MHT CET 2026 May 10 Shift 2 PCB Question Paper- Download PDF with Solutions

Ecological succession describes the process of change in the species structure of an ecological community over time. A lithosere is a type of succession that occurs on bare rock. The pioneer species are the first organisms to colonize this harsh environment. Lichens are symbiotic associations of fungi and algae or cyanobacteria. They are exceptionally well-adapted to colonize bare rock because:

1. Rock weathering: They secrete acids that can break down rock surfaces, initiating the formation of soil.

2. Nutrient acquisition: They can absorb moisture and nutrients directly from the atmosphere and rainwater.

3. Survival: They can tolerate extreme temperatures and desiccation.


Phytoplanktons are pioneer species in aquatic succession (hydrosere). Mosses and herbs typically colonize later, after lichens have created a thin layer of soil and altered the microenvironment.




Final Answer: \(\boxed{B}\) Quick Tip: \textbf{Key Exam Tip:}
In ecological succession, pioneer species are the first to colonize a new habitat. For lithoseres (bare rock), lichens are the pioneers due to their unique ability to break down rock and survive extreme conditions.

Detritivores are organisms that consume dead organic matter (detritus). Earthworms are a prime example. Their ecological role involves several processes in decomposition:

Fragmentation: Earthworms ingest dead organic matter, physically breaking it down into smaller pieces through chewing and churning in their digestive tracts. This increases the surface area available for microbial decomposition.

Leaching: This is the removal of soluble compounds from detritus by water. Earthworms do not directly cause leaching.
Catabolism: This refers to the metabolic breakdown of organic compounds within organisms to release energy. Earthworms perform catabolism for their own sustenance.

Humification: Earthworms play a significant role in this process. By ingesting detritus and soil, mixing them, and excreting them as nutrient-rich casts (vermicasts), they enhance the decomposition of organic matter and promote the formation of humus, a stable, dark, nutrient-rich organic material. This process is crucial for soil fertility.


While earthworms contribute to fragmentation and their own catabolism, humification is a key ecological process they actively facilitate, transforming raw detritus into a more stable and beneficial form for the ecosystem.




Final Answer: \(\boxed{D}\) Quick Tip: \textbf{Key Exam Tip:}
Earthworms are vital decomposers. They contribute to fragmentation and significantly aid in humification by processing detritus and mixing it with soil, forming humus.

Let's analyze each statement regarding ecological pyramids:
1. The pyramid of energy is always upright: This is TRUE. Energy transfer between trophic levels is inefficient (about 10%), so each successive level has less energy than the level below it.

2. The pyramid of biomass in the sea is generally inverted: This is TRUE. In many aquatic ecosystems, especially the open ocean, phytoplankton (producers) have a lower biomass at any given time than the zooplankton (primary consumers) that feed on them, leading to an inverted pyramid of biomass.

3. The pyramid of numbers in a grassland is upright: This is TRUE. Typically, a grassland has a large number of producers (grasses), fewer primary consumers (herbivores), and even fewer secondary consumers (carnivores), forming an upright pyramid of numbers.

4. Saprophytes occupy the top of every pyramid: This is FALSE. Saprophytes (decomposers) obtain nutrients from dead organic matter from all trophic levels. They are not positioned at the top consumer level of a pyramid but rather decompose organic matter from all levels.


Since statements 1, 2, and 3 are true, option (A) is correct.




Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
The pyramid of energy is invariably upright. Pyramids of numbers and biomass can be inverted in some cases (e.g., aquatic ecosystems, parasitic food chains). Decomposers are not placed at the apex of pyramids.

Statement I: Net Primary Productivity (NPP) is defined as Gross Primary Productivity (GPP) minus the energy lost through respiration by the producers (R). NPP represents the energy available for consumers. Thus, \(NPP = GPP - R\). This statement is correct.


Statement II: Energy flow in an ecosystem is typically unidirectional. Energy flows from producers to consumers and decomposers, with energy lost as heat at each transfer, following the second law of thermodynamics. It does not flow backward in a cyclic manner. Thus, energy flow is unidirectional, not bidirectional. This statement is incorrect.


Since Statement I is correct and Statement II is incorrect, option (C) is the correct answer.




Final Answer: \(\boxed{C}\) Quick Tip: \textbf{Key Exam Tip:}
Remember that NPP = GPP - R. Energy flow in ecosystems is unidirectional, not bidirectional.

Matching the interactions with their effects (+ benefit, - harm, 0 no effect):

i. Mutualism: Both species benefit (+,+). Match: i-b.

ii. Commensalism: One species benefits, the other is unaffected (+,0). Match: ii-a.

iii. Amensalism: One species is harmed, the other is unaffected (-,0). Match: iii-d.

iv. Competition: Both species are harmed (-, -). Match: iv-c.


The correct correspondence is i-b, ii-a, iii-d, iv-c.




Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Understand the effect of ecological interactions on the populations involved: Mutualism (+,+), Commensalism (+,0), Amensalism (-,0), Competition (-,-).

Statement I (Allen's Rule): This rule correctly states that endothermic animals living in colder climates tend to have shorter appendages (ears, limbs) compared to related animals in warmer climates. Shorter appendages reduce the surface area to volume ratio, thus minimizing heat loss. This is an adaptation for thermoregulation in cold environments. Thus, Statement I is correct.


Statement II: Opuntia has leaves modified into spines to reduce transpiration. Opuntia (prickly pear cactus) is adapted to arid conditions. Its leaves are modified into spines, which serve multiple purposes: reducing water loss by transpiration (due to lower surface area), deterring herbivores, and sometimes providing shade. Thus, Statement II is correct.


Since both statements are correct, option (A) is the answer.




Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Allen's Rule relates appendage size to climate for thermoregulation. Plant adaptations to arid conditions include leaf modifications like spines to reduce water loss.

Matching diseases with their causal organisms:

i. Amoebiasis: Caused by the protozoan *Entamoeba histolytica*. So, i-b.

ii. Filariasis: Caused by filarial worms like *Wuchereria bancrofti*. So, ii-a.

iii. Syphilis: Caused by the bacterium *Treponema pallidum*. So, iii-d.

iv. AIDS: Caused by the Human Immunodeficiency Virus (HIV). So, iv-c.


The correct matching is i-b, ii-a, iii-d, iv-c.



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Familiarize yourself with common diseases and their specific causative agents: Amoebiasis (protozoan), Filariasis (worm), Syphilis (bacterium), AIDS (virus).

Plasmodium is the parasite that causes malaria. Its life cycle involves both humans and mosquitoes. The infective stage for humans is the sporozoite. When an infected female Anopheles mosquito bites a human, it injects sporozoites from its salivary glands into the human bloodstream. These sporozoites then travel to the liver and initiate the infection cycle in humans. Merozoites are formed in the liver and infect red blood cells, causing symptoms. Trophozoites are the feeding stage within red blood cells. Gametocytes are the sexual stages that develop in red blood cells and are taken up by mosquitoes to continue the parasite's life cycle.




Final Answer: \(\boxed{C}\) Quick Tip: \textbf{Key Exam Tip:}
In the *Plasmodium* life cycle, sporozoites are the infective stage for humans, transmitted by mosquito bites.

Matching receptors with their stimuli:

i. Photoreceptors: Detect light. So, i matches with b.

ii. Gustatoreceptors: Detect taste. So, ii matches with a.

iii. Phonoreceptors: Detect sound. So, iii matches with d. (Note: "Phonoreceptors" is not a standard term; typically "mechanoreceptors" or "auditory receptors" are used for sound. However, given the context and options, this is the intended match.)

iv. Nociceptors: Detect painful stimuli. So, iv matches with c.


The correct matching is i-b, ii-a, iii-d, iv-c.



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Correctly associating receptors with their stimuli is key: Photoreceptors (light), Gustatoreceptors (taste), Phonoreceptors (sound), Nociceptors (pain).

Chlorobenzene reacts with magnesium metal in the presence of dry ether to form a Grignard reagent called phenylmagnesium chloride.

The reaction is: \[ C_6H_5Cl + Mg \xrightarrow[dry ether]{} C_6H_5MgCl \]

Grignard reagents are highly reactive organometallic compounds and behave as very strong bases as well as strong nucleophiles. They cannot exist in the presence of compounds containing acidic hydrogen atoms.

Ethanol (\(C_2H_5OH\)) contains an acidic hydrogen attached to oxygen. Therefore, when phenylmagnesium chloride reacts with ethanol, it abstracts the proton from ethanol and gets converted into benzene.

The reaction is: \[ C_6H_5MgCl + C_2H_5OH \rightarrow C_6H_6 + C_2H_5OMgCl \]

Thus, the final organic product formed is benzene.



Final Answer: \(\boxed{B}\) Quick Tip: \textbf{Key Exam Tip:}
Grignard reagents react immediately with compounds containing acidic hydrogen atoms such as water, alcohols, and carboxylic acids to form hydrocarbons.

The spin-only magnetic moment is calculated using the formula: \[ \mu = \sqrt{n(n+2)} BM \]
where \(n\) is the number of unpaired electrons.

We are given: \[ \mu = 2.84 BM \]

We check each configuration one by one.

Option (A): d⁵ in strong field

In a strong ligand field, electrons pair up in lower energy \(t_{2g}\) orbitals first.

Configuration: \[ t_{2g}^{5} \]

This arrangement contains 1 unpaired electron.

Thus: \[ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 BM \]

So, option (A) is incorrect.



Option (B): d³ configuration

Configuration: \[ t_{2g}^{3} \]

This contains 3 unpaired electrons.
\[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 BM \]

Hence, option (B) is incorrect.



Option (C): d⁴ in weak field

Weak field ligands produce high-spin complexes.

Configuration: \[ t_{2g}^{3}e_g^{1} \]

This contains 4 unpaired electrons.
\[ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 BM \]

So, option (C) is incorrect.



Option (D): d⁴ in strong field

Strong field ligands cause electron pairing.

Configuration: \[ t_{2g}^{4} \]

This arrangement has 2 unpaired electrons.
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 BM \]

This value matches the given magnetic moment of 2.84 BM.

Therefore, the correct answer is d⁴ in strong ligand field.



Final Answer: \(\boxed{D}\) Quick Tip: \textbf{Key Exam Tip:}
Always determine the number of unpaired electrons first. Strong field ligands cause pairing and reduce magnetic moment, while weak field ligands increase the number of unpaired electrons.

The acidity of carboxylic acids depends mainly on the stability of the conjugate base formed after loss of proton.

When a carboxylic acid loses \(H^+\), it forms a carboxylate ion: \[ RCOOH \rightarrow RCOO^- + H^+ \]

If the conjugate base is more stable, the acid becomes stronger.

Halogens such as fluorine, chlorine, and bromine are electron-withdrawing groups due to their negative inductive effect (\(-I\) effect). They pull electron density away from the carboxylate ion and stabilize the negative charge.

The electron-withdrawing power decreases in the order: \[ F > Cl > Br \]

Therefore: \[ FCH_2COOH > ClCH_2COOH > BrCH_2COOH \]

Acetic acid (\(CH_3COOH\)) contains a methyl group which shows a \(+I\) effect (electron donating effect). This destabilizes the carboxylate ion and reduces acidity.

Hence, the correct order is: \[ FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH \]



Final Answer: \(\boxed{C}\) Quick Tip: \textbf{Key Exam Tip:}
Electron-withdrawing groups increase acidity by stabilizing the conjugate base through the negative inductive effect.

Grignard reagents react with carbonyl compounds followed by hydrolysis to form alcohols.

The nature of alcohol formed depends upon the carbonyl compound used.


Reaction with formaldehyde (\(HCHO\)) gives primary alcohols.
Reaction with aldehydes gives secondary alcohols.
Reaction with ketones gives tertiary alcohols.


To prepare ethyl alcohol (\(CH_3CH_2OH\)), methyl magnesium halide reacts with formaldehyde.

Step 1: \[ CH_3MgX + HCHO \rightarrow CH_3CH_2OMgX \]

Step 2: Hydrolysis \[ CH_3CH_2OMgX \xrightarrow{H_3O^+} CH_3CH_2OH \]

Thus, formaldehyde is the correct reactant.



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Grignard reagent + HCHO gives primary alcohol. Grignard reagent + aldehyde gives secondary alcohol. Grignard reagent + ketone gives tertiary alcohol.

Acidity depends upon the ease with which a compound can lose a proton (H\(^+\)). Greater the stability of the conjugate base formed after removal of proton, greater is the acidity.

Alcohols such as benzyl alcohol and cyclohexanol are weak acids because the alkoxide ions formed after deprotonation are not highly stabilized.

Phenol is more acidic than ordinary alcohols because the phenoxide ion formed is resonance stabilized. The negative charge on oxygen gets delocalized over the benzene ring.

The resonance stabilization of phenoxide ion increases its stability and therefore increases acidity.

In m-chlorophenol, the chlorine atom exerts a strong electron-withdrawing inductive effect (−I effect). This effect pulls electron density away from the phenoxide ion and stabilizes the negative charge further.

Because of this additional stabilization, m-chlorophenol is more acidic than phenol.

Thus, the acidity order is: \[ m-chlorophenol > phenol > benzyl alcohol \approx cyclohexanol \]

Hence, the most acidic compound is m-chlorophenol.



Final Answer: \(\boxed{D}\) Quick Tip: \textbf{Key Exam Tip:}
Electron-withdrawing groups increase acidity by stabilizing the conjugate base through the negative inductive effect.

Ligands are ions or molecules that donate lone pairs of electrons to the central metal atom in coordination compounds.

Depending upon the number of donor atoms through which they coordinate with the metal ion, ligands are classified as monodentate, bidentate, or polydentate.

A monodentate ligand donates only one pair of electrons through a single donor atom.

Let us examine each option:


Cyanide ion (CN\(^-\)): It donates a lone pair through one donor atom (generally carbon). Therefore, it acts as a monodentate ligand.

Ethylenediamine (en): It contains two nitrogen donor atoms and forms two coordinate bonds with the metal ion. Hence, it is a bidentate ligand.

Oxalate ion (C\(_2\)O\(_4^{2-}\)): It coordinates through two oxygen atoms and therefore behaves as a bidentate ligand.

EDTA: Ethylenediaminetetraacetate contains six donor atoms (two nitrogen and four oxygen atoms). Therefore, it is a hexadentate ligand.


Thus, cyanide ion is the only monodentate ligand among the given options.



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Monodentate ligands bind through one donor atom, bidentate ligands through two donor atoms, and polydentate ligands through multiple donor atoms.

Polymers are classified on the basis of their structure into linear polymers, branched polymers, and cross-linked polymers.


Linear polymers consist of long straight chains placed closely together.

Branched polymers contain side chains attached to the main chain.

Cross-linked polymers contain extensive three-dimensional network structures.


High-density polythene (HDPE) contains long, unbranched chains of polyethylene molecules. Due to the linear arrangement, the chains can pack closely, resulting in high density and strength.

Low-density polythene (LDPE) contains branching in the polymer chains and therefore is not linear.

Bakelite and melamine are thermosetting polymers with highly cross-linked structures.

Therefore, high-density polythene is the linear polymer among the given options.



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
HDPE is linear and densely packed, whereas LDPE is branched. Bakelite and melamine are cross-linked thermosetting polymers.

The relationship between molar conductivity and conductivity is: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
where:


\(\Lambda_m\) = molar conductivity
\(\kappa\) = conductivity
\(C\) = concentration in mol L\(^{-1}\)


Rearranging the formula: \[ \kappa = \frac{\Lambda_m \times C}{1000} \]

Given: \[ \Lambda_m = 220\ \Omega^{-1} cm^2 mol^{-1} \] \[ C = 0.05 mol L^{-1} \]

Substituting the values: \[ \kappa = \frac{220 \times 0.05}{1000} \]
\[ \kappa = \frac{11}{1000} \]
\[ \kappa = 0.011\ \Omega^{-1} cm^{-1} \]

Therefore, the conductivity of the solution is: \[ 0.011\ \Omega^{-1} cm^{-1} \]



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Use the relation: \[ \kappa = \frac{\Lambda_m C}{1000} \] when concentration is given in mol L\(^{-1}\).

Elastomers are polymers that possess high elasticity. They can be stretched considerably and return to their original shape after removal of force.

This elasticity is due to weak intermolecular forces and slight cross-linking between polymer chains.

Let us analyze the given polymers:


Nylon 6,6: It is a polyamide used in fibres and plastics. It is strong and tough but not elastic.

Buna-S: It is a synthetic rubber formed by copolymerization of butadiene and styrene. It shows elastic properties and is therefore classified as an elastomer.

Terylene: It is a polyester fibre used in textiles and plastics.

Polythene: It is a thermoplastic polymer and not an elastomer.


Hence, Buna-S is the elastomer among the given options.



Final Answer: \(\boxed{B}\) Quick Tip: \textbf{Key Exam Tip:}
Synthetic rubbers such as Buna-S and neoprene are common examples of elastomers.

An electron moving in a circular orbit constitutes an electric current. The magnetic dipole moment associated with a current-carrying loop is given by:
\[ \mu = IA \]

where:

\(I\) = current produced by revolving electron
\(A\) = area of circular orbit


The electron completes \(f\) revolutions per second. Therefore, the current produced is:
\[ I = ef \]

where:

\(e\) = charge of electron
\(f\) = frequency of revolution


The area of the circular orbit is:
\[ A = \pi r^2 \]

Substituting these values in the formula:
\[ \mu = (ef)(\pi r^2) \]
\[ \mu = \pi efr^2 \]

Hence, the equivalent magnetic dipole moment is:
\[ \boxed{\pi efr^2} \]



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
For an electron revolving in a circular orbit: \[ I = ef \quad and \quad \mu = IA \] Always use area of circle = \(\pi r^2\).

A long straight current-carrying wire produces a magnetic field around it.

The magnitude of magnetic field at a distance \(d\) from the wire is:
\[ B = \frac{\mu_0 I}{2\pi d} \]

A proton moving with velocity \(v\) inside a magnetic field experiences magnetic force given by Lorentz force equation:
\[ F = qvB\sin\theta \]

where:

\(q=e\) for proton
\(v\) = velocity of proton
\(B\) = magnetic field
\(\theta\) = angle between velocity and magnetic field


The magnetic field due to the wire is perpendicular to the direction of motion of the proton.

Hence: \[ \theta = 90^\circ \]

Therefore:
\[ F = evB \]

Substituting value of magnetic field:
\[ F = ev\left(\frac{\mu_0 I}{2\pi d}\right) \]
\[ F = \frac{\mu_0 eIv}{2\pi d} \]

Using the right-hand rule, the force on the proton is directed towards the wire. Therefore, the force is attractive.

Hence, the correct option is:
\[ \boxed{Attractive, magnitude \dfrac{\mu_0 eIv}{2\pi d}} \]



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Magnetic field due to long straight wire: \[ B = \frac{\mu_0 I}{2\pi r} \] Use Fleming’s left-hand rule or right-hand rule to determine direction of force.

The electric field due to an infinite line charge at a perpendicular distance \(r\) is:
\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]

The potential difference between two points located at distances \(r_1\) and \(r_2\) from the line charge is:
\[ V = \frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right) \]

The work done in moving a charge \(q\) through potential difference \(V\) is:
\[ W = qV \]

Here: \[ r_1 = a \quad and \quad r_2 = 2a \]

Therefore:
\[ V = \frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{2a}{a}\right) \]
\[ V = \frac{\lambda}{2\pi\varepsilon_0}\ln 2 \]

Hence:
\[ W = q \times \frac{\lambda}{2\pi\varepsilon_0}\ln 2 \]
\[ W = \frac{q\lambda}{2\pi\varepsilon_0}\ln 2 \]

Thus, the required work done is:
\[ \boxed{\frac{q\lambda}{2\pi\varepsilon_0}\ln 2} \]



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
Potential due to infinite line charge varies logarithmically with distance: \[ V \propto \ln r \]

Let the original plate separation be \(d\).

Original capacitance:
\[ C = \frac{\varepsilon_0 A}{d} \]

A dielectric slab of dielectric constant \(K\) and thickness \(d/2\) is inserted.

The remaining air gap is also \(d/2\).

The effective separation becomes:
\[ d_{eff} = \frac{d}{2K} + \frac{d}{2} \]

Taking common factor:
\[ d_{eff} = \frac{d}{2}\left(\frac{1}{K}+1\right) \]

New capacitance:
\[ C' = \frac{\varepsilon_0 A}{d_{eff}} \]

Substituting effective separation:
\[ C' = \frac{\varepsilon_0 A}{\dfrac{d}{2}\left(\dfrac{1}{K}+1\right)} \]
\[ C' = \frac{2\varepsilon_0 A}{d\left(\dfrac{1+K}{K}\right)} \]
\[ C' = \frac{2K\varepsilon_0 A}{d(K+1)} \]

Since:
\[ C = \frac{\varepsilon_0 A}{d} \]

Therefore:
\[ C' = \frac{2K}{K+1}C \]

Hence, the new capacitance is:
\[ \boxed{\dfrac{2K}{K+1}C} \]



Final Answer: \(\boxed{A}\) Quick Tip: \textbf{Key Exam Tip:}
For partially filled dielectric capacitors, calculate effective separation first and then apply: \[ C = \frac{\varepsilon_0 A}{d_{eff}} \]