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Quiz 5
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Q.1
World Ozone Day is celebrated on
0%
\(16^{th}\) September
0%
\( 22^{nd}\) April
0%
\(5^{th}\) June
0%
\(21^{st}\) April
Explanation
World Ozone Day (the International Day for the Preservation of the Ozone Layer) is observed on 16th September each year.
Q.2
The experimental proof for semiconservative replication of DNA was first shown in a
0%
Plant
0%
Fungus
0%
Bacterium
0%
Virus
Explanation
The classic Meselson-Stahl experiment, which first experimentally confirmed DNA's semiconservative replication, was carried out using the bacterium E. coli.
Q.3
Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively?
0%
Inflammation of bronchioles; Decreased respiratory surface
0%
Increased number of bronchioles; Increased respiratory surface
0%
Decreased respiratory surface; Inflammation of bronchioles
0%
Increased respiratory surface; Inflammation of bronchioles
Explanation
Asthma involves inflammation (and narrowing) of the bronchioles, causing difficulty breathing, while emphysema involves destruction of alveolar walls, which reduces the total respiratory surface area available for gas exchange.
Q.4
Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system
0%
Amphibia
0%
Aves
0%
Reptilia
0%
Osteichthyes
Explanation
Birds (Aves) have a crop (for temporary food storage) and a gizzard (for grinding food, since birds lack teeth) as distinctive parts of their digestive tract.
Q.5
Match the columns
Column I
Column II
a. Glycosuria
(i) Accumulation of
uric acid in joints
b. Gout
(ii) Mass of crystallised
salts within the
kidney
c. Renal calculi
(iii) Inflammation in
glomeruli
d. Glomerular nephritis
(iv) Presence of glucose in urine
0%
a -- iii b -- ii c -- iv d -- i
0%
a -- iv b -- i c -- ii d -- iii
0%
a -- i b -- ii c -- iii d -- iv
0%
a -- ii b -- iii c -- i d --iv
Explanation
Glycosuria is the presence of glucose in urine (iv); gout is the accumulation of uric acid crystals in joints (i); renal calculi are masses of crystallised salts within the kidney, i.e. kidney stones (ii); and glomerular nephritis is inflammation of the glomeruli (iii).
Q.6
Which of the following is a secondary pollutant?
0%
\(CO\)
0%
\(SO_2\)
0%
\( O_3\)
0%
\(CO_2\)
Explanation
Ozone (O₃) is a secondary pollutant — it forms in the atmosphere through photochemical reactions (e.g. between NOₓ and volatile organic compounds in sunlight), rather than being directly emitted. CO, SO₂, and CO₂ are all primary pollutants, released directly from their sources.
Q.7
In which of the following forms is iron absorbed by plants?
0%
Free element
0%
Ferric
0%
Both ferric and ferrous
0%
Ferrous
Explanation
Plants absorb iron predominantly in its ferric (Fe³⁺) form.
Q.8
Which of the following is true for nucleolus?
0%
Larger nucleoli are present in dividing cells
0%
It takes part in spindle formation
0%
It is a membrane-bound structure
0%
It is a site for active ribosomal RNA synthesis
Explanation
The nucleolus is the site of active ribosomal RNA synthesis and ribosome subunit assembly. It has no surrounding membrane, plays no role in spindle formation, and (contrary to growing larger) tends to become less prominent or disappear during active cell division.
Q.9
The transparent lens in the human eye is held in its place by
0%
ligaments attached to the iris
0%
smooth muscles attached to the iris
0%
ligaments attached to the ciliary body
0%
smooth muscles attached to the ciliary body
Explanation
The lens is held in position by suspensory ligaments (zonule fibres) that attach it to the ciliary body — not directly to the iris, and it's the ligaments (whose tension the ciliary muscles adjust for focusing), not muscles, that physically suspend the lens.
Q.10
Niche is
0%
the range of temperature that the organism needs to live
0%
the functional role played by the organism where it lives
0%
all the biological factors in the organism's environment
0%
the physical space where an organism lives
Explanation
An organism's niche refers to its functional role within its habitat — what it does, what it eats, how it interacts with other species — distinct from its habitat, which is simply the physical place it lives.
Q.11
In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?
0%
Ascariasis
0%
Ringworm disease
0%
Elephantiasis
0%
Amoebiasis
Explanation
Elephantiasis (lymphatic filariasis), caused by filarial worms transmitted through mosquito bites, leads to chronic inflammation and blockage of the lymphatic vessels, producing the disease's characteristic swelling.
Q.12
Oxygen is not produced during photosynthesis by
0%
Nostoc
0%
Green sulphur bacteria
0%
Cycas
0%
Chara
Explanation
Green sulphur bacteria carry out anoxygenic photosynthesis, using hydrogen sulphide (rather than water) as their electron donor — so no oxygen is released. Nostoc, Cycas, and Chara are all oxygenic photosynthesisers that do produce oxygen.
Q.13
Conversion of milk to curd improves its nutritional value by increasing the amount of
0%
Vitamin A
0%
Vitamin E
0%
Vitamin \(B_{12}\)
0%
Vitamin D
Explanation
Fermenting milk into curd, via lactic acid bacteria, increases its vitamin B12 content compared to plain milk.
Q.14
Match the columns
Column I
Column II
a. Eutrophication
(i) UV-B radiation
b. Sanitary landfill
(ii) Deforestation
c. Snow blindness
(iii) Nutrient enrichment
d. Jhum cultivation
(iv) Waste disposal
0%
a. -- ii b. -- i c. -- iii d. -- iv
0%
a. -- iii b. -- iv c. -- i d. -- ii
0%
a. -- i b. -- ii c. -- iv d. -- iii
0%
a. -- i b. -- iii c. -- iv d. -- ii
Explanation
Eutrophication is nutrient enrichment of a water body (iii); a sanitary landfill is a method of waste disposal (iv); snow blindness results from excessive UV-B radiation exposure (i); and jhum (shifting) cultivation is a traditional farming practice associated with deforestation (ii).
Q.15
Match the columns
Column I
Column II
a. Tidal volume
(i) 2500 – 3000 mL
b. Inspiratory Reserve volume
(ii) 1100 – 1200 mL
c. Expiratory Reserve volume
(iii) 500 – 550 mL
d. Expiratory Reserve volume
(iv) 1000 – 1100 mL
0%
a -- iii b -- i c -- iv d -- ii
0%
a -- iii b -- ii c -- i d -- iv
0%
a -- iv b -- iii c -- ii d -- i
0%
a -- i b -- iv c -- ii d -- iii
Explanation
Standard adult lung volumes: tidal volume is about 500–550 mL (iii), inspiratory reserve volume is about 2500–3000 mL (i), and expiratory reserve volume is about 1000–1100 mL (iv). (Note: the source table mistakenly lists "Expiratory Reserve volume" twice for rows c and d — row d's value of 1100–1200 mL, per the marked answer, corresponds to what should be labelled residual volume.)
Q.16
What type of ecological pyramid would be obtained with the following data? Secondary consumer : 120 g Primary consumer : 60 g Primary producer : 10 g
0%
Upright pyramid of biomass
0%
Inverted pyramid of biomass
0%
Upright pyramid of numbers
0%
Pyramid of energy
Explanation
With biomass actually increasing at higher trophic levels (10g → 60g → 120g, rather than decreasing as usual), this produces an inverted pyramid of biomass — a pattern seen in some aquatic ecosystems where small, rapidly-reproducing producers support a much larger standing biomass of consumers.
Q.17
Which of the following organisms are known as chief producers in the oceans?
0%
Diatoms
0%
Dinoflagellates
0%
Euglenoids
0%
Cyanobacteria
Explanation
Diatoms are considered the chief primary producers in marine (ocean) ecosystems, forming a major part of oceanic phytoplankton.
Q.18
The contraceptive ‘SAHELI’
0%
is a post-coital contraceptive.
0%
is an IUD.
0%
increases the concentration of estrogen and prevents ovulation in females
0%
blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
Explanation
Saheli (centchroman) works by blocking estrogen receptors in the uterine lining, preventing a fertilised egg from implanting — rather than by directly altering estrogen levels or ovulation.
Q.19
Select the wrong statement :
0%
Mitochondria are the powerhouse of the cell in all kingdoms except Monera
0%
Mushrooms belong to Basidiomycetes
0%
Cell wall is present in members of Fungi and Plantae
0%
Pseudopodia are locomotory and feeding structures in Sporozoans
Explanation
Pseudopodia for locomotion and feeding are characteristic of Sarcodina (e.g. Amoeba), not Sporozoa — sporozoans (like Plasmodium) are largely non-motile or move by other means (like gliding), not via pseudopodia — making this the wrong statement.
Q.20
Iron carbonyl, \(\text{Fe(CO)}_5\) is
0%
monomuclear
0%
dinuclear
0%
tetranuclear
0%
trinnuclear
Explanation
Iron pentacarbonyl, Fe(CO)₅, contains a single iron atom bonded to five CO ligands — making it a mononuclear carbonyl complex.
Q.21
Consider the change in the oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
0%
\( HBrO\)
0%
\( Bro_{3}^{-}\)
0%
\( Bro_{4}^{-}\)
0%
\( Br_2\)
Explanation
In a Latimer diagram, a species is prone to disproportionation when its reduction potential going to a lower oxidation state (the step to its right) exceeds the potential for its own formation from a higher oxidation state (the step to its left). For HBrO: the step to Br₂ (to its right) is 1.595 V, while the step from BrO₃⁻ to HBrO (to its left) is only 1.5 V — since 1.595 > 1.5, HBrO satisfies the condition for spontaneous disproportionation.
Q.22
The correction factor ‘a’ to the ideal gas equation corresponds to
0%
the volume of the gas molecules
0%
Forces of attraction between the gas molecules
0%
the density of the gas molecules
0%
electric field present between the gas molecules
Explanation
In the van der Waals equation, the constant 'a' corrects for the attractive intermolecular forces between gas molecules, which reduce the pressure a real gas exerts compared to an ideal gas (the constant 'b' separately corrects for the finite volume of the molecules themselves).
Q.23
Which of the following statements is not true for halogens?
0%
Chlorine has the highest electron-gain enthalpy
0%
All but fluorine show positive oxidation states.
0%
All are oxidizing agents.
0%
All form monobasic oxyacids
Explanation
Fluorine is unique among the halogens in that it shows only the −1 oxidation state (being the most electronegative element, it never shows a positive oxidation state) — the other halogens (Cl, Br, I) do show various positive oxidation states in their oxoacids and oxides. (Chlorine genuinely does have the highest electron-gain enthalpy among halogens — an exception to the expected trend, caused by fluorine's small atomic size and electron-electron repulsion in its compact 2p subshell — and halogens are indeed generally good oxidising agents.)
Q.24
Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :
0%
a--iv b--i c--ii d--iii
0%
a--iii b--v c--i d--ii
0%
a--i b--ii c--iii d--iv
0%
a--iv b--v c--ii d--i
Explanation
Calculating unpaired electrons and spin-only magnetic moment (μ = √[n(n+2)] BM) for each ion: Co³⁺ ([Ar]3d⁶, high spin): 4 unpaired → μ = √24 BM (iv) Cr³⁺ ([Ar]3d³): 3 unpaired → μ = √15 BM (v) Fe³⁺ ([Ar]3d⁵): 5 unpaired → μ = √35 BM (ii) Ni²⁺ ([Ar]3d⁸): 2 unpaired → μ = √8 BM (i) So: a→iv, b→v, c→ii, d→i.
Q.25
Which of the following molecules represents the order of hybridization \(\text{sp}^2,\text{sp}^2,\text{sp,sp}\) from left to right atoms?
0%
\( CH_2=CH-C\equiv CH\)
0%
\( CH_2=CH-CH=CH_2\)
0%
\( HC\,\equiv C-C\equiv CH\)
0%
\( CH_3-CH=CH-CH_{_3}\)
Explanation
CH₂=CH−C≡CH: the first two carbons (CH₂=CH−) are part of a C=C double bond, giving them sp² hybridisation, while the last two carbons (−C≡CH) are part of a C≡C triple bond, giving them sp hybridisation — exactly matching sp², sp², sp, sp from left to right.
Q.26
In which case is the number of molecules of water maximum?
0%
0.00224 L of water vapors at 1 atm and 273 K
0%
18 mL of water
0%
\(10^{-3}\) mol of water
0%
0.18 g of water
Explanation
Converting each to moles: 0.00224 L vapour at STP = 0.00224/22.4 = 10⁻⁴ mol; 18 mL liquid water (density ≈1 g/mL) = 18 g = 1 mol; 10⁻³ mol given directly; 0.18 g = 0.18/18 = 0.01 mol. The 18 mL of liquid water (1 mol) contains by far the most molecules.
Q.27
On which of the following properties does the coagulating power of an ion depend?
0%
The magnitude of the charge on the ion alone
0%
The sign of charge on the ion alone
0%
Both magnitude and sign of the charge on the ion
0%
Size of the ion alone
Explanation
By the Hardy-Schulze rule, coagulating power depends on both the sign of the ion's charge (it must be opposite to the charge on the colloidal particles to cause coagulation) and the magnitude of that charge (higher charge magnitude gives much greater coagulating power).
Q.28
Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
0%
\(CH_3=CH_3\)
0%
\(CH\equiv CH\)
0%
\(CH_2=CH_2\)
0%
\(CH_4\)
Explanation
Wurtz reaction couples two alkyl halide molecules, doubling the carbon count. For the coupled product to still have fewer than 4 carbons, the starting alkyl halide must have just 1 carbon. Since the starting hydrocarbon reacts with bromine by SUBSTITUTION (not addition, which is what alkenes/alkynes would undergo), it must be a saturated hydrocarbon — matching methane (CH₄), which forms CH₃Br, then couples via Wurtz to ethane (C₂H₆).
Q.29
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
0%
In acidic (strong) medium aniline is present as anilinium ion.
0%
In absence of substituents nitro group always goes to m-position
0%
In electrophilic substitution reactions amino group is meta directive.
0%
In spite of substituents nitro group always goes to only m-position.
Explanation
In strongly acidic conditions, aniline's amino group gets protonated, forming the anilinium ion (C₆H₅NH₃⁺) — a positively charged, electron-withdrawing group that acts as a meta-director in electrophilic substitution, unlike neutral aniline's −NH₂ group, which is an ortho/para director.
Q.30
Which one of the following conditions will favor maximum formation of the product in the reaction, \[A_2\left( g \right) +B_2\left( g \right) \rightleftharpoons X_2\left( g \right) \Delta _rH=-X\,\,kJ\]
0%
High temperature and high pressure
0%
High temperature and Low pressure
0%
Low temperature and high pressure
0%
Low temperature and low pressure
Explanation
This reaction is exothermic (ΔH negative), so by Le Chatelier's principle, LOW temperature favours the forward (product-forming) direction. Since 2 moles of gaseous reactants form only 1 mole of gaseous product, HIGH pressure also favours the product side (fewer gas moles). So low temperature and high pressure together maximise product formation.
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