VITEEE PCBE Mock Test - 10 - JEE MCQ

• A. P-wave: Depolarization of atria leading to atrial contraction (R).
• B. QRS complex: Depolarization of ventricles (P).
• C. T-wave: Return of ventricles to resting state (Q).
• D. PR interval: Time taken for electrical impulse to travel from atria to ventricles (S).

Thus, the correct matching is Option 1: A - R, B - P, C - Q, D - S.

During the first pregnancy, the maternal immune system is not yet sensitized to the Rh antigen, so the mother produces antibodies against Rh only after exposure to the foetal blood during delivery. This is why the first pregnancy is generally unaffected.

• Statement I is correct: Uricotelic animals (e.g., birds, reptiles) excrete nitrogenous waste primarily as uric acid, which helps conserve water.
• Statement II is incorrect: Uric acid is not highly soluble in water, and it is typically excreted as solid or semi-solid crystals, not through the skin.

Thus, Statement I is correct, but Statement II is incorrect.

Red tides are caused by the rapid growth of dinoflagellates, which can release toxins that are harmful to marine life, including fish, shellfish, and other marine animals. These toxins can cause massive die-offs and can even be toxic to humans who consume contaminated seafood. Thus, Option 2 is correct, as it highlights the harmful effects of red tides as harmful algal blooms.

• Cytokinin is a class of plant hormones, but it is not classified as terpenes. It is derived from purine compounds.
• ABA (Abscisic acid) is related to carotenoids (Option 1).
• Ethylene is indeed a gaseous hormone (Option 2).
• Auxins are indole-based compounds (Option 3).

Sarcomere Components

• A-Band (I): Represents the length of the myosin filaments, including areas where actin and myosin overlap, and appears as the darker region.
• I-Band (II): Contains only actin filaments and is the lighter area that appears to shorten during muscle contraction.
• Sarcomere (III): The entire unit from one Z-line to the next.
• H-Zone (IV): The lighter region in the middle of the A-band where only myosin filaments are present when the muscle is relaxed.
• Myosin (V): The thick filament.
• Actin, Troponin, Tropomyosin (VI): Components of the thin filaments.
• Z-line (VII): The dark line to which actin filaments are attached, marking the boundary of each sarcomere.

Label Correspondence

• A: Appears to be the Z-line, marking the boundary of the sarcomere.
• B: Thick filaments (myosin) as seen in the center part of the sarcomere.
• C: Contains elements of both myosin and actin filaments, typical of the A-band.
• D: Shows the I-band, where only actin filaments are present.
• E: Another part of the A-band with overlapping actin and myosin filaments.
• F: The entire unit from one Z-line to the next Z-line, representing the sarcomere.
• G: The lighter middle region within the A-band where only myosin filaments are visible, representing the H-zone.

The structural proteins collagen and elastin and present in white collagen fibres and yellow elastic fibres respectively, which are connective tissue fibres. Neurogicalcells are specialized cells found in the brain and spinal cord supporting the neurons and their fibres. The muscles of biceps are voluntary and striated.

Large holes in ‘Swiss cheese’ are due to production of CO₂ by Propionibacterium sharmanii (a bacterium).

Golden rice is vitamin A rich variety developed by rDNA technology and used in the treatment of vitamin A deficiency.

'I' represent temporal bone in the given figure. The temporal lobe is one of the four major lobes of the cerebral cortex in the brain and located beneath the lateral fissure on both cerebral hemispheres of the mammalian brain. The temporal lobes are involved in processing sensory input into derived meanings for the appropriate retention of visual memories, language comprehension, and emotion association.

△x ⋍ (0.0127 – 0.0123) mm = 0.0004 mm = 4 × 10^-7 m
△p =
By uncertainty principle,


= 0.158 m/sec
= 0.16 m/sec

⇒ The SI unit of power is watt (W).
⇒ Ohm's law is valid only at a constant temperature.
⇒ Conventional current flows from positive terminal to negative terminal.
⇒ Electric current is the rate of flow of electric charges.

Focal length of the objective lens, f_o = 144 cmFocal length of the eyepiece, f_e = 6.0 cm
The magnifying power of the telescope is given as

The separation between the objective lens and the eyepiece is calculated as:
L = f₀ + f_e
L = 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24, and the separation between the objective lens and the eyepiece is 150 cm.

If M is the mass of the earth, the gain in potential energy is given by

Hence the correct choice is (2). Remember, the expression PE = mgh is an approximate one which is valid if h << R.

From the hysteresis curve of soft iron we know Soft iron has high retentivity and low coercive force therefore the loop (i) is for soft iron and the loop (ii) is for steel.

The tangential acceleration is along the tangent and a_c acts radially inwards. The resultant acceleration

 (from figure)

∴tan30° = a_c / at

The resistance of a conductor is, 

Here, V = Al is the volume of the wire. As both the wires are made of copper, their densities are same.
Hence, mass of the wire is directly proportional to the volume.
Let l₁ & l₂ be the lengths of the two wires and m₁ & m₂ be their masses, respectively.

Given that, 

From equation (1)1, we can write,

Consider a cube of length l which is heated, and the temperature is increased by ΔT. It has the coefficient of linear expansion α and the coefficient of volume expansion γ.

The volume of the cube is given by:
V = l³

Due to heating, the length of the cube increases and is given by:
Δl = αlΔT ...(1)

Now, the volume of the cube also increases and is given by:
ΔV = γVΔT ...(2)

The volume of the cube after expansion can also be given by:
V + ΔV = (l + Δl)³

From equation (1) and (2),
V + γVΔT = (l + αlΔT)³
= l³ + (αlΔT)³ + 3l²(αlΔT) + 3(αlΔT)²l

Ignoring α³l³ and α²l² terms, and using l³ = V, we get:
γ = 3α

D = coulomb / metre²

Dimensions 

D = M^₀L⁻²T¹A¹

Reduction of one mole of to Cr⁺³ involves the transfer of six moles of electrons.
Required charge = 6 F = 6 × 96500 C [ one F = 96500 C]

 does not have

 group required for iodoform test.

We know that antipyretics are those compounds that are applied to reduce the body temperature in fever.
Ex. Aspirin (acyl salicylic acid), paracetamol, phenacetin, novalgin, and analgin.

2KMnO₄ + 3H₂SO₄ + 5H₂S → K₂SO₄ + 2MnSO₄ + 8H₂O + 5S
KMnO₄  is a oxidising agent ; oxidises S^2- to S

An asymmetric carbon is one in which the carbon is bonded to four different atoms or groups.

• CH₃−CHCl−COOH: One chiral carbon
• CH₃−CH₂−COOH: No chiral carbons
• Cl−CH₂−CH₂−COOH: No chiral carbons
• Cl₂CH−CH₂−COOH: No chiral carbons

Hence, in the given options, only 2-chloropropanoic acid contains an asymmetric carbon.

The number of optically active forms of the above compound is two.

In BCl₃ molecule, boron has trigonal planar structure due to 3 bond pairs in the valence shell of boron. It's hybridisation is sp².

Whereas, NCl₃ has distorted tetrahedral structure due to one lone pair and three bond pair in the valence shell of nitrogen. It's hybridisation is sp³.

The meaning of the idiom/phrase 'Knit the brow' is 'To frown or look worried, angry or puzzled'.
Example-She knitted her brows as she listened to the strange story.
Thus, option A, that is, 'to frown' is the correct answer.

A man is facing South. He turns 135 degrees in the anticlockwise direction and then 180 degrees in the clockwise direction.

According to the question, the man turns in the anticlockwise direction, which means he is turning towards the northeast.
He turns 135°, which is  90° + 45.
On turning 90°90°, he would face east and on further turning 45°, he would face the exact centre line between north- east.
And Finally, he turns 180° in the clockwise direction, he would face the exact centre line in the South-west direction.
So, he is facing in the South-west direction at the final position.
Hence, this is the correct answer.

Prepare the eight directions diagram:

According to the question, prepare this diagram by taking the help of the directions:

We have to find out AB (Hypotenuse) using Pythagorean’s theorem. We need to first find BC (Base) and AC (perpendicular) as follows:

BC = BE + EC
= 2 + (FG − DH)
= 2 + (16 – 14)
= 4 km

AC = AD – CD
= 12 – (EF – HG)
= 12 – (16 – 14)
= 10 km

Now, in ΔACB, by Pythagoras' theorem:

AB² = AC² + BC²
= 10² + 4²
= 100 + 16
= 116

AB = √116
= √(2 × 2 × 29)
= 2√29 km