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RRB Technician Grade III 2024-25 26 December 2024 Shift 3 GK Questions – with answers & solutions

All 40 GK / GS questions asked in RRB Technician Grade III 2024-25 on 26 December 2024 Shift 3, with correct answers and short solutions, subject-wise: Biology, Chemistry, Physics.

📚 40 questions🗓️ 26 December 2024⏰ Shift 3
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Subjects: Biology (13) · Chemistry (14) · Physics (13)

Biology (13)

Control and Coordination · Brain - parts and functions

Q1. Which of the following actions are regulated by the medulla in the hind brain?

  1. (a) Riding a bicycle
  2. (b) Salivation
  3. (c) Maintaining balance and posture
  4. (d) Walking
Answer & solution

Answer: (b) Salivation

Correct Answer: Option 2 — Salivation
The medulla oblongata is the lowest part of the hind-brain; it joins the brain to the spinal cord. It controls the involuntary actions of the body — those that go on by themselves without our thinking. These include salivation, vomiting, swallowing, coughing, sneezing, blood pressure, heart rate and the rhythm of breathing.
Because the medulla governs such life-supporting functions, it is called the vital centre of the body, and any serious injury to it is usually fatal.
❌ Why the other options are incorrect
✗ Riding a bicycle — a learnt, voluntary skill needing precise coordination and balance; it is controlled by the cerebellum together with the cerebrum.
✗ Maintaining balance and posture — this is the special work of the cerebellum, which keeps the body steady and the movements smooth.
✗ Walking — a voluntary muscular activity initiated by the cerebrum (motor area) and smoothed by the cerebellum, not by the medulla.
📌 SSC & Railway Exam Facts
• The hind-brain has three parts — cerebellum, pons and medulla oblongata.
• The fore-brain has the cerebrum (thinking, memory, intelligence and the seat of all sensory areas).
• The mid-brain controls reflexes of the eye (pupil size) and of the ear.
• The hypothalamus in the fore-brain controls hunger, thirst, sleep and body temperature.
• The brain floats in cerebrospinal fluid (CSF) inside the cranium, which absorbs shocks; the covering membranes are the meninges.
Reproduction · Asexual reproduction (fission, budding, fragmentation, spores)

Q2. Which of the following does NOT belong to the category of asexual reproduction?

  1. (a) Vegetative propagation
  2. (b) Spore formation
  3. (c) Regeneration
  4. (d) Fertilisation
Answer & solution

Answer: (d) Fertilisation

Correct Answer: Option 4 — Fertilisation
Asexual reproduction needs only one parent, produces no gametes, and gives offspring that are genetically identical (clones) to the parent. Fertilisation, on the other hand, is the fusion of a male gamete with a female gamete to form a zygote — it is the central event of sexual reproduction, which normally needs two parents.
Because fertilisation mixes the DNA of two individuals, it creates genetic variation, which is the raw material of evolution. Asexual methods cannot do this.
❌ Why the other options are incorrect
✗ Vegetative propagation — new plants grow from a root, stem or leaf of the parent, as in potato, ginger and Bryophyllum; only one parent is involved.
✗ Spore formation — hundreds of thick-walled spores made inside a sporangium each germinate into a new individual, as in Rhizopus and ferns.
✗ Regeneration — a cut or broken piece of the body grows into a complete organism, as in Planaria and Hydra; again no gametes are formed.
📌 SSC & Railway Exam Facts
• Modes of asexual reproduction: binary fission, multiple fission, budding, fragmentation, regeneration, spore formation and vegetative propagation.
• Binary fission — Amoeba, bacteria; Multiple fission — Plasmodium; Budding — yeast, Hydra.
• In humans, fertilisation takes place in the fallopian tube (oviduct).
• In flowering plants double fertilisation occurs — one male gamete forms the zygote, the other forms the endosperm.
• Sexual reproduction gives variation; asexual reproduction gives uniform clones but is faster.
Cell - Structure and Function · Discovery of cell and cell theory

Q3. In which year were cells first discovered by Robert Hooke?

  1. (a) 1664
  2. (b) 1667
  3. (c) 1670
  4. (d) 1665
Answer & solution

Answer: (d) 1665

Correct Answer: Option 4 — 1665
In 1665, the English scientist Robert Hooke examined a thin slice of cork (dead bark tissue of a tree) under his own self-designed compound microscope. He saw a honeycomb-like arrangement of small box-shaped compartments and named each of them a 'cell', from the Latin word cellula, meaning 'a little room'.
Hooke published this observation in his famous book 'Micrographia' (1665). What he actually saw were only the dead cell walls of cork cells, but the name he gave became the foundation of cell biology, and the cell is today accepted as the structural and functional unit of life.
❌ Why the other options are incorrect
✗ 1664 — one year too early; Hooke's observation and his book Micrographia both belong to 1665.
✗ 1667 — no landmark cell discovery is dated to this year; it is only a distractor.
✗ 1670 — by this date microscopy had advanced, but the discovery of the cell had already been made in 1665.
📌 SSC & Railway Exam Facts
• Anton van Leeuwenhoek (1674) first saw free-living cells — bacteria, sperms and protozoa — and is called the Father of Microbiology.
• Robert Brown (1831) discovered the nucleus; Purkinje (1839) coined the term protoplasm.
• The Cell Theory was given by Schleiden (1838) and Schwann (1839); Rudolf Virchow (1855) added that 'Omnis cellula e cellula' — all cells arise from pre-existing cells.
• The largest cell is the ostrich egg; the longest cell is the nerve cell (neuron).
• The smallest cell is that of Mycoplasma (PPLO); RBCs of mammals have no nucleus.

Q4. When water enters into the guard cells, the guard cells ______ and the stomatal pore ______.

  1. (a) swell; opens
  2. (b) swell; shrinks
  3. (c) shrink; closes
  4. (d) shrink; opens
Answer & solution

Answer: (a) swell; opens

Correct Answer: Option 1 — swell; opens
Stomata are tiny pores in the epidermis of a leaf, each bounded by two bean-shaped (kidney-shaped) guard cells. They are the doors through which gaseous exchange and transpiration take place.
When the guard cells take in water by osmosis, they become turgid (swollen). The inner wall of a guard cell, facing the pore, is thick and inelastic while the outer wall is thin and elastic; so the swelling bulges the outer wall outwards, the guard cells curve apart, and the stomatal pore opens. When water is lost, the guard cells become flaccid, straighten and the pore closes — the way the plant checks excess water loss on a hot day.
❌ Why the other options are incorrect
✗ swell; shrinks — swelling of turgid guard cells can never make the pore shrink; turgidity always widens the pore.
✗ shrink; closes — true, but only for loss of water; the question specifically says water enters the guard cells.
✗ shrink; opens — self-contradictory; flaccid, shrunken guard cells come together and shut the pore.
📌 SSC & Railway Exam Facts
• Transpiration is the loss of water as vapour from the aerial parts of a plant, mostly through stomata.
• Transpiration creates the transpiration pull, which draws water up through the xylem — the main force of ascent of sap.
• It also cools the plant and helps the absorption and upward movement of minerals.
• Stomata are usually more numerous on the lower surface of a dicot leaf; in floating plants like the lotus they occur on the upper surface.
• Guard cells of stomata are the only epidermal cells that contain chloroplasts.
Cell - Structure and Function · Endoplasmic reticulum and ribosomes

Q5. Which of the following act(s) as a site for protein synthesis?

  1. (a) Vacuole
  2. (b) Cell membrane
  3. (c) Ribosomes
  4. (d) Plastid
Answer & solution

Answer: (c) Ribosomes

Correct Answer: Option 3 — Ribosomes
Ribosomes are the site of protein synthesis in every living cell. They are extremely small, dense, non-membrane-bound granules made of RNA (rRNA) and protein, which is why they are also called the protein factory of the cell.
Ribosomes may lie free in the cytoplasm (their proteins are used inside the cell) or remain attached to the Rough Endoplasmic Reticulum (RER) (their proteins are exported out of the cell). During synthesis the ribosome reads the codons of mRNA and links amino acids brought by tRNA into a polypeptide chain — a process called translation.
❌ Why the other options are incorrect
✗ Vacuole — a fluid-filled sac that stores water, cell sap, pigments and wastes and gives turgidity to plant cells; it makes nothing.
✗ Cell membrane — the selectively permeable outer boundary that controls the entry and exit of substances; it is a barrier, not a synthesis site.
✗ Plastid — a plant organelle; the chloroplast carries out photosynthesis and leucoplasts store starch, oil and protein — storage, not protein manufacture.
📌 SSC & Railway Exam Facts
• Ribosomes were discovered by George Palade; the name was given by Richard B. Roberts.
• Prokaryotic ribosome — 70S (50S + 30S); eukaryotic ribosome — 80S (60S + 40S).
• Ribosomes are also present inside mitochondria and chloroplasts, which are therefore semi-autonomous organelles.
• Organelle nicknames: mitochondrion — powerhouse; chloroplast — kitchen; lysosome — suicide bag; Golgi body — dispatching centre.
• The genetic message travels as DNA → RNA → Protein, called the central dogma of molecular biology.
Reproduction · Seed, fruit and germination

Q6. After the process of fertilisation which part of the flower undergoes transformation to develop into a seed?

  1. (a) Ovule
  2. (b) Stigma
  3. (c) Style
  4. (d) Ovary
Answer & solution

Answer: (a) Ovule

Correct Answer: Option 1 — Ovule
After pollination, the pollen grain germinates on the stigma and sends a pollen tube down through the style into the ovary, where the male gamete fuses with the egg cell inside the ovule. This fusion is fertilisation and it produces the zygote.
The zygote divides repeatedly to form the embryo, the integuments of the ovule harden into the seed coat, and food is stored as cotyledons or endosperm. Thus the whole ovule is transformed into the seed, while the ovary around it swells and ripens into the fruit.
❌ Why the other options are incorrect
✗ Stigma — the sticky top of the pistil that receives the pollen grain; after fertilisation it withers away along with the style, sepals and petals.
✗ Style — the tube-like stalk joining the stigma to the ovary; it only provides the path for the pollen tube and is shed later.
✗ Ovary — it develops into the fruit, not into the seed; in a mango the fleshy part is the ovary and the stone contains the seed.
📌 SSC & Railway Exam Facts
• Ovule → Seed; Ovary → Fruit; Zygote → Embryo; Integument → Seed coat.
• A fruit formed without fertilisation is called a parthenocarpic fruit (e.g. seedless banana and grapes).
• A false fruit develops from the thalamus, as in apple, pear and cashew.
• Angiosperms show double fertilisation: one male gamete + egg → zygote; the other + two polar nuclei → triploid endosperm.
• Gram, pea and bean are dicot seeds with two cotyledons; maize and wheat are monocot seeds with one cotyledon.

Q7. Select the INCORRECT statement.

  1. (a) Gonorrhea is a bacterial disease.
  2. (b) Warts are a protozoan disease.
  3. (c) AIDS is caused by the HIV virus.
  4. (d) Many diseases can be sexually transmitted.
Answer & solution

Answer: (b) Warts are a protozoan disease.

Correct Answer: Option 2 — Warts are a protozoan disease.
This statement is incorrect. Warts are small, rough growths on the skin or mucous membrane caused by the Human Papilloma Virus (HPV) — that is, they are a viral disease, not a protozoan one. Protozoan diseases are entirely different: malaria (Plasmodium), amoebic dysentery (Entamoeba), kala-azar (Leishmania) and sleeping sickness (Trypanosoma).
Both warts and the other sexually transmitted infections spread through close physical or sexual contact, which is why they are grouped together in the NCERT chapter on reproductive health.
❌ Why the other options are incorrect
✗ Gonorrhea is a bacterial disease — this is true; it is caused by the bacterium Neisseria gonorrhoeae and is curable with antibiotics.
✗ AIDS is caused by the HIV virus — true; the Human Immunodeficiency Virus destroys the helper T-cells, so immunity collapses.
✗ Many diseases can be sexually transmitted — true; syphilis, gonorrhoea, warts, AIDS and hepatitis-B all spread by sexual contact.
📌 SSC & Railway Exam Facts
• Bacterial STDs — gonorrhoea and syphilis; Viral STDs — AIDS, warts and genital herpes.
• Syphilis is caused by the bacterium Treponema pallidum.
• AIDS spreads by unprotected sex, infected blood transfusion, shared needles and from an infected mother to her child — never by touching, sharing food or mosquito bites.
• World AIDS Day is observed on 1 December; the HIV test commonly used is ELISA.
• Other viral diseases: polio, measles, chickenpox, rabies, dengue, influenza, COVID-19.
Reproduction · Asexual reproduction (fission, budding, fragmentation, spores)

Q8. In which of the following modes of asexual reproduction if an individual is cut or broken into many pieces, many of these pieces grow into separate individuals?

  1. (a) budding
  2. (b) regeneration
  3. (c) spore formation
  4. (d) fission
Answer & solution

Answer: (b) regeneration

Correct Answer: Option 2 — regeneration
Regeneration is the ability of a simple organism to rebuild its whole body from a cut or broken piece. If Planaria (flatworm) or Hydra is cut into many pieces, most of the pieces grow into complete new individuals.
Regeneration is carried out by specialised (differentiated) cells which proliferate — they divide rapidly and then differentiate in an organised sequence to make the missing tissues and organs. It is possible only in animals with a simple body organisation; it is not the normal method of reproduction for them, because a complete organism cannot be produced from every cell of the body in complex animals.
❌ Why the other options are incorrect
✗ budding — a bud grows out of one spot on the parent body and separates as a new individual, as in yeast and Hydra; the parent is not cut into pieces.
✗ spore formation — hundreds of thick-walled spores are released from a sporangium, as in Rhizopus; no cutting of the body is involved.
✗ fission — a unicellular organism splits into two (binary fission, Amoeba) or many (multiple fission, Plasmodium) daughter cells; it applies to a single cell, not to a cut-up multicellular body.
📌 SSC & Railway Exam Facts
• Planaria, Hydra, starfish and earthworm are classic examples of regeneration.
• A starfish can grow a whole body from a single arm with a part of the central disc.
• A lizard regenerating its lost tail is partial regeneration, not reproduction.
• In humans, the liver has a remarkable power of regeneration.
• Hydra shows both budding and regeneration; Planaria is studied for regeneration through the work of T. H. Morgan.

Q9. Which of the following is an example of a unicellular organism?

  1. (a) Lizard
  2. (b) House fly
  3. (c) Amoeba
  4. (d) Hydra
Answer & solution

Answer: (c) Amoeba

Correct Answer: Option 3 — Amoeba
A unicellular (single-celled) organism is one whose entire body is made of only one cell, and that single cell carries out all the life processes — nutrition, respiration, excretion, movement and reproduction. Amoeba is the classic example: it belongs to the kingdom Protista, has an irregular, ever-changing shape, and moves and captures food using finger-like pseudopodia (false feet).
Inside the one cell, food is digested in a food vacuole, extra water is thrown out by the contractile vacuole, and reproduction takes place by binary fission.
❌ Why the other options are incorrect
✗ Lizard — a multicellular vertebrate reptile with a backbone, lungs, a four-chambered heart (partly divided) and a well-developed nervous system.
✗ House fly — a multicellular insect of the phylum Arthropoda, with a jointed body, compound eyes and a complete life cycle.
✗ Hydra — though simple and capable of budding and regeneration, it is a multicellular animal of the phylum Cnidaria (Coelenterata) with tentacles and stinging cells.
📌 SSC & Railway Exam Facts
• Other unicellular organisms: Paramecium, Euglena, Chlamydomonas, yeast and all bacteria.
• Amoeba shows holozoic nutrition and engulfs food by phagocytosis.
• Euglena has both plant and animal characters, so it is called a connecting link; it moves by a flagellum.
• Paramecium moves with cilia and is slipper-shaped.
• Kingdom Protista was added by R. H. Whittaker in his five-kingdom classification (1969).
Nutrition and Digestion · Absorption and assimilation

Q10. Select the correct option related to the function of villi in the small intestine.

  1. (a) Secretion of minerals
  2. (b) Increases absorptive surface area
  3. (c) Protection from acid
  4. (d) Helps in excretion
Answer & solution

Answer: (b) Increases absorptive surface area

Correct Answer: Option 2 — Increases absorptive surface area
The inner lining of the small intestine carries millions of tiny finger-like projections called villi. Their single purpose is to increase the surface area for absorption of the digested food, so that maximum nutrients are taken in during the short time the food stays there.
Each villus is richly supplied with blood capillaries and a lymph vessel (lacteal). Glucose, amino acids, vitamins and minerals pass into the blood capillaries, while fatty acids and glycerol are absorbed by the lacteal. The blood then carries the nutrients to every cell of the body — this is assimilation.
❌ Why the other options are incorrect
✗ Secretion of minerals — minerals are absorbed from food, not secreted by villi; digestive juices are secreted by the glands of the gut wall, liver and pancreas.
✗ Protection from acid — the stomach is protected from HCl by a layer of mucus, not by villi; food entering the small intestine is already made alkaline by bile and pancreatic juice.
✗ Helps in excretion — excretion of nitrogenous waste is done by the kidneys; undigested food is egested through the large intestine and anus, which is not the same as excretion.
📌 SSC & Railway Exam Facts
• The small intestine is the longest part of the alimentary canal (about 6 metres) and the main site of digestion and absorption.
• It has three parts — duodenum, jejunum and ileum; villi are most numerous in the ileum.
• Bile from the liver (stored in the gall bladder) has no enzyme; it emulsifies fat and makes the food alkaline.
• Pancreatic juice contains trypsin (protein), amylase (starch) and lipase (fat).
• The large intestine mainly absorbs water and some salts; the appendix is a vestigial organ.
Tissues · Meristematic tissues

Q11. What type of plant tissue is classified as apical, lateral and intercalary?

  1. (a) Connective tissue
  2. (b) Simple permanent tissue
  3. (c) Meristematic tissue
  4. (d) Complex permanent tissue
Answer & solution

Answer: (c) Meristematic tissue

Correct Answer: Option 3 — Meristematic tissue
Meristematic tissue is the dividing tissue of a plant. Its cells are small, thin-walled, closely packed, with dense cytoplasm, a prominent nucleus and no vacuoles, and they divide continuously to add new cells. On the basis of position in the plant body, meristem is of three types:
• Apical meristem — at the tips of roots and shoots; it increases the length of the plant (primary growth).
• Lateral meristem (cambium) — along the sides of the stem and root; it increases the girth/thickness (secondary growth).
• Intercalary meristem — at the base of leaves or above the nodes; it allows grasses and sugarcane to regrow after grazing or cutting.
❌ Why the other options are incorrect
✗ Connective tissue — an animal tissue (bone, blood, cartilage, tendon); it does not exist in plants and is never classified in this way.
✗ Simple permanent tissue — parenchyma, collenchyma, sclerenchyma; classified by cell type, not by apical/lateral/intercalary position, and these cells cannot divide.
✗ Complex permanent tissue — xylem and phloem, conducting tissues made of several cell types; again divided by function, not by position.
📌 SSC & Railway Exam Facts
• Cells of meristem lose the power of division and become permanent tissue by differentiation.
• Xylem conducts water and minerals upwards; phloem conducts food in both directions (translocation).
• Annual rings are produced by the vascular cambium, a lateral meristem, and reveal a tree's age.
• Parenchyma stores food; chlorenchyma has chloroplasts; aerenchyma with air cavities helps aquatic plants float.
• Sclerenchyma gives hardness and strength — it forms the husk of coconut; its cells are dead and lignified.
Tissues · Meristematic tissues

Q12. Cells of the meristematic tissue are very active, but they lack:

  1. (a) prominent nuclei
  2. (b) vacuoles
  3. (c) dense cytoplasm
  4. (d) thin cellulose walls
Answer & solution

Answer: (b) vacuoles

Correct Answer: Option 2 — vacuoles
NCERT describes meristematic cells as very active, having dense cytoplasm, thin cellulose walls and prominent nuclei, but they lack vacuoles. A large vacuole would fill the cell with water and sap and push the cytoplasm to the sides — exactly the opposite of what a rapidly dividing cell needs.
Because there is no vacuole, the whole cell is packed with cytoplasm, nucleus and organelles, giving it the high metabolic activity needed for continuous cell division. Large vacuoles appear only later, when the cells stop dividing and mature into permanent tissue such as parenchyma.
❌ Why the other options are incorrect
✗ prominent nuclei — meristematic cells do have a large, prominent nucleus, because the nucleus controls the repeated divisions.
✗ dense cytoplasm — their cytoplasm is dense and abundant, a sign of very high metabolic activity.
✗ thin cellulose walls — their walls are indeed thin and made of cellulose, which lets the cells expand and divide easily.
📌 SSC & Railway Exam Facts
• Meristem types by position: apical (length), lateral or cambium (girth) and intercalary (grasses).
• Growth in plants is localised, unlike animals where it is spread over the whole body.
• In a mature plant cell, a single large central vacuole may occupy up to 90% of the cell volume.
• The vacuole is bounded by a membrane called the tonoplast and provides turgidity and rigidity.
• Parenchyma, collenchyma and sclerenchyma are simple permanent tissues formed from meristem by differentiation.

Q13. Which of the following statements is/are correct related to respiration?
(i) In alveoli, exchange of gases takes place.
(ii) The walls of alveoli contain an extensive network of blood vessel.
(iii) In human beings, the respiratory pigments is haemoglobin.

  1. (a) (i), (ii) and (iii)
  2. (b) Only (i)
  3. (c) Only (i) and (iii)
  4. (d) Only (i) and (ii)
Answer & solution

Answer: (a) (i), (ii) and (iii)

Correct Answer: Option 1 — (i), (ii) and (iii)
All three statements agree with NCERT:
• (i) is correct — the alveoli are the balloon-like air sacs at the end of the smallest bronchioles, and they are the actual site of gaseous exchange in the lungs. O₂ diffuses from alveolar air into the blood and CO₂ diffuses out.
• (ii) is correct — the wall of each alveolus is only one cell thick and is wrapped in an extensive network of blood capillaries, so the two gases have a very short distance to diffuse.
• (iii) is correct — the respiratory pigment of human beings is haemoglobin, a red, iron-containing protein present in the red blood cells; it has a very high affinity for oxygen.
❌ Why the other options are incorrect
✗ Only (i) — it leaves out (ii) and (iii), though both are true.
✗ Only (i) and (iii) — it wrongly rejects (ii); alveolar walls really are richly supplied with capillaries.
✗ Only (i) and (ii) — it wrongly rejects (iii); haemoglobin is indeed the human respiratory pigment.
📌 SSC & Railway Exam Facts
• Path of air: nostrils → nasal cavity → pharynx → larynx → trachea → bronchi → bronchioles → alveoli.
• Each pair of lungs has about 300 million alveoli, giving a surface area of nearly 80 m².
• Haemoglobin carries O₂ as oxyhaemoglobin; CO₂ is mostly carried as bicarbonate in plasma, because it is more soluble in water.
• Carbon monoxide (CO) binds haemoglobin about 200–250 times more strongly than oxygen, causing suffocation.
• The diaphragm and the intercostal muscles bring about breathing; the rate of breathing in an adult at rest is about 12–20 per minute.

Chemistry (14)

Environmental Chemistry · Ozone layer and its depletion

Q14. Which of the following statements is/are correct related to the eco-system?
(i) Ozone is a molecule formed by three atoms of oxygen.
(ii) Oxygen (O₂) is essential for all aerobic forms of life.
(iii) Ozone is non-poisonous.

  1. (a) Only (i)
  2. (b) Only (i) and (ii)
  3. (c) Only (i) and (iii)
  4. (d) (i), (ii) and (iii)
Answer & solution

Answer: (b) Only (i) and (ii)

Correct Answer: Option 2 — Only (i) and (ii)
Checking the three statements against NCERT:
• (i) is correct — ozone (O₃) is a molecule made of three atoms of oxygen, unlike the ordinary oxygen we breathe, which has two atoms (O₂).
• (ii) is correct — O₂ is essential for all aerobic forms of life, because aerobic respiration oxidises glucose in the mitochondria to release large amounts of ATP.
• (iii) is INCORRECT — ozone is a deadly poison. At ground level it is a harmful pollutant that irritates the eyes and lungs and damages crops. It is useful only high up in the stratosphere, where it filters out harmful UV radiation.
So statements (i) and (ii) alone are correct.
❌ Why the other options are incorrect
✗ Only (i) — it leaves out (ii), although oxygen really is essential for every aerobic organism.
✗ Only (i) and (iii) — statement (iii) is false, since ozone is poisonous at the surface.
✗ (i), (ii) and (iii) — it wrongly accepts (iii); ozone is not non-poisonous.
📌 SSC & Railway Exam Facts
• Ozone is formed in the stratosphere when UV rays split O₂ into free oxygen atoms, which then join O₂ molecules: O₂ → O + O; O + O₂ → O₃.
• Excess UV radiation causes skin cancer, cataract and damage to the immune system and crops.
• CFCs (chlorofluorocarbons) used in refrigerators and aerosol sprays are the main ozone-depleting substances.
• The ozone hole over Antarctica was reported in 1985; the Montreal Protocol (1987) phased out CFCs.
• 16 September is observed as the International Day for the Preservation of the Ozone Layer (World Ozone Day).
Carbon and its Compounds · IUPAC nomenclature

Q15. What would be the name of hydrocarbon having three carbon chain with a double bond?

  1. (a) Propyne
  2. (b) Propane
  3. (c) Propanone
  4. (d) Propene
Answer & solution

Answer: (d) Propene

Correct Answer: Option 4 — Propene
In the IUPAC system the name of a hydrocarbon is built from two parts — a root word that tells the number of carbon atoms, and a suffix that tells the kind of bond present.
• Three carbon atoms → root word prop-
• Only single bonds → alkane, suffix -ane
• One double bond → alkene, suffix -ene
• One triple bond → alkyne, suffix -yne
A chain of three carbons carrying one double bond is therefore prop + ene = propene, whose formula is C₃H₆. It belongs to the alkene family with general formula CₙH₂ₙ, so for n = 3 we get C₃H₆, and it is an unsaturated hydrocarbon.
❌ Why the other options are incorrect
✗ Propyne — the suffix -yne means a triple bond, so propyne is C₃H₄, an alkyne, not an alkene.
✗ Propane — the suffix -ane means all single bonds; propane is the saturated hydrocarbon C₃H₈, the common LPG gas.
✗ Propanone — the ending -one is the ketone functional group; propanone is CH₃COCH₃ (acetone), not a plain hydrocarbon.
📌 SSC & Railway Exam Facts
• Root words: meth- (1), eth- (2), prop- (3), but- (4), pent- (5), hex- (6).
• Alkanes CₙH₂ₙ₊₂, Alkenes CₙH₂ₙ, Alkynes CₙH₂ₙ₋₂.
• Ethene (C₂H₄) is used for the artificial ripening of fruits; ethyne (C₂H₂) is the oxy-acetylene welding gas.
• Functional group suffixes: -ol alcohol, -al aldehyde, -one ketone, -oic acid carboxylic acid.
• Unsaturated hydrocarbons decolourise bromine water; saturated ones do not.
Mixtures, Solutions and Separation · Suspensions and colloids

Q16. Which of the following statements is true regarding a suspension?

  1. (a) Particles are smaller than those in a solution.
  2. (b) Particles cannot be separated by filtration.
  3. (c) Particles settle down after some time.
  4. (d) Particles do not scatter light.
Answer & solution

Answer: (c) Particles settle down after some time.

Correct Answer: Option 3 — Particles settle down after some time.
A suspension is a heterogeneous mixture in which the solute particles are large enough to be seen with the naked eye — bigger than 100 nm (10⁻⁷ m) — and remain merely dispersed, not dissolved, in the liquid.
Because these particles are heavy, gravity pulls them down and, once the mixture is left undisturbed, they settle at the bottom as a separate layer. This settling is called sedimentation, and it is the single most characteristic property of a suspension. Chalk powder or sand shaken with water, muddy river water, and a bottle of paint are the standard examples.
Suspension particles are large enough to be retained by filter paper, and they do scatter a beam of light, making its path visible — the Tyndall effect.
❌ Why the other options are incorrect
✗ Particles are smaller than those in a solution — the reverse is true. Solution particles are less than 1 nm, colloid particles 1–100 nm, and suspension particles greater than 100 nm, the largest of the three.
✗ Particles cannot be separated by filtration — suspension particles are the only ones that are removed by ordinary filter paper; it is solution particles that pass straight through.
✗ Particles do not scatter light — suspensions do scatter light and show the Tyndall effect; it is a true solution that does not scatter light.
📌 SSC & Railway Exam Facts
• Particle size: true solution < 1 nm, colloid 1–100 nm, suspension > 100 nm.
• Solution — homogeneous, stable, no Tyndall effect; colloid — heterogeneous, stable, shows Tyndall effect; suspension — heterogeneous, unstable, shows Tyndall effect.
• Colloids are separated by centrifugation, suspensions by filtration.
• The Tyndall effect explains the visible beam of a torch in a misty forest and light entering a dusty room.
• Milk is an emulsion (colloid), muddy water is a suspension, and sugar in water is a true solution.
Matter and its States · Evaporation and its factors

Q17. In which state of matter does evaporation occur?

  1. (a) Liquid only
  2. (b) Both plasma and liquid
  3. (c) Solid only
  4. (d) Plasma only
Answer & solution

Answer: (a) Liquid only

Correct Answer: Option 1 — Liquid only
Evaporation is the change of a liquid into its vapour at any temperature below its boiling point. It is purely a surface phenomenon — only those particles lying at the surface of the liquid that happen to gain enough kinetic energy can overcome the force of attraction of their neighbours and escape into the air.
Since the process needs a free liquid surface from which particles can fly off, evaporation takes place only in the liquid state.
Evaporation is faster when the surface area is larger, the temperature is higher, the humidity is lower, and the wind speed is greater. The escaping particles carry away energy, so evaporation always causes cooling — that is why we sweat, why water stays cool in an earthen pot, and why a desert cooler works.
❌ Why the other options are incorrect
✗ Both plasma and liquid — plasma is the ionised, super-energetic fourth state of matter found in the Sun, stars and fluorescent tubes; its particles are already free ions and electrons, so evaporation does not apply to it.
✗ Solid only — a solid changing directly into vapour is called sublimation, not evaporation, as in camphor, naphthalene, iodine and dry ice.
✗ Plasma only — evaporation is entirely a property of liquids; plasma is created by supplying very high energy to a gas, an altogether different process.
📌 SSC & Railway Exam Facts
• Evaporation — at all temperatures, only at the surface, a slow and cooling process.
• Boiling — at a fixed temperature, throughout the bulk of the liquid, a rapid process.
• The five states of matter: solid, liquid, gas, plasma and Bose–Einstein condensate (BEC).
• Plasma is the most abundant state of matter in the universe; BEC was obtained by cooling to near 0 K.
• Latent heat of vaporisation of water = 22.5 × 10⁵ J/kg, which is why steam burns are worse than boiling-water burns.
Acids, Bases and Salts · Neutralisation reaction

Q18. When hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), which of the following products is formed?

  1. (a) Only water
  2. (b) Water and salt
  3. (c) Only salt
  4. (d) Carbon dioxide and water
Answer & solution

Answer: (b) Water and salt

Correct Answer: Option 2 — Water and salt
The reaction between an acid and a base which gives a salt and water is called a neutralisation reaction. Here hydrochloric acid (HCl), a strong acid, reacts with sodium hydroxide (NaOH), a strong base, to give the salt sodium chloride (NaCl) and water.
⚗️ The Reaction
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
• In ionic form the essence of the reaction is H⁺ + OH⁻ → H₂O.
• Na⁺ pairs with Cl⁻ to give the salt, so it is also a double displacement reaction.
• The reaction is exothermic — the mixture becomes noticeably warm.
• NaCl is the salt of a strong acid and a strong base, so its solution is neutral, pH = 7.
The general form worth memorising is Acid + Base → Salt + Water.
❌ Why the other options are incorrect
✗ Only water — water is only one of the two products; the metal of the base and the negative ion of the acid must also combine, forming the salt.
✗ Only salt — the H⁺ of the acid and the OH⁻ of the base necessarily combine, and water is always produced in a neutralisation.
✗ Carbon dioxide and water — CO₂ appears when an acid reacts with a carbonate or a bicarbonate, e.g. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. NaOH contains no carbon at all.
📌 SSC & Railway Exam Facts
• Acid + Base → Salt + Water (neutralisation); Acid + Metal → Salt + H₂; Acid + Metal carbonate → Salt + H₂O + CO₂.
• NaOH is caustic soda, KOH is caustic potash, Ca(OH)₂ is slaked lime.
• Antacids such as milk of magnesia, Mg(OH)₂ work by neutralising excess stomach HCl.
• Baking soda is used on a bee sting (acidic) and vinegar on a wasp sting (alkaline).
• NaCl solution is neutral (pH 7), Na₂CO₃ solution is basic, NH₄Cl solution is acidic.
Mixtures, Solutions and Separation · Solutions, solubility and concentration

Q19. What is the term used to describe a solution that cannot dissolve any more solute at a given temperature?

  1. (a) Dilute solution
  2. (b) Unsaturated solution
  3. (c) Heterogeneous solution
  4. (d) Saturated solution
Answer & solution

Answer: (d) Saturated solution

Correct Answer: Option 4 — Saturated solution
A saturated solution is one in which no more solute can be dissolved at a given temperature. Any extra solute added simply settles at the bottom as undissolved solid.
The amount of solute needed to saturate 100 g of the solvent at that temperature is called the solubility of the solute. Solubility is therefore a property that is always quoted at a stated temperature, because it changes with temperature.
For most solids the solubility increases on heating — a saturated solution becomes unsaturated when warmed, and more solute can then be dissolved. On cooling the solution back down, the excess solute separates out as crystals, which is exactly how the technique of crystallisation purifies a solid. For gases, however, solubility decreases with a rise in temperature.
❌ Why the other options are incorrect
✗ Dilute solution — this only means the solution contains a relatively small quantity of solute; it can still dissolve a great deal more.
✗ Unsaturated solution — this is the exact opposite: a solution that can still dissolve more solute at that temperature.
✗ Heterogeneous solution — a contradiction in terms. A true solution is always homogeneous; heterogeneous mixtures are suspensions and colloids.
📌 SSC & Railway Exam Facts
• Solubility = grams of solute that saturate 100 g of solvent at a given temperature.
• Solution = solute + solvent; the solvent is present in the larger amount.
• Solubility of solids increases with temperature; solubility of gases decreases with temperature and increases with pressure.
• Fizz escapes from a cold drink bottle when opened because lowering the pressure lowers CO₂ solubility — Henry's law.
• Crystallisation is used to purify copper sulphate, alum and common salt.
Chemical Reactions and Equations · Oxidation, reduction and redox reactions

Q20. Which of the following is NOT an example of oxidation reaction?

  1. (a) Rancidity of oils
  2. (b) Corrosion
  3. (c) Formation of zinc from zinc oxide
  4. (d) Formation of magnesium oxide from magnesium
Answer & solution

Answer: (c) Formation of zinc from zinc oxide

Correct Answer: Option 3 — Formation of zinc from zinc oxide
Oxidation is the gain of oxygen or the loss of hydrogen or of electrons by a substance. Reduction is exactly the opposite — the loss of oxygen or the gain of hydrogen or of electrons.
When zinc oxide (ZnO) is converted into zinc (Zn), oxygen is removed from the compound. That is a loss of oxygen, so this change is a reduction, not an oxidation. It is therefore the required answer.
⚗️ The Reaction
ZnO + C → Zn + CO
• ZnO loses oxygen → it is reduced (ZnO is the oxidising agent).
• C gains oxygen → it is oxidised (carbon is the reducing agent).
• Both processes occur together, so the reaction is a redox reaction. This step is the smelting of zinc.
❌ Why the other options are incorrect (they ARE oxidations)
✗ Rancidity of oils — fats and oils are oxidised by the oxygen of air, giving the foul smell and taste of stale food. It is prevented by antioxidants and by flushing packets with nitrogen gas.
✗ Corrosion — a metal is slowly oxidised by air and moisture, as in the rusting of iron to hydrated iron(III) oxide, Fe₂O₃·xH₂O.
✗ Formation of magnesium oxide from magnesium — magnesium gains oxygen, 2Mg + O₂ → 2MgO, a textbook oxidation and combination reaction.
📌 SSC & Railway Exam Facts
• OIL RIG — Oxidation Is Loss of electrons, Reduction Is Gain of electrons.
• The substance oxidised is the reducing agent; the one reduced is the oxidising agent.
• Rusting needs both air (oxygen) and water; it is prevented by painting, greasing, galvanisation and electroplating.
• Galvanisation is coating iron with a layer of zinc.
• Chips packets are filled with nitrogen to stop rancidity.
Carbon and its Compounds · Ethanol and ethanoic acid

Q21. Which of the following compounds is used in tincture iodine?

  1. (a) Ethanoic acid
  2. (b) Propanol
  3. (c) Ethanol
  4. (d) Methanol
Answer & solution

Answer: (c) Ethanol

Correct Answer: Option 3 — Ethanol
Tincture iodine is a solution of iodine in alcohol, specifically a mixture of iodine dissolved in ethanol and water. It is a well-known antiseptic, applied to cuts and wounds to kill germs.
Ethanol (C₂H₅OH) is chosen as the solvent because iodine is only slightly soluble in water but dissolves readily in ethanol, and ethanol itself is a good antiseptic and is not poisonous in the small amounts involved. A "tincture" in pharmacy always means a medicinal solution prepared in alcohol.
Ethanol is a colourless liquid with a boiling point of 78 °C, is the alcohol present in all alcoholic drinks, and belongs to the alcohol family with the −OH functional group.
❌ Why the other options are incorrect
✗ Ethanoic acid — this is CH₃COOH, acetic acid; its 5–8% solution in water is vinegar, used to preserve food, not as the solvent of tincture iodine.
✗ Propanol — C₃H₇OH is a laboratory and industrial solvent, but tincture iodine is standardised with ethanol, not propanol.
✗ Methanol — CH₃OH, wood spirit, is highly poisonous; even small amounts cause blindness and death, so it can never be used on skin or in medicine.
📌 SSC & Railway Exam Facts
• Ethanol C₂H₅OH — b.p. 78 °C; Methanol CH₃OH — b.p. 65 °C, extremely toxic.
• Denatured alcohol is ethanol made unfit for drinking by adding methanol and a blue dye.
• Ethanol + conc. H₂SO₄ at 170 °C → ethene, a dehydration reaction.
• Ethanol + Na → sodium ethoxide + H₂; it is the test that shows the −OH group.
• Ethanoic acid has a melting point of 17 °C, so it freezes in cold weather and is called glacial acetic acid.
Everyday and Industrial Chemistry · Fuels and calorific value (LPG, CNG, coal, petroleum)

Q22. What is the full form of CNG?

  1. (a) Compressed Natural Gas
  2. (b) Combustible Nitrogen Gas
  3. (c) Carbonated Natural Gas
  4. (d) Chemical Natural Gas
Answer & solution

Answer: (a) Compressed Natural Gas

Correct Answer: Option 1 — Compressed Natural Gas
CNG stands for Compressed Natural Gas. It is natural gas stored under a high pressure of about 200–250 bar so that a large quantity can be packed into a small cylinder and carried in a vehicle.
Its main constituent is methane, CH₄ (about 80–90%), which makes it a clean fuel: it burns almost completely and releases far less carbon monoxide, unburnt hydrocarbons and particulate matter than petrol or diesel. It also produces no sulphur dioxide, so it does not contribute to acid rain.
Because of this, the Supreme Court ordered all public transport in Delhi to switch to CNG, and CNG is now used widely in buses, autos and cars, and as a domestic and industrial fuel (PNG — piped natural gas).
❌ Why the other options are incorrect
✗ Combustible Nitrogen Gas — nitrogen is not combustible at all; it is in fact an inert gas used to fill food packets precisely because it does not support burning.
✗ Carbonated Natural Gas — "carbonated" means dissolved CO₂, as in soft drinks; CO₂ is a non-supporter of combustion and is used in fire extinguishers, so such a fuel is impossible.
✗ Chemical Natural Gas — not a real term; every gas is a chemical, so the expansion carries no meaning.
📌 SSC & Railway Exam Facts
• CNG — Compressed Natural Gas, mainly methane (CH₄).
• LPG — Liquefied Petroleum Gas, mainly butane (C₄H₁₀) with propane; the smell is from ethyl mercaptan added as a warning.
• The calorific value of CNG ≈ 35,000–50,000 kJ/kg; that of LPG ≈ 55,000 kJ/kg; of hydrogen ≈ 150,000 kJ/kg, the highest of all fuels.
• A good fuel has a high calorific value, low ignition temperature, low ash and causes no pollution.
• Hydrogen is the cleanest fuel because burning it gives only water.
Carbon and its Compounds · Soaps and detergents

Q23. Which of the following statements is correct about soaps and detergents?

  1. (a) Soaps molecules have two ends – one is hydrophilic and the other end is hydrophobic.
  2. (b) Hydrophobic interacts with water molecules.
  3. (c) Hydrophilic ends interact with hydrocarbons.
  4. (d) Soap molecules have no ends.
Answer & solution

Answer: (a) Soaps molecules have two ends – one is hydrophilic and the other end is hydrophobic.

Correct Answer: Option 1 — Soaps molecules have two ends – one is hydrophilic and the other end is hydrophobic.
A soap molecule is the sodium or potassium salt of a long-chain fatty acid, for example C₁₇H₃₅COONa (sodium stearate). Every such molecule has two distinctly different ends:
• A long hydrocarbon tail which is hydrophobic (water-repelling) but oil-loving — it dissolves in grease and dirt.
• An ionic head, −COO⁻Na⁺, which is hydrophilic (water-loving) — it dissolves in water.
This dual nature is precisely what makes cleaning possible. When soap is added to dirty cloth, the hydrophobic tails bury themselves in the oily dirt while the hydrophilic heads stay in the water. The soap molecules arrange themselves into a spherical cluster called a micelle, with the dirt trapped inside. Agitation then lifts the micelles away and the dirt is rinsed off.
❌ Why the other options are incorrect
✗ Hydrophobic interacts with water molecules — the opposite is true; "hydrophobic" literally means water-fearing, and this end interacts with oil and grease.
✗ Hydrophilic ends interact with hydrocarbons — the hydrophilic ionic head interacts with water; it is the hydrocarbon tail that mixes with hydrocarbons.
✗ Soap molecules have no ends — if a soap molecule had no two distinct ends, no micelle could form and soap could not clean at all.
📌 SSC & Railway Exam Facts
• Soap = sodium/potassium salt of a long-chain carboxylic (fatty) acid; the process of making it is saponification.
• Soaps do not lather in hard water because Ca²⁺ and Mg²⁺ form an insoluble scum; detergents work even in hard water.
• Detergents are ammonium or sulphonate salts of long-chain hydrocarbons.
• The micelle forms in water but not in ethanol, since the tails would dissolve in ethanol.
• Detergents with branched chains are non-biodegradable and cause water pollution and foaming in rivers.
Acids, Bases and Salts · Properties of bases and alkalis

Q24. Which of the following substances can act as a source of OH⁻ ions when dissolved in water?

  1. (a) Sodium hydroxide
  2. (b) Carbon dioxide
  3. (c) Hydrochloric acid
  4. (d) Nitric acid
Answer & solution

Answer: (a) Sodium hydroxide

Correct Answer: Option 1 — Sodium hydroxide
According to the Arrhenius concept, a base is a substance that produces hydroxide ions, OH⁻, when dissolved in water, while an acid is a substance that produces hydrogen ions, H⁺ (as H₃O⁺), in water.
Sodium hydroxide (NaOH) is a strong base; it dissolves in water and ionises almost completely, flooding the solution with OH⁻ ions.
⚗️ The Reaction
NaOH(aq) → Na⁺(aq) + OH⁻(aq)
• Because ionisation is nearly complete, NaOH is a strong alkali — a base that is soluble in water.
• Its solution turns red litmus blue, phenolphthalein pink and has a pH well above 7 (about 14 for 1 M NaOH).
• Dissolving NaOH in water is highly exothermic, so the acid/base must always be added to water, never the reverse.
❌ Why the other options are incorrect
✗ Carbon dioxide — an acidic oxide; dissolving in water it gives carbonic acid, CO₂ + H₂O → H₂CO₃, which releases H⁺, not OH⁻.
✗ Hydrochloric acid — a strong acid, ionising as HCl → H⁺ + Cl⁻; it is the acid present in gastric juice.
✗ Nitric acid — also a strong acid, HNO₃ → H⁺ + NO₃⁻, used in making fertilisers and explosives.
📌 SSC & Railway Exam Facts
• Common bases: NaOH (caustic soda), KOH (caustic potash), Ca(OH)₂ (slaked lime), NH₄OH, Mg(OH)₂ (milk of magnesia).
• All alkalis are bases, but all bases are not alkalis — only water-soluble bases are alkalis.
• pH < 7 acidic, pH = 7 neutral, pH > 7 basic; the pH scale was given by S. P. L. Sørensen.
• Turmeric and china rose are natural indicators; onion and vanilla are olfactory indicators.
• Ca(OH)₂ + CO₂ → CaCO₃ + H₂O — the lime-water test that turns lime water milky.
Metals and Non-metals · Reactivity series

Q25. Which of the following metals is LEAST reactive towards dilute hydrochloric acid?

  1. (a) Zn
  2. (b) Mg
  3. (c) Al
  4. (d) Fe
Answer & solution

Answer: (d) Fe

Correct Answer: Option 4 — Fe
A metal reacts with a dilute acid only if it lies above hydrogen in the reactivity series, displacing H⁺ to liberate hydrogen gas. The higher a metal stands in the series, the more vigorous the reaction.
The relevant order in the activity series is:
Mg > Al > Zn > Fe > (H) > Cu
Among the four metals given, iron (Fe) sits lowest, immediately above hydrogen. It therefore reacts with dilute HCl most slowly — the effervescence is gentle and the solution slowly turns pale green, whereas magnesium reacts almost explosively.
⚗️ The Reactions
Mg + 2HCl → MgCl₂ + H₂↑ — very vigorous, the test tube becomes hot
2Al + 6HCl → 2AlCl₃ + 3H₂↑ — vigorous once the oxide layer dissolves
Zn + 2HCl → ZnCl₂ + H₂↑ — brisk, the standard laboratory method for making H₂
Fe + 2HCl → FeCl₂ + H₂↑ — the slowest of the four
❌ Why the other options are incorrect
✗ Zn — it lies above iron, so it reacts faster than Fe; zinc with dilute HCl is the usual laboratory preparation of hydrogen.
✗ Mg — the most reactive of the four; it reacts so fast that the mixture heats up strongly.
✗ Al — more reactive than both zinc and iron; it is protected only by a thin oxide layer, which the acid removes.
📌 SSC & Railway Exam Facts
• Reactivity series: K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au > Pt.
• Metals below hydrogen — Cu, Ag, Au — do not displace H₂ from dilute acids.
• Dilute HNO₃ does not usually give H₂ with metals because it is a strong oxidising agent; Mg and Mn are the exceptions.
• Aqua regia (3 : 1 conc. HCl : conc. HNO₃) dissolves even gold and platinum.
• The burning splinter test — hydrogen burns with a pop sound.
Mixtures, Solutions and Separation · Suspensions and colloids

Q26. Which of the following is a suspension?

  1. (a) Solids dispersed in solids
  2. (b) Gases dispersed in solids
  3. (c) Gases dispersed in liquids
  4. (d) Solids dispersed in liquids
Answer & solution

Answer: (d) Solids dispersed in liquids

Correct Answer: Option 4 — Solids dispersed in liquids
A suspension is a heterogeneous mixture in which insoluble solid particles are spread throughout a liquid without dissolving. The solid is the dispersed phase and the liquid is the dispersion medium, and the particles are larger than 100 nm, big enough to be seen and to settle out on standing.
Everyday examples are chalk powder in water, sand in water, muddy river water, paint, and a bottle of medicine marked "shake well before use" — the instruction itself exists because the solid keeps settling.
Suspension particles scatter light (Tyndall effect), are retained by filter paper, and the mixture is unstable.
❌ Why the other options are incorrect
✗ Solids dispersed in solids — this is a solid sol colloid, as in coloured gemstones and alloys; it is not a suspension.
✗ Gases dispersed in solids — this is a solid foam, as in sponge, foam rubber and pumice stone.
✗ Gases dispersed in liquids — this is a foam, as in soap lather, shaving cream and whipped cream.
📌 SSC & Railway Exam Facts
• Colloid types: sol (solid in liquid — ink, paints), gel (liquid in solid — cheese, butter, jelly), emulsion (liquid in liquid — milk, cream), aerosol (solid/liquid in gas — smoke, fog, clouds), foam (gas in liquid — shaving cream).
• Smoke = solid in gas; fog and clouds = liquid in gas; milk = liquid in liquid.
• Suspension particles settle on standing; colloid particles do not settle.
• Suspensions are separated by filtration or sedimentation and decantation.
• Butter and cheese are gels; milk and cream are emulsions.
Acids, Bases and Salts · pH scale and its importance

Q27. Which of the following ion concentrations is measured by the pH scale?

  1. (a) Na⁺ ion concentration
  2. (b) Cl⁻ ion concentration
  3. (c) H⁺ ion concentration
  4. (d) F⁻ ion concentration
Answer & solution

Answer: (c) H⁺ ion concentration

Correct Answer: Option 3 — H⁺ ion concentration
The pH scale measures the concentration of hydrogen ions (H⁺) present in a solution. In fact the "p" in pH stands for the German "potenz", meaning power, so pH literally means the "power of hydrogen".
Mathematically, pH = −log[H⁺], where [H⁺] is the hydrogen-ion concentration in mol/L. The scale runs from 0 to 14:
• pH < 7 → acidic, because the H⁺ concentration is high (a lower pH means a stronger acid).
• pH = 7 → neutral, as in pure water, where [H⁺] = [OH⁻].
• pH > 7 → basic/alkaline, because the OH⁻ concentration exceeds the H⁺ concentration.
The scale was introduced by the Danish chemist S. P. L. Sørensen in 1909. A change of one pH unit means a tenfold change in hydrogen-ion concentration.
❌ Why the other options are incorrect
✗ Na⁺ ion concentration — sodium ions are spectator ions; they do not decide whether a solution is acidic or basic.
✗ Cl⁻ ion concentration — chloride ions likewise take no part; a 1 M NaCl solution is neutral even though it is full of Cl⁻.
✗ F⁻ ion concentration — fluoride content matters for dental health and fluorosis, but it is not what the pH scale measures.
📌 SSC & Railway Exam Facts
• pH = −log[H⁺]; pH + pOH = 14 at 25 °C.
• Human blood pH = 7.35–7.45; saliva ≈ 6.8; gastric juice ≈ 1.2–3.0; pure water = 7; milk ≈ 6.6.
• Rain below pH 5.6 is acid rain, caused by SO₂ and NO₂.
• Tooth decay begins when the mouth pH falls below 5.5; toothpastes are basic to neutralise this.
• Plants and crops grow best in a narrow pH range of soil, close to neutral.

Physics (13)

Gravitation · Numericals on gravitation

Q28. If an object on the surface of the moon weighs 3 N, the weight of the same object on surface of the Earth is ______.

  1. (a) 18 N
  2. (b) 3 N
  3. (c) 12 N
  4. (d) 24 N
Answer & solution

Answer: (a) 18 N

Correct Answer: Option 1 — 18 N
📖 Given
• Weight of the object on the moon, W(moon) = 3 N
• Acceleration due to gravity on the moon is one-sixth of that on the Earth, g(moon) = g(earth)/6
• The mass m of the object stays the same everywhere
✍️ Formula
W = mg, so W(earth)/W(moon) = g(earth)/g(moon) = 6
🔢 Solution
• W(earth) = 6 × W(moon)
• W(earth) = 6 × 3
• W(earth) = 18 N
So the same object that presses down with only 3 N on the moon presses down with 18 N on the Earth, because the Earth pulls six times as strongly.
❌ Why the other options are incorrect
✗ 3 N — this assumes weight is the same everywhere; that is true of mass, not of weight.
✗ 12 N — this comes from multiplying by 4 instead of 6.
✗ 24 N — this comes from multiplying by 8 instead of 6.
📌 SSC & Railway Exam Facts
• g(earth) = 9.8 m s⁻²; g(moon) ≈ 1.63 m s⁻², about 1/6 of the Earth's value.
• Mass is measured in kg and is constant; weight is a force measured in newton (N).
• Newton's law of gravitation: F = Gm₁m₂/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
• The value of g is maximum at the poles and minimum at the equator.
• In free fall (and inside an orbiting satellite) a body feels weightlessness.
Work, Energy and Power · Numericals on work, energy and power

Q29. A force of 20 N acts on a body in a direction opposite to the displacement of the body. If under the action of this force the displacement is 2m, the work done will be:

  1. (a) −40 J
  2. (b) 40 J
  3. (c) 0 J
  4. (d) −20 J
Answer & solution

Answer: (a) −40 J

Correct Answer: Option 1 — −40 J
📖 Given
• Force, F = 20 N
• Displacement, s = 2 m
• The force acts opposite to the displacement, so the angle θ = 180°
✍️ Formula
W = F s cos θ
🔢 Solution
• W = 20 × 2 × cos 180°
• cos 180° = −1
• W = 20 × 2 × (−1)
• W = −40 J
The minus sign means the force is taking energy away from the body instead of giving energy to it — exactly what friction or a brake does to a moving object.
❌ Why the other options are incorrect
✗ 40 J — this is the value you get if you forget the direction and take θ = 0°, i.e. treat the force as helping the motion.
✗ 0 J — work would be zero only if the displacement were zero or the force were perpendicular to the displacement (θ = 90°).
✗ −20 J — this uses only the force and drops the displacement; you must multiply F by s.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit of work is the joule (J), and 1 J = 1 N × 1 m.
• Work is positive at θ = 0°, negative at θ = 180° and zero at θ = 90°.
• Work and energy are scalar quantities even though force and displacement are vectors.
• CGS unit of work is the erg; 1 J = 10⁷ erg.
• Power = work/time, SI unit watt (W); 1 horsepower = 746 W.
Electricity · Resistance, resistivity and their factors

Q30. Which of the following factors do NOT affect the resistance of a conductor (at constant temperature):
(i) Area of cross-section of the conductor
(ii) Current flowing through the conductor
(iii)Length of the conductor

  1. (a) Both (i) and (ii)
  2. (b) Both (i) and (iii)
  3. (c) Only (iii)
  4. (d) Only (ii)
Answer & solution

Answer: (d) Only (ii)

Correct Answer: Option 4 — Only (ii)
The resistance of a metallic conductor is given by R = ρl/A, where ρ is the resistivity of the material, l the length and A the area of cross-section.
• Length (iii) — R ∝ l. A longer wire offers more resistance, so length does affect R.
• Area of cross-section (i) — R ∝ 1/A. A thicker wire offers less resistance, so area does affect R.
• Current (ii) — resistance is a property of the conductor itself. At a constant temperature, changing the current does not change R; by Ohm's law V = IR, a larger current simply means a larger potential difference across the same resistance.
Hence the only factor that does not affect resistance is (ii), the current.
❌ Why the other options are incorrect
✗ Both (i) and (ii) — area of cross-section certainly affects resistance, so (i) cannot be included.
✗ Both (i) and (iii) — both length and area are the two geometrical factors that decide resistance.
✗ Only (iii) — length is one of the strongest factors: double the length and the resistance doubles.
📌 SSC & Railway Exam Facts
• Resistance depends on length, area of cross-section, nature of material and temperature.
• SI unit of resistance is the ohm (Ω); of resistivity, the ohm metre (Ω m).
• Ohm's law: V = IR, valid at constant temperature.
• Silver has the lowest resistivity; nichrome has a high resistivity and is used in heaters.
• For a metal, resistance increases with rise in temperature.
Magnetism and Magnetic Effects of Current · Solenoid and electromagnet

Q31. What is the nature of the magnetic field inside a current-carrying solenoid?

  1. (a) Strong non-uniform magnetic field
  2. (b) Weak uniform magnetic field
  3. (c) Strong uniform magnetic field
  4. (d) Weak non-uniform magnetic field
Answer & solution

Answer: (c) Strong uniform magnetic field

Correct Answer: Option 3 — Strong uniform magnetic field
A solenoid is a long cylindrical coil made of many closely wound circular turns of insulated copper wire. When a current flows, each turn produces its own magnetic field and all these fields add up along the axis.
Inside the solenoid the field lines are parallel straight lines of the same direction and the same closeness everywhere. Equally spaced parallel lines mean the field has the same magnitude and same direction at every point — that is, it is uniform. Because the contributions of hundreds of turns add together, the field is also strong.
Its strength is B = μ₀nI, where n is the number of turns per unit length. Outside, the solenoid behaves exactly like a bar magnet, with one end acting as the north pole and the other as the south pole.
❌ Why the other options are incorrect
✗ Strong non-uniform magnetic field — the field is strong, but inside a solenoid it is not non-uniform; non-uniform fields occur near the ends and outside.
✗ Weak uniform magnetic field — uniform is right, but the many turns make the field strong, not weak.
✗ Weak non-uniform magnetic field — both parts are wrong for the region inside the solenoid.
📌 SSC & Railway Exam Facts
• Field inside a solenoid: B = μ₀nI; SI unit of magnetic field is the tesla (T).
• Placing a soft iron core inside a solenoid makes a strong electromagnet.
• Direction of the field is given by the right-hand thumb rule.
• Hans Christian Oersted (1820) discovered the magnetic effect of electric current.
• Around a straight current-carrying wire, field lines are concentric circles.
Light - Reflection and Mirrors · Concave mirror - image formation and uses

Q32. A concave mirror has a radius of curvature of 4.00 m. If any object is located at 2.00 m from this mirror, find the position of the image.

  1. (a) Zero
  2. (b) Infinity
  3. (c) Between pole and focus point
  4. (d) At centre of curvature
Answer & solution

Answer: (b) Infinity

Correct Answer: Option 2 — Infinity
📖 Given
• Radius of curvature, R = 4.00 m
• Object distance from the mirror, u = 2.00 m
• Mirror is concave
✍️ Formula
f = R/2 and 1/v + 1/u = 1/f
🔢 Solution
• f = R/2 = 4.00/2 = 2.00 m
• So the object distance u = 2.00 m is exactly equal to the focal length f — the object is placed at the principal focus F.
• Using the mirror formula: 1/v = 1/f − 1/u = 1/2.00 − 1/2.00
• 1/v = 0
• v = infinity
When an object is kept at the focus of a concave mirror, all the reflected rays become parallel to one another, so they meet only at infinity. The image is real, inverted and highly enlarged.
❌ Why the other options are incorrect
✗ Zero — v = 0 would mean the image is formed on the mirror itself, which happens for no ordinary object position.
✗ Between pole and focus point — an image in the P–F region belongs to a convex mirror, which always forms a virtual, diminished image between its pole and focus. A concave mirror never forms a real image there.
✗ At centre of curvature — the image forms at C only when the object is also placed at C (u = R), here that would need u = 4.00 m.
📌 SSC & Railway Exam Facts
• For a spherical mirror, R = 2f.
• Object at F → image at infinity; object at infinity → image at F.
• This is the working principle of a torch, headlight and searchlight, where the bulb is placed at the focus.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u.
• A concave mirror is used in solar cookers, shaving mirrors and dentists' mirrors.
Work, Energy and Power · Power and its units (watt, horsepower)

Q33. The rate of doing work can be defined as ______.

  1. (a) force
  2. (b) power
  3. (c) work
  4. (d) energy
Answer & solution

Answer: (b) power

Correct Answer: Option 2 — power
Power is defined as the rate of doing work, that is, the work done per unit time:
P = W/t
It can also be written as the rate of transfer of energy, P = E/t. Its SI unit is the watt (W), where 1 watt = 1 joule per second.
Power tells us how fast a job is done, not how big the job is. Two cranes may both lift the same load to the same height — so both do the same work — but the crane that finishes sooner has the greater power.
When a constant force F moves a body with a steady velocity v, power can also be written as P = F × v.
❌ Why the other options are incorrect
✗ force — force is a push or pull, F = ma, measured in newton (N); it has nothing to do with time.
✗ work — work is the product of force and displacement in its direction, W = Fs cos θ, measured in joule (J); it is the job itself, not the rate of doing it.
✗ energy — energy is the capacity to do work, also measured in joule (J); dividing energy by time gives power.
📌 SSC & Railway Exam Facts
• Power P = W/t = F v; SI unit watt (W), named after James Watt.
• 1 kilowatt = 1000 W; 1 horsepower (hp) = 746 W.
• Commercial unit of electrical energy is the kilowatt-hour (kWh); 1 kWh = 3.6 × 10⁶ J.
• Electric power P = VI = I²R = V²/R.
• Power is a scalar quantity.
Light - Reflection and Mirrors · Mirror formula and magnification

Q34. A positive and negative sign in the magnification indicates that the images are:

  1. (a) real and virtual, respectively
  2. (b) virtual and real, respectively
  3. (c) both real in nature
  4. (d) both virtual in nature
Answer & solution

Answer: (b) virtual and real, respectively

Correct Answer: Option 2 — virtual and real, respectively
The magnification of a mirror is m = h′/h = −v/u, where h′ is the image height and h the object height. The sign of m carries the information about the nature of the image:
• Positive magnification (m > 0) — the image is erect, and an erect image formed by a mirror or lens is always virtual.
• Negative magnification (m < 0) — the image is inverted, and an inverted image is always real.
So a positive sign means a virtual image and a negative sign means a real image — in that order. A plane mirror always gives m = +1 (virtual, erect, same size), while a concave mirror with the object beyond F gives a negative m (real, inverted).
❌ Why the other options are incorrect
✗ real and virtual, respectively — this reverses the rule; positive goes with virtual, not with real.
✗ both real in nature — then every image would be inverted, which is false for a plane mirror and a convex mirror.
✗ both virtual in nature — then no image could ever be caught on a screen, but real images certainly can be.
📌 SSC & Railway Exam Facts
• m = h′/h = −v/u for mirrors; m = h′/h = v/u for lenses.
• |m| > 1 → enlarged, |m| < 1 → diminished, |m| = 1 → same size.
• A convex mirror always forms a virtual, erect and diminished image, so m is positive and less than 1.
• A real image can be taken on a screen; a virtual image cannot.
• Mirror formula: 1/v + 1/u = 1/f, with sign conventions measured from the pole.
Electricity · Numericals on electricity

Q35. A current of 2A flows through a wire when the two ends of the wire are maintained at a potential difference of 4 V. The resistance of the wire will be:

  1. (a) 8 Ω
  2. (b) 2 Ω
  3. (c) 1/2 Ω
  4. (d) 4 Ω
Answer & solution

Answer: (b) 2 Ω

Correct Answer: Option 2 — 2 Ω
📖 Given
• Current through the wire, I = 2 A
• Potential difference across its ends, V = 4 V
• Temperature is constant
✍️ Formula
Ohm's law: V = IR, so R = V/I
🔢 Solution
• R = V/I
• R = 4/2
• R = 2 Ω
So this wire opposes the flow of charge with a resistance of 2 ohms: every 2 volts applied across it drives 1 ampere of current through it.
❌ Why the other options are incorrect
✗ 8 Ω — this comes from multiplying V by I (4 × 2) instead of dividing; V × I gives power in watts, not resistance.
✗ 1/2 Ω — this is I/V = 2/4, the ratio taken upside down; that quantity is the conductance in siemens.
✗ 4 Ω — this is simply the voltage value copied as the answer, ignoring the current altogether.
📌 SSC & Railway Exam Facts
• Ohm's law: V = IR, given by Georg Simon Ohm; valid at constant temperature.
• 1 Ω = 1 V/A; conductance G = 1/R, unit siemens (S).
• A V–I graph for an ohmic conductor is a straight line through the origin, its slope giving 1/R.
• Ammeter is connected in series, voltmeter in parallel.
• Series resistance: R(eq) = R₁ + R₂; parallel: 1/R(eq) = 1/R₁ + 1/R₂.
Motion · Numericals on motion

Q36. A bus accelerates uniformly from 12 m s⁻¹ to 30 m s⁻¹ in 30 s. Calculate the distance covered by the bus in that time.

  1. (a) 175 m
  2. (b) 400 m
  3. (c) 630 m
  4. (d) 1900 m
Answer & solution

Answer: (c) 630 m

Correct Answer: Option 3 — 630 m
📖 Given
• Initial velocity, u = 12 m s⁻¹
• Final velocity, v = 30 m s⁻¹
• Time taken, t = 30 s
• Acceleration is uniform
✍️ Formula
a = (v − u)/t and s = ut + ½at²
🔢 Solution
• a = (30 − 12)/30
• a = 18/30 = 0.6 m s⁻²
• s = (12 × 30) + ½ × 0.6 × (30)²
• s = 360 + (0.3 × 900)
• s = 360 + 270
• s = 630 m
The same answer comes straight from the average-velocity shortcut: s = ((u + v)/2) × t = ((12 + 30)/2) × 30 = 21 × 30 = 630 m.
❌ Why the other options are incorrect
✗ 175 m — does not follow from any correct step of this problem; it is far too small for a bus moving at 12–30 m s⁻¹ for half a minute.
✗ 400 m — this is roughly what you get if only the change in velocity (18 m s⁻¹) is used with a wrong time factor.
✗ 1900 m — this comes from wrongly multiplying the final velocity by a much larger time, ignoring that the bus started at 12 m s⁻¹ and ran for only 30 s.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• For uniform acceleration, average velocity = (u + v)/2.
• Area under a velocity–time graph gives the distance travelled.
• Slope of a velocity–time graph gives the acceleration.
• SI unit of acceleration is m s⁻²; it is a vector quantity.
Human Eye and the Colourful World · Prism and dispersion of light

Q37. When a white light falls on a glass prism, ________ colour undergoes maximum deviation.

  1. (a) Violet
  2. (b) Blue
  3. (c) Yellow
  4. (d) Red
Answer & solution

Answer: (a) Violet

Correct Answer: Option 1 — Violet
When white light passes through a glass prism it splits into seven colours — VIBGYOR. This splitting is called dispersion of light, first explained by Isaac Newton.
Dispersion happens because the refractive index of glass is different for different colours. Violet light has the shortest wavelength, and for it the refractive index of glass is the highest. A higher refractive index means the ray bends more, so violet is deviated the most.
Red light has the longest wavelength and the lowest refractive index in glass, so it bends the least. That is why on the screen violet appears at the bottom of the spectrum and red at the top.
❌ Why the other options are incorrect
✗ Blue — blue is deviated a lot, but still less than violet, because its wavelength is longer than violet's.
✗ Yellow — yellow lies in the middle of the spectrum and suffers only a moderate deviation.
✗ Red — red suffers the least deviation of all seven colours, the exact opposite of what is asked.
📌 SSC & Railway Exam Facts
• Order of deviation: Violet > Indigo > Blue > Green > Yellow > Orange > Red.
• Order of wavelength is the reverse: Red (≈ 700 nm) > … > Violet (≈ 400 nm).
• Dispersion was explained by Isaac Newton, who recombined the colours with a second prism.
• A rainbow is formed by refraction, total internal reflection and dispersion in water droplets.
• The sky is blue because of the scattering of light (Rayleigh scattering), not dispersion.
Work, Energy and Power · Conservation and transformation of energy

Q38. Which of the following statements related to the law of conservation of energy is true?

  1. (a) Energy cannot be converted from one form to another.
  2. (b) Energy can be created or destroyed.
  3. (c) The total energy before and after a transformation process are different.
  4. (d) The law of conservation of energy is valid in all situations and for all kinds of transformations.
Answer & solution

Answer: (d) The law of conservation of energy is valid in all situations and for all kinds of transformations.

Correct Answer: Option 4 — The law of conservation of energy is valid in all situations and for all kinds of transformations.
The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy before and after the transformation always remains the same.
This law has been tested in mechanics, heat, light, sound, electricity, chemistry and nuclear physics, and no exception has ever been found. That is why the correct statement is that it is valid in all situations and for every kind of transformation.
Everyday examples: in a bulb, electrical energy → light + heat; in a falling body, potential energy → kinetic energy; in a loudspeaker, electrical energy → sound energy.
❌ Why the other options are incorrect
✗ Energy cannot be converted from one form to another. — this is exactly the opposite of the truth; every machine works by converting one form of energy into another.
✗ Energy can be created or destroyed. — the law says precisely that energy can be neither created nor destroyed.
✗ The total energy before and after a transformation process are different. — the total energy always stays the same; what looks like a loss is usually energy turned into heat or sound.
📌 SSC & Railway Exam Facts
• Energy can neither be created nor destroyed, only transformed — total energy is always conserved.
• Mechanical energy = KE + PE; for a freely falling body the sum stays constant.
• In nuclear reactions mass and energy are linked by E = mc² (Einstein).
• Hydroelectric plant: PE of water → KE → electrical energy.
• SI unit of energy is the joule (J); 1 calorie = 4.2 J.
Motion · Numericals on motion

Q39. A motorcycle starting from rest attains a velocity of 30 m s⁻¹ in 10 s. Assuming that the acceleration is uniform, find the distance travelled by the motorcycle to attain this velocity.

  1. (a) 250 m
  2. (b) 150 m
  3. (c) 270 m
  4. (d) 300 m
Answer & solution

Answer: (b) 150 m

Correct Answer: Option 2 — 150 m
📖 Given
• Initial velocity, u = 0 (starts from rest)
• Final velocity, v = 30 m s⁻¹
• Time taken, t = 10 s
• Acceleration is uniform
✍️ Formula
a = (v − u)/t and s = ut + ½at²
🔢 Solution
• a = (30 − 0)/10 = 3 m s⁻²
• s = (0 × 10) + ½ × 3 × (10)²
• s = 0 + (1.5 × 100)
• s = 150 m
The average-velocity shortcut gives the same result: s = ((0 + 30)/2) × 10 = 15 × 10 = 150 m. So the motorcycle covers 150 metres while building up to 30 m s⁻¹.
❌ Why the other options are incorrect
✗ 250 m — not obtained from any correct step; it would need an acceleration of 5 m s⁻² over 10 s.
✗ 270 m — this is the answer to a different problem (u = 12 m s⁻¹ type data); it does not fit u = 0.
✗ 300 m — this is v × t = 30 × 10, which wrongly assumes the motorcycle moved at the full 30 m s⁻¹ for the whole 10 s; in fact it started from rest, so its average speed was only 15 m s⁻¹.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• Starting from rest means u = 0, which simplifies s = ½at².
• For uniform acceleration, average velocity = (u + v)/2.
• Area under the velocity–time graph = distance travelled.
• Distance is a scalar; displacement is a vector.
Work, Energy and Power · Work and its sign

Q40. A book is lying on a table and Rahul moves the book in the direction of the force for some distance D. The work done by the force is:

  1. (a) undetermined
  2. (b) zero
  3. (c) positive
  4. (d) negative
Answer & solution

Answer: (c) positive

Correct Answer: Option 3 — positive
Work done by a force is W = F s cos θ, where θ is the angle between the force and the displacement.
Here the book is moved in the direction of the force through a distance D, so θ = 0° and cos 0° = +1. Therefore
W = F × D × 1 = +FD
which is a positive quantity. Positive work means the force is giving energy to the book, and that energy shows up as the motion of the book across the table.
This is the standard case of positive work: a horse pulling a cart forward, a person pushing a trolley ahead, or gravity acting on a freely falling stone.
❌ Why the other options are incorrect
✗ undetermined — the work is perfectly well defined; F, D and θ = 0° are all known in principle, so W = FD.
✗ zero — work would be zero only if the displacement were zero or the force acted perpendicular to the displacement (θ = 90°).
✗ negative — negative work needs the force to act opposite to the displacement (θ = 180°), as friction does; here the force and displacement point the same way.
📌 SSC & Railway Exam Facts
• W = F s cos θ; SI unit joule (J), 1 J = 1 N m.
• θ = 0° → positive work, θ = 90° → zero work, θ = 180° → negative work.
• Work done against friction is always negative.
• A coolie carrying a load on his head on level ground does zero work against gravity.
• Work–energy theorem: net work done = change in kinetic energy.
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