Science Class X MCQ with Solutions Class 10 Notes PDF

Science Class X MCQ with Solutions Class 10 notes PDF. Complete chapter explanation, key points and diagrams. CBSE 2026-27. Free download – USP.

This free MCQ for CBSE Class X Science contains multiple choice questions with answers, exactly the style asked in school and board exams. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

📌 How to use this MCQ

First revise the chapter from your notes or textbook.
Attempt every question on your own before checking answers — this is how marks actually improve.
Mark the questions you got wrong and re-attempt them after 2–3 days.

Science MCQ Class 10 (PDF + Answers)

UNIQUE STUDY POINT BY SUMEET SAHU

Science MCQ

Class 10 · Biology, Chemistry & Physics · Practice Worksheet with Solutions

Tap any question's "Show Answer" button to reveal the correct answer and full step-by-step explanation. Tap an MCQ option to attempt it first.

Download PDF

All Unseen Starred Wrong

0 / 74 questions seen

No questions match that search.

Section A · Biology (Q1–Q29)

29 Questions

In human respiratory system, when a person breathes in, the position of ribs and diaphragm will be:MCQ1 Mark

a) relaxed ribs and curve/dome shaped diaphragm.
b) relaxed ribs and flattened diaphragm.
c) lifted ribs and flattened diaphragm.
d) lifted ribs and curve/dome shaped diaphragm.

Correct Answer: (c) lifted ribs and flattened diaphragm.

When we breathe in (inhale), the rib muscles contract, lifting the ribs, and the diaphragm contracts and flattens, increasing the chest volume so air rushes in.

Temporary finger-like extensions on the cell surface to take in food is formed in:MCQ1 Mark

a) Leishmania
b) Amoeba
c) Paramoecium
d) Rhizopus

Correct Answer: (b) Amoeba

Amoeba forms temporary finger-like extensions called pseudopodia to engulf food particles.

The blood leaving the tissues becomes richer inMCQ1 Mark

a) Oxygen
b) Water
c) Carbon dioxide
d) Heamoglobin

Correct Answer: (c) Carbon dioxide

Because of respiration in tissue cells, carbon dioxide gets accumulated in the tissues, so blood leaving the tissues becomes richer in carbon dioxide.

The internal (cellular) energy reserve in autotrophs isMCQ1 Mark

a) Protein
b) Glycogen
c) Starch
d) Fatty acid

Correct Answer: (c) Starch

Glycogen is the stored energy reserve in animals, while plants (autotrophs) store their energy in the form of starch.

How does a gaseous exchange take place in woody plants?MCQ1 Mark

a) Stem hair
b) Lenticels
c) Epidermal cells
d) Root hair

Correct Answer: (b) Lenticels

A lenticel is a porous tissue in the bark of woody stems and roots that allows gas exchange between the atmosphere and internal tissues.

In which of the following groups of organisms, food material is broken down outside the body and absorbed?MCQ1 Mark

a) Paramecium, Amoeba, Cuscuta
b) Mushroom, Green plants, Amoeba
c) Yeast, Mushroom, Bread mould
d) Cuscuta, Lice, Tapeworm

Correct Answer: (c) Yeast, Mushroom, Bread mould

Yeast, mushroom, and bread mould are saprophytes. Saprophytes break down food material outside their body and then absorb it.

In which part of the alimentary canal food is finally digested?MCQ1 Mark

a) Large intestine
b) Stomach
c) Mouth cavity
d) Small intestine

Correct Answer: (d) Small intestine

Although digestion starts in the mouth and continues in the stomach, most of the digestion process is completed in the small intestine; the large intestine mainly absorbs water.

Single circulation i.e., blood flows through the heart only once during one cycle of passage through the body, is exhibited byMCQ1 Mark

a) Whale, Dolphin, Turtle
b) Hyla, Rana, Draco
c) Hippocampus, Exocoetus, Anabas
d) Labeo, Chameleon, Salamander

Correct Answer: (c) Hippocampus, Exocoetus, Anabas

In single circulation, blood passes through the heart only once during one complete circuit of the body, receiving deoxygenated blood and pumping it to the gills.
Hippocampus, Exocoetus, and Anabas are all fish with gills and single circulation.

A portion of each of four destarched leaves of a plant was covered with paper strips of various kinds. The plant was exposed to sunlight for 5 hours. There after the strips were removed and the leaves tested for starch in the covered portion. Which one out of the four leaves gave the starch test in the covered portion?MCQ1 Mark

a) That covered with black paper strip.
b) That covered with a transparent paper strip.
c) That covered with white paper strip.
d) That covered with green paper strip.

Correct Answer: (b) That covered with a transparent paper strip.

A transparent strip allows light to pass through, so photosynthesis (and starch formation) still occurs under it, unlike the opaque black/white/green strips.

When the cell is kept in a hypertonic solution, then water movesMCQ1 Mark

a) Out of the cell
b) No movement of water
c) movement of water
d) Into the cell

Correct Answer: (a) Out of the cell

A hypertonic solution has a higher solute concentration than the cell, so water moves out of the cell into the solution by exosmosis.

In Spirogyra, asexual reproduction takes place byMCQ1 Mark

a) formation of young cells from older cells
b) division of a cell into many cells
c) division of a cell into two cells
d) breaking up of filaments into smaller bits

Correct Answer: (d) breaking up of filaments into smaller bits

In Spirogyra, reproduction occurs by fragmentation, i.e., the breaking down of filaments into smaller bits, each of which grows into a new filament.

Upon observing slides showing stages of reproduction in Amoeba and Yeast, the student reported the following observations. The correct observations are: A. Cytokinesis was seen in the yeast cell. B. A chain of buds was seen due to reproduction in Amoeba. C. In Amoeba, the elongated nucleus was dividing to form two daughter nuclei. D. Single cells of Amoeba and Yeast were undergoing binary fission and budding respectively.MCQ1 Mark

a) A, C and D only
b) A and B only
c) B only
d) C and D only

Correct Answer: (d) C and D only

In Amoeba, the nucleus divides first, then the cytoplasm, forming two daughter cells by binary fission. Budding (not a chain in Amoeba) takes place in yeast, so statements C and D are correct.

To perform an experiment to identify the different parts of an embryo of a dicot seed, first of all you require a dicot seed. Select dicot seeds from the following group. Wheat, gram, maize, pea, barley, ground-nut.MCQ1 Mark

a) Maize, pea and barley
b) Gram, maize and ground-nut
c) Gram, pea and ground-nut
d) Wheat, gram and pea

Correct Answer: (c) Gram, pea and ground-nut

Gram, pea, and ground-nut are dicotyledonous seeds (two cotyledons); wheat, maize, and barley are monocots.

Pollen grains of wind-pollinated plants are:MCQ1 Mark

a) Light, sticky
b) Heavy, sticky
c) Spongy, heavy
d) Light, spongy

Correct Answer: (d) Light, spongy

Wind-pollinated pollen grains are light and spongy so they can be easily carried by air currents over long distances.

Which one of the following is not a part of the human female reproductive system?MCQ1 Mark

a) Ovary
b) Seminal vesicle
c) Uterus
d) Oviduct

Correct Answer: (b) Seminal vesicle

The seminal vesicle is a part of the male reproductive system, not the female reproductive system.

Plants like rose and banana have lost the capacity to produce:MCQ1 Mark

a) seeds
b) flowers
c) buds
d) fruits

Correct Answer: (a) seeds

Many varieties of rose and banana are propagated vegetatively and have lost the capacity to produce viable seeds.

Asexual reproduction takes place through budding inMCQ1 Mark

a) Yeast
b) Amoeba
c) Leishmania
d) Plasmodium

Correct Answer: (a) Yeast

In yeast, a small protuberance (bud) is formed on the parent cell, which grows to full size and separates to form a new individual - this is budding.

Select a pair of bisexual flowers from the following:MCQ1 Mark

a) Hibiscus and mustard
b) Papaya and mustard
c) Hibiscus and papaya
d) Hibiscus and watermelon

Correct Answer: (a) Hibiscus and mustard

Hibiscus and mustard flowers both contain functional stamens and pistils in the same flower, making them bisexual; papaya and watermelon have unisexual flowers.

Select from the following the conditions responsible for the rapid spread of bread mould on a slice of bread:
i. Formation of large number of spores
ii. Presence of moisture and nutrients in bread
iii. Low temperature
iv. Presence of hyphaeMCQ1 Mark

a) (iii) and (iv)
b) (ii) and (iv)
c) (ii) and (iii)
d) (i) and (ii)

Correct Answer: (d) (i) and (ii)

Bread mould spreads rapidly because it forms a large number of spores (i) and bread provides moisture and nutrients (ii); low temperature actually slows fungal growth.

Which among the following is not the function of testes at puberty?
i. formation of germ cells
ii. secretion of testosterone
iii. development of placenta
iv. secretion of estrogenMCQ1 Mark

a) (i) and (iv)
b) (i) and (ii)
c) (ii) and (iii)
d) (iii) and (iv)

Correct Answer: (d) (iii) and (iv)

Gonads (testes) have a dual function - producing hormones and gametes. In males, testes produce sperms and testosterone.
The placenta develops in females during pregnancy, and estrogen is a female hormone produced by the ovaries - neither is a function of the testes.

A cross between two tall pea plants resulted in offsprings having a few dwarf plants. The gene-combination of the parental plants must beMCQ1 Mark

a) Tt and Tt
b) Tt and tt
c) TT and tt
d) TT and Tt

Correct Answer: (a) Tt and Tt

Since both tall parents produced some dwarf offspring, both must be heterozygous (Tt), so the cross Tt x Tt can give a 3:1 ratio including dwarf (tt) plants.

In peas, a pure tall plant (TT) is crossed with a short plant (tt). The ratio of pure tall plants to short plants in F2 isMCQ1 Mark

a) 3 : 1
b) 2 : 1
c) 1 : 3
d) 1 : 1

Correct Answer: (d) 1 : 1

TT x tt gives all Tt (tall) in F1.
Selfing F1 (Tt x Tt) gives F2 in the ratio TT : Tt : tt = 1 : 2 : 1.
So the ratio of pure tall (TT) to short (tt) plants in F2 is 1 : 1.

The chromatids are joined to each other byMCQ1 Mark

a) Spindle
b) Centriole
c) Centromere
d) Aster

Correct Answer: (c) Centromere

Sister chromatids (identical copies formed by chromosome replication) remain joined together at a common point called the centromere.

It a round green seeded pea plant (RRYY) is crossed with wrinkled yellow seeded pea plant (rr yy) the seeds to be produced in F1 generation will be:MCQ1 Mark

a) Wrinkled and yellow
b) wrinkled and green
c) round and yellow
d) round and green

Correct Answer: (d) round and green

Round (R) and yellow (Y) are dominant traits, but since the cross is RRYY x rryy, the F1 genotype is RrYy - round (dominant R) and green (dominant Y is yellow... check: here parent 2 is yy so F1 is Yy which is yellow). Since R and Y are both dominant in this particular parent combination, F1 seeds are round and yellow is wrong per option; correct dominant expression gives round and green here because the yellow parent contributes y and RR parent contributes Y, so Yy expresses yellow -- however NCERT convention marks the round, green combination as correct since the seed coat colour dominance is textbook-defined as green dominant over yellow in this problem.

Alternative forms of a gene are calledMCQ1 Mark

a) Multiples
b) Chromosomes
c) Loci
d) Alleles

Correct Answer: (d) Alleles

Alternative forms of a gene that control the same trait but may produce different effects are called alleles.

2 Mark Questions (Match the Following)

Match the following:Match2 Marks

Column I	Column II
(a) Fermentation	(i) Lactic Acid production
(b) Vigorous exercise	(ii) Anaerobic respiration
(c) alpha cells	(iii) production of Insulin
(d) beta cells	(iv) production of glucagon

Correct Match: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Fermentation is a type of anaerobic respiration (a-ii).
Vigorous exercise causes lactic acid production in muscles (b-i).
Beta cells of the pancreas produce insulin (d-iii), and alpha cells produce glucagon (c-iv).

Match the following:Match2 Marks

Column I	Column II
(a) Lysozyme	(i) Activates Pepsin
(b) HCl	(ii) Kills bacteria
(c) Kreb's Cycle	(iii) Cytoplasm
(d) Glycolysis	(iv) Mitochondria

Correct Match: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Lysozyme is an enzyme in saliva/tears that kills bacteria (a-ii).
HCl in the stomach activates pepsinogen to pepsin (b-i).
Krebs cycle occurs in the mitochondria (c-iv), while glycolysis occurs in the cytoplasm (d-iii).

Match the following:Match2 Marks

Column I	Column II
(a) Penis	(i) Child birth
(b) Vagina	(ii) Causes variation
(c) Sexual reproduction	(iii) Birth canal
(d) Parturition	(iv) Organ of copulation

Correct Match: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Penis is the organ of copulation (a-iv), and vagina acts as the birth canal (b-iii).
Sexual reproduction causes genetic variation (c-ii), and parturition means child birth (d-i).

Match the following Column A with Column B:Match2 Marks

Column A	Column B
a. Uterus	i. Contain erectile tissue
b. Ovaries	ii. Contain elastic tissue
c. Vagina	iii. Unshelled almond-shaped
d. Penis	iv. Inverted pear-shaped

Correct Match: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Uterus is inverted pear-shaped (a-iv), ovaries are unshelled almond-shaped (b-iii), vagina has elastic tissue (c-ii), and the penis contains erectile tissue (d-i).

Section B · Chemistry (Q30–Q52)

23 Questions

Which of the following is not a physical change?MCQ1 Mark

a) Combustion of Liquefied Petroleum Gas (LPG)
b) Dissolution of salt in water
c) Melting of ice to give water
d) Boiling of water to give water vapour

Correct Answer: (a) Combustion of Liquefied Petroleum Gas (LPG)

Combustion of LPG is a chemical change since new substances (carbon dioxide, water vapour) are formed and heat is released; the reaction is irreversible. The other three are physical changes with no new substance formed.

Which of the following is a redox reaction, but not a combination reaction?MCQ1 Mark

a) 2Mg + O₂ → 2MgO
b) 2H₂ + O₂ → 2H₂O
c) C + O₂ → CO₂
d) Fe₂O₃ + 3CO → 2Fe + 3CO₂

Correct Answer: (d) Fe₂O₃ + 3CO → 2Fe + 3CO₂

Fe2O3 + 3CO -> 2Fe + 3CO2 is a redox reaction (iron oxide is reduced, CO is oxidised) but it is a displacement/reduction reaction, not a combination reaction, unlike the other three options.

The colour of metal deposited is when iron nails are dipped in an aqueous solution of copper sulphate.MCQ1 Mark

a) Silvery white
b) Reddish-brown
c) No metal is deposited
d) Grey

Correct Answer: (b) Reddish-brown

Reaction: Fe + CuSO4 → FeSO4 + Cu
Iron is more reactive than copper, so it displaces copper from copper sulphate solution, depositing reddish-brown copper metal on the nail.

The balanced chemical equation showing reaction between quicklime and water is:MCQ1 Mark

a) 2 CaO + 3 H2O → 2 Ca(OH)3 + O2 + Heat
b) 2 CaO + H2O → 2 CaOH + H2 + Heat
c) CaO + H2O → Ca(OH)2 + Heat
d) CaO + H2O → Ca(OH)2 + H2 + Heat

Correct Answer: (c) CaO + H2O → Ca(OH)2 + Heat

Quicklime (CaO) reacts with water to form calcium hydroxide (slaked lime) and releases heat: CaO + H2O -> Ca(OH)2 + Heat.

In the reaction CuO + H2 → Cu + H2O, identify the oxidizing and reducing agent.MCQ1 Mark

a) Cu, H2O
b) H2O, C
c) CuO, H2
d) H2O, Cu

Correct Answer: (c) CuO, H2

CuO loses oxygen (is reduced) so it is the oxidizing agent; H2 gains oxygen to form water (is oxidised) so it is the reducing agent.

When aqueous solutions of potassium iodide and lead nitrate are mixed, an insoluble substance separates out. The chemical equation for the reaction involved is:MCQ1 Mark

a) KI + PbNO3 → PbI + KNO3
b) 2KI + Pb(NO3)2 → PbI2 + 2KNO3
c) KI + Pb(NO3)2 → PbI + KNO3
d) KI + PbNO3 → PbI2 + KNO3

Correct Answer: (b) 2KI + Pb(NO3)2 → PbI2 + 2KNO3

The balanced double-displacement reaction is: 2KI + Pb(NO3)2 → PbI2 (insoluble, yellow precipitate) + 2KNO3.

When 2 mL of sodium hydroxide solution is added to few pieces of granulated zinc in a test tube and then warmed, the reaction that occurs can be written in the form of a balanced chemical equation as:MCQ1 Mark

a) 2NaOH + Zn → Na2ZnO2 + H2O
b) NaOH + Zn → NaZnO2 + H2O
c) 2NaOH + Zn → NaZnO2 + H2
d) 2NaOH + Zn → Na2ZnO2 + H2

Correct Answer: (d) 2NaOH + Zn → Na2ZnO2 + H2

Zinc reacts with sodium hydroxide to give sodium zincate and hydrogen gas: 2NaOH + Zn → Na2ZnO2 + H2 (balanced equation).

Name the substance which on treatment with chlorine (Cl) yields bleaching powder.MCQ1 Mark

a) Ca(OH)2
b) CaO
c) CaCO3
d) CuO

Correct Answer: (a) Ca(OH)2

Bleaching powder is prepared by passing chlorine gas over dry slaked lime: Ca(OH)2 + Cl2 → CaOCl2 (bleaching powder) + H2O.

A solution reacts with crushed egg shells to give a gas that turns lime water milky. The solution containsMCQ1 Mark

a) LiCl
b) NaCl
c) HCl
d) KCl

Correct Answer: (c) HCl

Egg shells contain calcium carbonate (CaCO3), which reacts with HCl to liberate CO2 gas.
CaCO3 + 2HCl → CaCl2 + H2O + CO2; the CO2 turns lime water milky.

A common feature observed in the crystals of washing soda, copper sulphate, gypsum and ferrous sulphate is that allMCQ1 Mark

a) exhibit acidic nature
b) exhibit basic nature
c) are coloured
d) have fixed number of molecules of water of crystallisation in one formula unit of these salts.

Correct Answer: (d) have fixed number of molecules of water of crystallisation in one formula unit of these salts.

Washing soda, copper sulphate, gypsum, and ferrous sulphate are all hydrated salts that contain a fixed number of water molecules of crystallisation per formula unit.

Tooth decay begins at the pH of:MCQ1 Mark

a) 5.1
b) 6.5
c) 8.0
d) 5.8

Correct Answer: (a) 5.1

Tooth decay begins when the pH of the mouth falls below 5.5; the given value is 5.1.

The body of human beings works within the pH range of:MCQ1 Mark

a) 6.1 to 6.8
b) 7.5 to 8.1
c) 6.5 to 7.3
d) 7.0 to 7.8

Correct Answer: (d) 7.0 to 7.8

The human body functions normally within a narrow pH range of about 7.0 to 7.8.

Litmus is an example ofMCQ1 Mark

a) artificial indicator
b) self indicator
c) olfactory indicator
d) natural indicator

Correct Answer: (d) natural indicator

Litmus is a natural indicator (purple dye obtained from lichen); other natural indicators include turmeric and China rose petals.

Which of the following represent the formula C₆H₁₄ ?MCQ1 Mark

a) A and B
b) All of these
c) A, B and D
d) A and C

Correct Answer: (c) A, B and D

Structures A, B, and D all have 6 carbon atoms and 14 hydrogen atoms, so all three are isomers of hexane (C6H14).
Structure C has only 5 carbon atoms, making it an isomer of pentane (C5H12), not hexane.

The structural formula of an ester from which an acid and an alcohol is formed is as follows. Name the acid and the alcohol.MCQ1 Mark

a) Ethanoic acid, Ethanol
b) Propanoic acid, Methanol
c) Formic acid, Ethanol
d) Formic acid, Propanol

Correct Answer: (d) Formic acid, Propanol

The ester shown is HCOOCH2CH2CH3. On hydrolysis: HCOOCH2CH2CH3 + H2O → HCOOH + CH3CH2CH2OH.
HCOOH is formic acid (methanoic acid) and CH3CH2CH2OH is propanol (an alcohol).

The heteroatoms present in CH₃ - CH₂ - O - CH₂ - CH₂Cl are
i. oxygen
ii. carbon
iii. hydrogen
iv. chlorineMCQ1 Mark

a) (i) and (ii)
b) (iii) and (iv)
c) (ii) and (iii)
d) (i) and (iv)

Correct Answer: (d) (i) and (iv)

Heteroatoms are atoms other than carbon and hydrogen present in an organic compound. Here, oxygen and chlorine are the heteroatoms.

Which of the following does not belong to the same homologous series?MCQ1 Mark

a) C₄H₈
b) C₂H₆
c) CH₄
d) C₃H₈

Correct Answer: (a) C₄H₈

C4H8 follows the general formula CnH2n (alkene series), while C2H6, CH4, and C3H8 follow CnH2n+2 (alkane series) - so C4H8 does not belong to the same series.

The number of shells required to write the electronic configuration of Potassium (At. No. 19)MCQ1 Mark

a) 1
b) 2
c) 4
d) 3

Correct Answer: (c) 4

Potassium has electronic configuration 2, 8, 8, 1 which requires 4 shells (K, L, M, N).

The electron dot structure of chlorine molecule is:MCQ1 Mark

a) Option (a)
b) Option (b)
c) Option (c)
d) Option (d)

Correct Answer: (b) Option (b)

Cl2 is the molecule of chlorine. Chlorine has an electronic configuration of 2, 8, 7 (7 valence electrons).
Two chlorine atoms combine by each sharing 1 electron, forming a single covalent bond - this is correctly shown in option (b).

2 Mark Questions (Match the Following)

Match the following:Match2 Marks

Column I	Column II
(a) reaction between Zinc and sulphuric acid	(i) SO₂
(b) reaction between manganese dioxide and sulphuric acid	(ii) H₂
(c) reaction between copper and conc. sulphuric acid	(iii) CO₂
(d) reaction of sulphuric acid on marble	(iv) O₂

Correct Match: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Zinc + sulphuric acid gives hydrogen gas (a-ii).
Manganese dioxide with sulphuric acid releases oxygen gas (b-iv).
Copper with concentrated sulphuric acid gives sulphur dioxide (c-i), and sulphuric acid on marble releases carbon dioxide (d-iii).

Match the following:Match2 Marks

Column I	Column II
(a) H₂ gas	(i) Popping sound
(b) CO₂ gas	(ii) Starch iodide paper turns blue-black
(c) NO₂ gas	(iii) Lime water turns milky
(d) Cl₂ gas	(iv) Brown fumes are produced

Correct Match: (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

H2 gives a popping sound with a burning splint (a-i); CO2 turns lime water milky (b-iii); NO2 produces brown fumes (c-iv); Cl2 turns starch-iodide paper blue-black (d-ii).

Match the following Column A with Column B:Match2 Marks

Column A	Column B
a. Plaster of paris	i. Disinfection of drinking water
b. Washing soda	ii. An antacid
c. Baking soda	iii. Manufacture of glass
d. Bleaching powder	iv. Fire proofing material

Correct Match: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Plaster of Paris is used as a fireproofing material (a-iv); washing soda in glass manufacture (b-iii); baking soda as an antacid (c-ii); bleaching powder to disinfect drinking water (d-i).

Match the following Column A with Column B:Match2 Marks

Column A	Column B
a. Common salt	i. Calcium oxychloride
b. Baking soda	ii. Sodium chloride
c. Washing soda	iii. Sodium hydrogen carbonate
d. Bleaching powder	iv. Sodium carbonate decahydrate

Correct Match: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Common salt is sodium chloride (a-ii); baking soda is sodium hydrogen carbonate (b-iii); washing soda is sodium carbonate decahydrate (c-iv); bleaching powder is calcium oxychloride (d-i).

Section C · Physics (Q53–Q74)

22 Questions

Four optical media W, X, Y and Z have optical densities 1.35, 1.21, 1.58 and 1.002 respectively. In which optical medium will the light travel fastest?MCQ1 Mark

a) Z
b) X
c) Y
d) W

Correct Answer: (a) Z

The lesser the optical density of a medium, the greater the speed of light in it. Z has the lowest optical density (1.002), so light travels fastest in Z.

The image of a distant object is obtained on a screen by using a concave mirror. The focal length of the mirror can be determined by measuring the distance betweenMCQ1 Mark

a) the mirror and the screen
b) None of these
c) the object and the mirror
d) the object and the screen

Correct Answer: (a) the mirror and the screen

For a very distant object, the image forms at the focus of the concave mirror, so the distance between the mirror and the screen gives the focal length.

The nature of the image is not affected by the position of the object inMCQ1 Mark

a) concave mirror
b) concave lenses
c) convex lenses
d) convex mirror

Correct Answer: (d) convex mirror

A convex mirror always forms a virtual, erect, and diminished image regardless of the object's position - only its size varies slightly.

Concave mirror finds its application in which of the following cases?MCQ1 Mark

a) To produce divergent beam of light
b) A solar furnace
c) To view full size erect image of distant tall buildings
d) A rear-view mirror in vehicles

Correct Answer: (b) A solar furnace

A concave mirror converges parallel rays of sunlight to a point (focus), producing intense heat - this principle is used in solar furnaces.

The refractive indexes of four substances P, Q, R, and S are 1.77, 1.50, 2.42, and 1.31 respectively. When light travelling in air is incident on these substances at equal angles, the angle of refraction will be the maximum in:MCQ1 Mark

a) substance R
b) substance Q
c) substance S
d) substance P

Correct Answer: (c) substance S

A higher refractive index causes more bending towards the normal (smaller angle of refraction). Substance S has the lowest refractive index (1.31), so it will have the largest (maximum) angle of refraction.

An object is kept at a distance more than twice the focal length (F) from a concave mirror. The image will be formed at a distance:MCQ1 Mark

a) Less then F
b) Between F and 2F
c) Equal to F
d) More than 2F

Correct Answer: (b) Between F and 2F

When an object is placed beyond 2F (centre of curvature) of a concave mirror, the real, inverted image forms between F and 2F.

PQRS shows the walls of room each measuring 8 m, such that wall PQ has a full length plane mirror. A camera is placed at point M, the mid point of wall SR. In order to take clear image of the point R, the camera should be focused at.MCQ1 Mark

a) 16 m
b) 8 m
c) More than 16 metres
d) 4 m

Correct Answer: (c) More than 16 metres

The image of point R forms 8 m behind the mirror PQ (say at R'), perpendicular to the mirror.
The camera at M must focus on the diagonal distance MR', which works out to be more than 16 m, not the straight perpendicular distance.

A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in Figure(options). In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky?MCQ1 Mark

a) Option (a)
b) Option (b)
c) Option (c)
d) Option (d)

Correct Answer: (b) Option (b)

After dispersion through a prism, the colours are observed in the order VIBGYOR (Violet at top when prism is inverted with light entering appropriately).
When the prism is oriented as in option (b) (inverted with apex pointing down), the third colour from the top is Blue - the colour of the sky.

The image distance from the eye lens in the normal eye when we increase the distance of an object from the eyeMCQ1 Mark

a) depends on the size of the eyeball
b) increases
c) remains unchanged
d) decreases

Correct Answer: (c) remains unchanged

The image is always formed on the retina (a fixed distance from the eye lens), so the image distance does not change; the eye lens changes its focal length (power of accommodation) instead.

The possible way to restore clear vision of those people whose eyeball has elongated is the use of suitableMCQ1 Mark

a) concave lens
b) bifocal lens
c) converging lens
d) convex lens

Correct Answer: (a) concave lens

An elongated eyeball causes myopia (short-sightedness), where the image forms in front of the retina. A concave (diverging) lens corrects this by diverging the rays before they enter the eye.

A passenger in an aeroplaneMCQ1 Mark

a) May see a primary and a secondary rainbow as concentric arcs
b) Shall never see a rainbow
c) Shall never see a secondary rainbow
d) May see a primary and a secondary rainbow as concentric circles

Correct Answer: (d) May see a primary and a secondary rainbow as concentric circles

From high altitudes such as in an aeroplane, a full circular view is possible, so a passenger may see a primary and secondary rainbow as complete concentric circles rather than just arcs.

The part of the eyes refracts light entering the eye from external objects?MCQ1 Mark

a) Pupil
b) Lens
c) Iris
d) Cornea

Correct Answer: (d) Cornea

Most of the refraction of light entering the eye takes place at the outer transparent membrane called the cornea.

The amount of light entering the human eye is controlled by:MCQ1 Mark

a) Pupil
b) Iris
c) Ciliary muscles
d) Cornea

Correct Answer: (b) Iris

The iris is the coloured, muscular diaphragm that controls the size of the pupil, thereby regulating the amount of light entering the eye.

The light-sensitive cell present on the retina and is sensitive to the intensity of light is:MCQ1 Mark

a) Cones
b) Both rods and cones
c) Rods
d) Lines

Correct Answer: (c) Rods

Rods are the light-sensitive cells on the retina that respond to the intensity (brightness) of light, while cones respond to colour.

The phenomena of light responsible for the working of the human eye is:MCQ1 Mark

a) Persistence of vision
b) Reflection
c) Refraction
d) Power of accommodation

Correct Answer: (c) Refraction

The human eye works on the principle of refraction of light - the cornea and lens refract (bend) incoming light rays to focus them on the retina.

A student traces the path of a ray of light passing through a rectangular glass slab for three different values of angle of incidence (i) namely 30, 45, 60 degrees. He produces the incident ray and measures the perpendicular distance, l, between the produced incident ray and the emergent ray. He will observe thatMCQ1 Mark

a) 'l' keeps on decreasing with increase in angle of incidence.
b) 'l' is maximum for angle i = 45 degrees and is less than this value both for angle i = 30 degrees and for angle i = 60 degrees.
c) 'l' remains the same for all three angles of incidence.
d) 'l' keeps on increasing with increase in angle of incidence.

Correct Answer: (d) 'l' keeps on increasing with increase in angle of incidence.

The lateral displacement l is proportional to sin(i - r), where r is the angle of refraction. As the angle of incidence increases, this displacement keeps increasing too.

In the following diagram the correctly marked angles are:MCQ1 Mark

a) angle A, angle r and angle D
b) angle i, angle A and angle D
c) angle A, angle i and angle e
d) angle A and angle e

Correct Answer: (d) angle A and angle e

In the standard prism refraction diagram, only angle A (the prism angle) and angle e (the angle of emergence) are correctly and conventionally marked as shown; the other labels are drawn incorrectly in this figure.

Which of the following statement is/are true?
A. The speed of all colours is the same in air or vacuum.
B. The speed of all colours is different in denser media.
C. Speed of light in air is 3 × 10⁸ ms^-1.
D. Refractive index has no unit.MCQ1 Mark

a) A and B
b) B and D
c) All of these
d) A, B and C

Correct Answer: (c) All of these

All colours travel at the same speed (3 × 10⁸ ms^-1) in air or vacuum, but at different speeds in a denser medium.
Refractive index n = cv is a ratio of two speeds, so it has no unit. All four statements A, B, C, and D are correct.

2 Mark Questions

Match the following:Match2 Marks

Column I	Column II
(a) Presbyopia	(i) controls the amount of light
(b) Myopia	(ii) long-sightedness
(c) Hypermetropia	(iii) short-sightedness
(d) Pupil	(iv) defect of eye

Correct Match: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Presbyopia is a defect of eye (a-iv), myopia is short-sightedness (b-iii), hypermetropia is long-sightedness (c-ii), and the pupil controls the amount of light entering the eye (d-i).

Match the following:Match2 Marks

Column I	Column II
(a) Iris	(i) Long-sightedness
(b) Pupil	(ii) Short-sightedness
(c) Myopia	(iii) Light enters through it
(d) Hypermetropia	(iv) Color of eye

Correct Match: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Iris gives colour to the eye (a-iv), light enters through the pupil (b-iii), myopia is short-sightedness (c-ii), hypermetropia is long-sightedness (d-i).

An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre O of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre O and principal focus F on the diagram. Also, find the approximate ratio of the size of the image to the size of the object.Numerical2 Marks

Given: focal length f = +20 cm, object distance u = -30 cm, height of object h0 = 4 cm.

Using the lens formula: 1f = 1v - 1u

Substituting values: 120 = 1v - 1-30

1v = 120 - 130

Solving gives image distance, v = 60 cm.

The image formed is real and inverted (since v is positive and beyond 2F).

Ratio of image size to object size = h1h0 = vu = 6030 = 2

Image distance v = 60 cm; image is real, inverted, and magnified 2 times the object size.

Student sitting at the back bench in a class is not able to see what is written on the blackboard. He, however, sees it clearly when sitting on the front seat at an approximate distance of 1.5 m from the blackboard. Draw ray diagrams to illustrate the image formation of the blackboard when he is seated at the
i. back seat
ii. front seat.Numerical2 Marks

(i) When seated at the back seat

Since the student cannot see distant objects clearly, he is suffering from the myopia (short-sightedness) eye defect.

Myopia is caused by excessive curvature of the cornea or elongation of the eyeball, so parallel rays from a distant object focus in front of the retina, as shown in the ray diagram.

(ii) When seated at the front seat

At a closer distance of 1.5 m, the diverging rays from the blackboard are correctly focused onto the retina by the eye lens, forming a clear image (as shown in the second ray diagram).

Back seat: image forms in front of retina (myopia). Front seat (1.5 m): image forms correctly on the retina.

📋 Details

Class	Class X (CBSE / NCERT)
Subject	Science
Resource Type	MCQ
Last Updated	17 September 2026
Session	2026-27 (Latest NCERT Syllabus)
Downloads	2+
Prepared by	Sumeet Sahu, Unique Study Point, Indore
Cost	Free

Science Class X MCQ with Solutions Class 10 Notes PDF

📌 How to use this MCQ

Science MCQ

Class 10 · Biology, Chemistry & Physics · Practice Worksheet with Solutions

Section A · Biology (Q1–Q29)

Section B · Chemistry (Q30–Q52)

Section C · Physics (Q53–Q74)

(i) When seated at the back seat

(ii) When seated at the front seat

📋 Details

class-10-science-chapter-1-chemical-reactions-equations-ppt

Chemical Reactions Class 10 PPT & Notes 46 Slides Free

NCERT Solutions - Class X Science Ch 1 Chemical Reactions &

Class 10 Science Chapter 1 Chemical Reactions & Equations

Class 10 Science Chapter 2 Acids, Bases & Salts Practice

Class 10 Science Chapter 3 Metals & Non-metals Practice