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Full Mock Test & Solutions: VITEEE PCBE Mock Test - 2 (125 Questions)

You can boost your JEE 2026 exam preparation with this VITEEE PCBE Mock Test - 2 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 150 minutes
  • - Total Questions: 125
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Biology, Physics, Chemistry, English, Aptitude

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VITEEE PCBE Mock Test - 2 - Question 1

Cardiac Muscle Tissue What unique structural feature of cardiac muscle tissue enhances its functionality in the heart?

Detailed Solution: Question 1

Cardiac muscle cells feature intercalated discs that allow them to contract in a synchronized manner, critical for the heart's pumping action, making Option B correct. Option A is incorrect as it falsely states that tight junctions prevent communication. Option C is incorrect because cardiac muscles are indeed striated, and Option D is incorrect as cardiac muscle cells do not operate independently.

VITEEE PCBE Mock Test - 2 - Question 2

How do bacteriophages differ from most plant-infecting viruses?

Detailed Solution: Question 2

Bacteriophages, which are viruses that infect bacteria, typically contain double-stranded DNA. In contrast, the majority of viruses that infect plants usually carry single-stranded RNA. This structural difference in genetic material is significant for their replication processes and host interactions, making Option C the correct answer. Option A is incorrect as bacteriophages infect bacteria, not animals. Option B is incorrect because all viruses, including bacteriophages, contain genetic material either DNA or RNA. Option D is incorrect as neither bacteriophages nor plant viruses are considered living organisms in traditional biological classifications.

VITEEE PCBE Mock Test - 2 - Question 3

ICBN stands for

Detailed Solution: Question 3

ICBN (International Code of Botanical Nomenclature) is one of the codes of nomenclature which is independent of zoological and bacteriological nomenclature. The foundations of ICBN were found in Philosophia Botanica, a book written by C. Linnaeus.

VITEEE PCBE Mock Test - 2 - Question 4

Assertion (A): Slime molds form fruiting bodies that can disperse spores.
Reason (R): The spores of slime molds are capable of surviving extreme conditions.

Detailed Solution: Question 4

Assertion A is true as slime molds indeed form fruiting bodies to disperse spores. Reason R is also true and is the correct explanation for A because the ability of spores to survive extreme conditions is crucial for the effective dispersal and lifecycle continuation of slime molds in varying environmental conditions.

VITEEE PCBE Mock Test - 2 - Question 5

Solanum, Panthera, Homo are examples of:

Detailed Solution: Question 5

Solanum, Panthera, Homo are examples of Genera.
Example:

  1. Potato Genus: Solanum; 
    Species: tuberosum
  2. Human Genus: Homo;
    Species: sapiens
  3. Lion Genus: Panthera; 
    Species: leo

Hence, the Correct Answer is C

NCERT Reference: Topic “Genus” from Chapter "The Living world" of NCERT

VITEEE PCBE Mock Test - 2 - Question 6

Viruses have:

Detailed Solution: Question 6

  • A virus is a small collection of genetic code, either DNA or RNA, surrounded by a protein coat. A virus cannot replicate alone. Viruses must infect cells and use components of the host cell to make copies of themselves.
  • All viruses are nucleoproteins (Nucleic acid + Protein) in the structure. The nucleic acid (DNA and RNA) is the genetic material.
  • In a particular virus, either DNA or RNA is genetic material never both are present in a virus.
  • Single-stranded RNA or ssRNA - Tobacco mosaic virus (TMV) Virus envelope is known as capsid. The capsid is composed of protein subunits called capsomeres.

Hence, the correct option is A

NCERT Reference: Page no. 26 of topic “2.6 VIRUSES, VIROIDS, PRIONS AND LICHENS” of chapter 2 of NCERT

VITEEE PCBE Mock Test - 2 - Question 7

Which of the following is correctly matched?

Detailed Solution: Question 7

Coelenterates have radial symmetry. Aschelminthes are pseudocoelomates. Molluscs do not show metamerism. Sponges are diploblastic.

VITEEE PCBE Mock Test - 2 - Question 8

The given figure shows an angiogram of the coronary blood vessel. Which one of the following statements correctly describes, what is being done?

Detailed Solution: Question 8

Angiogram is an X-ray diagnostic procedure used to visualize the blood vessels following injection of a contrast 16 substance into an artery. Coronary angiography is used to
detect obstruction in the coronary arteries, due to presence of atherosclerotic plaques which can lead to heart attack, To clear the passage, a very small balloon tipped catheter is inserted into the coronary artery under X-ray observation, Then the balloon is inflated with air to destroy the plaques, thereby clearing the lumen of the blood vessel for blood flow, This procedure is known as coronary artery angioplasty.

VITEEE PCBE Mock Test - 2 - Question 9

In china rose the flower are

Detailed Solution: Question 9

China rose flowers are – Actinomorphic – Hypogynous – Twisted aestivation

VITEEE PCBE Mock Test - 2 - Question 10

Read the following statements about the vascular bundles:
(a) In roots, xylem and phloem in a vascular bundle are arranged in an alternate manner along the different radii.
(b) Conjoint closed vascular bundles do not possess cambium
(c) In open vascular bundles, cambium is present in between xylem and phloem
(d) The vascular bundles of dicotyledonous stem possess endarch protoxylem
(e) In monocotyledonous root, usually there are more than six xylem bundles present
Choose the correct answer from the options given below :

Detailed Solution: Question 10

(a) True. In the roots of many plants, the vascular bundles are arranged radially, where xylem and phloem are alternately positioned along different radii.

(b) True. Conjoint closed vascular bundles are those where cambium is absent. These are typically found in monocots.

(c) True. In open vascular bundles, cambium is present between xylem and phloem. This cambium is meristematic tissue that allows for the secondary growth of the plant by adding layers to xylem and phloem.

(d) True. In the vascular bundles of dicotyledonous stems, the protoxylem (the first-formed xylem) is oriented towards the center, which is known as endarch.

(e) True. Monocotyledonous roots often feature a polyarch condition, where there are multiple (usually more than six) xylem bundles.

VITEEE PCBE Mock Test - 2 - Question 11

Which of the following statements are correct?
(A). Depolarization of an axonal membrane is caused due to rise in stimulus-induced permeability to Na⁺ and its rapid influx into axoplasm.
(B). Diffusion of K⁺ outside the axonal membrane restores the resting potential of the membrane.
(C). Sodium-potassium pump maintains active transport of 2 Na⁺ outwards for 3 K⁺ into the axoplasm across the resting membrane.

Detailed Solution: Question 11

Depolarization of the axonal membrane happens due to an increase in Na⁺ permeability, which allows Na⁺ ions to rapidly enter the axoplasm, making Statement A correct. Statement B is also correct, as the diffusion of K⁺ out of the membrane restores the resting potential. Statement C is incorrect because the sodium-potassium pump moves 3 Na⁺ out for every 2 K⁺ in, not the other way around. Hence, the correct statements are A and B.
Topic in NCERT: Generation and Conduction of Nerve Impulse.
Line in NCERT: "The rise in the stimulus-induced permeability to Na* is extremely short-lived. It is quickly followed by a rise in permeability to K*. Within a fraction of a second, K+ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation."

VITEEE PCBE Mock Test - 2 - Question 12

Refer to the given diagram of the structure of a neuron and identify A, B and C.

Detailed Solution: Question 12


Topic in NCERT: NEURON AS STRUCTURAL AND FUNCTIONAL UNIT OF NEURAL SYSTEM
Line in NCERT: " A neuron is a microscopic structure composed of three major parts, namely, cell body, dendrites and axon."

VITEEE PCBE Mock Test - 2 - Question 13

Which organelle is involved in the formation of cilia and flagella in animals?

Detailed Solution: Question 13

Basal body is also known as a modified centriole which serves both to create and connect the cilia and flagella to the cell.

Topic in NCERT: Cilia and Flagella

Line in NCERT: "Both the cilium and flagellum emerge from centriole-like structure called the basal bodies."

VITEEE PCBE Mock Test - 2 - Question 14

The most common fungal partners of mycorrhiza are _________ species.

Detailed Solution: Question 14

  • Mycorrhiza is a symbiotic association between fungus and angiosperms.
  • The most common fungal partners of mycorrhiza are the Glomus species.

Hence, the correct option is D.
NCERT Reference: Topic- MICROBES AS BIOFERTILISERS” of chapter "Microbes in Human Welfare" of NCERT.  

Topic in NCERT: MICROBES AS BIOFERTILISERS

Topic in NCERT: Microbes as biofertilisers

Line in NCERT: "many members of the genus glomus form mycorrhiza."

VITEEE PCBE Mock Test - 2 - Question 15

A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles

Detailed Solution: Question 15

The charging battery is removed. Therefore, q = constant
The distance between the plates is increased. Therefore, C decreases.
Now, V = q/C, q is constant and C is decreasing.
Therefore, V should increase.
U = again q is constant and C is decreasing.
Therefore U should increase.

VITEEE PCBE Mock Test - 2 - Question 16

A proton moving with a speed u along the positive x-axis enters at y = 0, a region of uniform magnetic field B = B0 which exists to the right of y-axis as shown in the figure. The proton leaves the region after some time with a speed v at coordinate y. Then,

Detailed Solution: Question 16

When the proton enters the region of the magnetic field, it will experience a force F given by: F = q (u B); where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it does not do any work. Hence, the magnitude of the velocity of the particle will remain unchanged; only the direction of the velocity changes. Hence, v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along positive x-axis and B is directed out of the page; the proton will move in a circle in the x-y plane in the clockwise direction. Hence, its y-coordinate will be negative, when it leaves the region. Thus, the correct choice is (4).

VITEEE PCBE Mock Test - 2 - Question 17

Ampere-hour is the unit of

Detailed Solution: Question 17

Ampere is the unit of current, so we can write it as Q/t
Hour is the unit of time, so we can write it as t.
Ampere-hour is = Q. Hence, ampere-hour is the unit of electric charge.

VITEEE PCBE Mock Test - 2 - Question 18

A current of 2 A flows through a system of resistors, as shown in the given figure. What is the value of 'V'?

Detailed Solution: Question 18

I = 2 A
Total resistance = R = 3/2Ω


As V = IR, V = volts = 3 volts

VITEEE PCBE Mock Test - 2 - Question 19

A magnet of length 10 cm and magnetic moment 1 Am2 is placed along the side AB of an equilateral triangle ABC. If the length of side AB is 10 cm, then the magnetic field at point C is

Detailed Solution: Question 19

Let m be the pole strength of each pole of the magnet figure. The magnetic field at C due to the N-pole is given by

direction a long AC away from C. The magnetic induction at C due to the S-pole is given by


directed along CB towards B. Since AC = BC, B1 = B2.
The resultant magnetic induction at C is given by




Given: M = 1 A m2, a = 10 cm = 0.1 m. Also μ0 =
4π × 10-7 T A-1 m. Substituting these values in (1),
we get B = 10-4 T, which is choice (4).

VITEEE PCBE Mock Test - 2 - Question 20

For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is

Detailed Solution: Question 20

RC = 2 kΩ and V0 = 4 V
IC = = 2 mA
β = IC/IB = 100
 IB = IC/100 = 2 × 10–5 A
Vin = IBRi = 2 × 10–5 × 1 kΩ = 20 mV
Hence, the input signal voltage is 20 mV
.

VITEEE PCBE Mock Test - 2 - Question 21

Which of the following solutions will show positive deviation from ideal behaviour?

Detailed Solution: Question 21

Solution of ethyl alcohol and water will have a positive deviation from ideal behaviour.
This is because the interactive forces after mixing are less strong than the interactive forces before mixing.

VITEEE PCBE Mock Test - 2 - Question 22

For an octahedral complex MX4Y2 (M = a transition metal, X and Y are monodentate achiral ligands), the correct statement, among the following, is

Detailed Solution: Question 22

MX4Y2 → Octahedral complex
Key point - If there is a plane of symmetry in a complex, then it will be achiral and optically inactive.
Geometrical isomers means they exist in cis and trans forms.


Hence, there are two geometrical isomers both of which are achiral.

VITEEE PCBE Mock Test - 2 - Question 23

An amine (X) reacts with benzenesulphonyl chloride and the product thus obtained is soluble in KOH.
The amine (X) is

Detailed Solution: Question 23

Since the amine (X) reacts with benzenesulphonyl chloride and forms a product which is soluble in KOH, so amine (X) must be a primary amine.
Secondary amine forms a product which is insoluble in KOH.
Tertiary amine does not react with benzenesulphonyl chloride.

VITEEE PCBE Mock Test - 2 - Question 24

Which of the following combinations in an aqueous medium will give a red colour or precipitate?

Detailed Solution: Question 24

Fe+3 + 3SCN- Fe(SCN)3
Ferric thiocyanate forms a a blood red colouration.

VITEEE PCBE Mock Test - 2 - Question 25

The reagent(s) used for the conversion of benzene diazonium hydrogensulfate to benzene is/are:

Detailed Solution: Question 25

VITEEE PCBE Mock Test - 2 - Question 26

The pair of products formed during the reaction of yellow phosphorous with aqueous potash solution is

Detailed Solution: Question 26

P4 + 3KOH + 3H2O → 3KH2PO2 + PH3.

VITEEE PCBE Mock Test - 2 - Question 27

Choose the word/group of words which is the most similar in meaning to the word/group of words printed in underline as used in the passage.
Pounding

Detailed Solution: Question 27

'Pounding' means 'repeated and heavy striking'. 'Beating', in context of the passage, means to stir (cooking ingredients) vigorously to make a smooth or frothy mixture. None of the other terms are used in relation to food items. Thus, at the midnight Punna just finished pounding (beating into a smooth mixture) rice for the next day's meal.

VITEEE PCBE Mock Test - 2 - Question 28

Abhijit purchased a TV set for ₹18000 and a DVD player for ₹4000. He sold both the items together for ₹26400
₹26400. How much percent profit did he make?

Detailed Solution: Question 28

The profit percentage is calculated using the formula:

Profit % = (SP - CP) / CP × 100

where SP is the total selling price and CP is the total cost price for the TV set as well as the DVD player.

Given Data:
Selling Price (SP) = ₹26,400
Cost Price (CP) = ₹18,000 + ₹4,000 = ₹22,000
Calculation:
Profit % = (26,400 - 22,000) / 22,000 × 100
= 4,400 / 22,000 × 100
= 20%

Conclusion: The profit percentage is 20%.

VITEEE PCBE Mock Test - 2 - Question 29

A person walking along a straight road observes a pole from two points that are 1 km (1000 m) apart. The angles of elevation of the pole from these two points are 30° and 45°. The height of the pole is:

Detailed Solution: Question 29

A person walking along a straight road observes a pole from two points that are 1 km (1000 m) apart.

The height of the pole is given by:

h = 1000 × (cot 30° - cot 45°)

Since cot 30° = √3 and cot 45° = 1, we get:

h = 1000 × (√3 - 1)

On rationalizing, we get:

h = 500 × (√3 + 1) m

Thus, the correct answer is option C: 500(√3 + 1) m.

VITEEE PCBE Mock Test - 2 - Question 30

If a3 + b3 = 28 and a + b = 4, then what is the value of ab?

Detailed Solution: Question 30

(a + b)3 = a3 + b3+ 3ab(a + b)
43 = 28 +3ab × 4
64 = 28 + 12ab
64 - 28 = 12ab
36 = 12ab
ab = 3
This is the correct answer.

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