This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The incorrect accounting equation is …(a) Assets = Liabilities + Capital(b) Assets = Capital + Liabilities (c) Liabilities = Assets + Capital (d) Capital = Assets – Liabilities |
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Answer» (c) Liabilities = Assets + Capital |
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| 2. |
For one mole of an ideal gas, which of these statements must be true ? (a) U and H each depends only on temperature (b) Compressibility factor z is not equal to 1 (c) CP,m – CV,m = R (d) dU = CVdT for any process (1) (a), (c) and (d) (2) (b), (c) and (d) (3) (c) and (d) (4) (a) and (c) |
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Answer» (1) (a), (c) and (d) For ideal Gas # U = f(T), H = f(T) # Z = 1 # CP – CV = R # dU = CVdT |
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| 3. |
Consider the following reaction N2O4 (g) ⇌ 2NO2 (g) ; △Ho = +58 kJ For each of the following cases (a, b), the direction in which the equilibrium shifts is: (a) Temperature is decreased (b) Pressure is increased by adding N2 at constant T (1) (a) towards reactant, (b) no change (2) (a) towards product, (b) towards reactant (3) (a) towards product, (b) no change (4) (a) towards reactant, (b) towards product |
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Answer» (1) (a) towards reactant, (b) no change △Ho > 0 T↓ equation shifts back ward. N2 is treated as inert gas in this case hence no effect on equilibrium. |
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| 4. |
J.F means ……(a) Ledger page number (b) Journal page number (c) Voucher number (d) Order number |
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Answer» (b) Journal page number |
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| 5. |
Which one of the following graphs is not correct for ideal gas ?d = Density, P = Pressure, T = Temperature(1) II (2) III (3) I (4) IV |
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Answer» Answer is (1) II PM = dRT ⇒ 1 d ∝ \(\frac{1}{T}\) PM = dRT for constant P(pressure), dT = constant {M ( Molar mass) and R ( universal gas constant) are constant for a gas} so Graph II is not valid Correct option (1) |
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| 6. |
The process of finding the net amount from the totals of debit and credit columns in a ledger is known as ……(a) Casting (b) Posting(c) Journalising (d) Balancing |
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Answer» (d) Balancing |
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| 7. |
An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules : (1) leaves the vapour increases (2) leaves the solution increases (3) leaves the solution decreases (4) leaves the vapour decreases |
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Answer» With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure |
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| 8. |
_________, totals/adds data arranged in an array—that is, a group of cells with labels for columns and/or rows Which step one must should follow before using the Subtotal option? a. Consolidate b. Rename Data c. Filter Data d. Subtotal |
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Answer» Correct answer is d. Subtotal d- subtotal
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| 9. |
A ______ hyperlink contains the full address of the destination file or web page. a. Relative b. Absolute c. Mixed d. Address |
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Answer» Correct answer is b. Absolute |
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| 10. |
Each video on Facebook has an id which shown in the______. |
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Answer» Each video on Facebook has an id which shown in the URL or Uniform Resource Locator. |
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| 11. |
Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils. |
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Answer» Total number of pens = 1001. Total number of pencils = 910. Maximum number of students who get the same number of pens and same number of pencils = HCF (1001, 910). Using prime factorization, 910 = 2 × 5 × 7 ×13. 1001 = 7 × 11 × 13. Hence, HCF (1001, 910) = 7 × 13 = 91 Hence, Maximum number of students who get the same number of pens (11) and same number of pencils (10) is 91. |
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| 12. |
The wavelength of the first Lyman lines of hydrogen `He^(+)` and `Li^(2+)` ions and `lamda_(1),lamda_(2)& lamda_(3)` the ratio of these wavelengths isA. `1:4:9`B. `9:4:1`C. `36:9:4`D. `6:3:2` |
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Answer» Correct Answer - C `(1)/(lamda)=RZ^(2)(1-(1)/(4))` for lyman line |
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| 13. |
The binding energy per nucleon of `._(5)B^(10)` is 8 MeV and that of `._(5)B^(11)` is 7.5 MeV. The energy required to remove a neutron from `._(5)B^(11)` is (mass of electron and proton are `9.11 xx 10^(-21) kg` and `1.67 xx 10^(-27) kg` respectively.A. `2.5 MeV`B. `8.0 MeV`C. `0.5 MeV`D. `7.5 MeV` |
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Answer» Correct Answer - A `10 xx 8 = 80` `11 xx 7.5 = 82.5` Energy required `= 82.5 -80 = 2.5` |
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| 14. |
Files with .voc extension are the ___files. 1. Image 2. Sound 3. Video 4. Text |
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Answer» Correct option: 2. Sound |
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| 15. |
The wavelength of the first balmer line caused by a transition from the `n = 3` level to the n = 2 level in hydrogen is `lamda_(1)`. The wavelength of the line caused by an electronic transition from `n =5` to `n =3` isA. `(375)/(128) lamda_(1)`B. `(125)/(64) lamda_(1)`C. `(64)/(125) lamda_(1)`D. `(128)/(378) lamda_(1)` |
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Answer» Correct Answer - B `(lamda_(1))/(lamda_(2))=((1)/(2^(2))-(1)/(3^(2)))/((1)/(3^(2))-(1)/(5^(2)))=((1)/(4)-(1)/(9))/((1)/(9)-(1)/(25))=(125)/(64)` `lamda_(2)=(125)/(64)lamda_(1)`. |
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| 16. |
JPEG stands for _____. a. Joint Photographic Explorer Group b. Joint pictures export graphic c. Joint photographic experts group d. Joint pixel export group |
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Answer» Correct option: c. Joint Photographic Experts Group |
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| 17. |
What is the full form of JPEG? |
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Answer» JPEG- Joint Photographic Experts Group |
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| 18. |
Two parallel plate capacitors of capacitance C each are connected in series with a battery of emf `epsilon`. Then, one of the capacitors is filled with a dielectric of dielectric constane K. i. Find the change in electric field in the two capacitors, if any ii. What amount of charge flows through the battery? iii Find the change in the energy stored in the circuit, if any.A. `(k-1)/(2(k-1))CE`B. `(k-1)/(2(k+1))CE`C. `(k-2)/(k+2)CE`D. `(K+2)/(k-2)CE` |
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Answer» Correct Answer - B `Q=(C)/(2)E` `C_(eq)=(C.KC)/(C+KC)=(C^(2)K)/(C(1+K))=(CK)/(1+K)` `Q=(CK)/(1+k)E` `Delta Q=((CK)/(1+K)-(C)/(2))E =(2CK-C(K-1))/(2(1+K))` `=(2CK-CK-C)/(2(1+K))=(CK -C)/(2(1+K))`. |
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| 19. |
Mention any two taglines of TV advertisement. |
Answer»
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| 20. |
___ is a free video editor for fast and lossless AVI editing. |
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Answer» AVI Trimmer is a free video editor for fast and lossless AVI editing. |
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| 21. |
What is celebrity endorsement? |
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Answer» Celebrity endorsement is a form of advertising campaign used by brands, companies or a non- profit organization which involves celebrities or a well known person using their social status or their fame to promote a product or a service or even raise awareness on social or environmental matters. |
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| 22. |
What is a text? |
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Answer» The text in any form can be developed in to multimedia software. Text can be edited using any standard text editor. However to give special effects, one needs graphic software. Text can be in varied in fonts, size, color and style to add value to the multimedia presentation. |
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| 23. |
What is meant by ‘frame’ in still photography? |
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Answer» There are several places where the word “framing” appears in photography, from the act of framing an image in the viewfinder of a camera, to the act of putting your favorite print in a frame for display. |
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| 24. |
What is a video clip? |
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Answer» Video clips are short clips of video, usually part of a longer recording. The video clips may be in various forms such as just visuals, visuals and sounds interlocked together, sounds and graphics etc. |
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| 25. |
What is the sound card used for? |
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Answer» A computer can be used to compose music and for recognition of speech and synthesis. Therefore, a sound card is used to convert the conventional sound signals to digital signals. |
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| 26. |
Explain two non-destructive and non-linear tools for video editing. |
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Answer» Two Non-destructive and non-linear tools for video editing are : a. Zwei-Stien Video Editor: It makes cutting and joining footage easy and includes a wide range of built-in adjustable effects that can be used separately or in combination b. AVI Tricks Video Editor : It makes cutting and joining footage easy and includes a wide range of built-in adjustable effects that can be used separately or in combination. |
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| 27. |
Give any one example of sound card. |
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Answer» Example of sound card- Asus, MSI, Razor |
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| 28. |
What are the tools for editing image? |
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Answer» Tools for editing image:- 1. Rectangular Marquee Tool 2. Move Tool 3. Polygon Lasso Tool 4. Magic Wand Tool 5. Crop Tool 6. Slice Tool 7. Healing Brush Tool 8. Brush Tool 9. Clone Stamp Tool 10. History Tool |
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| 29. |
Who are the target buyers? |
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Answer» Every product belongs to a sector of business. And the buyers are distributed as per their needs and income levels. These buyers are called Targeted Buyers. |
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| 30. |
What is quota sampling? |
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Answer» Quota Sampling is a method for selecting survey participants. In quota sampling, a population is first segmented into mutually exclusive sub- groups, just as stratified proportion. |
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| 31. |
Which compound show both G.I. and O.I. ? (AA = Symetric bidentate ligand) (AB = Asymetric bidentate ligand) M = MetalA. `"M(AA)"_(3)`B. `M(AB)_(3)`C. (1)& (2)D. None |
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Answer» Correct Answer - B `2CH_(3)-COOH overset(Ca(OH)_(2))rarr(A) overset(Delta)rarr(B)overset((i)OH^(Theta))underset((ii)Delta)rarr (C)` |
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| 32. |
In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?(1) N2 → N+2 (2) NO → NO+(3) O2 → O2-2(4) O2 → O+2 |
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Answer» The correct option (2) NO → NO+ Explanation:
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| 33. |
The C–C bond length is maximum in : (1) C60 (2) diamond (3) graphite (4) C70 |
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Answer» Correct option (2) diamond Explanation: C–C in C60 = 1.4Å C–C in C70 = 1.37 to 1.46 Å Diamond = l > 1.54Å Graphite = 1.54 Å |
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| 34. |
`CH_(2)-CH-underset(OH)underset("| ")"CH"-underset(OH)underset("| ")"CH"_(2)overset(MnO_(2))rarr` Product is :-A. `CH_(2)=CH-underset(O)underset(||)C-underset(O)underset(||)C-H`B. `CH_(2)=CH-CH_(2)-underset(OH)underset("| ")"CH"_(2)`C. `CH_(2)=CH-underset(O)underset(||)C-CH_(3)`D. `CH_(2)=CH-underset(O)underset(||)C-underset(OH)underset("| ")"CH"_(2)` |
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Answer» Correct Answer - D `MnO_(2)` oxidised only `1^(@)` Allylic acid and Benzylic Alcohol into Aldehyde. `2^(@)` Allylic and Benzylic Alcohol into Ketone. |
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| 35. |
Aryl halides are extermely less reactive towards nucleophilic substitution. Predictand explain the order of reactivity of the compounds towards nucleophilic substitution. A. a gt b gt c gt dB. d gt c gt a gt bC. b gt c gt a gt dD. c gt b gt d gt a |
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Answer» Correct Answer - C Rate of Nucleophilic sub. `prop-m.-I` order is = (b) gt (c) gt (a) gt (d) |
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| 36. |
`105 mL` of pure water at `4^(circ)C` saturated with `NH_(3)` gas yielded a solution of density `0.9g mL^(-1)` and containing `30%NH_(3)` by mass. Find out the volume of `NH_(3)` solution resulting and the volume of `NH_(3)` gas at `4^(circ)C` and `775 mm` of `Hg`, which was used to saturate water.A. 66.67 mlB. 166.67 mlC. 133.33 mlD. 266.67 ml |
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Answer» Correct Answer - B Mass of water =105x1=105 g let wt. of `NH_3` is x given `x/(x+105)xx100=30` 10x=3x+315 7x=315 x=45 Total mass =45+105=150 g Volume of solution =mass of solution /density of solution =`150/0.9=1500/9=166.67` ml |
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| 37. |
Name the major product formed when nitrous acid is treated with aniline at low temperature. |
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Answer» In a aniline, the lone pair of electrons present on nitrogen are involved in the delocalisation with n electrons of the benzene ring due to resonance. Hence, the lone pair of electrons present on nitrogen are less available in aniline. In addition, the anilinium ion formed is destabilised by the benzene ring. Hence, aniline is a weaker base than ammonia. |
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| 38. |
Name the major product formed when nitrous acid is treated with methylamine. |
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Answer» Methylamine reacts with nitrous acid to give methyl alcohol liberating nitrogen. CH3NH2 + ONOH → CH3OH + N2 + H2O. |
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| 39. |
Acetaldehyde does not undergo Cannizzaro’s reaction. Why? |
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Answer» Acetaldehyde contains three α-hydrogen. Hence it does not answer Cannizzaro’s reaction. |
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| 40. |
If `F^(ө)` attacks on `BF_(3)` then the co`-`ordinate Bond is formed. Find out the approximate change in Bond Angle with respect to unhybridised orbitalA. `11^(@)`B. `27^(@)`C. `19^(@)`D. `42^(@)` |
| Answer» Correct Answer - C | |
| 41. |
Read the following pie diagram. Distribution of students at Graduate level in Seven Institutes named P, Q, R, S, T, M & N :Now answer the following questions. 1) What does the above pie diagram show?2) How many institutes are there in total?3) In which two institutes maximum number of students studied?a) N & P b) R & M c) Q & S4) In which two institutes least number of students studied?a) Q & P b) M & R c) N & P5) The number of students of which college are exactly half of the sum of students from Q & Ra) P b) Q c) T |
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Answer» 1. The pie-chart shows the distribution of students at the Graduate level in seven institutes named P Q, R, S, T, M, N. 2. There are seven institutes in total. 3. b) R & M 4. c) N & P 5. c) T |
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| 42. |
An inspector of the Alaska Pipeline has the task of comparing the reliability of two pumping stations. Each station is susceptible to two kinds of failure: pump failure and leakage. When either (or both) occurs, the station must be shut down. The data at hand indicate that the following probabilities prevail:StationP (Pump Failure)P (Leakage)P (Both)10.070.10020.090.120.06Which station has the higher probability of being shut down? |
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Answer» Let A = Event which indicates pump failure. B = Event which indicates pump is leakaged For station 1, P(A) = 0.07, P(B) = 0.1 P(A ∪ B) = 0 ∴ P(A ∪ B)1 = P(A) + P(B) - P(A ∪ B) = 0.07 + 0.1 - 0 = 0.17 For station 2, P(A) = 0.09, P(B) = 0.12 P(A ∪ B) = 0.06 ∴ P(A ∪ B)2 = P(A) + P(B) - P(A ∪ B) = 0.09 + 0.12 - 0.06 = 0.21 - 0.06 = 0.15 ∴ P(A ∪ B)1 > P(A ∪ B)2 Thus, station 1 has the higher probability of being shut down. |
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| 43. |
Read the following table. Number of Engineering students at Institutes of different kind.College1988 – 19891989- 19901990 – 1991Private Engineering college120001800025000Government Engineering colleges80001200013000Regional Engeering colleges4000750010000IIT300040008000Now answer the following questions.1) What does the table show?2) What is the general trend of Engineering education?3) What was the total number of engineering students in 1989 – 90?a) 38500 b) 41500 c) 42500 4) In which category of Engineering colleges highest number of students are studying Engineering?a) IITs b) Govt. Engineering colleges c) Private Engineering college5) In which category of colleges lowest number of students are studying Engineering?a) IITs b) Govt. Engineering colleges c) Regional Engineering colleges |
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Answer» 1. The table shows the number of engineering students studying at different institutes during the period 1989-89 to 1990-91. 2. The general trend of engineering education has shown consistent growth during the given period. 3. b) 41500 4. c) Private Engineering college 5. 5) In which category of colleges lowest number of students are studying Engineering? a) IITs b) Govt. Engineering colleges c) Regional Engineering colleges |
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| 44. |
A study showed that 60% of manager had some business education and 55% had some engineering education. Furthermore, 25% of the mangers had some business education but no engineering education. What is the probability that a manager has some business education, given that he has some engineering education? |
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Answer» Let E1 = Manger has some business education and E2 = Manager has some engineering education P(E1) = 60% = 60/100 = 3/5 and P(E2) = 55% = 55/100 = 11/20 Also, given that 25% managers has some business education but no engineering education. \(\therefore\) P(E1 - E2) = 25% = 25/100 = 1/4 Now, P(E1) = P(E1 - E2) + P(E1 ∩ E2) \(\therefore\) P(E1 ∩ E2) = P(E1) - P(E1 - E2) = 3/5 - 1/4 = (12 - 5)/20 = 7/20. Now, Probability that a manager has some business education given that he has some engineering education is P(E1/E2) = P(E1 ∩ E2)/P(E2) = \(\frac{7/20}{11/20}\) = 7/11 |
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| 45. |
The substance undergoes first order decomposition.The decomposition follows two parallel first order reactons as : `K_1=10^(-2) sec^(-1) and K_2=10^(-2)sec^(-1)` If the corresponding activation energies of parallel reaction are 100 and 120 kJ `"mol"^(-1)` then the net activation energy of A is/are:A. 120 kJ `"mol"^(-1)`B. 116 kJ `"mol"^(-1)`C. 100 kJ `"mol"^(-1)`D. 150 kJ `"mol"^(-1)` |
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Answer» Correct Answer - B `E=(k_1E_1+k_2E_2)/(k_1+k_2)=(10^(-2)xx100+120xx4xx10^(-2))/(10^(-2)+4xx10^(-2))=116 kJ mol^(-1)` |
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| 46. |
A certain reaction obeys the rate equation (in the integrated form) `[C^(1-n)-C_(@)^(1-n)]=(n-1) kt` where `[email protected]` is the initial concentration of C is the concentration after time, t Then:A. The unit of k for n=1 is `"sec"^(-1)`B. The unit of k for n=2 is litre `"mol"^(-1) "sec"^(-1)`C. The unit of k for n=3 is mol `"litre"^(-1) "sec"^(-1)`D. The unit of k for n=3 is `"lite"^2"mol"^2"sec"^(-1)` |
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Answer» Correct Answer - A,B,D Unit of `k=("mol"/"litre")^(1-n) xx s^(-1)` for n=1 is `sec^(-1)`, n=2 is litre `mol^(-1) sec^(-1)`, n=3 is `"litre"^2 mol^(-2) sec^(-1)` |
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| 47. |
A bag 'A' contains 4 black and 6 red balls and bag 'B' contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag 'A' is chosen, otherwise bag 'B'. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black. |
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Answer» The event can happen in two possible ways - the first ball chosen is red and the second one is black or vice versa. For bag A, the probability that first ball is red and second ball is black is 6/10 x 4/9 =4/15 For bag A, the probability that the first ball is black and the second ball is red is 4/10 x 6/9=4/15. The total probability for bag A is therefore 4/15+4/15=8/15 Similarly the total probability for bag B can be found to be 7/30+7/30=7/15. The probability that bag A is selected from throwing the die is 1/3 and the probability that bag B is selected is 2/3 So, the total probability of the occurence of this event is 1/3 x 8/15 x 2/3 x 7/15= 22/45 |
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| 48. |
satisfying the given condition:11. \( \left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x ; y=1 \) when \( x=0 \) |
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Answer» Toppers village ..... you tube channel |
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| 49. |
Here are three reactions to the language policy followed in India. Give an argument and an example to support any of these positions.Sangeeta: The policy of accommodation has strengthened national unity.Arman: Language-based States have divided us by making everyone conscious of their language.Harish: This policy has only helped to consolidate the dominance of English over all other languages. |
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Answer» Sangeeta’s reaction is better than those of the other two. Unlike Sri Lanka (where the language of the majority has been promoted), the Indian polity has given equality of status to all the major languages spoken in the country. This has led to the avoidance of social conflict on linguistic basis. The policy of accommodation has made administration of States easier. It has also ensured a larger participation in the government’s activities by people who speak various languages. |
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| 50. |
differential equation (2x + 3y) dx + (4x + y) dy = 0? |
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Answer» The given differential equation is (4x – 3y)dx + (2y – 3x)dy = 0 4xdx + 2ydy – 3(xdy + ydx) = 0 4xdx + 2ydy – 3d(xy) = 0 Integrating, we get 2x2 + y2 – 3xy = c which is the required solution of the given differential equation= |
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