CBSE EXAMINATION PAPER-2024
NEET-ZOOLOGY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 200 questions. All questions are compulsory.
- This question paper is divided into 1 sections.
- Section A – questions number 1 to 200 are multiple choice questions
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
Match List I with List II. Choose the correct answer from the options given below
CH₃–CH₃ : 1σ, CH₂ = CH₂ : 1σ, 1π C₂ : 2π, CH= CH :1σ, 2π
The Henry’s law constant (K_H) values of three gases (A,B,C) in water are 145, 2×10⁻⁵ and 35 kbar, respectively. The solubility of these gases in water follows the order:
According to Henry's Law: P = K_H x Thus, a lower value of K_H indicates a higher value of x (mole fraction/concentration) for a given pressure and therefore higher solubility. Since, K_H : B < C < A ; Solubility : B > C > A
Statement I: The boiling points of hydrides of Group 16 elements follow the order H₂O > H₂Te > H₂Se > H₂S
Statement II: On the basis of molecular mass, H₂O is expected to have a lower boiling point than the other members of the group, but due to the presence of extensive H-bonding in H₂O, it has a higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
H₂O has the highest boiling point due to extensive intermolecular H-bonding, and H₂Te comes next due to its highest molecular mass. Thus, both statement I and II are correct.
Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane
Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
The boiling point of isomeric alkanes decreases with branching due to a decrease in the exposed surface area, which decreases the strength of intermolecular forces. Thus, the boiling point order for isomers of pentane is: n-pentane > iso-pentane > neo-pentane Thus, both statement I and II are correct.
Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II: Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements,
choose the correct answer from the options given below:
Aniline does not undergo Friedel–Crafts reactions as the –NH₂ group reacts with acidic groups to form salts instead.
The –NH₃⁺ group acts as a strong deactivating group. Thus, statement I is true.
Aromatic primary amines (aniline here) cannot be prepared by Gabriel–phthalimide synthesis as aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Thus, statement II is true.
Match List I with List II. Choose the correct answer from the options given below:
Match List I with List II. Choose the correct answer from the options given below:
NH₃ (1 lp, 3 bp) → Trigonal pyramidal (shape) BrF₅ (1 lp, 5 bp) → Square pyramidal (shape) XeF₄ (2 lp, 4 bp) → Square planar (shape) SF₆ (0 lp, 6 bp) → Octahedral shape (geometry)
In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C. 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
D. Cl₂(g) → 2Cl(g)
Choose the correct answer from the options given below:
Entropy represents degree of Randomness. Thus, it increases for those processes that involve a decrease in the order of arrangement or stability of the system.
Process A involves change of state from liquid to gas which is less ordered, so entropy increases. Process B involves lowering of T which makes the system more stable so entropy decreases.
Process C involves formation of gaseous products from solid reactant, so entropy increases.
Process D involves an increase in the number of gaseous moles so disorderliness or entropy increases.
Which reaction is NOT a redox reaction?
Match List I with List II. Choose the correct answer from the options given below:
n = Principal quantum number → Size of orbital/shell
l = Azimuthal quantum number → Shape of orbital
mₗ = Magnetic orbital quantum number → Orientation of orbital mₛ = Magnetic spin quantum number → Spin of electron
‘Spin only’ magnetic moment is same for which of the following ions?
A. Ti³⁺
B. Cr²⁺
C. Mn²⁺
D. Fe²⁺
E. Sc³⁺
Choose the most appropriate answer from the options given below:
The highest number of helium atoms is in
For helium: 4 g of He/4 g mol⁻¹ = 1 mol
2.271 L of He = 2.271/22.7 = 0.1 mol
Thus, 4 g of He has the highest number of atoms.
The reagents with which glucose does NOT react to give the corresponding test/products are
A. Tollen’s reagent
B. Schiff’s reagent
C. HCN
D. NH₂OH
E. NaHSO₃
Choose the correct options from the given below:
Glucose does not give Schiff's test and does not form a hydrogen sulphite addition product with NaHSO₃ due to its cyclic structure (absence of a free –CHO group). Glucose reacts with NH₂OH, but its pentaacetate does not react with NH₂OH due to the absence of a free –CHO group.
Identify the correct reagents that would bring about the following transformation.
-
(A)
-
(B)
-
(C)
-
(D)
Hydroboration - oxidation gives Anti-Markovnikoff product (alcohol) which is then oxidized to aldehyde by a mild oxidizing agent, which is Pyridinium chlorochromate (PCC) here.
A compound with a molecular formula of C₆H₁₄ has two tertiary carbons. Its IUPAC name is:
Match List I with List II. Choose the correct answer from the options given below:
Isothermal ⇒ Constant T, Isochoric ⇒ Constant V ,Isobaric ⇒ Constant P ,Adiabatic ⇒ Constant heat/no exchange of heat
The pair of lanthanoid ions which are diamagnetic is
Ce⁴⁺ = [Xe] 4f⁰ Yb²⁺ = [Xe] 4f¹⁴ Ce³⁺ = [Xe] 4f¹ Eu²⁺ = [Xe] 4f⁷ Gd³⁺ = [Xe] 4f⁷ Er4³⁺ = [Xe] 4f⁶ Pm³⁺ = [Xe] 4f⁴ Sm³⁺ = [Xe] 4f⁵ Thus, Ce⁴⁺ and Yb²⁺ are diamagnetic due to all electrons being paired.
Statement I: [Co(NH₃)₆]³⁺ is a homoleptic complex whereas [Co(NH₃)₄Cl₂]⁺ is a heteroleptic complex. Statement II: Complex [Co(NH₃)₆]³⁺ has only one kind of ligands but [Co(NH₃)₄Cl₂]⁺ has more than one kind of ligands. In the light of the above statements, choose the correct answer from the options given below.
Homoleptic complex ⇒ has only one type of ligand. [Co(NH₃)₆]³⁺ Heteroleptic complex ⇒ has different types of ligands. [Co(NH₃)₄Cl₂]⁺ Thus, Statement I and II, both are correct.
Mass in grams of copper deposited by passing 9.6487 A current through a voltameter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol⁻¹, 1 F = 96487 C)
t = 100 seconds, I = 9.6487 A.
Charge (Q) = I × t = 9.6487 × 100 = 964.87 C
Now, Cu⁺²(aq) + 2e⁻ → Cu(s)
Thus, 2 moles of electrons = 2 × 96487 C of electricity give 1 mole (63 g) of Cu.
Therefore, 964.87 C of electricity will give:
M = 63 × 964.87/2 × 96487
M = 0.315g of Cu
The products A and B obtained in the following reactions, respectively, are 3ROH + PCl₃ →RCl + A ROH + PCl₅ → RCl + HCl + B
3ROH + PCl₃ → (3RCl , A ) + H₃PO₃
ROH + PCl₅ → RCl + (HCl ,B )+ POCl₃
The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given: R=8.314 J K⁻¹ mol⁻¹, log4=0.6021
Using Arrhenius equation:
Log k₂ /k₁ =Eₐ / 2.303R [ T₂-T ₁/ T₁ T ₂ ]
Log 4k ₁ /k ₁ = log (4) =E ₐ /2.303(8.314)
[330-300/300×330]
0.6021= Eₐ×(1.582×10⁻⁵) ⇒ Eₐ = 38040 j = 38.04kj
During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe²⁺ ion?
Dilute sulphuric acid (H₂SO₄) is added to aqueous solutions of Mohr’s salt to prevent the hydrolysis of ferrous sulphate.
This is in accordance with Le-Chatelier’s principle, according to which the addition of more SO₄²⁻ ions to the solution will favour the formation of more FeSO₄ to decrease the concentration of SO₄²⁻ ions and thus prevent hydrolysis.
Concentrated H₂SO₄ is not added because it is an oxidizing agent and might oxidize Fe²⁺ to Fe³⁺ and also does not dissociate completely.
Identify the correct answer.
Ozone has only two resonance structures. BF₃ is a non-polar molecule due to zero dipole moment. The dipole moment of NF₃ is less than that of NH₃ due to F being more electronegative than H and pulling electrons towards itself, resulting in cancellation of bond dipoles. Carbonate ion (CO₃²⁻) has three canonical structures, with each oxygen atom forming a double bond with carbon in one of the structures.
Identify the major product C formed in the following reaction sequence:
Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI.
A. Al³⁺
B. Cu²⁺
C. Ba²⁺
D. Co²⁺
E. Mg²⁺
Choose the correct answer from the options given below.
Al³⁺ = Group III, Cu²⁺ = Group II, Ba²⁺ = Group V, Co²⁺ = Group IV, Mg²⁺ = Group VI Thus, the correct increasing order is B→ A→ D→C→E
Consider the following reaction in a sealed vessel at equilibrium with concentrations of N₂ = 3.0 × 10⁻³ M, O₂ = 4.2 × 10⁻³ M and NO = 2.8 × 10⁻³ M. 2NO(g) ⇌ N₂(g) + O₂(g) If 0.1 mol L⁻¹ of NO(g) is taken in a closed vessel, what will be the degree of dissociation (α) of NO(g) at equilibrium?
A compound X contains 32% of A, 20% of B and remaining percentage of C. Then the empirical formula of X is: (Given atomic masses of A = 64; B = 40; C = 32 u)
Let us take 100 g of this compound.
⇒ Masses present: A = 32 g B = 20 g C = 48 g
⇒ Number of moles: A = 32/64 = 0.5 B = 20/40 = 0.5 C = 48/32 = 1.5 Dividing these moles by the smallest mole value: A:B:C = 1:1:3
⇒ Empirical Formula = ABC₃
In a vernier callipers, (N+1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (cm) is:
Vernier Constant V.C. = MSD – VSD …(i) where MSD = Main scale division and VSD = Vernier scale division Given : (N+1) VSD = N MSD VSD = (N/N+1)MSD ...(ii) From eq. (i) and (ii), V.C. = MSD - N/N+1(MSD) = MSD(1 - N/N+1) = MSD/N+1 = 0.01/N+1 = 1/100(N+1)
A logic circuit provides the output Y as per the following truth table:
The expression for the output Y is :
As per given truth table, output is independent of input A. ∴ Output Y = B̅
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 × 10⁻⁶ kg m². If the magnitude of magnetic moment of the needle is 1 × 10⁻⁵ A m², then the value of ' x' is:
Time period of oscillation, T = 2π√(I/MB) ⇒ 1/4 = 2π(√9.8×10⁻⁶/M×0.049) ⇒1/16 = 4π² ×9.8 × 10⁻⁶/ M × 49 × 10⁻³
M = 4π² × 9.8 × 10⁻⁶ /49 × 10⁻³ × 16
= 4π² × 9.8 × 16 × 10⁻³/49
= 12.8π² × 10⁻³ × 10⁻² × 10² = 1280π² × 10 ⁻⁵ Am²
∴ x = 1280 π²
A wire of length 'l' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
Resistance R = ρl/A Divided into 10 parts ∴ Resistance of each part = 5Ω Resistance in series, Rₛ = 5 × 10 = 50 Ω Resistance in parallel, Rₚ = 10/5= 2Ω ∴ R_eq = Rₛ +Rₚ =52Ω
A horizontal force 10 N is applied to a block A as shown in the figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
Force (F) = (M₁+M₂)a ⇒ a = F/M₁+M₂ or a = 10/2+3 = 2 ms⁻² ∴Force exerted by block A on block B
F′ = M₂(2) = 3×2 = 6 N
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below.
Statement I is correct as atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II is incorrect as atom of most of the elements but not every atom are stable and emit characteristic spectrum. This statement is not true for every atom.
A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. If the tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes:
Consider the following statements A and B and identify the correct answer: A. For a solar cell, the I–V characteristics lies in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the current measured in μA is due to majority charge carriers.
In a reverse biased pn junction diode, the current measured in μA is due to minority charge carriers. Solar cell: The I–V characteristics is as shown in figure.
A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path be is:
Path bc is an isochoric process, i.e., volume is constant. ΔV=0
∴ Work done by gas along path bc,
W_bc = P. ΔV = 0
If c is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is E=hν.
B. The velocity of a photon is c.
C. The momentum of a photon, p= hν/c
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
(A) The energy of a photon E = hv
(B) Velocity of photon is equal to velocity of light i.e. c.
(C) Since λ = h/p ⇒ p = h/λ ⇒p = hv/c
(D) In photon-electron collision both total energy and total momentum are conserved.
A light ray enters through a right-angled prism at point P with the angle of incidence 30° as shown in the figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is:
In the following circuit, the equivalent capacitance between terminal A and terminal B is :
Given circuit is balanced Wheatstone bridge so equivalent circuit is as shown ∴ Equivalent capacitance C_AB = (1/2 + 1/2) + (1/2 + 1/2) = 2μF
The quantities which have the same dimensions as those of solid angle are:
Solid angle, dΩ = dA/r² has dimensions [M⁰L⁰T⁰] Strain, Strain=Δℓ/ℓ has dimensions [M⁰L⁰T⁰] Angle measured in radians, θ = ℓ/r is also dimensionless, [M⁰L⁰T⁰] Solid angle, strain, and angle are dimensionless .
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 10⁸ N m⁻² and 2 × 10¹¹ N m⁻², is:
For maximum elongation, Stress = Elastic limit
Maximum elongation
= σelastic × L / Young's modulus = 8 × 10⁸ ×1/ 2 × 10¹¹ = 4 × 10⁻³ m =4 mm
In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South.
The mass of a planet is 1/10that of the Earth and its diameter is half that of the Earth. The acceleration due to gravity on that planet is:
Acceleration due to gravity g = GM/R² So acceleration due to gravity on the surface of the planet g' = GM'/R'² = GM/10(R/2)² or g' = 4/10.GM/R² = 0.4 × g = 0.4 × 9.8 ∴ g' = 3.92 m s⁻²
Match List I with List II.Choose the correct answer from the options given below:
Energy of transition, ΔE = hc/λ ∴ λ ∝ 1/ΔE As, (ΔE)₆₋₂ > (ΔE)₅₋₂ > (ΔE)₄₋₂ >(ΔE)₃₋₂
⇒ λ₆₋₂< λ₅₋₂ < λ₄₋₂ < λ₃₋₂ A-III, B-IV, C-II, D-I
An unpolarised light beam strikes a glass surface at Brewster’s angle. Then
According to Brewster’s law when an unpolarised light beam strikes a glass at Brewster’s angle, then reflected rays are completely polarized and refracted rays are partially polarized.
Match List I with List II. Choose the correct answer from the options given below:
The susceptibility of different magnetic materials are given below
Two bodies A and B of same mass undergo completely inelastic one-dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁:v₂ is:
By conservation of linear momentum, Pᵢ = P_f ⇒ mv₁ = mv₂ + mv₂ ⇒ mv₁ = 2mv₂
∴ v₁/v₂ = 2/1
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are
If x = 5sin (πt +π/3) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
x = 5 sin (πt +π/3) m
Amplitude, A = 5 m
Time perind T = 2π/ω = 2π/π = 2 s
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m⁻¹, then the excess force required to take it away from the surface is:
Excess force = T × length = T × 2πR = 7/100 × 2 × 3.14 × 4.5/100 = 19.8 mN
The output (Y) of the given logic gate is similar to the output of an/a
Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector 𝐏⃗ of magnitude 4 × 10⁻⁶ Cm, is ±9 × 10³V.
(Take 1/4πε₀ = 9 × 10⁹ SI units)
Reason R: V = ± 2P/4πε₀r², where r is the distance of any axial point, situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:
The potential V at any point, at distance r from centre of dipole = 1/4πε₀.Pcosθ/r²
At axial point where θ = 0°; V = 1/4πε₀.P/r²
= 9 × 10⁹ × 4 ×10⁻⁶/2² = 9 × 10³ V
V = -1/4πε₀.P/r² = -9 × 10³ V
At the axial point where θ = 180°
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?
In the case of pure rolling, v = ωR ∴ vₚ = V+ ωR = 2v
v_Q = v - ωR = 0
Hence, point P moves faster than point Q.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
.
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
f₀ = 140 cm and fₑ = 5 cm
For distant object, the magnifying power is given by m = f₀/fₑ = 140/5 = 28 cm
The following graph represents the T–V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P₁, P₂ and P₃ compared with those of Charles’ law represented as dotted lines. Then the correct relation is:
Using Charles' law, T ∝ V
∴ T = Slope × V ...(i)
From the ideal gas equation, PV = nRT ⇒ T = (P/nR)V ...(ii)
From equations (i) and (ii), we get: Slope ∝ P
As, (slope)₁ > (slope)₂ > (slope)₃ ⇒ P₁ > P₂ > P₃
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is:
Time period of simple pendulum is T = 2π√(l/g) ⇒ T ∝ √l
⇒ T₂/T₁ = √(l₂/l₁) = 1/2 [∵ l₂ = l₁/2]
⇒ T₂ = 1/√2 T₁ = x/2 T₁(given)
∴ x = √2
A sheet is placed on a horizontal surface in front of a strong magnetic pole.
A force is needed to: A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar. Choose the correct statement(s) from the options given below:
A. A magnetic pole will repel or attract magnetic sheet, so force is need to hold the sheet.
B. If sheet is non-magnetic, then magnetic pole will not exert force on it.
C. If it is conducting, then there will be eddy current in sheet, which opposes the motion. So forces is needed to move sheet with uniform speed.
D. The non-conducting and non-polar sheet do not interact with magnetic poles of magnet.
A metallic bar of Young’s modulus 0.5 × 10¹¹ N m⁻² and coefficient of linear thermal expansion 10⁻⁵°C⁻¹, length 1 m and area of cross-section 10⁻³ m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is:
The compressive force developed in the metallic bar is F = YAαΔT = 0.5 × 10¹¹ × 10⁻³ × 10⁻⁵ × 100 = 50 × 10³ N
An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is :
The compressive force developed in the metallic bar is F = YAαΔT = 0.5 × 10¹¹ × 10⁻⁵ × 10⁻³ × 100 = 50 × 10³ N
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
When the plates of a parallel plate capacitor is connected to a battery, then potential difference remains constant.
V′=V=Constant
(i) C′ = ε₀A/d′, C= ε₀A/d
Since, d′ < d ⇒ C′ > C
Hence, the final capacitance is greater than the initial capacitance. (ii) U′ = 1/2 C′V² and U = 1/2CV²
⇒ U′ > U [∵ C′ > C] Hence, the final energy is greater than the initial energy.
(iii) Q′/V′ = C′ and Q/V = C ⇒ Q′/V′ ≠ Q/V
(iv) Product of charge and voltage: X′ = Q′V = C
′V² and X = QV = CV² ⇒ X′ > X
[∵ C′ > C]
A force defined by F = αt² + βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:
From principle of homogeneity [F] = [αt²] = [βt] ⇒ [αt] = [F]/[t] and [β] = [F]/[t] ∴ [αt]/[β] = [F⁰t⁰] ∴ αt/β = dimensionless
Hindi II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
Hind II was the first restriction endonuclease discovered. The functioning of restriction endonuclease depends on a specific DNA nucleotide sequence that was isolated. Hind II is a 6-base pair cutter.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-II, C-IV, D-I
Statement I : Bt toxins are insect group specific and coded by a gene cry IAc.
Statement II : Bt toxin exists as inactive protoxin in B. thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
Sol. Bt toxin exists as inactive protoxins in B. thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to alkaline pH of the insect gut.
Identify the set of correct statement: A. The flowers of Vallisneria are colourful and produce nectar.
B. The flowers of waterlily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below.
As flowers of Vallisneria are pollinated by water so they are not colourful and do not produce nectar.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-I , C-IV, D-II
The equation of Verhulst-Pearl logistic growth is dN/dt = rN[K-N/K]. From this equation, K indicates:
In the given equation K denotes the carrying capacity.
Statement I : Parenchyma is living but collenchyma is dead tissue.
Statement II : Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
Collenchyma is also a living tissue.
These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
Habitat loss and fragmentation , Over-exploitation , Alien species invasions ,and Co-extinctions are the major causes of biodiversity losses. So, correct options are A, B and D only.
What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism. B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
The DNA carrying only gene of interest which may get integrated into the genome of the recipient, hence multiply and be inherited along with the host DNA . The piece of DNA not be able to multiply itself in the progeny cells of the organism independently. It needs to be integrated into the genome of the recipient.
Tropical regions show greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below.
Tropical environment are less seasonal, and relatively more constant and predictable when compared to temperate or other environments.
Statement I: Chromosomes become gradually visible under light microscope during leptotene stage. Statement II: The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements,
choose the correct answer from the options given below:
Both the statement I and II are true.
A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
Both red as well as pink flowered plants are expected in the progeny as per the below cross: Pink colour flower in snapdragon : Rr Red flowered snapdragon : RR when both pink and red are crossed Phenotype Red : Pink : White , 2 : 2 : 0
So the progeny that we get are red and pink flowered plants only.
Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C. CO₂ D. ATP
E. NADPH
Choose the correct answer from the options given below:
Light and chlorophyll are required in light reaction not in dark reaction.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-IV, C-I, D-II
Match List I with List II. Choose the correct answer from the options given below:
A-II, B-IV, C-I, D-III
Match List I with List II. Choose the correct answer from the options given below:
A-II, B-I, C-IV, D-III
Read the following statements and choose the set of correct statements: In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin. Choose the correct answer from the options given below:
In Phaeophyceae sexual reproduction can occur by oogamy, isogamy or anisogamy . Hence correct statements are A, C, D and E.
In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is 100x (kcal m⁻² yr⁻¹), what would be the GPP (Gross Primary Productivity) of the third trophic level of the same ecosystem?
NPP of first trophic level can act as GPP for second trophic level and NPP of second trophic level can act as GPP for third trophic level. So, 100x (kcal/m²/yr) would be GPP of second trophic level and 100x × 10% (kcal/m²/yr) i.e., 10x (kcal/m²/yr) energy would be GPP of third trophic level.
Match List I with List II. Choose the correct answer from the options given below:
A-IV, B-I, C-II, D-III
Statement I: In C₃ plants, some O₂ binds to RuBisCO, hence CO₂ fixation is decreased
. Statement II: In C₄ plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
In C4 plants photorespiration does not occur
Which of the following are fused in somatic hybridization involving two varieties of plants?
Somatic hybridization involves the fusion of protoplasts isolated from two varieties of plants.
Following are the stages of pathway for conduction of an action potential through the heart
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below
The pathway that action potential follows when generated in heart is Sino Atrial Node > Atrio Ventricular Node > Atrio Ventricular Bundle > Bundle Branches >Purkinje Fibres
The flippers of the Penguins and Dolphins are the example of the
Similar function is performed by the flippers of the Penguins and Dolphins perform but structurally they are not similar. Hence these structures represents convergent evolution.
Adaptive radiation occurs when different species in a given geographical area starting from a point and literally radiating to the other areas of geography in terms of evolution.
According to natural selection is the key mechanism behind evolution.
Divergent evolution represents homology.
Given below are some stages of human evolution. Arrange them in correct sequence. (Past to Recent)
A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus
Choose the correct sequence of human evolution from the options given below:
The sequence of Human Evolution from past to recent is as follows: Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens
Following are the stages of cell division :
A. Gap 2 phase
B. Cytokinesis
C. Synthesisphase
D. Karyokinesis
E. Gap I phase
Choose the correct sequence of stages from the options given below:
The following sequence represents the stages of cell division in correct order:
Which of the following are Autoimmune disorders?
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
Myasthenia gravis, Rheumatoid arthritis and Systemic Lupus Erythematosus (SLE) are the examples of auto immune disorders. Muscular dystrophy is a genetic disorder which progressively affects the skeletal muscles,
Gout is the inflammation of joints due to deposition of uric acid crystals.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male.
Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being.
In the light of the above statements, choose the correct answer from the options given below :
FSH acts on ovarian follicles in females and stimulating the growth of follicles. In males LH affects Leydig cells leading to secretion of androgens.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-IV, C-I, D-II
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
In the light of the above statements,
choose the most appropriate answer from the options given below:
Both A and R are correct and R is the correct explanation of A.
Match List I with List II. Choose the correct answer from the options given below:
A-II, B-IV, C-I, D-III
Consider the following statements :
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below:
Poriferans are the acoelomates. Aschelminths are pseudocoelomates. Platyhelminths are acoelomates.
Statement I: In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II: The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements,
choose the correct answer from the options given below:
Descending limb of loop of Henle is impermeable to electrolytes and permeable to water. Simple cuboidal brush -border epithelium lines the proximal convoluted tubule and increases the surface area for reabsorption.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-I, C-II, D-IV
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above above statements,
choose the correct answer from the options given below :
It is not necessary that hymen is torn during the first coitus only it can be torn due to sudden jolt, insertion of a vaginal tampon, active participation in some sports and in some women the hymen persists even after coitus.
Statement I: Gause’s competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause’s principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below :
Gause’s competitive exclusion principle states that two closely related species competing for the same resources cannot exist indefinitely. Due to the competition the superior species will outnumber the inferior one leading to their elimination.
The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Postanal tail is absent.
Choose the most appropriate answer from the options given below:
Perforated pharyngeal gill slits and dorsal central nervous system are the characteristics of chordates.
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum. Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
Mid brain, pons and medulla oblongata forms the brain stem.
Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes. In the light of above statements,
choose the most appropriate answer from the options given below :
Both statement I and II are correct.
Rgarding catalytic cycle of an enzyme action, select the correct sequential steps :
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below:
The correct sequence of catalytic cycle of an enzyme action is: (1) The substrate binds to the active site of the enzyme. (2) The binding of the substrate causes the change in shape of the enzyme to alter its shape. (3) Chemical bonds break and Enzyme Product complex is formed. (4 )Product is released by enzyme and enzyme becomes ready for the next substrate.
Match List I with List II. Choose the correct answer from the options given below:
A-III, B-IV, C-I, D-II
Statement I: Mitochondria and chloroplasts both double membranes bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast.
In the light of the above statements, choose the m is appropriate answer from the options given below
The inner membrane of mitochondria is relatively more permeable, as compared to chloroplast.
The graph which shows the variation of (1/λ²) and its A kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
-
(A)
-
(B)
-
(C)
-
(D)
The velocity (v) — time (t) plot of the motion of a body is shown below:
The acceleration (a) — time (t) graph that best suits this motion is :
-
(A)
-
(B)
-
(C)
-
(D)
As acceleration is the slope of the v–t curve:
∴ Initially, a = 0
After some time, a = constant (+ve)
Again, a = 0
Again, a = constant (−ve)
∴ The corresponding a-t graph is:
Choose the correct circuit which can achieve the bridge balance.
-
(A)
-
(B)
-
(C)
-
(D)
For bridge balance,
R₁/R₂ = R₃/R₄ ⇒ 10/15 = 10/R₄
∴ R₄ = 15 Ω
As R₄ is a combination of 5 Ω resistance and diode resistance,
∴ The diode should be forward biased and connected in series with the 5 Ω resistance.
Intramolecular hydrogen bonding is present in
-
(A)
-
(B)
-
(C)
- (D) HF
o-Nitrophenol → Intramolecular hydrogen bonding.
p-Nitrophenol, m-Nitrophenol, HF → Intermolecular hydrogen bonding.
The compound that will undergo S_N1 reaction with the fastest rate is
-
(A)
-
(B)
-
(C)
-
(D)
Which one of the following alcoholreacts instantaneously with Lucas reagent?
- (A) CH₃ – CH₂ – CH₂ – CH₂OH
-
(B)
-
(C)
-
(D)
Reaction of alcohols with Lucas reagent (Anhydrous ZnCl₂ + conc. HCl) follows the S_N1 mechanism. Therefore, a 3° alcohol, which forms the most stable carbocation, reacts instantaneously with Lucas reagent.
The most stable carbocation among the following is :
-
(A)
-
(B)
-
(C)
-
(D)
is most stable as it is a tertiary carbocation stabilized by three + I groups. The stability follows the order: 1° < 2° < 3°,
Which plot of In k vs 1/T is consistent with Arrhenius equation?
-
(A)
-
(B)
-
(C)
-
(D)
For the given reaction:
'P' is
-
(A)
-
(B)
-
(C)
-
(D)
Acidic KMnO₄ oxidizes alkenes that are substituted to carboxylic acids or ketones.
Both doubly bonded carbon atoms having one H-atom will also get oxidized to –OH.
Thus,
Major products A and B formed in the following reaction sequence, are
-
(A)
-
(B)
-
(C)
-
(D)
Identify the correct description about the given figure:
The depicted diagram shows a wind pollinated plant showing flowers with well exposed stamens. Stamens are exposed to assist pollination by wind.
Three types of muscles are given as a, b and c. Identify the correct matching pair along with their location in the human body:
Name of muscle/location
Figure (a) depicts skeletal muscle fibres .These muscles are closely attached to skeletal bones. They are found in voluntary organs like triceps muscles .Figure (b) depicts smooth muscle fibres .They are found in walls of internal organs such as the blood vessels, stomach and intestine. Figure (c) depicts cardiac muscle fibres ,they are only found in heart.
The following diagram shows restriction sites in E. coli cloning vector pBR322. Find the role of 'X' and 'Y' genes:
T'X' in the given diagram represents the ori region which controls the copy number of plasmid and hence the copy number of linked DNA. 'Y' is rop region that codes for protein involved in replication of plasmid.
Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
(A) represents FSH (Follicle stimulating hormone) acts on sertoli cells in males.
(B) represents Leydig cells involved in synthesis of androgens.
(C) represents Sertoli cells plays vital role in secretion of factors inovolved in spermiogenesis.
(D) represents Spermiogenesis, the process of formation of sperms from spermatozoa.
which one of the following factors will not affect the hardy - weinberg equilibrium?
The following factors affects the Hardy-Weinberg equilibrium:
1. Genetic recombination
2. Genetic Drift
3. Gene Migration
4. Mutation
5. Natural Selection
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