MCQOPTIONS
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MCQOPTIONS
This section includes 5314 Mcqs, each offering curated multiple-choice questions to sharpen your Chemical Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 2301. |
What type of steam is produced in the steam boiler? |
| A. | Low pressure steam |
| B. | High pressure steam |
| C. | Saturated steam |
| D. | Unsaturated steam |
| Answer» D. Unsaturated steam | |
| 2302. |
What is the pH value at which the boiler is alkalized? |
| A. | 9 |
| B. | 5.4 |
| C. | 6.9 |
| D. | 5.7 |
| Answer» B. 5.4 | |
| 2303. |
What is the storage used to hold the condensed water? |
| A. | Reservoir |
| B. | Tarn |
| C. | Hot well |
| D. | Basin |
| Answer» D. Basin | |
| 2304. |
Assertion (A): DC servomotors have practically taken over from AC servomotors in most control application.Reason (R): Low brush commutator friction and still higher torque/inertia have been achieved in DC servomotors. |
| A. | Both A and R are true and R is correct explanation of A |
| B. | Both A and R are true but R is not correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» B. Both A and R are true but R is not correct explanation of A | |
| 2305. |
Speed of a permanent magnet DC motor: |
| A. | It is directly proportional to the armature voltage at a given load torque |
| B. | The speed torque characteristics are more flat than in a wound field motor |
| C. | It depends upon armature voltage and are more flat than in a wound field motor |
| D. | None of the mentioned |
| Answer» D. None of the mentioned | |
| 2306. |
DC motors are constructed using Permanent Magnet resulting in: |
| A. | Higher torque/inertia ratio |
| B. | Higher operating frequency |
| C. | No filed losses |
| D. | All of the mentioned |
| Answer» E. | |
| 2307. |
Commutation process of a DC motor can be accomplished by solid devices (transistors and SCR’s). |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2308. |
Assertion (A): Slotted armature type in this armature is placed in slotted armature with DC windings placed inside the slots.Reason (R): The construction is highly reliable and rugged but has high torque. |
| A. | Both A and R are true and R is correct explanation of A |
| B. | Both A and R are true but R is not correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» C. A is true but R is false | |
| 2309. |
Different type of construction in permanent type DC magnet are : |
| A. | Slotted armature type |
| B. | Surface wound armature type |
| C. | Surface wound rotating armature, stationary wound type. |
| D. | All of the mentioned |
| Answer» E. | |
| 2310. |
In constant electric field model, power dissipation per unit area is scaled by |
| A. | α |
| B. | β |
| C. | 1 |
| D. | β2 |
| Answer» D. β2 | |
| 2311. |
In constant field model, maximum operationg frequency is scaled by |
| A. | α |
| B. | β |
| C. | α2 |
| D. | β2 |
| Answer» B. β | |
| 2312. |
In constant voltage model, the saturation current is scaled by |
| A. | α |
| B. | β |
| C. | 1 |
| D. | β2 |
| Answer» D. β2 | |
| 2313. |
Power dissipation per unit area is scaled by |
| A. | 1/α |
| B. | 1/β |
| C. | β2/α2 |
| D. | α2/β2 |
| Answer» E. | |
| 2314. |
Carrier density is scaled by |
| A. | α |
| B. | β |
| C. | 1 |
| D. | α2 |
| Answer» D. α2 | |
| 2315. |
What is minimum amount of air required per m3 of gaseous fuel for complete combustion? |
| A. | 1/21 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel |
| B. | 1/100 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel |
| C. | 1/21 [(H2/2) + (CO/2) + 3C2H4] m3/m3 of fuel |
| D. | 1/100 [(H2/2) + (CO/2) + 3C2H4] m3/m3 of fuel |
| Answer» B. 1/100 [(H2/2) + (CO/2) + 2CH4 + 3C2H4] m3/m3 of fuel | |
| 2316. |
What is amount of minimum air required per kg of liquid fuel for complete combustion using carbon, oxygen, hydrogen and sulfur? |
| A. | 1/23 [8/3 C + 8(H – (O/8)) + S] |
| B. | 1/100 [8/3 C + 8(H – (O/8)) + S] |
| C. | 1/100 [8/3 C + 8(H – (0/8))] |
| D. | 1/23 [8/3 C + 8(H – (0/8))] |
| Answer» B. 1/100 [8/3 C + 8(H – (O/8)) + S] | |
| 2317. |
Which is the common method to relate higher calorific value to lower calorific value? |
| A. | HCV = LCV + HV (nH2O, out / nfuel, in) |
| B. | LCV = HCV + HV (nH2O, out / nfuel, in) |
| C. | HCV = LCV + HV (nfuel, in/ nH2O, out) |
| D. | LCV = HCV + HV (nfuel, in/ nH2O, out) |
| Answer» B. LCV = HCV + HV (nH2O, out / nfuel, in) | |
| 2318. |
Loss of activity which may be restored by the removal of inhibitor is referred to as ___________ |
| A. | reversible inhibition |
| B. | irreversible inhibition |
| C. | competitive inhibition |
| D. | mixed inhibition |
| Answer» B. irreversible inhibition | |
| 2319. |
The following diagram represents _______________ inhibition. |
| A. | competitive |
| B. | non-competitive |
| C. | product |
| D. | uncompetitive |
| Answer» C. product | |
| 2320. |
Inhibition of lactase by galactose is an example of which kind of inhibition? |
| A. | Uncompetitive inhibition |
| B. | Mixed inhibition |
| C. | Competitive inhibition |
| D. | Substrate inhibition |
| Answer» D. Substrate inhibition | |
| 2321. |
The rate equation for _______________ inhibition is given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{[P]}{K_p})+[S]}\). |
| A. | substrate |
| B. | non-competitive |
| C. | product |
| D. | competitive |
| Answer» D. competitive | |
| 2322. |
The following rate equation is given by ___________ inhibition.\(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i’}}\right ) [S]}{\left (\frac{K_m}{1+\frac{[I]}{K_i’}}\right )+[S]}\) |
| A. | mixed |
| B. | competitive |
| C. | uncompetitive |
| D. | non-competitive |
| Answer» D. non-competitive | |
| 2323. |
Mercury causes irreversible inhibition. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2324. |
If I = Ki = Ki‘, then what will happen to Vmax and Km when inhibitor acts non-competitively? |
| A. | Lowers to 0.5 Vmax and 0.5 Km |
| B. | Vmax is unchanged and Km increases 2Km |
| C. | Lowers to 0.5 Vmax and Km remains unchanged |
| D. | Lowers to 0.67 Vmax and Km increases to 2Km |
| Answer» D. Lowers to 0.67 Vmax and Km increases to 2Km | |
| 2325. |
What happens to Vmax and Km in case of uncompetitive inhibition when I = Ki‘? |
| A. | Lowers to 0.5 Vmax and 0.5 Km |
| B. | Vmax is unchanged and Km increases 2Km |
| C. | Lowers to 0.5 Vmax and Km remains unchanged |
| D. | Lowers to 0.67 Vmax and Km increases to 2Km |
| Answer» B. Vmax is unchanged and Km increases 2Km | |
| 2326. |
In a drained test, the drainage is permitted during the application of ______ |
| A. | No stress |
| B. | Deviator stress |
| C. | Cell pressure |
| D. | Cell pressure and deviator stress |
| Answer» E. | |
| 2327. |
The test results from the dynamic cone penetration test can be related to CBR. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2328. |
Determine the sensitivity of the soil if the undisturbed undrained shear strength is 52.78 kN/m2 and the disturbed undrained shear strength is 23.54 kN/m2. |
| A. | Less sensitive |
| B. | Sensitive |
| C. | Highly sensitive |
| D. | Quick |
| Answer» B. Sensitive | |
| 2329. |
In a CBR test, the load values corresponding to different deflection values are noted. |
| A. | True |
| B. | False |
| Answer» C. | |
| 2330. |
Which of the below soil samples has better shear strength at the plastic limit? |
| A. | Soil A |
| B. | Soil B |
| C. | Soil C |
| D. | Soil DView Answer |
| Answer» B. Soil B | |
| 2331. |
A vane of 80 mm diameter and 160 mm height was pushed into the soil using a torque of 80 Nm. What will be the undrained shear strength of the soil? |
| A. | 4.263×10-5 kN/m2 |
| B. | 4.263×105 kN/m2 |
| C. | 42.63 kN/m2 |
| D. | 4.263 kN/m2 |
| Answer» D. 4.263 kN/m2 | |
| 2332. |
The diameter of the standard plate used to find the value of modulus of subgrade is ______ |
| A. | 7.5 cm |
| B. | 75 mm |
| C. | 75 cm |
| D. | 750 cm |
| Answer» D. 750 cm | |
| 2333. |
What is the liquidity index of soil if the plastic limit is 24%, the liquid limit is 49% and the natural water content is 35%. |
| A. | -54% |
| B. | -44% |
| C. | 44% |
| D. | 54% |
| Answer» D. 54% | |
| 2334. |
Which of the below facts about the particle size distribution graph is false? |
| A. | Uniformity coefficient can be obtained from the graph |
| B. | Particle size is plotted on the X-axis |
| C. | Helps in understanding the physical properties of soil |
| D. | It is also called a semi-log graph |
| Answer» E. | |
| 2335. |
What is the cohesion for a soil sample that fails under axial stress of 140 kN/m2 when subjected to unconfined compression test and has an angle of internal friction of 17°? |
| A. | 38.33 kN/m2 |
| B. | 0 kN/m2 |
| C. | 38 kN/m2 |
| D. | 40 kN/m2 |
| Answer» B. 0 kN/m2 | |
| 2336. |
The cone penetration test is also called ______ test. |
| A. | Swiss cone |
| B. | Dutch cone |
| C. | Dynamic cone penetration |
| D. | Static cone penetration |
| Answer» C. Dynamic cone penetration | |
| 2337. |
What is the CBR value of the soil if the CBR value at 2.5 mm penetration is found to be 4.5% and that at 5 mm penetration is found to be 4.3%? |
| A. | 4.5% |
| B. | 4.4% |
| C. | 4.6% |
| D. | 4.3% |
| Answer» B. 4.4% | |
| 2338. |
Which test setup does the figure represent? |
| A. | Direct shear |
| B. | Plate bearing |
| C. | CBR |
| D. | Unconfined compression |
| Answer» C. CBR | |
| 2339. |
Determine the deviator stress for a soil sample subjected to triaxial test having the following details:Cohesion = 15 kN/m2Angle of shearing stress = 30°Cell pressure = 260 kN/m2 |
| A. | 1091.96 kN/m2 |
| B. | 571.96 kN/mm2 |
| C. | 1091.96 kN/mm2 |
| D. | 571.96 kN/m2 |
| Answer» E. | |
| 2340. |
______ method is used to determine water content as well as the specific gravity of soil. |
| A. | Measuring flask |
| B. | Sand bath |
| C. | Oven drying |
| D. | Pycnometer |
| Answer» E. | |
| 2341. |
In a dark image, the components of histogram are concentrated on which side of the grey scale? |
| A. | High |
| B. | Medium |
| C. | Low |
| D. | Evenly distributed |
| Answer» D. Evenly distributed | |
| 2342. |
Histograms are the basis for numerous spatial domain processing techniques. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2343. |
The histogram of a digital image with gray levels in the range [0, L-1] is represented by a discrete function: |
| A. | h(r_k)=n_k |
| B. | h(r_k )=n/n_k |
| C. | p(r_k )=n_k |
| D. | h(r_k )=n_k/n |
| Answer» B. h(r_k )=n/n_k | |
| 2344. |
At synchronous speed, relative speed between rotating air-gap flux and damper is? |
| A. | zero |
| B. | Ns |
| C. | lesser than Ns |
| D. | more than zero |
| Answer» B. Ns | |
| 2345. |
Damper winding should have ___________ |
| A. | high resistance |
| B. | low starting torque |
| C. | low resistance |
| D. | low reactance |
| Answer» D. low reactance | |
| 2346. |
Turbo generators do not use damper winding due to ___________ |
| A. | solid steel rotor core |
| B. | laminated armature |
| C. | soft iron core |
| D. | small current flow in pole shoes |
| Answer» B. laminated armature | |
| 2347. |
The damper winding are placed in ___________ |
| A. | pole shoes |
| B. | series with armature |
| C. | series with field |
| D. | rotor slots |
| Answer» B. series with armature | |
| 2348. |
Damper windings are made of ___________ |
| A. | copper |
| B. | iron |
| C. | silicon |
| D. | cast iron |
| Answer» B. iron | |
| 2349. |
Employment of damper winding will reduce hunting. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2350. |
Using a _____ can be used to limit hunting. |
| A. | Flywheel |
| B. | DC Exciter |
| C. | Large load |
| D. | Any of the mentioned |
| Answer» E. | |