VITEEE PCBE Mock Test - 2 - JEE MCQ

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.

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.

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

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.

(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.

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."

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."

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

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

• 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

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

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).

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.

I = 2 A
Total resistance = R = 3/2Ω

As V = IR, V = volts = 3 volts

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).

R_C = 2 kΩ and V₀ = 4 V
I_C = = 2 mA
β = I_C/I_B = 100
⇒ I_B = I_C/100 = 2 × 10^–5 A
V_in = I_BR_i = 2 × 10^–5 × 1 kΩ = 20 mV
Hence, the input signal voltage is 20 mV.

The resistance of a conductor is given as,

R = ρlA

Where l is the length measured along the direction of the current.

A is the area of cross-section measured perpendicular to the direction of the current.

The cross-sectional area of the A−A face is,
2x × 4x = 8x²

The resistance between A−A (top and bottom face) is,

The cross-sectional area of the B−B face is,
x × 4x = 4x²

The resistance between B−B is,

The cross-sectional area of the C−C face is,
x × 2x = 2x²

The resistance between C−C is,

From the above three relations, it is clear that the maximum electrical resistance is,
R_c−c

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.

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.

P₄ + 3KOH + 3H₂O → 3KH₂PO₂ + PH₃.

Beryllium (Be) has a higher ΔiH (ionization enthalpy) than boron (B). In both cases, the electron to be removed belongs to the same principal shell.

In Be (Z = 4): 1s², 2s², the electron to be removed is a 2s electron.
In B (Z = 5): 1s², 2s², 2p¹, the electron to be removed is a 2p electron.
The 2s electron penetrates closer to the nucleus compared to the 2p electron. This means 2s electrons experience a stronger attraction from the nucleus than 2p electrons. As a result, a higher amount of energy is required to remove a 2s electron compared to a 2p electron. Thus, Be has a higher ionization enthalpy than B.

Ionization Enthalpy of Oxygen vs. Nitrogen and Fluorine
Oxygen (O) has a lower ionization enthalpy than nitrogen (N) and fluorine (F).

Electronic configurations:
Nitrogen (N, Z = 7): 1s², 2s², 2pₓ¹, 2pᵧ¹, 2p��¹
Oxygen (O, Z = 8): 1s², 2s², 2pₓ², 2pᵧ¹, 2p��¹
Fluorine (F, Z = 9): 1s², 2s², 2pₓ², 2pᵧ², 2p��¹
Across a period, ionization enthalpy generally increases from left to right due to the decrease in atomic size and increase in nuclear charge.

However, the ionization enthalpy of nitrogen is greater than that of oxygen. This is because nitrogen has a more stable half-filled p-orbital configuration (2p³), which provides extra stability. Removing an electron from this stable configuration requires more energy.

Thus, oxygen has a lower ionization enthalpy than nitrogen and fluorine.

This problem involves conceptual mixing of preparation of styrene and addition polymerisation of styrene. Students are advised to follow these steps.

• Complete the reaction first using information provided in question. (use elimination reaction)
• Then, using the product of above reaction as starting material identify the correct product.

Preparation of styrene

Styrene is obtained due to elimination reaction of 1-chloro-2 phenyl ethane in presence of strong base KO But
Polymerisation of styrene

'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.

'Coarse' means 'of textures that are rough to the touch or substances consisting of relatively large particles'. So, 'refined' is the correct answer. When the Buddha passed by Punna's house, she only had an unrefined/coarse pancake to offer to the Buddha.

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%.

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.

(20 × 16) women can complete the work in 1 day.
1 day's work of 1 woman = 1/320
(16 × 15) men can complete the work in 1 day.
1 day's work of 1 man = 1/240
So, required ratio = 1/240 : 1/320 : = 1/3 : 1/4   
= 4 : 3 (cross-multiplied)