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.

What is the API gravity of a substance with specific gravity 1?(a) 131.5(b) 141.5(c) 10(d) 50

Answer» Right choice is (c) 10

To explain I would say: API = (141.5/sp.gr) – 131.5 = 141.5/1 – 131.5 = 10.
2.

What is the specific gravity of 10 Kg of water occupied in 5 m3 with respect to 100 g/m^3?(a) 1(b) 5(c) 10(d) 20

Answer» Correct choice is (d) 20

For explanation I would say: Specific gravity = (10/5)/0.1 = 20.
3.

What is the mass of a cone of radius 1 m and height 3 m having specific gravity 0.1? (Density of water = 1000 kg/m^3)(a) 314.16 kg(b) 425.24 kg(c) 136.16 kg(d) 325.24 kg

Answer» Correct choice is (a) 314.16 kg

The explanation is: Volume of cone = (1/3) π1^2*3 = 3.1416 m^3, density of cone = 0.1*1000 = 100 kg/m^3, => mass of cone = 100*3.1416 = 314.16 kg.
4.

API gravity is necessary for the calculation of specific gravity of petroleum products, because(a) Volume of the petroleum products changes with temperature(b) Mass of petroleum products changes with temperature(c) Potential energy of petroleum products changes with temperature(d) None of the mentioned

Answer» Right choice is (a) Volume of the petroleum products changes with temperature

The best I can explain: Since volume of petroleum products changes significantly with temperature, that means density of petroleum products changes significantly with temperature, so that’s why specific gravity of petroleum products changes with temperature, and that’s why API gravity is required.
5.

A cylinder completely filled with water has density = 10 kg/m^3 and volume 5 m^3 , now a cube with side 1 m and density 25 kg/m^3 is dipped into the cylinder with some water dropping out, what is the final density of cylinder?(neglect the mass of cylinder)(a) 13 kg/m^3(b) 15 kg/m^3(c) 17 kg/m^3(d) 19 kg/m^3

Answer» Right choice is (a) 13 kg/m^3

Easy explanation: Initial mass of cylinder = 10*5 = 50 kg, mass of cube = 25*1 = 25 kg, => mass of water coming out = 10*1 = 10 kg, => final density of cylinder = (50 + 25 – 10)/5 = 13 kg/m^3.
6.

What is the unit of specific gravity?(a) Kg/m^3(b) N/m^3(c) m/s^2(d) Dimensionless

Answer» The correct answer is (d) Dimensionless

The explanation: Specific gravity of a substance is the ratio of density of the substance and density of a reference substance, so it is dimensionless.
7.

It is necessary to mention the temperature at which specific gravity is calculated, because(a) Mass of the substance changes with temperature(b) Rigidity of the substance changes with temperature(c) Density of the substance changes with temperature(d) None of the mentioned

Answer» The correct choice is (c) Density of the substance changes with temperature

Explanation: Since the volume of the substance changes with temperature, which means density of the substance changes with temperature, that’s why it is required to mention the temperature at which specific gravity is calculated.
8.

What is the specific gravity of a substance with density 100 kg/m^3 with respect to reference substance of density 100 lb/m^3?(a) 1.1(b) 2.2(c) 3.3(d) 4.4

Answer» Right answer is (b) 2.2

To elaborate: Density of the substance = 100*2.2 = 220 lb/m^3, Specific gravity = (100*2.2)/100 = 2.2.
9.

What is the specific gravity of N2 at 27oC and 6 Pa with respect to 0.1 Kg/m^3?(a) 0.16(b) 0.33(c) 0.51(d) 0.89

Answer» Correct option is (b) 0.33

Easiest explanation: Specific gravity = (6*14/8.314*300)/0.1 = 0.33.
10.

What is the specific gravity of O2 at 27oC and 60 Pa with respect to 1 Kg/m^3?(a) 0.15(b) 0.28(c) 0.54(d) 0.76

Answer» Correct option is (d) 0.76

Easy explanation: Specific gravity = (60*32/8.314*300)/1 = 0.76.
11.

Mass of a substance is changing with respect to its volume as m = -v^3 + 3v^2 -2v, what is the maximum specific gravity of the substance between v = 1 m^3 and v = 2 m^3? (Density of reference substance = 0.25 kg/m^3)(a) 1(b) 2(c) 3(d) 4

Answer» Right option is (a) 1

The explanation: Density of substance D = m/v = -v^2 + 3v – 2, => Specific gravity of substance S = D/0.25 = 4(-v^2 + 3v – 2), => dS/dv = 4(-2v + 3) = 0, => v = 1.5, => S is maximum at v = 1.5, => maximum specific gravity = 4(-1.5^2 + 3*1.5 – 2) = 1.
12.

A vessel of volume `3V` contains a gas at pressure `4 P_(0)` and another vessel of volume `2V` contains same gas at pressure `1.5 P_(0)`. Both vessels have same temperature. When both vessels are connected by a tube of negligible volume, the equilibrium is `IP_(0)`, where `I` is an integer. Find the value of `I`.A. `P_(0)`B. `2P_(0)`C. `3P_(0)`D. `4P_(0)`

Answer» Correct Answer - C
No change of temperature
So, `(n_(1) + n_(2))_(i) = (n_(1) + n_(2))_(f)`
`(P_(1)V_(1))/(RT) + (P_(2)V_(2))/(RT) = (PV_(1))/(RT) + (PV_(2))/(RT)`
So, `P = (P_(1)V_(1) + P_(2)V_(2))/(V_(1) + V_(2))`
`= ((4P_(0)) (3v) + (1.5 P_(0)) (2V))/(5V)`
`= 3 P_(0)`
13.

A force `vec(F) == b (-y hat(i) + x hat(j))/(x^(2) + y^(2)) N` (b is a constant) acts on a particle as it undergoes counterclock-wise circular motion in a circle : `x^(2) + y^(2) = 16`. The work done (in J) by the force when the particle undergoes one complete revolution is (x, y are in m)A. ZeroB. `2 pi b`C. `2b`D. None of these

Answer» Correct Answer - B
`W = oint vec(F) .d vec(r)`
`= oint (bd vec(r))/(|d vec(r)|R).d vec(r)`
`= (b)/(R) oint |d vec(r)| = (2 pi Rb)/(R) = 2 pi bJ`
14.

The end of a chain of length L and mass per unit length `rho`, which is piled up on a horizontally platform is lifted vertically with a constant velocity u by a variable force F. Find F as a function of height x of the end above platform:A. `rho (gx + 2u^(2))`B. `rho (2 gx + rho u^(2))`C. `rho (gx + u^(2))`D. `rho (u^(2) - gx)`

Answer» Correct Answer - C
Force is rate of change of momentum in chain plus the weight of chain at x
`F = rho xg + rho u^(2)`
15.

In development, neruvous system (brain,spinal cord,nerve cells) isA. EctodermalB. EndodermalC. EctomesodermalD. Endomesodermal

Answer» Correct Answer - A
(A) In development, nervous system (brain, spinal cord, nerve cells) is ectodermal.
16.

The electric current in a discharging R-C circuit is given by `i=i_0e^(-t/RC) where i_0` R and C are constant parameters and t is time. Let `i_0=2.00A, R=6.00xx10^5 ohm and C=0.500 muF`. a. Find the current at t=0.3 s. b. Find the rate of change of current at t=0.3 s. Find approximately the current at t=0.31 s.

Answer» Correct Answer - (a)2/e (b)-20/3e (c)5.8/3e
Equation `i =i_0.e^(-t/RC), where
i_0=2A, R=6x10^5 ohm
C=0.0500xx10^(-6)F
=5x10^(-7)F
=5xx10^(-7)F
i=2.0 e^)-t/0.3)
a. i=2xxe^(-1)= 2/e amp.
b.di/dt=(-i_0)/RC.e^(-t/RC)
when t=0.3 sec, then di/dt=2/(0.3) e^)(-0.3)/(0.3))(-20)/(3e) amp/sec
c. At t= 0.31 st
i= 2e((-0.3)/0.3), 5.8/(3e) amp (approx.)
17.

If alpha, beta are roots of the equation ax2 + 3x + 2 = 0 (a< 0), thenAlpha square by beta +beta square by alpha greater than

Answer» Since a << 0, therefore discriminant D=9−8a>0D=9-8a>0.
So, αandβαandβ are real.
We have, α+β=−3aandαβ=2aα+β=-3aandαβ=2a
∴ α2β+β2α=α3+β3αβ=(α+β)3−3α(α+β)αβ∴ α2β+β2α=α3+β3αβ=(α+β)3-3α(α+β)αβ
⇒ α2β+β2α=(α+β)3αβ−3(α+β)=−272a2+9a<0 [∵a<0]
18.

 Root development is promoted by (1) Auxin (2) Gibberellin (3) Ethylene (4) Absicisic acid

Answer»

Root development is promoted by Auxin.

19.

One of the commonly used plant growth hormone in tea plantations is(1) Absicisic acid (2) Zeatin (3) Indole-3-acetic acid (4) Ethylene

Answer»

One of the commonly used plant growth hormone in tea plantations is Zeatin.

20.

Oxygen molecule is(a) diamagnetic with no-unpaired electron(s)(b) diamagnetic with two unpaired electrons(c) paramagnetic with two unpaired electrons(d) paramagnetic with no unpaired electron(s)

Answer» (c) It is paramagnetic with two unpaired electrons
21.

The number of electrons that are paired in oxygen molecule are(a) 16 (b) 12 (c) 14 (d) 7

Answer»

(c) Total number of electrons in O2 is 16. It has 2 unpaired electrons, the rest 14 are paired.

22.

My body is a gametophyte and I grow under moist and shaded conditions. My group is commonly known as amphibians of the plant kingdom. Name my group.

Answer»

Group is Bryophyta

23.

What is the difference between Ectoderm and Endoderm?

Answer»

• Ectoderm is the outermost layer of the primary germ cells, but the endoderm is the innermost layer of the early embryo.

• Both cell layers line some common as well as separate organs but endoderm never lines any exteriorly exposed organ.

• Few genes are required to form the ectoderm, but most of the genes of the genome are required to form the endoderm.

• The endoderm cells are mostly columnar shaped while there is no particular shape or has almost all the shapes of cells in ectodermic cells after differentiation.

24.

Which is completely absent in cestodaa) digestive systemb) alimentary canalc) reproductive systemd) both a and b

Answer» a) Digestive System :)
25.

Plants that have a foliar root system

Answer»

Foliar roots: They arise from petiole (e.g., Pogostemon, rubber plant etc.) or veins of leaf due to some injury. These can also be induced by application of hormones. Some foliar buds can produce foliar roots, e.g., Bryophyllum, Begonia etc.

26.

Duck-billed-platypus isa) oviparousb) viviparousc) ovoviviparousd) vivioviparous

Answer»

Duck-billed-platypus is 

a) oviparous 

b) viviparous

c) ovoviviparous 

d) vivioviparous

they are one of the mammals who lay eggs rather than giving birth to young ones. thus they are oviparous.

27.

Choose the correct combination (a) Aves and Chordata – Classes (b) Annelida and Porifera – Phyla (c) Mollusca and Hydrozoa – Classes (d) Oligochaeta and Arthropoda – Phyla

Answer»

(b) Annelida and Porifera – Phyla 

28.

What is meant by holoblastic cleavage?

Answer»

Once an egg is fertilized, it rapidly divides into many smaller cells. This is called cleavage. But just as there are many different animals in the world, there are also many cleavage patterns. Here you'll learn about the patterns specific to holoblastic, or complete cleavage.

It is characterized by complete cleavage that divides the whole egg into distinct and separate blastomeres — compare meroblastic :)

29.

Five kingdom classification was proposed by (a) Woese (b) Haeckel (c) Darwin (d) Whittaker

Answer»

(d) Whittaker was most active in the areas of plant community analysis, succession, and productivity. He also first proposed the five-kingdom taxonomic classification of the world's biota into the Animalia, Plantae, Fungi, Protista, and Monera.

30.

‘Sanjeevani booti’ is (a) Selaginella kraussiana (b) Selaginella chrysocaulos (c) Selaginella bryopteris (d) None of the above

Answer»

(c) Selaginella bryopteris 

31.

The unique feature of bryophytes being member of kingdom plantae is that (a) they lack roots. (b) they produce spores. (c) they lack vascular tissue. (d) their sporophyte is attached to gametophyte.

Answer»

(c) they lack vascular tissue. 

32.

____ is useful in urinary disease. (A) Lobaria (B) Usnea (C) Parmelia (D) Peltigera

Answer»

Usnea is useful in urinary disease.

33.

Non-motile and thin walled spored of algae are known asA. ZygosporesB. ZoosporesC. AplanosporesD. Hypnospores

Answer» Correct Answer - C
Non-motile and thin walled spores of algae are known as aplanospores.
34.

Inducer molecule in lac-operon of E.coli is chemically- (a) Disaccharides (b) Amino acid (c) Protein (d) RNA

Answer»

Inducer molecule in lac-operon of E.coli is chemically Disaccharides.

35.

Define decomposition and describe the processes and products of decomposition

Answer»

Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients. The various processes involved in decomposition are as follows:

  •  Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.
  •  Leaching: It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.
  •  Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.
  •  Humification: The next step is humification which leads to the formation of a darkcoloured colloidal substance called humus, which acts as reservoir of nutrients for plants.
  •  Mineralization: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization.

Decomposition produces a dark coloured, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO2, water, and other nutrient in the soil.

36.

Give an example for:(a) An endothermic animal(b) An ectothermic animal(c) An organism of benthic zone

Answer»

(a) Endothermic animal: Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such as bears, cows, rats, rabbits, etc. are endothermic animals.
(b) Ectothermic animal: Fishes such as sharks, amphibians such as frogs, and reptiles such as tortoise, snakes, and lizards are ectothermic animals.
(c) Organism of benthic zone: Decomposing bacteria is an example of an organism found in the benthic zone of a water body

37.

Identify which of the following shows +1 and -I effect?(I) -NO2 (II) -SO3H (III) -I (IV) -OH (V) CH3O- (VI) CH3-

Answer»
+ I - effect-1 - effect
(v) CH3O(i) -NO2
(vi) CH3(ii) - SO3H
 (iii) -I
(iv) -OH

38.

One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?

Answer»

Given that [PCl5]initial = \(\frac{1 mol}{dm^3}\)

[Cl2]eq = 0.6 mol dm-3 

PCl5 ⇄ PCl3 + Cl2 

[PCl5]eq = 0.6 mole dm-3 

[PCl5]eq = 0.4 mole dm-3

∴ KC = \(\frac{0.6 \times 0.6}{0.4}\)

KC = 0.9

39.

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer»

Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy. 

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.

(Gas + Solvent ⇄ Solution + Heat)

40.

Read the following lines and answer the questions that follow. Gainst death and all-oblivious enmityShall you pace forth; your praise shall still find room, Even in the eyes of all posterity That wear this world out to the ending doom. So, till the judgement that yourself arise, You live in this, and dwell in lovers’ eyes. (i) What shall the rhyme pace forth against?(ii) Where shall its praise still find room?(iii) Where will posterity wear this world out to?(iv) Until the judgement, where does the poem dwell?

Answer»

(i) Against death and all-oblivious enmity.

(ii) In the eyes of all posterity.

(iii) It will wear the world out to the ending doom.

(iv) In lovers’ eyes.

41.

Read the following lines and answer the questions that follow. When wasteful war shall statues overturn, And broils root out the work of masonry, Nor Mars his sword nor war’s quick fire shall burn The living record of your memory.(i) What is the effect of war?(ii) What roots out the work of masons?(iii) What things cannot affect the living record referred to here ?(iv) What is beyond burning by war?

Answer»

(i) War overturns statues.

(ii) Broils root out the work of masons.

(iii)  War, broils, and Mars’ sword cannot destroy the living record.

(iv) The living record of the poem’s memory.

42.

Find the word which is out of the logic list:A) expect B) await C) disappoint D) wait for

Answer»

Correct option is C) disappoint

43.

Sunil often reaches his office late. He takes regular leaves due to bad health and mismanagement. Sunil is unable to work properly due to stress. He is also suffering from lower back pain and BP. Can the situation be changed through yoga? Write about yogic management for the same?

Answer»

Sunil suffers from BP, lower back pain and stress,These diseases can be treated by yoga. 

Yogic Management 

1. Kriyas: Jalaneti, 

2. Om chanting and prayer 

3. Selected practices of sukshmavyayama: Uccaranasthalatatha Visuddha chakra shuddi.

4. Yogasanas: Tadasana, Katichakrasana, Urdhwahastottanasana, Gomukhasana, Uttanamandukasana, Bhujanasana, Markatasana, Shavasana. 

5. Pranayama: Nadishodhana Pranayama, Bhramari, Bhastrika. 

6. Special Practice: Yoganindra 

7. Dhyana: Meditation

44.

Find the word which is out of the logic list:A) entirely B) partly C) quite D) completely

Answer»

Correct option is B) partly

45.

Find the word which is out of the logic list:A) impress B) astonish C) stun D) astound

Answer»

Correct option is A) impress

46.

Find the word which is out of the logic list:A) conclude B) question C) ask D) inquire

Answer»

Correct option is A) conclude

47.

I used to smoke at one time but now I ______ it up. A) have given B) gave C) had given D) am given

Answer»

Correct option is A) have given

48.

Someone ______ my bike last night. ______ you ever ______ your bike stolen? A) stole / Did / have B) stolen / Have / had C) was stolen / Have / had D) stole / Have / had

Answer»

Correct option is D) stole / Have / had

49.

The doctor ______ the patient with a smile and told him that he ______ better. A) approached / is looking B) approached / was looking C) approaches / is looking D) had approached / looked

Answer»

Correct option is B) approached / was looking

50.

Find the word which is out of the logic list:A) plainly B) evidently C) rapidly D) obviously

Answer»

Correct option is C) rapidly