq = 3.7 x 1010 Ci |
| C. | Curie and Roentgens 1 Ci = 1000 R |
| D. | Roentgen and Becquerel 1 R = 1000 Bq |
| Answer» B. Curie and Becquerel 1 Bq = 3.7 x 1010 Ci | |
| 267. |
What is the relation between 1 Rad, 1 Rem and 1 R? |
| A. | 1 Rad ≈ 1.5 Rem ≈ 1000 R |
| B. | 1 Rad ≈ 10 Rem ≈ 1.8 R |
| C. | 1 Rad ≈ 1 Rem ≈ 1 R |
| D. | 1 Rad ≈ 10 Rem ≈ 100 R |
| Answer» D. 1 Rad ≈ 10 Rem ≈ 100 R | |
| 268. |
For what all purposes is diathermy principal used? |
| A. | Surgical and Therapeutic |
| B. | Therapeutic and Diagnostic |
| C. | Diagnostic and surgical |
| D. | Diagnostic and rehabilitative |
| Answer» B. Therapeutic and Diagnostic | |
| 269. |
If this does not solve the problem then new S.P.C. techniques that are being introduced in the pulp mill would probably decrement process variability of the pulp going to the bleach plant from the pulp mill. What is the topic of discussion? |
| A. | Control chart |
| B. | Pareto charts |
| C. | Student’s |
| D. | SPC |
| Answer» B. Pareto charts | |
| 270. |
One can quickly see if the process control variable is within the control area. If the actual y values are consistently within the boundaries, the process is said to be in control. This allows an operator to make changes when appropriate, but not to try to make changes when no action is called for. What is the topic of discussion? |
| A. | Control chart |
| B. | Pareto charts |
| C. | Student’s |
| D. | SPC |
| Answer» B. Pareto charts | |
| 271. |
Upper and lower boundaries of the “desired” values of y are plotted as lines. These are the _________ and _________ The average value (X, said as “X-bar”) is centred b/w the control limits. The range of the control limits, R, depends on the statistical variation of the process. |
| A. | Higher limit and lower limit |
| B. | Upper control limit and lower control limit |
| C. | Max. limit and Min. Limit |
| D. | Upper point and lower point |
| Answer» C. Max. limit and Min. Limit | |
| 272. |
It is very important to collect useful data before trying to analyze it by charting and statistical analysis. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 273. |
Castor requires a mapping configuration file to know how to map objects to and from XML documents. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 274. |
A marshalling endpoint requires:- |
| A. | marshaller |
| B. | unmarshaller |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 275. |
For some marshalling APIs, the object model must be generated by them so that they can insert marshalling-specific information. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 276. |
Spring-WS supports various XML marshalling APIs. |
| A. | Castor |
| B. | JAXB 1.0 |
| C. | XMLBeans |
| D. | All of the mentioned |
| Answer» E. | |
| 277. |
Marshaller for XStream:- |
| A. | org.springframework.oxm.jaxb.Jaxb1Marshaller |
| B. | org.springframework.oxm.jaxb.Jaxb2Marshaller |
| C. | org.springframework.oxm.xstream.XStreamMarshaller |
| D. | org.springframework.oxm.jibx.JibxMarshaller |
| Answer» D. org.springframework.oxm.jibx.JibxMarshaller | |
| 278. |
Marshaller for JAXB 2.0:- |
| A. | org.springframework.oxm.jaxb.Jaxb1Marshaller |
| B. | org.springframework.oxm.jaxb.Jaxb2Marshaller |
| C. | org.springframework.oxm.castor.CastorMarshall |
| D. | org.springframework.oxm.xmlbeans.XmlBeansMarshaller |
| Answer» C. org.springframework.oxm.castor.CastorMarshall | |
| 279. |
Spring-WS supports using XML marshalling technology to marshal and unmarshal objects to and from XML documents. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 280. |
Ray command draws lines diverging at a specified angle. |
| A. | True |
| B. | False |
| Answer» C. | |
| 281. |
RAY icon is shown _______ |
| A. | on the draw panel |
| B. | on the home tab |
| C. | on the expanded draw panel of the home tab |
| D. | nowhere |
| Answer» D. nowhere | |
| 282. |
If one specifies three nonlinear points P1, P2, P3, the ray passes through _________ |
| A. | P1 and P2 |
| B. | P1 and P3 |
| C. | P2 and P3 |
| D. | All three points |
| Answer» C. P2 and P3 | |
| 283. |
If one specified point P1 first and then point P2, the line passing through P2 to infinity and starting from P1 is called as ________ |
| A. | Line |
| B. | Segment |
| C. | Ray |
| D. | 2-point line |
| Answer» D. 2-point line | |
| 284. |
For drawing perspective view firstly draw the __________ |
| A. | Perspective view |
| B. | Top view |
| C. | Side view |
| D. | Front view |
| Answer» C. Side view | |
| 285. |
A simple pendulum is vibrating in an evacuated chamber. It will oscillate with ___________ |
| A. | Constant amplitude |
| B. | Increasing amplitude |
| C. | Decreasing amplitude |
| D. | First increasing amplitude and then decreasing amplitude |
| Answer» B. Increasing amplitude | |
| 286. |
A hollow spherical pendulum is filled with mercury has time period T. If mercury is thrown out completely, then the new time period ___________ |
| A. | Increases |
| B. | Decreases |
| C. | Remains the same |
| D. | First increases and then decreases |
| Answer» D. First increases and then decreases | |
| 287. |
A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the increase in the time period of the pendulum of increased length is? |
| A. | 10% |
| B. | 30% |
| C. | 21% |
| D. | 50% |
| Answer» B. 30% | |
| 288. |
The time period of a simple pendulum is 2 sec. If its length is increased by 4 times, then its period becomes ___________ |
| A. | 16 sec |
| B. | 8 sec |
| C. | 12 sec |
| D. | 4 sec |
| Answer» E. | |
| 289. |
Time period of a simple pendulum will be double if we ___________ |
| A. | Decrease the length 2 times |
| B. | Decrease the length 4 times |
| C. | Increase the length 2 times |
| D. | Increase the length 4 times |
| Answer» E. | |
| 290. |
To show that a simple pendulum executes simple harmonic motion, it is necessary to assure that ___________ |
| A. | Length of the pendulum is small |
| B. | Amplitude of oscillation is small |
| C. | Mass of the pendulum is small |
| D. | Acceleration due to gravity is small |
| Answer» C. Mass of the pendulum is small | |
| 291. |
A particle executes simple harmonic motion of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy? |
| A. | 0.51A |
| B. | 0.71A |
| C. | 0.61A |
| D. | 0.81A |
| Answer» C. 0.61A | |
| 292. |
A particle having potential energy 1/3 of the maximum value at a distance of 4 cm from mean position. Amplitude of motion is ___________ |
| A. | 4√3 |
| B. | 6/√2 |
| C. | 2/√6 |
| D. | 2√6 |
| Answer» B. 6/√2 | |
| 293. |
A particle executing simple harmonic motion has amplitude 0.01 and frequency 60Hz. The maximum acceleration of the particle is ___________ |
| A. | 144 π2 m/s2 |
| B. | 80 π2 m/s2 |
| C. | 120 π2 m/s2 |
| D. | 60 π2 m/s2 |
| Answer» B. 80 π2 m/s2 | |
| 294. |
The maximum velocity and maximum acceleration of a body moving in a simple harmonic motion are 2m/s and 4m/s2 respectively. What will be the angular velocity? |
| A. | 4 rad/sec |
| B. | 3 rad/sec |
| C. | 2 rad/sec |
| D. | 8 rad/sec |
| Answer» D. 8 rad/sec | |
| 295. |
The magnitude of acceleration of particle executing simple harmonic motion at the position of maximum displacement is? |
| A. | Zero |
| B. | Minimum |
| C. | Maximum |
| D. | Infinity |
| Answer» D. Infinity | |
| 296. |
A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is ___________ |
| A. | 10 cm/s |
| B. | 12 cm/s |
| C. | 16√16 cm/s |
| D. | 16 cm/s |
| Answer» C. 16√16 cm/s | |
| 297. |
A body is executing the simple harmonic motion with an angular frequency of 2rad/sec. Velocity of the body at 20m displacement, when amplitude of motion is 60m, is ___________ |
| A. | 90 m/s |
| B. | 118 m/s |
| C. | 113 m/s |
| D. | 131 m/s |
| Answer» D. 131 m/s | |
| 298. |
A particle executes simple harmonic motion, its time period is 16s. If it passes through the centre of oscillation, then its velocity is 2 m/s at time 2s. The amplitude will be ___________ |
| A. | 7.2m |
| B. | 4cm |
| C. | 6cm |
| D. | 0.72m |
| Answer» B. 4cm | |
| 299. |
A particle in simple harmonic motion is described by the displacement function x(t)=Acos(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s. |
| A. | 1 cm |
| B. | √2 cm |
| C. | 2 cm |
| D. | 2.5 cm |
| Answer» C. 2 cm | |
| 300. |
Vibratory feeder or toothed belt feeder as used to measure the quantity of the abrasive particles before mixing. |
| A. | True |
| B. | False |
| Answer» B. False | |