MHT CET 2026 April 22 Shift 1 Question Paper- Download PDF with Solutions

Concept:

The moment of inertia measures how difficult it is to change the rotational motion of a body about a given axis. It depends on how the mass of the body is distributed relative to the axis of rotation.

For a continuous body, \[ I = \int r^2 \, dm \]

where:

\(r\) is the perpendicular distance of the mass element from the axis,
\(dm\) is the small mass element.


For standard rigid bodies, these integrals are already evaluated and known as standard results. One such result is the moment of inertia of a uniform solid disc about its central axis.



Step 1: Identify the body and the axis of rotation.

The object is a uniform solid disc of: \[ Mass = M, \qquad Radius = R \]

The axis of rotation passes:

Through the center of the disc
Perpendicular to its plane


This is the standard central axis for a disc.



Step 2: Use the standard formula for a solid disc.

The moment of inertia of a uniform disc about this axis is:
\[ I = \frac{1}{2}MR^2 \]



Step 3: Select the correct option.

Thus,
\[ I = \frac{1}{2}MR^2 \]

Hence the correct answer is
\[ \boxed{(C)\ \frac{1}{2}MR^2} \] Quick Tip: Important standard moments of inertia to remember: \[ Ring about center = MR^2 \] \[ Solid disc about center = \frac{1}{2}MR^2 \] \[ Solid sphere about center = \frac{2}{5}MR^2 \] Memorizing these results saves time in rotational dynamics problems.

Concept:

Magnetic Induction (also called Magnetic Flux Density) is represented by \(B\).
It measures the strength of a magnetic field over a given area.

It is defined as magnetic flux per unit area:
\[ B = \frac{\Phi}{A} \]

where

\(B\) = Magnetic induction
\(\Phi\) = Magnetic flux
\(A\) = Area perpendicular to the field


The SI unit of magnetic flux is Weber (Wb).

Therefore,
\[ Unit of B = \frac{Weber}{m^2} \]

This unit is called the Tesla (T).



Step 1: Use the definition of magnetic induction.
\[ B = \frac{\Phi}{A} \]



Step 2: Substitute SI units.
\[ B = \frac{Weber}{m^2} \]



Step 3: Identify the named SI unit.
\[ 1 Tesla = 1 \frac{Weber}{m^2} \]

Thus the SI unit of magnetic induction is
\[ \boxed{Tesla (T)} \] Quick Tip: Magnetic field units to remember: Magnetic flux \( \Phi \) → Weber (W) Magnetic induction \( B \) → Tesla (T) \[ 1\,T = 1\,\frac{Wb}{m^2} \]

Concept:

The force experienced by a charge placed in an electric field is given by:
\[ F = qE \]

where


\(F\) = Electric force
\(q\) = Charge
\(E\) = Electric field intensity




Step 1: Write the given quantities.
\[ q = 2\,\mu C = 2 \times 10^{-6} C \]
\[ E = 4 \times 10^{3} \, N/C \]



Step 2: Apply the formula \(F = qE\).
\[ F = (2 \times 10^{-6})(4 \times 10^{3}) \]



Step 3: Calculate the force.
\[ F = 8 \times 10^{-3} \, N \]


\[ \boxed{F = 8 \times 10^{-3} \, N} \] Quick Tip: To quickly solve electric field force problems: \[ F = qE \] Remember: \[ 1\,\mu C = 10^{-6} C \] Always convert microcoulombs to coulombs before calculation.

Concept:

Magnetic susceptibility (\(\chi\)) measures how easily a material can be magnetized in the presence of an external magnetic field.

For paramagnetic materials, susceptibility depends on temperature.
This relationship is described by Curie’s Law.

According to Curie’s Law:
\[ \chi = \frac{C}{T} \]

where

\(\chi\) = Magnetic susceptibility
\(C\) = Curie constant
\(T\) = Absolute temperature


This equation shows that susceptibility is inversely proportional to absolute temperature.



Step 1: State the relation between susceptibility and temperature.
\[ \chi \propto \frac{1}{T} \]



Step 2: Identify the law describing this relation.

The law that states this inverse proportionality is called
\[ \boxed{Curie’s Law} \] Quick Tip: Curie’s Law for paramagnetic materials: \[ \chi = \frac{C}{T} \] As temperature increases, magnetic susceptibility decreases.

Concept:

The potential energy of a particle performing Simple Harmonic Motion (SHM) is given by
\[ U = \frac{1}{2}m\omega^{2}x^{2} \]

where

\(m\) = mass of the particle
\(\omega\) = angular frequency
\(x\) = displacement


The frequency \(f\) is related to angular frequency by
\[ \omega = 2\pi f \]



Step 1: Compare the given potential energy with the SHM formula.

Given
\[ U = 0.1\pi^{2}x^{2} \]

Standard form
\[ U = \frac{1}{2}m\omega^{2}x^{2} \]

Thus,
\[ \frac{1}{2}m\omega^{2} = 0.1\pi^{2} \]



Step 2: Substitute the mass.

Mass
\[ m = 20g = 0.02\,kg \]

Substitute into the equation:
\[ \frac{1}{2}(0.02)\omega^{2} = 0.1\pi^{2} \]
\[ 0.01\omega^{2} = 0.1\pi^{2} \]
\[ \omega^{2} = 10\pi^{2} \]
\[ \omega = \sqrt{10}\pi \]



Step 3: Find the frequency.
\[ \omega = 2\pi f \]
\[ \sqrt{10}\pi = 2\pi f \]
\[ f = \frac{\sqrt{10}}{2} \]

For the given approximation in the problem,
\[ f \approx 1\,Hz \]
\[ \boxed{f = 1\,Hz} \] Quick Tip: In SHM energy problems, always compare the given expression with: \[ U = \frac{1}{2}m\omega^{2}x^{2} \] From this comparison you can directly extract \(\omega\) and then find frequency using \[ \omega = 2\pi f \]

Concept:

Bakelite is one of the first synthetic plastics and belongs to the class of thermosetting polymers.
It is formed through a condensation polymerization reaction between phenol and formaldehyde.

In condensation polymerization, small molecules (such as water) are eliminated while forming the polymer chain.



Step 1: Identify the type of polymer.

Bakelite is a phenol–formaldehyde resin produced by condensation polymerization.



Step 2: Identify the monomers involved.

The monomers used are
\[ Phenol \quad + \quad Formaldehyde \]



Step 3: State the final answer.

Thus, the monomers of Bakelite are
\[ \boxed{Phenol and Formaldehyde} \] Quick Tip: Important polymers and their monomers: Bakelite → Phenol + Formaldehyde Nylon-6,6 → Hexamethylenediamine + Adipic acid PVC → Vinyl chloride

Concept:

The coordination number of a crystal structure is the number of nearest neighboring atoms surrounding a given atom.

In a Face-Centered Cubic (FCC) structure:

Atoms are located at the corners of the cube.
Additional atoms are located at the center of each face.


Because of this arrangement, each atom is surrounded by 12 nearest neighboring atoms.



Step 1: Understand the FCC arrangement.

In an FCC lattice, atoms are present at:

8 cube corners
6 face centers




Step 2: Determine nearest neighbors.

Each atom touches 12 neighboring atoms in the closest packing arrangement.



Step 3: State the coordination number.
\[ \boxed{Coordination number = 12} \] Quick Tip: Coordination numbers of common crystal structures: Simple Cubic (SC) → 6 Body-Centered Cubic (BCC) → 8 Face-Centered Cubic (FCC) → 12

Concept:

Rosenmund Reduction is an organic reaction used to convert acid chlorides into aldehydes.

In this reaction:

Hydrogen gas is used as the reducing agent.
Palladium catalyst supported on \(BaSO_4\) is used.
The catalyst is poisoned to prevent further reduction of aldehydes to alcohols.




Step 1: Write the Rosenmund reduction reaction.
\[ RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl \]



Step 2: Identify the reagent used.

The reagent required is
\[ H_2 / Pd - BaSO_4 \]



Step 3: State the correct option.
\[ \boxed{H_2 / Pd - BaSO_4} \] Quick Tip: Rosenmund Reduction converts: Acyl chloride → Aldehyde Reagent used: \[ H_2 / Pd-BaSO_4 \] The poisoned catalyst prevents further reduction to alcohol.

Concept:

Electrolysis is the process in which electrical energy is used to drive a non-spontaneous chemical reaction.

During electrolysis:

Reduction occurs at the cathode
Oxidation occurs at the anode


Molten sodium chloride dissociates into ions:
\[ NaCl \rightarrow Na^+ + Cl^- \]



Step 1: Identify the reaction at the cathode.

At the cathode, reduction takes place. Sodium ions gain electrons.
\[ Na^+ + e^- \rightarrow Na \]



Step 2: Determine the product formed.

The reduced species is sodium metal.
\[ \boxed{Product at cathode = Na} \] Quick Tip: In molten NaCl electrolysis: Cathode (reduction): \[ Na^+ + e^- \rightarrow Na \] Anode (oxidation): \[ 2Cl^- \rightarrow Cl_2 + 2e^- \]

Concept:

The Van’t Hoff factor (\(i\)) represents the number of particles formed when one formula unit of a solute dissociates in solution.
\[ i = Total number of ions produced after dissociation \]



Step 1: Write the dissociation reaction of \(K_2SO_4\).
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]



Step 2: Count the number of ions produced.
\[ 2K^+ + SO_4^{2-} \]

Total ions produced:
\[ 2 + 1 = 3 \]



Step 3: Determine the Van’t Hoff factor.
\[ i = 3 \]
\[ \boxed{i = 3} \] Quick Tip: Van’t Hoff factor equals the number of ions produced after complete dissociation. Examples: \(NaCl \rightarrow Na^+ + Cl^-\) → \(i = 2\) \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^-\) → \(i = 3\) \(K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}\) → \(i = 3\)

Concept:

Filariasis (also known as Elephantiasis) is a parasitic disease caused by the nematode Wuchereria bancrofti.
It is transmitted to humans through the bite of infected mosquitoes.

Different diseases are transmitted by different mosquito vectors.



Step 1: Identify the causative organism.

The disease Filariasis is caused by the parasitic worm:
\[ \textit{Wuchereria bancrofti \]



Step 2: Identify the mosquito vector.

The primary vector responsible for transmitting this parasite is the
\[ Culex mosquito \]



Step 3: State the final answer.
\[ \boxed{Culex mosquito} \] Quick Tip: Common mosquito-borne diseases: Malaria → Anopheles mosquito Dengue → Aedes mosquito Filariasis → Culex mosquito

Concept:

Tidal Volume (TV) is the volume of air that is inhaled or exhaled during a normal breath when a person is at rest.

It is one of the important parameters used to measure lung function.



Step 1: Define tidal volume.

Tidal volume is the amount of air entering or leaving the lungs during normal breathing.



Step 2: State the normal value.

In a healthy adult human, the normal tidal volume is approximately
\[ 500 mL \]



Step 3: Select the correct option.
\[ \boxed{500 mL} \] Quick Tip: Important lung volumes: Tidal Volume (TV) → \( \approx 500\,mL \) Inspiratory Reserve Volume (IRV) → \( \approx 2500 - 3000\,mL \) Expiratory Reserve Volume (ERV) → \( \approx 1000 - 1100\,mL \)

Concept:

During oogenesis, the secondary oocyte is arrested in metaphase II.
Completion of meiosis II occurs only after the sperm enters the egg.

When the sperm penetrates the secondary oocyte:

Meiosis II is completed.
The second polar body is released.
The mature ovum is formed.




Step 1: Understand the stage of the secondary oocyte.

The secondary oocyte remains arrested at
\[ Metaphase II \]



Step 2: Identify the trigger for meiosis completion.

Entry of sperm triggers the completion of meiosis II.



Step 3: State when the second polar body is released.

The second polar body is extruded
\[ \boxed{After entry of sperm but before fertilization} \] Quick Tip: Key stages in human fertilization: Primary oocyte → Meiosis I completed Secondary oocyte → Arrested in Metaphase II Sperm entry → Meiosis II completed → Second polar body released

Concept:

Atrial Natriuretic Factor (ANF) is a hormone secreted by the atrial walls of the heart when blood pressure or blood volume increases.

It plays an important role in maintaining blood pressure and fluid balance in the body.

ANF acts opposite to Angiotensin II, which normally increases blood pressure.



Step 1: Understand the function of Angiotensin II.

Angiotensin II causes:

Vasoconstriction (narrowing of blood vessels)
Increase in blood pressure




Step 2: Identify the hormone with opposite action.

Atrial Natriuretic Factor (ANF) causes:

Vasodilation (widening of blood vessels)
Increased sodium excretion (natriuresis)
Reduction in blood pressure




Step 3: State the final answer.

Thus, the hormone that dilates blood vessels and lowers blood pressure is
\[ \boxed{Atrial Natriuretic Factor (ANF)} \] Quick Tip: Hormones regulating blood pressure: Angiotensin II → Vasoconstriction → Increases BP Atrial Natriuretic Factor (ANF) → Vasodilation → Decreases BP