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 cancer? How is a cancer cell different from normal cell? How do normal cells attain cancerous nature?

Answer»

An abnormal and uncontrolled division of cells is termed as cancer.

S.No.Cancer cellNormal cell
(i)Cancer cells divide in an uncontrolled manner.Normal cells divide in a controlled manner.
(ii)These cells do not show contact inhibition.These cells show contact inhibition.
(iii)Lifespan is indefinite.Lifespan is definite.

In our body, the growth and differentiation of cells is highly controlled and regulated. The normal cells show a property called contact inhibition. The surrounding cell inhibits uncontrolled growth and division of a cell. The normal cells when lose this property, become cancerous, giving rise to masses of cells called tumours. Transformation of normal cells into cancerous cells is induced by some physical, chemical or biological agents (carcinogens).

2.

State any four methods to reduce friction.

Answer»

Friction can be reduced by using polished surfaces, using lubricants, using grease and using ball bearings.

3.

Why is haemophillia called as bleeder's disease?

Answer»

In haemophilic persons, clotting of blood takes an hour to 24 hours. Such persons bleed excesivelly form even minor wounds. Continuous bleeding may cause death. So it is also called bleeder's desease.

4.

Which nuclei fuse to give endosperm ?

Answer»

Polar nuclei.

5.

Egg apparatus of angiosperms consist of(a) one synergid and two egg cells(b) two synergid and one egg cells(c) one central cell, two synergid, and three antipodal cells(d) one egg cell, two polar nuclei, and three antipodal cells

Answer»

(b) two synergid and one egg cells

6.

List the general characteristics of the pollen grains of wind pollinated plants.

Answer»

The pollen grains are smooth, dry, light, non-sticky and may be winged.

7.

Pollen grains in angiosperms ace not motile. They have to be carried to the stigma by external agents.1. Name the process of transfer of pollen grains.2. Write the name of three such agents.3. Explain how it takes place in vallisnaria.

Answer»

1. Transfer of pollen grains from the anther to the stigma is called pollination.

2. Wind water and animals

3. In vallisneria the female flowers reach the surface of water by long stalk and the male flowers or pollen grains are released to the surface of water and are carried by water currents to the female flower and fertilization takes place.

8.

In many grasses seeds are formed only after fertilization. There are reports that in some grasses, seeds are formed without fertilization. Explain the phenomenon.

Answer»

The phenomenon of formation of seeds without fertilization is called Apomixis. Apomixis is a form of asexual reproduction that mimics sexual reproduction. In this phenomenon , the diploid egg cell is formed without reduction division and develops into the embryo without fertilization.

9.

Egg cell formation in angiosperms involves me- gasporogenesis and female gametophyte development.a) Briey write the various steps involved in female gametophyte development.b) Mature angiosperm embryosac at maturity, though 8 nucleated is 7 celled.What is your explanation related to this statement? Explain.

Answer»

(a) Single megaspore mother cell (MMC) in the micropylar region of the nucellus undergoes meiotic division results in the production of four megaspores. In a majority of flowering plants, one of the megaspores is functional while the other three degenerate. Only one functional megaspore develops into the female gametophyte (embryo sac).

(b) 2 polar nuclei are situated below the egg apparatus in the large central cell. Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus consists of two synergids and one egg cell. The synergids have special cellular thickenings at the micropylar tip called filiform apparatus, which play an important role in guiding the pollen tubes into the synergid. Three cells are at the chalazal end and are called the antipodals.

10.

Given below are the components related to simplified model of mineral cycling in a terrestrial ecosystem. Construct a flow chart.(Hint: Weathering of rock)

Answer»

Producers consumers -> detritus -> soil solution -> Minerals in rock

11.

In market, seedless grapes are high priced than seeded grapes. Give the method used to develop the former.

Answer»

The seedless grapes are developed through induced parthenocarpy.

12.

Given below in the diagram showing the transfer of pollen grains.(i) Identify a & b with technical terms.(ii) Critically evaluate a & b.

Answer»

(i) (a) Autogamy

(b) Geitonogamy 

(ii) (a) Transfer of pollen from antherto the stigma of the same ower is called autogamy.

(b) Transfer of pollen from anther to the stigma of a dierent ower on the same plant.

13.

Give the technical terms for the following1. Produce seeds without fertilization2. Persistent nucellus3. Cotyledons of grass family4. Protective sheath of radical.

Answer»

1. Parthenocarpy

2. Perisperm

3. Scutellum 

4. Coleorhizae

14.

What is a false fruit ? Give an example.

Answer»

A fruit which is formed by any floral parts of the flower other than ovary, eg. apple, pear, cashew nut, etc. 

15.

Banana is a true fruit and also a parthenocarpic fruit. Justify.

Answer»

Since banana fruit is formed from ovary it is true fruit. It is parthenocarpic because the ovary develops into fruit without fertilization.

16.

Development of fruit without fertilization and are seedless known as .............(a) Polyembryony (b) Apomixix (c) Parthenocarpy (d) Parthenogenesis

Answer»

Parthenocarpy

17.

Wind pollinated (anemophilous) plants

Answer»

Cannabis, Coconut

18.

What technical term is applied to fruits formed without fertilization ?

Answer»

Parthenocarpy

19.

The thick protective covering of the fruit is known as ......

Answer»

The thick protective covering of the fruit is known as Pericarp

20.

When does a human body elicit an anamnestic response?

Answer»

Anamnestic response is the secondary response which is elicited when our body encounters with the same antigen to which the body has previously encountered.

21.

Name some water pollinated plats.

Answer»
  • Fresh water – vallesneria
  • Hydrolla Aquatic – Zoster
22.

Explain the importance of syngamy and meiosis in a sexual life cycle of an organism.

Answer»

Syngamy  :  Restoration (2n) chromosome number/ diploidy / zygote formation / variations (due to syngamy). 

Meiosis :  Gamete formation / reduction of (n) chromosome number/haploidy / variation (due to crossing over).

23.

Name the protective substance present on the pollen envelop to tide over adverse condition. 

Answer»

Sporopollenin.

24.

Explain the process of emasculation and bagging of flowers. State their importance in breeding experiments.

Answer»

Emasculation : If the female parent bears bisexual flowers, removal of anthers from the flower's bud before the anther dehiscence, using a pair of forceps is referred to as emasculation.

Bagging  : Emasculated flowers have to be covered with a bag of suitable size, generally made up of butter paper, to prevent contamination of its stigma with unwanted pollen. This process is called bagging. 

Importance : When the stigma of bagged flower attains receptivity, mature pollen grains collected {rom anthers of the male parents are dusted on the stigma and the flowers are rebagged, and the fruits are allowed to develop. If the female parent produces unisexual female flowers, there is no need of emasculation. The female flower buds are bagged before the flowers open. When the stigma becomes receptive, pollination is carried out using the desired pollen and the florver is rebagged.

25.

What is apomixos and what is its importance ?

Answer»

Apomixis in the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also by sowing hybrid seeds, it is difficult to maintain hybrid . characters that segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also it is a cost effective method for producing seeds.

26.

Why is apple called a false fruit ? Which part(s) of the flower forms fruit ?

Answer»

Fruits derived not from the ovary but from other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits in which fleshy part develop from ovary, but don’t consists of the thalamus or any other floral parts. In an apple the fleshy receptacle forms the main edible part. Hence it is a false fruit.

27.

Define Emasculation.

Answer»

Removal of anthers from the flower bud of a bisexual flower before the anther dehisces using a pair of forceps.

28.

A small telescope has an objective lens of focal length 150cm and eye piece of focal length 5cm.  If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25cm away from eyepiece.

Answer»

Given, Focal length of objective, fo​=150 cm

Focal length of eye-piece, fe​=5 cm

Height of tower, H=100 m

Distance of tower, u=3 km

Magnification of telescope is given by:

m=−fe​fo​​(1+Dfe​​)

m=−5150​(1+255​)=−36

Also, m=tanαtanβ​

tanα=H/u=100/3000=1/30

tanβ=−3036​

tanβ=DH′​

Height of the image of the tower, H′=−30−36×25​=−30cm

29.

Why the land has a higher temperature than the ocean during the day but a lower temperature at night ?

Answer»

Specific Heat of water is more than land (earth). Therefore for given heat change in temp. of land is more than ocean (water).

30.

A hotter gas implies higher average value of (A) internal energy (B) total energy (C) kinetic energy (D) heat constant.

Answer»

Answer is (C) kinetic energy

The term hot refers to higher temperature and temperature depends on kinetic energy.

31.

The number of degrees of freedom for translatory motion are (A) same for all types of molecules.(B) less for multi atomic molecules. (C) more for multi atomic molecules. (D) dependent upon the nature of translatory motion.

Answer»

Answer is (A) same for all types of molecules.

32.

Pure water vapour is trapped in a vessel of volume 10 cm3 . The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

Answer»

RH=VP/SVP

The point where the vapour starts condensing, VP = SVP
We know P1V1 = P2V2
RH SVP × 10 = SVP × V2 => V2 = 10RH => 10 × 0.4 = 4 cm3

33.

A 2m long rectangular bar of 7.5 cm × 5 cm is subjected to an axial tensile load of 1000kN. Bar gets elongated by 2mm in length and decreases in width by 10 × 10–6 m. Determine the modulus of elasticity E and Poisson's ratio of the material of bar.

Answer»

Given:

L = 2m; 

B = 7.5cm = 0.075m; 

D = 5cm = 0.05m 

P = 1000kN

\(\delta\)L = 2mm = 0.002m 

\(\delta\)b = 10 × 10-6 m.

Longitudinal strain 

eL = et = \(\delta\)L/L = 0.002/2 = 0.001 

Lateral strain

\(\delta\)b/b = 10 × 10-6/0.075 = 0.000133

Tensile stress (along the length)

\(\sigma\)t = P/A = (1000 × 1000)/(0.075 × 0.05) = 0.267 × 109 N/m2

Modulus of elasticity, 

E = \(\sigma\)t /et = 0.267 × 109 /0.001 = 267 × 109 N/m2

Poisson’s ratio = Lateral strain/ Longitudinal strain 

= (\(\delta\)b/b)/(\(\delta\)L/L) = 0.000133/0.001

34.

What is meant by Rocket propulsion ? 

Answer»

It is an example of momentum conservation in which the large backward momentum of the ejected gases imparts an equal forward momentum to the rocket. Due to the decrease in mass of the rocket-fuel system, the acceleration of the rocket keeps on increasing.

35.

A solid steel cylinder 500 mm long and 70 mm diameter is placed inside an aluminium cylinder having 75 mm inside diameter and 100 mm outside diameter. The aluminium cylinder is 0.16 mm. longer than the steel cylinder. An axial load of 500kN is applied to the bar and cylinder through rigid cover plates as shown in Fig.. Find the stresses developed in the steel cylinder and aluminium tube. Assume for steel, E = 220 GN/m2 and for Al E = 70 GN/m2

Answer»

Since the aluminium cylinder is 0.16 mm longer than the steel cylinder, the load required to compress this cylinder by 0.16 mm will be found as follows : 

E = stress/ strain = P.L/A.\(\delta\)

Or P = E.A.\(\delta\)L/L 

= 70 × 109 × \(\pi\)/4 (0.12 - 0.0752) × 0.00016/0.50016 = 76944 N 

When the aluminium cylinder is compressed by its extra length 0.16 mm, the load then shared by both aluminium as well as steel cylinder will be, 

500000 – 76944 = 423056 N

Let es = strain in steel cylinder 

ea = strain in aluminium cylinder 

\(\sigma\)s = stress produced in steel cylinder 

\(\sigma\)a = stress produced in aluminium cylinder 

Es = 220 GN/m2 

Ea = 70 GN/m2

As both the cylinders are of the same length and are compressed by the same amount

es = e

\(\sigma\)s /Es =\(\sigma\)a /Ea 

or; \(\sigma\)s = Es/Ea.\(\sigma\)a= (220 × 109 /70 × 109 ). 

\(\sigma\)a = (22/7). \(\sigma\)a

Also Ps + Pa = P 

or; \(\sigma\)s As + \(\sigma\)a. Aa = 423056 (22/7). 

 As + \(\sigma\)a Aa = 423056 ...(i) 

As = \(\pi\)/4 (0.072 ) = 0.002199 m2 

Aa = \(\pi\)/4 (0.12 – 0.0752 ) = 0.003436 m2

Putting the value of As and Aa in equation (i) we get

\(\sigma\)a = 27.24 × 106 N/m2 = 27.24 MN/m

\(\sigma\)S = 22/7 × 27.24 = 85.61 MN/m2

Stress in the aluminium cylinder due to load 76944 N

= 76944/ \(\pi\)/4 (0.12 - 0.0752) 

= 23.39 × 109 N/m2 = 22.39 MN/m2

Total stress in aluminium cylinder 

= 27.24 + 22.39 = 49.63 MN/m2 

and stress in steel cylinder = 85.61 MN/m2

36.

A piece of steel 200 mm long and 20 mm x 20 mm cross section is subjected to a tensile force of 40 kN in the direction of its length. Calculate the change in volume. Take 1/m = 0.3. E = 2.05 x 105 N/mm2 .

Answer»

e1\(\frac{40000}{(20\times20)(2.05\times10^5)}\) = 4.88 x 10-4

e2 = e3 = - \(\frac{e_1}m\) = - 4.88 x 0.3 x 10-4 = -1.464 x 10-4

δV/V = e1 + e2 + e3 = [4.88 - (1.464 x 2)]10-4 = 1.952 x 10-4

V = 200 x 20 x 20 = 80000 mm3 

\(\therefore\) δV = 1.952 x 80000 x 10-4 = 15.62 mm3

37.

Stability of interhalogen compounds follows the order A. BrF > IBr > ICl > ClF > BrCl B. IBr > BeF > ICl > ClF > BrCl C. ClF > ICl > IBr > BrCl > BrF D. ICl > ClF > BrCl > IBr > BrF

Answer»

Correct answer is

C. ClF > ICl > IBr > BrCl > BrF

38.

A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens ? Give reason.

Answer»

As a diverging lens.

Light rays diverge on going from a rarer to a denser medium.

39.

(i) What is the relation between critical angle and refractive index of a material ?(ii) Does critical angle depend on the colour of light ? Explain

Answer»

(i) Refractive index (μ) = 1/sin C

where C is the critical angle.

(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths (colours) of light.

40.

 Why does sun appear bigger during sunset or sunrise?

Answer»

During sunset or sunrise, the rays of light from sun travel through a long distance through air. Thus, the apparent image of sun is formed closer to eye, which in turn appears bigger. 

41.

Magnification m = ……………A) \(\cfrac{v}u\)B) \(\cfrac{u}v\)C) \(\cfrac{h_o}{h_i}\)D) \(\cfrac{h_i}{h_o}\)

Answer»

 D) \(\cfrac{h_i}{h_o}\)

42.

Emission of electrons by the absorption of heat energy is called ……… emission(a) photoelectric (b) field (c) thermionic (d) secondary

Answer»

Correct answer is (c) thermionic

43.

Paramagnetism is not exhibited byA. `CuSO_(4)5H_(2)O`B. `CuCl_(2)5H_(2)O`C. `Cul`D. `NiSO_(4).6H_(2)O`

Answer» Correct Answer - C
44.

State the appropriate concept for the given statement.Law treats all citizens equally.

Answer»

Law treats all citizens equally : Equality before the law

45.

Chlorofluoro carbons are used in: (a) Air conditioners (b) Aerosol sprays (c) Refrigerators (d) All of these

Answer»

(b) Aerosol sprays

46.

State the appropriate concept for the given statement.Application of the abstract concept of justice through the implementation of the law.

Answer»

Application of the abstract concept of justice through the implementation of the law : Legal justice

47.

In which year Ganga Action Plan was launched?(a) 1980 (b) 1984 (c) 1982 (d) 1985

Answer»

Ganga Action Plan was launched in 1985

48.

State whether the following statements are true or false with reasons.Social justice is essentially reformative and distributive.

Answer»

This statement is True. 

  • The concept of social justice implies equal social opportunities for every individual to progress to the fullest possible extent. Social justice is reformative i.e., it aims at a revision of the social order and involves the eradication of existing social evils.
  •  Social justice is also distributive i.e., available resources should be equitably distributed to ensure social welfare.
49.

Bhopal gas tragedy was caused due to(a) Air pollution (b) Emission of poisonous gas (c) Water pollution (d) Leakage of poisonous gas

Answer»

(d) Leakage of poisonous gas

50.

The disease ‘Minamata’ is caused by poisoning(a) Lead (b) Zinc (c) Mercury (d) Copper

Answer»

The disease ‘Minamata’ is caused by poisoning Mercury