Explore topic-wise InterviewSolutions in Current Affairs.

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.

Avogardo's number(N) is equal to(a) 6.023 x 1024(b) 6.023 x 1023(c) 6.023 x 10-23(d) 11.2

Answer»

Answer (b) 6.023 x 1023

2.

Modern periodic table is given by(a) Debonair(b) Mendeleef(c) Mendel(d) none of them

Answer»

Answer (d) none of them

none of them
3.

Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is(a) Bi2O5 (b) BiF5 (c) BiCl5 (d) Bi2S5

Answer»

 The compound is (b) BiF5 

4.

Nitrogen is relatively inactive element because(a) its atom has a stable electronic configuration(b) it has low atomic radius(c) its electronegativity is fairly high(d) dissociation energy of its molecule is fairly high

Answer»

(d) N2 molecule contains triple bond between N atoms having very high dissociation energy (946 kJ mol–1) due to which it is relatively inactive.

5.

Which of the following has the highest pπ – pπ bonding tendency ?(a) N (b) P (c) As (d) Sb

Answer»

 (a) Nitrogen due to small size is able to show  pπ -pπ lateral overlap forming N = N, rest elements due to bigger size are not able to show pπ -pπ lateral overlap

6.

Nitrogen forms N2, but phosphorus is converted into P4 from P, the reason is(a) Triple bond is present between phosphorus atom(b) pπ – pπ bonding is strong(c) pπ – pπ bonding is weak(d) Multiple bond is formed easily

Answer»

(c) Nitrogen form N2 (i.e. N = N) but phosphorus form P4 , because in P2 , pπ — pπ bonding is present which is a weaker bonding

7.

Which of the following statements is not correct for nitrogen?(a) Its electronegativity is very high(b) d-orbitals are available for bonding(c) It is a typical non-metal(d) Its molecular size is small

Answer» (b) In case of nitrogen, d-orbitals are not available
8.

The three important oxidation states of phosphorus are(a) –3, +3 and +5 (b) –3, +3 and –5(c) –3, +3 and +2 (d) –3, +3 and +4

Answer»

(a) –3, +3 and +5

9.

Pick out the wrong statement.(a) Nitrogen has the ability to form pπ -pπ bonds with itself.(b) Bismuth forms metallic bonds in elemental state.(c) Catenation tendency is higher in nitrogen when compared with other elements of the same group.(d) Nitrogen has higher first ionisation enthalpy when compared with other elements of the same group

Answer» (c) Catenation tendency is higher in phosphorus when compared with other elements of same group
10.

Collectively the elements of group 15 are called –(a) pnicogens (b) pnicopens(c) nicopen (d) None of these

Answer» (a) Collectively these elements are called pnicogens and their compound pniconides
11.

Which one of the following elements is most metallic ?(a) P (b) As (c) Sb (d) Bi

Answer» (d) Metallic character increases down the group, Bi is most metallic
12.

Which of the follow group 15 element forms metallic bonds in elemental state ?(a) As (b) P (c) Sb (d) B

Answer» (d) Bismuth forms metallic bonds in elemental state
13.

Which of the following statement is incorrect for group 15 elements ?(a) Order of ionization enthalpies is ΔiH1 < ΔiH2 < ΔiH3(b) The boiling point and melting point increases from top to bottom in the group(c) Dinitrogen is a gas while all others are solids(d) All statements are correct

Answer» (b) The melting point in group 15 increases upto arsenic and then decreases upto bismuth.
14.

Maximum covalence of nitrogen is '4' but the heavier elements of group 15 show covalence greater than '4' Why ?

Answer»

[Hint : Nitrogen is not having vacant d- orbitals in its octet, hence cannot expand its octet, while heavier members have vacant d-orbitals to expand their octet.]

15.

How does ammonia solution react with Ag+ (aq) ? Write the balanced chemical equation.

Answer»

[Hint : Ag+ (aq) + 2NH3 (aq) => [Ag(NH3)2]+ (aq)]

16.

Choose the weakest acid among the following :A. `F_(3)"CCOOH"`B. `FCH_(2)COOH`C. `CH_(3)COOH`D. `(CH_(3))_(2)CHCOOH`

Answer» Correct Answer - D
17.

Among the following which one is most basic in aqueous solution ?A. `NH_(3)`B. `CH_(3)NH_(2)`C. `(CH_(3))_(2)NH`D. `(CH_(3))_(3)N`

Answer» Correct Answer - C
18.

The airport is A LONG WAY FROM the centre of the city. A) far from B) not far from C) in the distance D) remote E) distant

Answer»

Correct option is B) not far from

19.

His parents were very RELIGIOUS and the boy had to sing at church services. A) devoted B) faithful C) reluctant D) atheistic E) competent

Answer»

Correct option is D) atheistic

20.

Which of the following represent a set of nucleophiles?(a) BF3 , H2O, NH2- (b) AlCl3 , BF3 , NH3 (c) CN- , RCH2- , ROH (d) H+ , RNH3+ , CCl2

Answer»

(c) CN- , RCH2- , ROH 

21.

Consider the following statements:(I) E2 reaction is a bimolecular elimination reaction of second order (II) E2 reaction takes place in two steps. (III) E2 reaction generally Jakes place in primary alkyl halides.Which of the above statements is/are not correct? (a) (I) only (b) (II) only (c) (III) only (d) (I) & (III)

Answer»

(II) E2 reaction takes place in two steps.

22.

Which structure does NOT play a part in the motion of cells?a) microvilli b) cilia c) flagella d) pseudopodia

Answer»

 MICROVILLI does NOT play a part in the motion of cells.

23.

Which structure does NOT play a part in the motion of cells?a) microvilli b) cilia c) flagella d) pseudopodia

Answer»

Which structure does NOT play a part in the motion of cells  MICROVILLI

24.

Which type of muscle is a syncytium?a) skeletal b) cardiac c) smooth d) all of the above

Answer»

SKELETAL muscle is a syncytium.

25.

When the potential difference across a membrane of a neuron equals the threshold, what results?a) movement of the membrane b) action potential c) relaxation d) contraction

Answer»

 ACTION POTENTIAL

26.

Which of the following is NOT a type of neuron?a) sensoryb) motorc) associationd) stimulatory

Answer»

 STIMULATORY.

27.

Which of the following are NOT part of a neuron? a) synapse b) axon c) Nissl bodies d) dendrite

Answer»

SYNAPSE are NOT part of a neuron.

28.

From which grandparent or grandparents did you inherit your mitochondria , Is it your:a) mother's parents b) paternal grandfather c) grand mothers d) maternal grandmother

Answer»

maternal grandmother

29.

Environmental pollution affectsA. Biotic componentsB. Plants onlyC. Man onlyD. Biotic and abiotic components of environment

Answer» Correct Answer - D
(D) Environmental pollution affects biotic and abiotic components of environment.
30.

Name the outer portion of a stem or root, bounded externally by the epidermis, and internally by the cells of the pericycle.

Answer»

Cortex is the outer portion of a stem or root, bounded externally by the epidermis, and internally by the cells of the pericycle.

31.

Deforestation is not a major environmental problem in w) the Amazon x) Borneo y) Oregon z) West Germany 

Answer»

WEST GERMANY .

32.

In most species of Paramecium there are how many contractile vacuoles? Is it:a) one b) two c) three d) four

Answer»

In most species of Paramecium there are Tow contractile vacuoles.

33.

POPULATION GROWTHA. EmigrationB. MortalityC. NatalityD. All the above

Answer» Correct Answer - D
(D) Population growth is indicated by emigration, mortality and natality.
34.

The major fibrous proteins are:a) peptone and edestin b) glutelin and leucinec) valine and lysine d) myosin and actin

Answer»

Myosin and Actin.

35.

How many statements are correct? (a) Hind II always cut DNA molecules at particular point by recognizing a specific sequence of 6 bases. (b)After the restriction endonuclease action, single stranded portions which are overhanging stretches called sticky ends because they form nucleotide bonds with their complementary cut counter part. (c)The separated bands of DNA are cut out from the agarosegel and extracted from gel piece. This step is known as spooling. (d)Taq polymerase is used between annealing and extension of r-DNA technology.A. 1B. 2C. 0D. 3

Answer» Correct Answer - C
(C) (a) Hind-II always cut DNA molecules at particular point by recognising a specific sequence of 6 base pairs.
(b) After the restriction endonuclease action, single stranded portions which are overhanging stretches called sticky ends because they form hydrogen bonds with their complementary cut counter part.
(c) The separated bands of DNA are cut out from the agarose gel and extracted from gel piece. This step is known as elution.
(d) Taq polymerase is used between annealing and extension of PCR.
36.

Which of the following is not a major pollutant from automobiles.w) carbon monoxide x) unburned hydrocarbons y) nitrous oxide z) sulfur dioxide

Answer»

 SULFUR DIOXIDE .

37.

What are the advantages of using natural gas (or CNG) as a fuel?

Answer»

CNG Run vehicles have low maintenance cost as compared to other vehicles this significantly less emission of pollutants like carbon dioxide oxides of Nitrogen and Sulphur and other better than petrol and diesel therefore the use of CNG can help in reducing global warming.

38.

Which of the following is believed to be responsible for the hole in the ozone layer over Antarctic?w) carbon dioxide x) compounds containing sulfur y) radioactivity z) compounds containing chlorine 

Answer»

 COMPOUNDS CONTAINING CHLORINE 

39.

What are the major products (or fractions) of petroleum refining? Give one use of each petroleum product.

Answer»

Petroleum refinery is an industrial process plant where crude oil is transformed and refined into more useful products such as petroleum, diesel fuel, asphalt base, heating oil, kerosene, liquefied petroleum gas, jet fuel and fuel oils.etc.
The use of liquefied petroleum gas

It is used in domestic household work such as cooking equipments and fuels

40.

The blastula develops into the: a) gastrulab) morula c) endoderm d) zygote

Answer»

 The blastula develops into the GASTRULA .

41.

A layer of dead skin cells is found in the:a) subcutaneous tissue b) dermis c) epidermis d) no dead cells are in the skin

Answer»

 A layer of dead skin cells is found in the EPIDERMIS.

42.

The nervous system develops from which germ layer?a) ectoderm b) mesoderm c) endoderm d) none of the above

Answer»

The nervous system develops from ECTODERM germ layer.

43.

The primary role of aldosterone is1. Conserve sodium, promote potassium excretion2. Conserve potassium, promote sodium excretion3. Conserve sodium and potassium4. Promote excretion of sodium and Potassium

Answer» Correct Answer - Option 1 : Conserve sodium, promote potassium excretion

Explanation:

Aldosterone

  • Aldosterone, a steroid hormone secreted by the adrenal glands.
  • Aldosterone serves as the principal regulator of the salt and water balance of the body and thus is categorized as a mineralocorticoid.
  • It also has a small effect on the metabolism of fats, carbohydrates, and proteins.
  • Aldosterone’s primary function is to act on the late distal tubule and collecting duct of nephrons in the kidney, directly impacting sodium absorption and potassium excretion.
  • It also indirectly affects the excretion of hydrogen ions by changing the amount of potassium in the lumen of the nephron, causing downstream consequences on alpha-intercalated cells.
  • It affects blood pressure by regulating the amount of sodium (and the chloride that diffuses with sodium across the membranes) by increasing or decreasing the total amount of volume in the extracellular fluid (ECF).
  • This is not to be confused with the effect of anti-diuretic hormone (ADH). ADH is often released simultaneously with aldosterone.
  • This allows for blood pressure control by causing the release and fusion of aquaporin channels into the membrane of the principal cells. Water will then be reabsorbed into the ECF.
  • Together these two hormones can cause an increase in the amount of water taken up through the nephron, therefore increasing blood pressure.
44.

The Sub Centre is renamed as 1. Health & sciences Centre2. Health Centre3. Health and wellness Centre4. Health sciences treatment and palliative centre

Answer» Correct Answer - Option 3 : Health and wellness Centre

Explanation:

Health and Wellness Centres (HWCs)

  • In February 2018, the Government of India's announced the creation of 1,50,000 Health and Wellness Centres (HWCs) by transforming existing Sub Centres and Primary Health Centres as the base pillar of Ayushman Bharat.
  • These centres would deliver Comprehensive Primary Health Care (CPHC) bringing healthcare closer to the homes of people covering both maternal and child health services and non-communicable diseases, including free essential drugs and diagnostic services.
  • Health and Wellness Centers, are envisaged to deliver and expanded range of services to address the primary health care needs of the entire population in their area, expanding access, universality and equity close to the community.
  • The emphasis of health promotion and prevention is designed to bring focus on keeping people healthy by engaging and empowering individuals and communities to choose healthy behaviours and make changes that reduce the risk of developing chronic diseases and morbidities.
  • The delivery of Universal Comprehensive Primary Health Care, through HWCs will increase the health system responsiveness to people by bringing services closer to the communities and being able to address the needs of most marginalized, through the Primary Health Care team.
45.

The (∂E/∂T)p of different types of half cells are as follows : (Where E is the electromotive force) Which of the above half cells would be preferred to be used as reference electrode ? (A) A (B) B (C) C (D) D

Answer»

Correct option is (C) C

A cell with less variation in EMF with temperature is preferred as reference electrode because it can be used for wider range of temperature without much derivation from standard value so a cell with less (∂E/∂T)p is preferred.

46.

Before birth, the alveoli in a baby’s lungs are expanded and filled with1. Air2. Fluid3. Blood4. Meconium

Answer» Correct Answer - Option 2 : Fluid

Explanation:
Changes in the newborn at birth refer to the changes an infant's body undergoes to adapt to life outside the womb.
Lungs, Heart, and Blood Vessels

  • The mother's placenta helps the baby "breathe" while it is growing in the womb.
  • Oxygen and carbon dioxide flow through the blood in the placenta. Most of it goes to the heart and flows through the baby's body.
  • At birth, the baby's lungs alveoli are filled with fluid. They are not inflated.
  • The baby takes the first breath within about 10 seconds after delivery.
  • This breath sounds like a gasp, as the newborn's central nervous system reacts to the sudden change in temperature and environment.

Once the baby takes the first breath, a number of changes occur in the infant's lungs and circulatory system:

  • Increased oxygen in the lungs causes a decrease in blood flow resistance to the lungs.
  • Blood flow resistance of the baby's blood vessels also increases.
  • Fluid drains or is absorbed from the respiratory system.
  • The lungs inflate and begin working on their own, moving oxygen into the bloodstream and removing carbon dioxide by breathing out (exhalation).
47.

Complete the equations for the following nuclear processes: (a)`._(17)^(35) Cl + ._(1)^(0)n rarr... + ._(2)^(4)He` (b) `._(92)^(235)U + ._(0)^(1) n rarr ...+ ._(54)^(137)Xe + 2 _(0)^(1)n` (c) `._(13)^(27) Al + ._(2)^(4) He rarr ... + ._(0)^(1) n` (d) `...(n,p) ._(16)^(35) S` (e) `._(94)^(239) Pu (alpha, beta^(-))...`

Answer» Correct Answer - `{:(a.``._(15)P^(32),,b. ``._(38)Sr^(97),,c. ``._(15)P^(30)),(d.``._(17)Cl^(35),,e. ``._(97)P^(243),,):}`
`a.` `._(17)Cl^(35)+``._(0)n^(1)rarr ``._(15)P^(32)+``._(2)He^(4)`
`b.` `._(92)U^(235)=``._(0)n^(1)rarr ``._(38)Sr^(97)+``._(54)Xe^(137)+2``._(0)n^(1)`
`c.` `._(13)Al^(27)+``._(2)He^(4)rarr``._(15)P^(30)+``._(0)n^(1)`
`d.` `._(17)Cl^(35)+``._(0)n^(1) rarr ``._(16)S^(350+``._(1)H^(1)`
`e.` `._(94)Pu^(239)+``._(2)He^(4)rarr ``._(97)Bk^(243)+``._(-1)e^(0)`
48.

The first line of drugs for the treatment of tuberculosis are:1. Isoniazid, Rifampicin, Pyrazinamide, Ethambutol and streptomycin2. Rifampicin, INH, streptomycin, PAS and Amikacin3. Rifampicin, INH, streptomycin, Kanamycin and Ethambutol4. Rifampicin, INH, Streptomycin, Kanamycin and PAS

Answer» Correct Answer - Option 1 : Isoniazid, Rifampicin, Pyrazinamide, Ethambutol and streptomycin

Explanation:

Treatment for TB Disease

  • When TB bacteria become active (multiplying in the body) and the immune system can’t stop the bacteria from growing, this is called TB disease.
  • TB disease will make a person sick. People with TB disease may spread the bacteria to people with whom they spend many hours.
  • It is very important that people who have TB disease are treated, finish the medicine, and take the drugs exactly as prescribed.
  • If they stop taking the drugs too soon, they can become sick again; if they do not take the drugs correctly, the TB bacteria that are still alive may become resistant to those drugs.
  • TB that is resistant to drugs is harder and more expensive to treat.
  • TB disease can be treated by taking several drugs for 6 to 9 months. 

The first-line anti-TB agents that form the core of treatment regimens are:

  1. Isoniazid (INH)
  2. Rifampicin (RIF)
  3. Ethambutol (EMB)
  4. Pyrazinamide (PZA)
  5. Streptomycin
49.

Calculate the binding energy per nucleon for `C^(12),N^(14), O^(16)`, and comment on their relative magnitudes. Masses of proton and neutron are `1.0078` and `1.0087m_(u)`, respectively. `(m_(u)=931MeV)`

Answer» Correct Answer - `C^(12)=7.68MeV`
`N^(14)=7.68MeV`
`O^(16)=7.68MeV`
Stability is same.
`a. C^(12),` proton `=6`, neutron `=6`
`Delta_(m)=(6xx1.0078+6xx1.0087)-12`
`=(6.0468+6.0522)-12`
`=12.099-12=0.099m _(u)`
`BE` per nucleus
`=(0.099xx931)/(12)MeV=7.68MeV`
`b.` `N^(14), proton =7`, neutron `=7`
`Delta m=(7xx1.0078+7xx1.0087)-14`
`=(7.0546+7.0609)-14`
`=14.1155-14=0.1155m_(u)`
`BE` per nucleus
`=(0.1155xx931)/(14)MeV=7.68m_(u)`
`c.` `O^(16),` proton `=8` neutron `=8`
`Delta_(m)=(8xx1.0078+8xx1.0087)-16`
`=(8.0624+8.0696)-16`
`=0.132m_(u)`
`BE` per nucleus
`=(0.132xx931)/(16)MeV=7.68MeV`
50.

Why do radioactive element decay?

Answer» The stable nuclei are found to have `n//p` ratio in the range `1` to `1.5` . The nuclei whose `n//p` ratio lies outside this range `i.e.,(lt1)` or `(gt1.5)` lose `alpha-`or `beta-`particles so that their `n//p` ratio shifts into the stability belt.