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.

The electric potential `V(z, y,z)` for a planar charge distribution is given by: `V(x, y, z)={(0,,"for "x lt -d),(-V_(0)(1+x/d)^(2),,"for "-d le x le 0),(-V_(0)(1+2x/d),,"for "0 le x lt d),(-3V_(0),,"for " x gt d):}` where `-V_(0)` is the potential at the origin and d is a distance. Graph of electric field as a function of position is given asA. B. C. D.

Answer» Correct Answer - A
`E=(-dv)/(dx)`
2.

Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown below. Calculate the acceleration and the frictional force of the floor on the ball?

Answer»

Acceleration =  v - u / t = 80 – 0 / 8 = 10 ms-2

Force = m a×= 50 /1000 x 10 = 0.5 N

3.

Why step farming is common in hills?

Answer»

Hint— This is practiced to check soil erosion through water currents on the slopes.

4.

The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in a saturated salt solution, if the density of the solution is 1.2 g cm–3 ? What will be the mass of the solution displaced by this packet?

Answer»

Density of the sealed packet = m / v = 500/350 = 1.4 g cm-3

As the density of the packet is greater than that of the saturated salt solution, the packet will sink.

Mass of the solution displaced = Volume of the packet × Density of the solution.

= 350 × 1.2 = 420 g.

5.

The most pupular electrochemical cell, daniel cell was originally developed by the English chemist john F daniel the general assembly of the cell is as given below An undergraduate student made a Daniel cell using `100cm^(3)` of `0.100MCuSO_(4)` ad `0.100M` `ZnSO_(4)` solution respectively. The two compartments are connected by suitable salt bridge. [given `E_((Cu^(2+)//Cu))^(@)=0.34V,E_((Zn^(2+)//Zn))=-0.76V,(2.30RT)/(F)=0.06,log2=0.3]` Q. A labmate of the student asked her for some solid `CuCl_(2)`. while she was lifting the bottle from a shelf, the lid of the bottle slipped and some amount of `CuCl_(2)` fell in the `CuSO_(4)` compartment at constant volume. She measured the emf of the cell again and found that it has increase by 9 mV. She used this data to calculate the amount of `CuCl_(2)` that had spilled in the compartment. Calculate the mass of `CuCl_(2)` that had spilled into the daniel cell? (molar mass of `CuCl_(2)=135(g)/(mol))`A. `1.35g`B. `13.5g`C. `27g`D. `2.7g`

Answer» Correct Answer - A
`1.109=1.100-(0.06)/(2)log(0.1)/([Cu^(2+)])`
`[Cu^(2+)]=0.2M`
`Deltan=0.2xx0.1-0.1xx0.1`
`=0.01`
`w=0.01xx135=1.35` gm
6.

How many 4-letter words can be formed out of the letters of 'EGOIST' having at least one vowel? (repetition is not allowed)1. 12 × 4!2. 15 × 4!3. 21 × 4!4. 24 × 4!

Answer» Correct Answer - Option 2 : 15 × 4!

Concept:

  • The ways of arranging n different things = n!
  • The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
  • The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
  • The number of ways for selecting r from a group of n (n > r) = nCr 


Calculation:

There are 3 vowels O, I, E

Required number of ways N = 1 vowel and 3 consonants + 2 vowel and 2 consonants + 3 vowel and 1 consonants

⇒ N = [\(\rm\left(^3C_1×{^3C_3}\right) + \left({^3C_2}×{^3C_2}\right) + \left({^3C_3}×{^3C_1}\right)\)] × 4!

⇒ N =  (3 + 9 + 3) × 4!

⇒ N = 15 × 4! words

7.

From a group of 7 men and 6 women , a committee of 5 person with more males than females is to be formed . In how many ways can this be done ?1. 5642. 6453. 7354. 756

Answer» Correct Answer - Option 4 : 756

Formula Used:

Number of ways of selection of  r items out of  n items: nCr   = n!/((n-r)! × r!)

Concept:

Probable combinations of selection could be as follows:

Cases

Required number of Males

Required number of Females

I

3

2

II

4

1

III

5

0

 

Calculation:

Number of ways: 7C× 6C2 + 7C4 × 6C1 + 7C5 × 6C0

⇒ 35× 15 + 35× 6 + 21× 1

⇒ 756

Number of ways in which the following activities can be done = 756

8.

A team of 11 members is to be formed from 8 males and 5 females. If A is the number of ways the committee must have at least 6 males and B is the number of ways the committee must have at least 3 females then;1. A = B2. A = 2B3. A = 4B4. 2A = B5. 4A = B

Answer» Correct Answer - Option 1 : A = B

GIVEN:

Team members = 11

Male = 8

Female = 5

CALCULATION:

Case: A

Males = At Least 6 (6, 7, 8)

Females = (3, 4, 5)

Possible ways = 8C8 x 5C3 + 8C7 x 5C4 + 8C6 x 5C5

Possible ways = A = 78

Case: B

Females = At Least 3 (3, 4, 5)

Males = (6, 7, 8)

Possible ways = 8C8 x 5C3 + 8C7 x 5C4 + 8C6 x 5C5

Possible ways = B = 78

A = B

9.

Without saturation of transport proteins and under the same concentration gradient how can the speed of simple diffusion be compared to the speed of facilitated diffusion?

Answer»

The action of facilitator proteins in facilitated diffusion makes this type of diffusion faster than simple diffusion under equal concentration gradients of the moved substance.

10.

What are the three main types of passive transport? 

Answer»

The three main types of passive transport are simple diffusion, osmosis and facilitated diffusion.

11.

How does the intensity of facilitated diffusion vary in relation to the concentration of the moved substance? What is the limiting factor?

Answer»

Like simple diffusion facilitated diffusion is more intense when the concentration a gradient of the substance increases and less intense when the gradient lessens. In facilitated diffusion, however, there is a limiting factor: the quantity of the permeases that facilitate the transport through the membrane. Even in a situation in which the concentration a gradient of the diffusing substance increases, if there are not enough permeases to perform the transport there will be no increase in the intensity of the diffusion. This situation is called saturation of the transport proteins and it represents the point at which the maximum transport capacity of the substance across the membrane is achieved.

12.

How does the intensity of simple diffusion vary in relation to the concentration gradient of the moved substance?

Answer»

The higher the concentration gradient of a substance the more intense its simple diffusion will be. If the concentration gradient diminishes the intensity of simple diffusion diminishes too.

13.

What is the energy source used in active transport through biological membranes?

Answer»

The energy necessary for active transport (against the concentration gradient of the transported substance) to occur comes from ATP molecules. The active transportation uses chemical energy from ATP. 

14.

What is the difference between simple and facilitated diffusion? Facilitated by which type of molecule does the term “facilitated” mean?

Answer»

Simple diffusion is the direct passage of substances across the membrane in favour of their concentration gradient. In facilitated diffusion the movement of substances are also in favour of their concentration gradient but the substances move bound to specific molecules that act as “permeabilizers”, i.e., facilitators of their passage through the membrane.

15.

What are some examples of biological activities in which osmosis plays an important role?

Answer»

Hemolysis (destruction of red blood cells) by the entrance of water, the hydric regulation in plants and the entrance of water in the xylem of vascular plants are all examples of biological phenomena caused by osmosis.

Excessive dilution of the blood plasma causes, by osmosis, the entrance of too much water into red blood cells and then the destruction of these cells (hemolysis). Osmosis is also the main process for maintenance of the flaccid, turgid or plasmolytic states of plant cells. Osmosis is one of the forces responsible for the entrance of water into plant roots since root cells are hypertonic in comparison to the soil.

16.

What is the difference between osmosis and diffusion?

Answer»

Osmosis is the phenomenon of movement of solvent particles (in general, water) from a region of lower solute concentration to a region of higher solute concentration. Diffusion, on the other hand, is the movement of solutes from a region of higher solute concentration to a region of lower solute concentration.

One can consider osmosis as movement of water (solvent) and diffusion as movement of solutes, both concentration gradient-driven.

17.

Why are salt and sugar used in the production of dried meat and dried fruits?

Answer»

Substances that maintain a highly hypertonic environment, like sugar and salt, are used in the production of dried meat, fruits or fish (for example, cod) because the material to be conserved is then dehydrated and the resulting dryness prevents the growth of populations of decomposer beings (since these beings also lose water and die).

18.

What are vestigial organs? How they are different from atavistic organs?

Answer»

The organs of the body of the organisms which are non-functional but were functional in their early ancestors are called as the vestigial organs. eg. Coccyx, third eyelid, vermiform appendix in man. 

Atavistic organs are those vestigial organs which have become functional eg. Birth of human baby with a small tail, development of power of moving pinna in same persons and presence of additional mammae in some human beings.

19.

Three layered embryo is called--------- a. Blastula b. Gastrula c. Morula d. All of them

Answer»

Three layered embryo is called Gastrula.

20.

Calculate the physical fitness index using short formula for a 12 year old boy having completed Harvard Step Test for a duration of 3 minutes and a pulse rate of 54 beats for 1 to 1.5 minutes. 

Answer»

The athlete’s fitness index score is calculated with the help of following formula. Fitness index score = (100 × test duration in seconds) divided by (2 × sums of heartbeats in recovery period). 

(100 × 180 Sec)/ 2 × 54 = 500/3 = 166.66

21.

In germinating seeds Amylase, Proteases, Lipases are stimulated by)(1) Auxin (2) Gibberellin (3) Cytokinin (4) Ethylene

Answer»

(2) Gibberellin

22.

A pigment concerned with both floral induction and seed germination is(1) Florigen(2) Chlorophyll (3) Plastocyanin (4) Phytochrome

Answer»

(4) Phytochrome

23.

The olfactory umpulses are transmitted to the ………… lobe of brain.(a) Parietal (b) Temporal (c) Occipital (d) Frontal

Answer»

Correct Answer is :(d) Frontal

24.

Inflammation of joints due to accumulation of uric acid crystals is called as ………(a) Myasthenia gravis (b) Gout(c) Osteoporosis (d) Osteomalacia

Answer»

Correct Answer is : (b) Gout

25.

Select the non-narcotic drug, which is a paraaminophenol derivative: a) Analgin b) Aspirinc) Baclophen d) Paracetamol

Answer»

d) Paracetamol 

26.

Non-narcotic analgesics are all of the following drugs EXCEPT: a) Paracetamol b) Acetylsalicylic acid c) Butorphanol d) Ketorolac

Answer»

c) Butorphanol  

27.

Indicate the non-narcotic analgesic, which lacks an anti-inflammatory effect: a) Naloxone b) Paracetamol c) Metamizole d) Aspirin 

Answer»

b) Paracetamol

28.

Which one of the following non-narcotic agents inhibits mainly cyclooxygenase (COX) in CNS? a) Paracetamol b) Ketorolac c) Acetylsalicylic acid d) Ibuprofen

Answer»

Paracetamol non-narcotic agents inhibits mainly cyclooxygenase (COX) in CNS.

29.

Which bone is remnant of tall in human beings ?

Answer» Coccyx bone is the remnant of tail in human beings.
30.

Name the epoch when mammals were at the height of evolution.

Answer» Correct Answer - Miocene epoch.
31.

What is meant by coacervate?

Answer» Coacervate is the term given by Haldane to the substances present on the primitive earth.
32.

Read the following passage and answer the questions that follow:Vitamin C is an essential ingredient for human beings. It is required for the synthesis of intercellular cement “collagen”, which is responsible for keeping various cells of the human body together. As an oxidant, it acts like a cop in the body, apprehending the rowdy molecules called free radicals that are naturally formed during various chemical reactions in the body and have the potential to damage. Vitamin C is also required for dentine formation in the teeth and it also aids in the absorption of iron from the body.Rats have the capacity to synthesise vitamin C with their body whereas human beings are required to provide it through diet. The richest source of vitamin C is amla. The other sources are lemon, orange, papaya, pineapple, cabbage, cauliflower, green peas and tomato. While cooking above 70-degree centigrade vitamin C gets destroyed!Scurvy is a disease caused by the deficiency of this vitamin. Due to its deficiency, gums are swollen or bleeding, bones are weak and readily fractured, delayed wound healing and poor teeth formation in children result. A balanced diet with vitamin C is the requirement to avoid scurvy. For pregnant or lactating mothers vitamin C is the remedy.Each adult is required to consume 75 mg per day; an infant 30 mg per day; a pregnant woman 100 mg per day; a lactating mother 150 mg per day. Doctors use vitamin C for the treatment of infections, healing of ulcers, burns and trauma, quick healing of fractures, etc.However, taking vitamin C tablets on a long-term basis is not without side-effects, like the reformation of oxalates, kidney stones, diarrhoea, abnormal heart rhythm, damage to the outer. a layer of teeth and rebound scurvy if tablets are stopped abruptly. Amla with honey is a supertonic, particularly when treated with solar rays since honey is an ingredient that provides instant energy without consuming insulin for its absorption in the body.A newly born child in India is first provided with honey to purify his/her internal system since it acts as antibiotic and also without side – effects. Most of us are not aware that during World War II a lot of food material was sent to the British army fighting for and on behalf of the British Government in different parts of the world.In this food material, Amla was one of the main ingredients to protect fighting forces from the scurvy disease since during day-time they had to live in bunkers and during night-time, they had to fight. Therefore, they were generally deprived of sunlight which is a major source of vitamins for the human body. Therefore, vitamin C in the form of amla is the best for human health.1. Why is vitamin C important for us?2. What are the sources of vitamin C?3. What is the cause of the disease scurvy?4. How can the disease scurvy be prevented?5. What are the side effects of taking vitamin C tablets on long term basis?6. Why was Amla one of the main food ingredients for army people during world war II?Choose from the passage the words that mean…7. A situation when there is not enough of the people or things that are needed8. Suddenly.9. A woman producing milk from the breasts to feed a baby.

Answer»

1. Vitamin C is important because it is required for the synthesis of intercellular cement “collagen”, which is responsible for keeping various cells of the human body together.

2. Amla, lemon, orange, pineapple, cabbage, cauliflower, green peas and tomato, etc. are the sources of vitamin C.

3. The disease scurvy is caused due to deficiency of vitamin C.

4. The disease scurvy can be prevented by taking a balanced diet enriched with vitamin C.

5. The side effects of taking vitamin C tablets on a long term basis are – the reformation of oxalates, kidney stones, diarrhoea, abnormal heart rhythm, etc.

6. Amla was one of the main ingredients to save fighting forces from scurvy disease during World War II.

7. Deficiency

8. Abruptly

9. Lactating

33.

Which period is dominant for Amphibia?

Answer» Carboniferous is dominant period for Amphibia.
34.

A normal vision female whose father is colour-blind marries a normal vision male. What would be the probability of her sons and daughters to be colour blind?(a) 25% sons colour blind and all daughters with phenotypically normal vision(b) 50% sons colour blind and 50% daughter normal(c) 50% sons colour blind and 50% daughter colour blind(d) All sons normal and 50% colour blind daughter

Answer»

Answer (a) 25% sons colour blind and all daughters with phenotypically normal vision

35.

What volume of oxygen gas `(O_(2))` measured at `0^(@)C` and 1 atm is needed to burn completely `1 L` of propane gas `(C_(3)H_(8))` measured under the same condition?A. 7 LB. 6 LC. 5 LD. 10 L

Answer» Correct Answer - C
`{:(C_(3)Hg,+,5O_(2),to,3CO_(2),+,4H_(2)O),(22.4L,,5xx22.4L,,,,):}`
For the combustion of 22.4 L propane , oxygen required `=5xx22.4 L`
For the combustion of 1 L of propane
Oxygen required `=(5xx22.4)/(22.4)L= 5 L`
36.

Why `NF_3` cannot be hydrolysed, while `NCl_3` can be readily hydrolysed?

Answer» In `NH_3` atomic sizes of N and F are quite comparable and thus N-F bond is quite strong and `NF_3` is inert towards hydrolysis. Whereas atomic size of N and Cl vary much and `N-Cl` is much weak and `NCl_3` undergoes hydrolysis.
37.

How many of the following halides are singnificantly hydrolysed under the normal conditions: `BF_3 , NCl_3, NBr_3 ,BCl_3, PCl_5, C Cl_4, NF_3, SiCl_4`

Answer» Correct Answer - 6
`NCl_3, NBr_3, BCl_3, PCl_3, PCl_5, SiCl_4` are singnificantly hydrolysed under the normal conditionsl.
38.

Most stable trihalide from NF3, NBr3, NCl3 & Nl3 is (1) NF3(2) NCl3(3) NBr3(4) Nl3

Answer»

Correct option is (1) NF3

NFis stable while NCl3, NBr3, Nl3 are explosive.

39.

Which one of the following is least basic ?(a) NCl3(b) NBr3(c) NI3  (d) NF3 

Answer»

 NI3 is least basic

40.

The most stable trihalide of nitrogen is: (A) NF3 (B) NCl3 (C) NBr3 (D) NI3

Answer»

Correct option is (A) NF3

Order of stability:-

NF3 > NCl3 > NBr3 > NI3

41.

When SnCl2 solution is added excess into a solution of HgCl2 , a white ppt is turning grey. Why?  

Answer»

When SnCl2 solution is added to HgCl2 solution, initially white ppt, of Hg2Cl2 is formed which further reacts with SnCl2 forming metallic grey Hg

2HgCl2 + SnCl2 → SnCl4 + Hg2Cl2 (white)

Hg2Cl2 + SnCl2 → SnCl4 + 2Hg (grey)

42.

Which is a planar molecule ?(a) XeO4 (b) XeF4 (c) XeOF4 (d) XeO2F2

Answer»

 (b) XeF4 is planar 

43.

Four moles of an ideal diatomic gas `(gamma = 1.4)` at `300 K` and `12 atm` expanded irreversibly & adiabatically to a final pressure of `2.4 atm` against a constant pressure of `2 atm`. Calcualate `W` & `DeltaH` (magnitude only) . Report your answer as `(|X + Y|)/(10)`, where `W = X xx 10R, DeltaH = Y xx 10R` and `R` is gas constant.

Answer» Correct Answer - 9
`q = 0 rArr DeltaU = w`
`nC_(v)(T_(2) - T_(1)) = - P_(ext.)(V_(2) - V_(1))`
`C_(v)(T_(2) - 300) = -2 ((R xx T_(2))/(2.4) - (R xx 300)/(12))`
`rArr (5)/(2)(T_(2) - 300) = (360)/(6) - (T_(2))/(1.2)`
`5T_(2) - 1500 = 100 - (5T_(2))/(3)`
`(20T_(2))/(3) = 1600 :. T_(2) = 240 K`
`W = 4 xx (5R)/(2)(-60) = - 600R`
`DeltaH = gammaDeltaU`
or `DeltaH = (7)/(5) xx -600 R = -840 R`
44.

`1 gH_(2)` gas `STP` is expanded so that the volume is doubled. Hence, work done isA. (a)22.4 L-atmB. (b)5.6 L-atmC. (c )11.2 L-atmD. (d)44.8 L-atm

Answer» Correct Answer - C
`V_(1) ("volume of 1 g" H_(2))=11.2 L` at STP
`V_(2) ("volume of g" H_(2))=22.4 L`
`:. W=PDeltaV=11.2 L atm`
45.

A reaction occurs spontanecously if `:-`A. `TDeltaSoltDeltaH`and both `DeltaH,DeltaS` are `+ve`B. `TDeltaSgtDeltaH` and `DeltaH=+ve,DeltaS=-ve`C. `TDeltaSgtDeltaH` and both `DeltaH,DeltaS`are `+ve`D. `TDeltaS=DeltaH` and both `DeltaH,DeltaS` are `+ve`

Answer» Correct Answer - C
46.

Molecular shape of ` SF_4, CF_4` and ` XeF_4` are :A. The same with `2, 0` and `1` lone pair of electron respectively.B. The same with `1, 1` and `1` lone pair of electron respectively.C. Different with `0, 1` and `2` lone pair of electrons respectively.D. Different with `1, 0` and `2` lone pair of electrons respectively.

Answer» Correct Answer - D
`SF_4` has `sp^3` d-hybridisation with one lone pair, `CF_4` has `sp^3-"hybridisation"` with no lone pair and `XeF_4` has `sp^3 d^2-"hybridisation"` with two lone pair.
47.

Considering the above data, the standard electrode potential (E°) for Fe3+ + e → Fe2+   is(a) + 1.2 V(b) 0.404 V(c) 0.771 V(d) – 0.40 V

Answer»

Considering the above data, the standard electrode potential (E°) for Fe3+ + e → Fe2+   is 0.771 V .

48.

In coordination chemistry there are a variety of methods applied to find out the structure of complexes.One method involves treating the complex with known reagents and from the nature of reaction, the formula of the complex can be predicted.An isomer of the complex `Co(en)_2(H_2O)Cl_2Br`,on reaction with concentrated `H_2SO_4` (dehydrating agent) it suffers loss in weight and on reaction with `AgNO_3` solution it gives a white precipitate which is soluble in `NH_3`(aq). If all the ligands in the coordination sphere of the above complex be replaced by `F^(bar)`, then the magnetic moment of the complex ion (due to spin only) will be :A. 2.8 BMB. 5.9 BMC. 4.9 BMD. 1.73 BM

Answer» Correct Answer - C
With `F^(-)` ligand, metal ion has electron configuration , `t_(2g)^(2,1,1)e_(g)^(1.1)`
So, `mu=sqrt(4(4+2))~~4.9` B.M , where n=number of unpaired electrons
49.

In coordination chemistry there are a variety of methods applied to find out the structure of complexes.One method involves treating the complex with known reagents and from the nature of reaction, the formula of the complex can be predicted.An isomer of the complex `Co(en)_2(H_2O)Cl_2Br`,on reaction with concentrated `H_2SO_4` (dehydrating agent) it suffers loss in weight and on reaction with `AgNO_3` solution it gives a white precipitate which is soluble in `NH_3`(aq). The correct formula of the complex is :A. `[CoClBr(en)_2] H_2O`B. `[CoCl(en)_2 (H_2O)]BrCl`C. `[Co Br(en)_2 (H_2O)]Cl_2`D. `[CoBrCl(en)_2]Cl. H_2O`

Answer» Correct Answer - D
Water present in ionisation sphere is absorbed by conc. `H_2SO_4`(dehydrating agent).If `Cl^(-)` present as counter ion then reacts with `AgNO_3` to give white precipitate of AgCl
50.

`Co^(2+)(aq)+SCN^(-)(aq)to`Complex (X) `Ni^(2+)(aq.)+"Dimethylglyoxime"overset(NH_4OH)to`Complex (Y). The coordination number of cobalt and nickel in compleses X and Y are four. The IUPAC names of the complexes (X) and (Y) are respectively :A. Tetrathiocyanato-S-cobalt(II) and Bis(dimethylglyoximate) nickel(II)B. Tetrathiocyanato-S-cobaltate(II) and Bis(dimethylglymato) nickel(II)C. Tetrathiocyanato-S-cobaltate(II) and Bis(dimethylglymato) nickelate(II)D. Tetrathiocyanato-S-cobaltate(III) and Bis(dimethylglymato) nickel(II)

Answer» Correct Answer - B
`(X)=[Co(SCN)_4]^(2-),(Y)=[Co(dmg)_2]`