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.

ComprehensionDirections: Please read the following passage and answer the questions.Tropical cyclones are closed low - pressure systems. generally about 650 km in diameter. which bring violent winds, torrential rainfall and thunderstorms. Generally. they contain a central region, the eye. with a diameter of about ten kilometres with light winds and more or less lightly clouded sky. In order for cyclones to develop various conditions need to be fulfilled. There must be a plentiful supply of moisture, and for this reason distribution of cyclones is closely related to those regions where the highest sea - surface temperatures are found. i.e. the western portions of the tropical oceans in late summer with a temperature of at least 27°C. This moisture provides the necessary latent heat to drive the storm and to provide the rainfall. Tropical cyclones tend to occur mainly in between the Tropic of Capricon and Tropic of cancer except the equatorial region i.e.5° on both side of the equator because closer than that to the Equator the Coriolis parameter approaches Zero. The main cyclone activity in the Northern Hemisphere is in late summer and autumn during the time of the equatorial trough's northern displacement. 1. Willy Willy2. Hurricane3. Typhoon4. Cyclone

Answer» Correct Answer - Option 2 : Hurricane
  • The regional name of a tropical cyclone in the West Indies is Hurricane.
  • They are known as willy willies in Australia, cyclones in Indian and typhoon in the western Pacific, eastern Asia.
2.

Which of the elemetns of group 15 are metalloids?

Answer» Group 15 elements which are metalloids are arsenic (As) and antimony (Sb).
3.

Consider the amplitude modulated (AM) signal AC cos\(\omega\)ct+2 cos\(\omega\)m t cos\(\omega\)ct For demodulating the signal using envelope detector, the minimum value of AC should be (a) 2 (b) 1 (c) 0.5(d) 0

Answer»

(a) 2 

Modulated signal is given as

\(\theta\)A(t) = [Ac+ 2os\(\omega\)mt] cos\(\omega\)ct

Note that for envelope detection the modulation should not go beyond full modulation i.e. \(\mu\) = 1, so amplitude of baseband signal has to be less than the carrier amplitude (Ac)

|f(t)|max ≤  Ac

|2 osmt|max= 2 ≤ Ac

Or A >= 2

4.

What is the nature of the compounds of group 15 elements?

Answer» Compounds of group 15 elemets are mostly covalent in nature.
5.

If modulation index of an AM wave is increased from 1.5 to 2, then the transmitted power..........a) remains same b) increases by 20% c) increases by 41% d) increases by 50%

Answer»

c) increases by 41% 

When m =1.5, transmitted power

(pt) = pc (1 + \(\cfrac{1.5^2}2\))= 2.125 pc and when m = 2, pt = pc ( 1 + \(\cfrac{2^2}2\)) = 3pc

so increase = \(\cfrac{3p_c-2.125p_c}{2.125p_c}\) = 0.41

6.

What is the number of unpaired electrons present in the valency shell of group elements?

Answer» The number of unpaired electrons present in the valence shell of group 15 elements is three.
7.

A wave is modulated by two sin waves having modulation indices of 0.3 and 0.5. Find the total modulation index? a) 0.1 b) 0.7 c) 0.58 d) 0.35

Answer»

c) 0.58 

Given that m1 = 0.3 and m2 = 0.5. Total modulation index will be equal to \(\sqrt{m_1^2+m_2^2}\).

By substituting values we have \((\sqrt{0.3^2+0.5^2)}\) which is equal to 0.58.

8.

Power of carrier wave is 500W and modulation index is 0.25. Find its total power? a) 500W b) 415W c) 375W d) 516W

Answer»

d) 516W

Total power, Pt = 500(1+(0.25)2⁄2) = 516W.

9.

For 100% modulation, power in each sideband is of that of carrier? a) 50% b) 70% c) 60% d) 25%

Answer»

d) 25%

Modulation index = 1. Power in sidebands may be calculated as

\(\cfrac{m^2p_c}4\).

so, power in each sideband is \(\cfrac{1^2p_c}4\) i.e 25%

10.

Calculate power in each sideband, if power of carrier wave is 176W and there is 60% modulation in amplitude modulated signal? a) 13.36W b) 52W c) 67W d) 15.84W

Answer»

d) 15.84W

Modulation index = 0.6 and Pc = 176W. Power in sidebands may be calculated as (m2Pc)/4 = 15.84 W

11.

A 50 V battery is supplying current of 10 A when connected to a resistor. If the efficiency of the battery at this current is 25%, then the internal resistance of the battery isA. `2.5 Omega`B. ` 3.75 Omega`C. `1.25 Omega`D. `5 Omega`

Answer» Correct Answer - B
b. `50 = 10[R+r]`
or `R + r = 5 Omega`
` eta = R/(R+r) or 0.25 = R/(R+r)`
`R + r = 4R `
or `r = 3R`
Then `R = 5//4 = 1.2 Omega and r = 3.75 Omega` .
12.

Which of the following circuits gives the correct value of resistance, when computed by using `R = (V//I)` where V and I are voltmeter and ammeter reading, respectively? The meters are not ideal.A. B. C. D. none of these

Answer» Correct Answer - D
d. In first two circuit diagrams, corrections have to be made while
third one is absolutely wrong. So, none of the circuit diagrams
is giving correct value of R.
13.

A cell of emf `epsilon` and internal resistance r is charged by a current I, thenA. the cell stores chemical energy at the rate of `epsiloni`B. the cell stores chemical energy at the rate of `i^2r`C. the cell stores chemical energy at the rate of `epsilon I - i^2r `D. the storage of chemical energy rate cannot be calculated

Answer» Correct Answer - C
c. From work-energy theorem, we get `Delta = epsiloni - i^2r`.
14.

Two ideal batteries having emf `E_1 and E_2` are connected as shown in Figure .The values of resistances are chosen in such a way that ammeter reading is zero. The reading of voltmeter will be (consider the meters to be ideal) A. `E_1`B. `E_2`C. In between `E_1 and E_2`D. nothing can be predicted about voltmeter reading from the given information.

Answer» Correct Answer - B
b. The ammeter is not showing any reading. It means the potential
drop across CB due to `E_1` is equal to that of the emf `E_2` . And this is the reading of the voltmeter.
15.

Statement I: In the meter bridge experiment shown in figure, the balance length AC corresponding to null deflection of the galvanometer is x. If the radius of the wire AB is doubled, the balanced length becomes 4x. Statement II: The resistance of a wire is inversely proportional to the square of its radius. A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.

Answer» Correct Answer - D
d. The condition for no deflection of the galvanometer is
`R_1/R_2 = R_(AC)/R_(CB)`
where `R_(AC) and R_(CB)` are the resistances of the bridge wire of
length AC and CB, respectively. If the radius of the wire AB is
doubled, the ratio `R_(AC)//R_(CB)` will remain unchanged. Hence, the
balanced length will remain the same.
16.

Statement I: In the potentiometer circuit shown in figure. `E_1 and E_2` are the emf of cells `C_1 and C_2`, repsectively, with `E_1gtE_2`. Cell `C_1` has negligible internal resistance. For a given resistor R, the balance length is x. If the diameter of the potentiometer wire AB is increased, the balance length x will decrease. Statement II: At the balance point, the potential difference between AD due to cell `C_1 is E_2` , the emf of cell `C_2`.A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.

Answer» Correct Answer - D
d. If the diameter of wire AB is increased, its resistance will
decrease. Hence, the potential difference between A and B
due to cell `C_1` will decrease. Therefore, the null point will be
obtained at a higher value of x.
17.

Statement I: When an external resistor of resistance R (connected across a cell to internal resistance r ) is varied, power consumed by resistance R is maximum when R = r. Statement II: Power consumed by a resistor of constant resistance R is maximum when current through it is maximum.A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.

Answer» Correct Answer - B
b. Both statements I and II are true. In Statements I, R is varied
while in Statement II, R is kept constant. Hence, both statements
are independent.
18.

STATEMENT 1: The current density `vec J` at any point in ohmic resistor is in direction of electric field `vec E` at that point . STATEMENT 2: A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False.

Answer» Correct Answer - C
From relation `vecJ=sigmavecE`, the current density `vecJ` at any point in ohmic resistor is in direction of electric field `vecE` at the point. In space having non-uniform electric field, charges released from rest may not move along ELOF.
19.

STATEMENT 1: The current density `vec J` at any point in ohmic resistor is in direction of electric field `vec E` at that point . STATEMENT 2: A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.

Answer» Correct Answer - C
c. From relation `vecJ = sigma vecE`, the current density `vecJ` at any point in
ohmic resistor is in direction of electric field `vecE` at that point. In
space having nonuniform electric field, charges released from
rest may not move along EL. Hence, Statement I is true while
Statement II is false.
20.

AM broadcast station transmits modulating frequency upto 6KHz. If transmitting frequency is 810KHz, then maximum and lower sidebands are...........a) 816KHz and 804KHz b) 826KHz and 804KHz c) 916KHz and 904KHz d) Not possible

Answer»

a) 816KHz and 804KHz 

Maximum frequency = 810 + 6 = 816KHz and Minimum frequency = 810 – 6 = 804KHz. Moreover it has a bandwidth of (816 – 804) = 12KHz.

21.

How does electronegativity vary down the group `17` and why? How does it vary from left to right in period? Name an element having highest electronegativity.

Answer» Electronegativity decreases down the group due to increase in atomic size. It increases along the period from left to right due to decreases in atomic size.
Fluorine `(F)` has the highest electronegativity.
22.

Name the element of group 15 which has highest electronegativity.

Answer» The elements of group 15 which has highest electronegativity is nitrogen.
23.

What is maximum covalency of nitrogen?

Answer» Maximum covalency of nitrogen is four.
24.

What are common oxidation states of group 15 elements?

Answer» Common oxidation states of group 15 elements are `-3,+3,+5`
25.

What is the relation between oxidation number of the central atom and covalent nature of the compound?

Answer» The covalent nature of the compound increases as the oxidation number of the central atom increases.
26.

What happens at the synapse between two neurons?

Answer»

The neurons lie end-to-end in a chain to transmit the impulses in an animal body. There occurs a very minute gap between the terminal portion of axon of one neuron and dendron of the other neuron. This gap is called synapse. At the end of the axon, the electrical impulse sets off the release of some chemicals. These chemicals cross the gap (synapse) and start a similar electrical impulse in the dendrite of the next neuron.

27.

(a) Which hormone is responsible for the changes noticed infemales at puberty?(b) Dwarfism results due to deficiency of which hormone?(c) Blood sugar level rises due to deficiency of which hormone?(d) Iodine is necessary for the synthesis of which hormone?

Answer» (a) Oestrogen
(b) Growth hormone
(c) Insulin
(d) Thyroxin
28.

What are the major parts of the brain? Mention the functions of different parts.

Answer»

The three major parts of the brain are :

(1) Fore-brain, 

(2) Mid-brain and 

(3) Hind-brain. 

The major parts of the brain and their functions are as follows:  

(1). Fore-brain - It is the main thinking part of the brain. It has three main parts-the olfactory lobes, the cerebrum and the diencephalon.  

(a) The olfactory lobes - These are a pair of very small club-shaped bodies which are fully covered by the cerebrum. They act as a center of smell.  

(b) The cerebrum - It is the largest and most prominent part of the brain. It is divided into right and left cerebral hemisphere by a deep groove. It is the center of consciousness, thoughts, memory and analytical thinking. It also controls voluntary actions.  

(c) Diencephalon - It mainly consists of pituitary gland, hypothalamus and thalamus. It posses control unit of thirst, hunger, temperature, sleep, etc.

(2). Mid-brain - It acts as a coordinating unit between forebrain and hindbrain. It also controls some involuntary actions.  

(3). Hind-brain - It has three main centers - Cerebellum, Pons and Medulla oblongata.  

(a) Cerebellum- It is responsible for precision of voluntary actions and maintaining the posture and balance of the body.  

(b) Pons - It is the center through which nerve impulses travels to and from the cerebellum, spinal cord and other parts of the brain. It also helps in respiration.  

(c) Medulla oblongata - It is the lowermost part of the brain. It contains vital centres for controlling blood pressure, respiration, swallowing, sneezing, coughing, salivation and vomiting.

Hints— Fore brain
Mid brain
Hind brain
Give its functions
29.

what is strong acid 

Answer»

A strong acid is one that is completely dissociated or ionized in an aqueous solution. It is a chemical species with a high capacity to lose a proton, H+. In water, a strong acid loses one proton, which is captured by water to form the hydronium ion: HA(aq) + H2O → H3O+(aq)+ A(aq).
30.

What is intracellular digestion?

Answer»

Intracellular digestion, or cellular digestion is the breaking in the interior of the cell of big molecules coming from outside or even from its own cell metabolism into smaller molecules. Products and residues of the intracellular digestion are used by the cell or excreted.

Intracellular digestion is classified into two types: heterophagic intracellular digestion and autophagic intracellular digestion.

31.

What is the main cell organelle involved in cell digestion? What are the properties of that organelle that enable it to do the task?

Answer»

The organelles responsible for intracellular digestion are the lysosomes. Lysosomes are vesicles that contain digestive enzymes capable of breaking big molecules into smaller ones. These vesicles fuse with others that carry the material to be digested and then digestion takes place. 

32.

What is autophagic intracellular digestion? Why is this type of intracellular digestion intensified in an organism undergoing starvation?

Answer»

Autophagic intracellular digestion is the cellular internal digestion of waste and residual materials. In general, it is done by lysosomes.

Autophagic intracellular digestion is intensified in situations of starvation because in such condition the cell tries to obtain from its own constituent materials the nutrients necessary to stay alive.

33.

Write the Uses of CaCO3.

Answer»

Some common uses of calcium carbonate(CaCO3) are given below.

  • Calcium carbonate plays an important role in construction, be it as a building material such as marble or as an ingredient in cement.
  • Calcium carbonate is in use in medicinal industries. Industries manufacture antacids, tablets made of base materials etc.
  • It is in use as calcium supplements.
  • Calcium carbonate is in use in the manufacture of paints, paper, plastics, etc.
34.

Explain Cannizzaro reaction with an example.

Answer»

Aldehydes with no α-hydrogen atom when treated with concentrated alkali (NaOH or KOH), undergo self oxidation and reduction (disproportionation) to give carboxylic acid salt and alcohol. This is called Cannizzaro’s reaction. 

2HCHO + Conc.KOH → HCOOK + CH3OH

35.

Name the gas liberated when Lanthanoids (Ln) react with acids.

Answer»

Hydrogen gas or H2 gas.

36.

Noble gases are chemically inert give one reason?

Answer»

The orbitals and of noble gases are completely filled they generally have 8 valence electrons and have stable electronic configuration hence they are chemically inert.

37.

Colonies of yeast fail to multiply in water, but multiply in sugar solution. Give one reason for this.

Answer»

Sugar provides energy for sustaining all life activities in yeasts. In water, it fails to reproduce because of inadequate energy in its cells.

38.

Write the unit of molality of a solution?

Answer»

mol/kg-1 or mol/kg

39.

What are thermo setting polymers?

Answer»

These polymers are cross linked molecules which undergo permanent change on heating. They become hard and infusible on heating and cannot be softed again.

40.

Name the monomers of Nylon-6, 6.

Answer»

Tetra methyline dicarboxylic acid (Adipic acid) and hexa methylene diamine.

41.

What are ferromagnetic substances? Give one example?

Answer»

The substances which are strongly affected by magnetic field are called ferromagnetic substances and the phenomenon is known as ferromagnetism.

Ex: Iron or Nickel, or Cobalt or CrO2 any other.

42.

Zr and Hf have almost identical radii give reason?

Answer»

It is due to Lanthanoid contraction.

43.

Calculate the no of particles (atoms) per unit cell in a FCC crystal lattice?

Answer»

No.of atoms per unit cell of FCC 

= 8 corner atoms × \(\frac{1}{8}\) atom + 6 face centred atoms × \(\frac{1}{2}\)

i.e. 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) =1 + 3 = 4

44.

What is “Chirality”?

Answer»

Chirality is a geometric property of some molecules and ions a chiral molecule or ion is non super posable on its mirror image the presence of asymmetric carbon atom is one of several structural features that induce chirality in organic an dinorganic molecules.

45.

Zr and Hf have almost identical radii. Give reason.

Answer»

Due to Lanthanoid contraction.

46.

Which hormone regulates the sugar level in the blood?

Answer»

Insulin or glucagon.

47.

Why is the use of iodised salt advisable?

Answer»

Iodine stimulates the thyroid gland to produce thyroxin hormone. It regulates carbohydrate, fat, and protein metabolism in our body. Deficiency of this hormone results in the enlargement of the thyroid gland. This can lead to goitre, a disease characterized by swollen neck. Therefore, iodised salt is advised for normal functioning of the thyroid gland.

48.

Name the hormone which regulates blood sugar level in the body.

Answer»

The hormone which regulates blood sugar level in the body is Insulin.

49.

Name any two factors affecting the rate of a reaction.

Answer»

1. Concentration of the reactants 

2. Temperature of the reactants 

3. Nature of the reacting substances

50.

Consider the following statements:Statement I: Chloroform is stored in dark brown bottles. Statements II: This is due to avoid aerial oxidation.Of these statements:(a) Both the statement are true and Statement II is the correct explanation of Statement I(b) Both the statement are true, but statement II is not the correct explanation of Statement I(c) Statement I is true, but Statement II is false(d) Statement I is false. but Statement II is true

Answer»

Answer (a) Both the statement are true and Statement II is the correct explanation of Statement I