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What is Y in the above reaction?
for the above reaction at 298 K, is found to be . If the concentration of at equilibrium is 0.040 M then concentration of in M is
NEET 2022 Jul 17 Question Paper with Solutions
All 200 questions from the NEET 2022 (Jul 17) paper — Physics (50), Chemistry (50) and Biology (100) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2022Chemistry PYQs 2022Biology PYQs 2022
- Questions
- 200
- Physics
- 50
- Chemistry
- 50
- Biology
- 100
Physics50 questions
Q1Single correctDual Nature of Radiation and Matter
The graph which shows the variation of the de Broglie wavelength () of a particle and its associated momentum (p) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The rectangular hyperbola, lambda inversely proportional to p (drawn graph, option 4)
Approach:
The de Broglie relation links a particle's wavelength to its momentum; identifying the functional form fixes the shape of the graph.
Step 1:The de Broglie wavelength of a particle is inversely proportional to its momentum.
Step 2:A relation of the form plots as a rectangular hyperbola, with falling steeply at small p and approaching zero for large p.
Final answer: The rectangular hyperbola, lambda inversely proportional to p (drawn graph, option 4)
Q2Single correctCurrent Electricity
As the temperature increases, the electrical resistance
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Increases for conductors but decreases for semiconductors
Approach:
Resistance varies with temperature according to the dominant charge-transport mechanism, which differs between metals and semiconductors.
Step 1:In a conductor the carrier density is fixed and a temperature rise increases lattice vibrations, scattering electrons more often, so resistance increases.
Step 2:In a semiconductor a temperature rise frees many more charge carriers, and this growth in carrier density outweighs the increased scattering, so resistance falls.
Final answer: Increases for conductors but decreases for semiconductors
Q3Single correctAtoms
Let and be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to the Bohr's model of an atom, the ratio is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The Bohr energy of a level scales inversely with the square of its principal quantum number, so identifying the correct quantum numbers gives the ratio.
Step 1:The first excited state corresponds to and the second excited state to .
Step 2:Taking the ratio of energies and cancelling the common constant leaves the inverse squares of the quantum numbers.
Final answer:
Q4Single correctSystems of Particles and Rotational Motion
Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
The centre of mass of two point masses on a rod lies along the rod at a position weighted by the masses; placing the origin at the 10 kg mass gives its distance directly.
Step 1:Place the origin at the 10 kg mass, so , and the 20 kg mass at m.
Step 2:Substituting the masses and positions into the centre-of-mass expression gives the distance from the 10 kg mass.
Final answer: m
Q5Single correctMotion in a Straight Line
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The distance covered during the nth second of free fall from rest follows a fixed expression, and forming the ratio for successive seconds reveals the pattern.
Step 1:For a body released from rest, and , so the distance in the nth second is proportional to .
Step 2:Evaluating for gives the sequence of odd numbers.
Final answer:
Q6Single correctSystems of Particles and Rotational Motion
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The radius of gyration about each axis follows from the corresponding moment of inertia, and their ratio reduces to the square root of the ratio of moments of inertia.
Step 1:The moment of inertia about the central normal axis is twice that about a diameter for a thin disc.
Step 2:Since the radius of gyration is the square root of the moment of inertia per unit mass, the ratio of radii equals the square root of the ratio of moments of inertia.
Final answer:
Q7Single correctSystems of Particles and Rotational Motion
The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/ is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Angular acceleration is the change in angular speed over time; converting the speeds from rpm to rad/s allows a direct calculation.
Step 1:Convert the initial and final speeds from revolutions per minute to radians per second.
Step 2:Dividing the change in angular speed by the elapsed time gives the angular acceleration.
Final answer:
Q8Single correctThermodynamics
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
On a P-V diagram the four processes are distinguished by their slopes; the adiabatic curve is steeper than the isothermal one through the same point.
Step 1:Curve 4 is horizontal (constant pressure, isobaric) and the vertical line is isochoric; the two falling curves are isothermal and adiabatic.
Step 2:At the common point the adiabatic slope exceeds the isothermal slope by the factor , so the steeper of the two falling curves, curve 2, is adiabatic.
Final answer:
Q9Single correctElectrostatic Potential and Capacitance
Two hollow conducting spheres of radii and () have equal charges. The potential would be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2More on smaller sphere
Approach:
The surface potential of a charged conducting sphere depends on its charge and radius, so comparing the two spheres at equal charge fixes which has the higher potential.
Step 1:With equal charge on both spheres, the potential is inversely proportional to the radius.
Step 2:Since , the smaller sphere of radius has the higher potential.
Final answer: More on smaller sphere
Q10Single correctElectromagnetic Waves
When light propagates through a material medium of relative permittivity and relative permeability , the velocity of light, v is given by (c-velocity of light in vacuum)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The speed of an electromagnetic wave in a medium follows from the medium's permittivity and permeability, which can be written in terms of their vacuum values and relative factors.
Step 1:Express the medium's permittivity and permeability as the product of vacuum values and relative factors.
Step 2:Factoring out the vacuum speed of light leaves the relative permittivity and permeability under the root.
Final answer:
Q11Single correctMoving Charges and Magnetism
A solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 T
Approach:
The axial field inside a long solenoid depends only on the number of turns per unit length and the current; substituting the given values yields the field.
Step 1:Convert the winding density of 100 turns per mm to turns per metre.
Step 2:Substituting n, the current A and into the solenoid field expression gives the central field.
Final answer: T
Q12Single correctAlternating Current
The peak voltage of the ac source is equal to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 times the rms value of the ac source
Approach:
The peak and rms values of a sinusoidal ac voltage are related by a fixed factor, which fixes the correct statement.
Step 1:For a sinusoidal source the rms voltage is the peak voltage divided by the square root of two.
Step 2:Rearranging makes the peak voltage equal to times the rms value.
Final answer: times the rms value of the ac source
Q13Single correctWork, Energy and Power
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 m. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : ( m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
At constant speed the motor force balances gravity and friction; the power is the product of this total force and the speed.
Step 1:The motor must overcome both the weight of the load and the frictional force, since the lift moves at constant speed.
Step 2:Multiplying the total force by the constant speed gives the minimum power.
Final answer:
Q14Single correctWave Optics
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Within a fixed region of the screen the number of fringes is inversely proportional to the wavelength, so the new count scales accordingly.
Step 1:The length of the segment stays fixed, so the product of fringe count and wavelength is constant.
Step 2:Solving for the new fringe count with the smaller wavelength gives a larger number of fringes.
Final answer:
Q15Single correctCurrent Electricity
A copper wire of length 10 m and radius m has electrical resistance of 10 . The current density in the wire for an electric field strength of 10 (V/m) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 A/
Approach:
The current density equals the total current divided by the cross-sectional area; the current follows from the applied electric field, wire length and resistance.
Step 1:The potential difference across the wire is the field times its length, and the current then follows from Ohm's law.
Step 2:The cross-sectional area uses the given radius, and dividing the current by this area gives the current density.
Final answer: A/
Q16Single correctElectromagnetic Induction
The dimensions [ML] belong to the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Magnetic permeability
Approach:
Each listed quantity has a characteristic dimensional formula; deriving the dimension of permeability and comparing identifies the match.
Step 1:Rearranging the force-per-length relation for two parallel currents isolates the permeability in terms of force, distance and current.
Step 2:Simplifying the dimensional expression yields the given combination.
Final answer: Magnetic permeability
Q17Single correctWaves
If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The transverse wave speed on a string varies with the square root of the tension, so doubling the tension scales the speed by a known factor.
Step 1:At fixed linear mass density the wave speed is proportional to the square root of the tension.
Step 2:Substituting the doubled final tension gives the ratio of initial to final speed.
Final answer:
Q18Single correctSemiconductor Electronics
In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 Hz
Approach:
A half wave rectifier passes only one half of each input cycle, so the repetition rate of the output equals that of the input.
Step 1:In half wave rectification only one half-cycle per input cycle appears at the output, producing one pulse for every input cycle.
Step 2:With the input at 60 Hz, the output retains the same frequency.
Final answer: Hz
Q19Single correctMotion in a Straight Line
The displacement-time graphs of two moving particles make angles of 30 and 45 with the x-axis as shown in the figure. The ratio of their respective velocity is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
On a displacement-time graph the velocity is the slope, equal to the tangent of the angle the line makes with the time axis; the ratio of velocities is the ratio of these tangents.
Step 1:The velocity of each particle equals the tangent of its graph angle with the time axis.
Step 2:Forming the ratio of the two velocities gives the required result.
Final answer:
Q20Single correctMagnetism and Matter
A square loop of side 1 m and resistance 1 is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 weber
Approach:
Magnetic flux is the product of field strength and area when the loop plane is perpendicular to the field; the resistance is not needed for the flux.
Step 1:With the loop plane perpendicular to the field, the field is along the area normal, so , and the area of the square is one square metre.
Step 2:Multiplying the field by the area gives the flux through the loop.
Final answer: weber
Q21Single correctElectromagnetic Waves
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 J
Approach:
Energy radiated is the product of power and time, with the power and duration converted to base SI units.
Step 1:Convert the power to watts and the time to seconds.
Step 2:Multiplying power by time gives the radiated energy.
Final answer: J
Q22Single correctGravitation
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 N/kg
Approach:
Gravitational field intensity is the force per unit mass; converting the mass to kilograms and dividing gives the result.
Step 1:Convert the mass from grams to kilograms.
Step 2:Dividing the gravitational force by the mass gives the field intensity.
Final answer: N/kg
Q23Single correctElectromagnetic Waves
Choose the correct answer from the options given below
| List-I (Electromagnetic waves) | List-II (Wavelength) |
|---|---|
| (a). AM radio waves | (i). m |
| (b). Microwaves | (ii). m |
| (c). Infrared radiations | (iii). m |
| (d). X-rays | (iv). m |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Approach:
Each band of the electromagnetic spectrum has a characteristic wavelength range, so matching by decreasing wavelength order pairs the lists correctly.
Step 1:AM radio waves have the longest wavelengths, on the order of m, pairing (a) with (ii).
Step 2:Microwaves lie near m and infrared near m, while X-rays are shortest at m, giving the remaining pairs.
Final answer: (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Q24Single correctWork, Energy and Power
A shell of mass is at rest initially. It explodes into three fragments having mass in the ratio . If the fragments having equal mass fly off along mutually perpendicular directions with speed , the speed of the third (lighter) fragment is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Total momentum is conserved and starts at zero, so the third fragment's momentum balances the resultant of the two equal-mass fragments moving perpendicular to each other.
Step 1:Let the equal fragments have mass each and the light fragment mass . The two equal fragments move perpendicular to each other, so their combined momentum has magnitude times one of them.
Step 2:The light fragment carries equal and opposite momentum, so its speed follows by dividing by its mass.
Final answer:
Q25Single correctRay Optics and Optical Instruments
A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 D
Approach:
The lens maker's formula gives the focal length from the refractive index and radii of curvature, and the power is the reciprocal of the focal length in metres.
Step 1:For a biconvex lens cm and cm; substituting these with gives the focal length.
Step 2:The power is the reciprocal of the focal length expressed in metres.
Final answer: D
Q26Single correctMoving Charges and Magnetism
Given below are two statements
Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (Idl) of a current carrying conductor only.
Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q.
In light of above statements choose the most appropriate answer from the options given below
Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (Idl) of a current carrying conductor only.
Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q.
In light of above statements choose the most appropriate answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct and Statement II is incorrect
Approach:
Biot-Savart's law and Coulomb's law are compared on the nature of their respective sources to judge each statement.
Step 1:Biot-Savart's law gives the magnetic field contribution of an infinitesimal current element Idl, so Statement I is correct.
Step 2:In Coulomb's law the source is the charge q, which is a scalar, whereas in Biot-Savart's law the source is the current element Idl, which is a vector. Statement II reverses these natures and is therefore incorrect.
Step 3:With Statement I correct and Statement II incorrect,the corresponding combination is identified.
Final answer: Statement I is correct and Statement II is incorrect
Q27Single correctNuclei
In the given nuclear reaction, the element X is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Conservation of mass number and charge across the positron emission fixes the unknown nuclide X.
Step 1:In positron emission the positron carries charge +1 and mass number 0, while the neutrino carries no charge and no mass number.
Step 2:Equating mass numbers gives the mass number of X.
Step 3:Equating atomic numbers gives the atomic number of X.
Step 4:The nuclide with mass number 22 and atomic number 10 is neon-22.
Final answer:
Q28Single correctUnits and Measurements
Plane angle and solid angle have
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Units but no dimensions
Approach:
Plane angle and solid angle are expressed as ratios of like quantities, which determines whether they carry units and dimensions.
Step 1:Plane angle is the ratio of arc length to radius, both lengths, so its dimensions cancel.
Step 2:Solid angle is the ratio of an area to the square of a radius, so its dimensions also cancel.
Step 3:Although dimensionless, both quantities are assigned distinct units, the radian for plane angle and the steradian for solid angle.
Final answer: Units but no dimensions
Q29Single correctElectrostatic Potential and Capacitance
The angle between the electric lines of force and the equipotential surface is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The relation between the work done moving a charge on an equipotential surface and the electric force fixes the angle between field lines and that surface.
Step 1:An equipotential surface has the same potential everywhere, so moving a charge along it does zero work.
Step 2:Since the field strength and the displacement along the surface are non-zero, the dot product can vanish only when the field is perpendicular to the displacement.
Step 3:Electric lines of force, which point along the field, therefore meet the equipotential surface at a right angle.
Final answer:
Q30Single correctRay Optics and Optical Instruments
A light ray falls on a glass surface of refractive index , at an angle . The angle between the refracted and reflected rays would be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Snell's law gives the refraction angle, after which the reflected and refracted directions are combined about the normal to obtain the angle between them.
Step 1:Applying Snell's law with incidence angle 60 degrees and refractive index root three gives the refraction angle.
Step 2:The reflected ray makes 60 degrees with the normal on the incidence side, while the refracted ray makes 30 degrees with the normal on the transmission side.
Step 3:The angle between the reflected and refracted rays equals 180 degrees minus the two angles each ray makes with the normal.
Final answer:
Q31Single correctSemiconductor Electronics
In the given circuits (a), (b) and (c), the potential drop across the two - junctions are equal in

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both circuits (a) and (c)
Approach:
Each labelled circuit shows two p-n junctions in a series arrangement; the bias direction of the junctions determines whether the same potential drop appears across both.
Step 1:When two identical junctions face the same way and carry the same current, the drop across each is identical, giving equal potential drops.
Step 2:Circuits (a) and (c) place the two junctions in equivalent bias conditions so that the same potential develops across each, whereas circuit (b) does not.
Step 3:The symmetric configurations of (a) and (c) yield equal junction drops simultaneously.
Final answer: Both circuits (a) and (c)
Q32Single correctMechanical Properties of Fluids
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball () as a function of time () is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B
Approach:
A sphere released from rest in a viscous liquid first accelerates because gravity exceeds the opposing viscous drag, but the upward viscous force grows with speed until it balances the net downward force; thereafter the sphere moves at constant terminal speed. The speed therefore rises from zero with a steadily decreasing slope and levels off to a horizontal asymptote, which matches curve D.
Step 1:At the instant of release the speed is zero, so the viscous drag is zero and the acceleration takes its maximum value. The curve starts at the origin with a finite positive slope.
Step 2:As the speed increases the viscous force increases, reducing the net downward force and hence the acceleration. The slope of the speed–time curve decreases with time.
Step 3:When the viscous force balances the effective weight the acceleration vanishes and the speed becomes constant at the terminal value, giving a horizontal asymptote.
Step 4:The curve labelled B is the one that starts at the origin, rises with continuously decreasing slope and flattens to a constant value.
Final answer: B
Q33Single correctCurrent Electricity
Two resistors of resistance, and are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in to that in in a given time is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For resistors in parallel the voltage is common, so the heat dissipated in a given time is compared through the relation between power and resistance at fixed voltage.
Step 1:In a parallel combination the same voltage V appears across both resistors.
Step 2:For a common voltage the heat developed in a given time is inversely proportional to the resistance.
Step 3:Taking the ratio of the heats in the two resistors gives the inverse ratio of their resistances.
Final answer:
Q34Single correctDual Nature of Radiation and Matter
When two monochromatic lights of frequency, and are incident on a photoelectric metal, their stopping potential becomes and respectively. The threshold frequency for this metal is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Einstein's photoelectric equation is written for both frequencies, and the two stopping-potential relations are combined to eliminate the work function and obtain the threshold frequency.
Step 1:For frequency v with stopping potential Vs/2 the photoelectric equation reads as follows.
Step 2:For frequency v/2 with stopping potential Vs the photoelectric equation reads as follows.
Step 3:Multiplying the second equation by one half and equating both expressions for eVs/2 eliminates the stopping potential.
Step 4:Solving the combined relation for the threshold frequency yields its value.
Final answer:
Q35Single correctMechanical Properties of Fluids
If a soap bubble expands, the pressure inside the bubble
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Decreases
Approach:
The excess pressure inside a soap bubble is related to its radius, which determines how the inside pressure changes as the bubble expands.
Step 1:A soap bubble has two surfaces, so its excess pressure over the surroundings is four times the surface tension divided by the radius.
Step 2:When the bubble expands its radius increases, and since the excess pressure is inversely proportional to the radius, this excess pressure falls.
Step 3:With the outside pressure fixed, a smaller excess pressure means the pressure inside the bubble decreases.
Final answer: Decreases
Q36Single correctElectric Charges and Fields
Two point charges and are placed at a distance of , as shown in the figure. The magnitude of electric field intensity at a distance varies as:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The two equal and opposite charges separated by L form an electric dipole, whose far-field intensity follows the standard dipole dependence on distance.
Step 1:Equal and opposite charges +q and -q separated by L constitute an electric dipole of moment p equal to qL.
Step 2:At distances much larger than the separation, the dipole field magnitude is proportional to the dipole moment divided by the cube of the distance.
Step 3:Therefore the field intensity at distance R varies as one over R cubed.
Final answer:
Q37Single correctUnits and Measurements
The area of a rectangular field (in ) of length m and breadth m after rounding off the value for correct significant digits is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The product of the two measurements is computed and then rounded to the number of significant figures set by the least precise factor.
Step 1:Multiplying length and breadth gives the raw area value.
Step 2:The breadth 25 m carries only two significant figures, so the area must be rounded to two significant figures.
Step 3:Rounding 1382.5 to two significant figures gives 1400, written as fourteen hundred.
Final answer:
Q38Single correctSemiconductor Electronics
The truth table for the given logic circuit is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The gates in the drawn circuit are traced for every input pair. The two NAND gates feed an AND gate, so the output is the product of the two NAND outputs.
Step 1:A enters the upper NAND together with B, while an inverter turns A into its complement, which enters the lower NAND together with B.
Step 2:The final gate has no inversion bubble, so it takes the product of the two NAND outputs.
Step 3:Evaluating the complement of B for the four ordered input pairs gives the output column.
Final answer:
Q39Single correctMechanical Properties of Solids
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring.
Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring.
Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions.
In the light of the above statements, choose the most appropriate answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A) is true but (R) is false
Approach:
The Assertion and Reason are each judged against the elastic behaviour of a coil spring and the relative tensile strengths of copper and steel.
Step 1:When a coil spring is stretched, its turns undergo torsion, so the deformation is governed by the shear modulus of the material, making the Assertion true.
Step 2:Steel has a greater tensile strength than copper for the same dimensions, so the claim that copper has more tensile strength than steel is false, making the Reason false.
Step 3:With a true Assertion and a false Reason,the corresponding combination is identified.
Final answer: (A) is true but (R) is false
Q40Single correctMoving Charges and Magnetism
From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with dependence for the outside region.
Approach:
Ampere's circuital law is applied to amperian loops inside and outside the wire to determine how the magnetic field depends on distance in each region.
Step 1:Inside the wire the enclosed current grows with the square of the loop radius, so dividing by the loop circumference gives a field that rises linearly with distance.
Step 2:Outside the wire the full current is enclosed, so the field falls inversely with distance.
Step 3:The field therefore increases linearly up to the surface and then decreases with a one over r dependence.
Final answer: A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with dependence for the outside region.
Q41Single correctAlternating Current
A series LCR circuit with inductance H, capacitance , resistance is connected to an ac source of voltage, volt. If the resonant frequency of the LCR circuit is and the frequency of the ac source is , then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 Hz
Approach:
The source frequency is read from the supply expression and the resonant frequency from the LC values, then the two are compared.
Step 1:The supply voltage has angular frequency 100 rad per second, so the source frequency is 100 divided by 2 pi.
Step 2:The resonant frequency follows from the inductance and capacitance values.
Step 3:Both the source frequency and the resonant frequency equal 50 over pi hertz.
Final answer: Hz
Q42Single correctGravitation
Choose the correct answer from the options given below
| List-I | List-II |
|---|---|
| (a). Gravitational constant (G) | (i). |
| (b). Gravitational potential energy | (ii). |
| (c). Gravitational potential | (iii). |
| (d). Gravitational intensity | (iv). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)
Approach:
The dimensional formula of each gravitational quantity in List I is derived and matched to the corresponding entry in List II.
Step 1:The gravitational constant from Newton's law of gravitation has dimensions of M to the minus one, L cubed, T to the minus two, matching entry (ii).
Step 2:Gravitational potential energy is an energy, with dimensions M, L squared, T to the minus two, matching entry (iv).
Step 3:Gravitational potential is energy per unit mass, with dimensions L squared, T to the minus two, matching entry (i).
Step 4:Gravitational intensity is a field strength equal to acceleration, with dimensions L, T to the minus two, matching entry (iii).
Final answer: (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)
Q43Single correctOscillations
Two pendulums of length cm and cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The two pendulums realign when the shorter completes exactly one more oscillation than the longer, so the ratio of their periods, fixed by their lengths, gives the required count.
Step 1:The ratio of the periods equals the square root of the ratio of the lengths.
Step 2:While the shorter pendulum makes 11 oscillations, the longer pendulum makes 10, so the shorter gains one full oscillation and the two return to the same phase.
Step 3:The minimum number of vibrations of the shorter pendulum is therefore 11.
Final answer:
Q44Single correctElectromagnetic Induction
A big circular coil of turns and average radius m is rotating about its horizontal diameter at rad . If the vertical component of earth's magnetic field at that place is T and electrical resistance of the coil is , then the maximum induced current in the coil will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 A
Approach:
The peak emf of the rotating coil is obtained from the number of turns, the field, the coil area and the angular speed, and dividing by the resistance gives the maximum current.
Step 1:The coil area follows from its radius.
Step 2:Substituting the values into the peak emf expression gives its magnitude.
Step 3:Dividing the peak emf by the coil resistance gives the maximum induced current.
Final answer: A
Q45Single correctElectrostatic Potential and Capacitance
A capacitor of capacitance pF is charged fully by V battery as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance pF as shown in figure (b). The electrostatic energy stored by the system (b) is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 J
Approach:
The charge on the first capacitor is conserved and shared with the equal second capacitor, after which the energy of the combined system is computed at the common voltage.
Step 1:The initial charge on the first capacitor follows from its capacitance and the battery voltage.
Step 2:After connection the charge is shared by two equal capacitors in parallel, so the common voltage is half the original.
Step 3:The energy stored by the parallel combination follows from the total capacitance and the common voltage.
Final answer: J
Q46Single correctNuclei
A nucleus of mass number splits into two nuclei having mass number and . The ratio of radius of two daughter nuclei respectively is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Nuclear radius scales with the cube root of mass number, so the ratio of the two daughter radii is the ratio of the cube roots of their mass numbers.
Step 1:The ratio of the two radii equals the ratio of the cube roots of their mass numbers.
Step 2:Evaluating the cube roots gives five for 125 and four for 64.
Step 3:The ratio of the daughter radii is therefore five to four.
Final answer:
Q47Single correctCurrent Electricity
A wheatstone bridge is used to determine the value of unknown resistance by adjusting the variable resistance as shown in the figure. For the most precise measurement of , the resistances and

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Should be approximately equal and are small
Approach:
The balance condition of the Wheatstone bridge and the requirement of maximum galvanometer sensitivity together fix the desirable values of P and Q.
Step 1:At balance the ratio of P to Q equals the ratio of X to Y, so the bridge can be balanced for the unknown.
Step 2:The bridge is most sensitive when all four arms are of the same order of magnitude, which calls for P and Q to be nearly equal and not large.
Step 3:Therefore the most precise measurement is obtained when P and Q are nearly equal and small.
Final answer: Should be approximately equal and are small
Q48Single correctKinetic Theory
The volume occupied by the molecules contained in kg water at STP, if the intermolecular forces vanish away is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
If intermolecular forces vanish the water behaves as an ideal gas, so the number of moles is found from the mass and each mole occupies the molar volume at STP.
Step 1:The number of moles in 4.5 kg of water follows from its molar mass of 18 grams per mole.
Step 2:Treating the molecules as an ideal gas at STP, each mole occupies 22.4 litre.
Step 3:Converting litre to cubic metre gives the occupied volume.
Final answer:
Q49Single correctMotion in a Plane
A ball is projected with a velocity, m, at an angle of with the vertical direction. Its speed at the highest point of its trajectory will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
At the highest point only the horizontal velocity component survives, so the angle measured from the vertical is used to resolve the projection velocity.
Step 1:The angle of 60 degrees is measured from the vertical, so the horizontal component is the projection velocity times the sine of 60 degrees.
Step 2:Evaluating the sine gives the horizontal component.
Step 3:At the highest point the vertical velocity is zero, so the speed equals the horizontal component.
Final answer: m
Q50Single correctRay Optics and Optical Instruments
Two transparent media A and B are separated by a plane boundary. The speed of light in those media are m/s and m/s, respectively. The critical angle for a ray of light for these two media is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The refractive indices of the two media follow from the speeds of light, and the critical angle is obtained from the ratio of the speeds for light passing from the denser to the rarer medium.
Step 1:Medium A has the smaller speed of light, so it is the denser medium, while medium B is the rarer one.
Step 2:The sine of the critical angle equals the ratio of the speed in the denser medium to that in the rarer medium.
Step 3:The critical angle is therefore the inverse sine of 0.750.
Final answer:
Chemistry50 questions
Q51Single correctThe s-Block and p-Block Elements (Group trends)
Gadolinium has a low value of third ionisation enthalpy because of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2high exchange enthalpy
Approach:
The third ionisation enthalpy of gadolinium is examined in terms of the stability of the resulting electronic configuration.
Step 1:Gadolinium has the ground state configuration .
Step 2:Removal of the third electron from leaves , an exactly half-filled f sub-shell that carries large exchange energy and extra stability.
Step 3:Because the product ion gains exchange stabilisation from the half-filled set, the energy required for the third ionisation falls.
Final answer: high exchange enthalpy
Q52Single correctStates of Matter
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture
(where = partial pressure of gas, = mole fraction of gas in gaseous mixture, = pressure of gas in pure state)
(where = partial pressure of gas, = mole fraction of gas in gaseous mixture, = pressure of gas in pure state)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each relation is tested against Dalton's law, which states that the total pressure of a non-reacting gaseous mixture equals the sum of partial pressures.
Step 1:The relation expresses total pressure as the sum of individual partial pressures, the direct statement of Dalton's law.
Step 2:Writing each partial pressure as simply substitutes the ideal gas equation into that sum, which is valid.
Step 3:The relation gives the partial pressure of a component as its mole fraction in the mixture times the total pressure, which is the standard corollary of Dalton's law.
Step 4:The relation equates partial pressure to mole fraction times the pure-state pressure; this is Raoult's law for solutions, not Dalton's law of partial pressures.
Final answer:
Q53Single correctThe Solid State
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A):
In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells.
Reason (R):
In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality.
In the light of the above statements, choose the most appropriate answer from the options given below:
Assertion (A):
In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells.
Reason (R):
In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Approach:
Each statement is evaluated independently and then the explanatory link between them is checked.
Step 1:An ionic solid that stays electrically neutral while some cations are absent from their normal sites describes the Frenkel defect, where the missing cations sit in interstitial positions. The solid as a whole keeps equal positive and negative charge, so the assertion is correct.
Step 2:The reason correctly defines the Frenkel defect as dislocation of a cation from its lattice site to an interstitial site with overall neutrality preserved.
Step 3:The assertion centres on neutrality being maintained despite cation absence from lattice sites, while the reason states the mechanism of the Frenkel defect; both are true but the reason does not by itself explain why neutrality is preserved, so it is not the correct explanation.
Final answer: Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Q54Single correctEquilibrium
The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is
[Given of ]
[Given of ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 15.57
Approach:
The mixture of acetic acid and sodium acetate forms an acidic buffer, so the Henderson-Hasselbalch equation is applied.
Step 1:Equal volumes are mixed, so the ratio of salt to acid concentrations equals the ratio of their molarities.
Step 2:Substituting into the Henderson-Hasselbalch equation gives the pH.
Step 3:Evaluating the logarithm completes the calculation.
Final answer: 5.57
Q55Single correctThe s-Block Elements
Identify the incorrect statement from the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The oxidation number of K in is +4.
Approach:
Each statement about alkali metals is checked against known group properties, and the false one is identified.
Step 1:Alkali metals react vigorously with water to give the corresponding hydroxides and hydrogen, so statement 1 is correct.
Step 2:Potassium superoxide contains the superoxide ion . With potassium fixed at +1, the oxidation state requirement makes the claim of +4 for potassium false.
Step 3:Down the alkali metal group atomic size increases and the outer electron is held less tightly, so ionisation enthalpy decreases, making statement 3 correct.
Step 4:Lithium has the most negative standard electrode potential owing to its high hydration enthalpy, making it the strongest reducing agent, so statement 4 is correct.
Final answer: The oxidation number of K in is +4.
Q56Single correctAlcohols, Phenols and Ethers
Given below are two statements
Statement I:
The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group.
Statement II:
o-nitrophenol, m-nitrophenol and p-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I:
The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group.
Statement II:
o-nitrophenol, m-nitrophenol and p-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
The acidity effect of the nitro group on phenol and the dependence of acidity on the position of substitution are evaluated.
Step 1:The nitro group is strongly electron withdrawing and stabilises the phenoxide ion, so a nitrophenol is more acidic than phenol; Statement I is correct.
Step 2:Acid strength depends on the position of the nitro group. The ortho and para isomers stabilise the phenoxide by resonance and are more acidic, while the meta isomer relies only on the inductive effect, so the three isomers differ in acidity.
Final answer: Statement I is correct but Statement II is incorrect.
Q57Single correctSome Basic Concepts of Chemistry
What mass of 95% pure will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
[Calculate upto second place of decimal point]
[Calculate upto second place of decimal point]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 21.32 g
Approach:
The moles of HCl are found, converted to moles of pure calcium carbonate by stoichiometry, and the mass is corrected for 95% purity.
Step 1:Moles of HCl are obtained from its molarity and volume.
Step 2:The reaction consumes 2 mol HCl per mol , so the moles of carbonate are half the moles of acid.
Step 3:Mass of pure carbonate follows from its molar mass of 100 g/mol.
Step 4:Since the sample is only 95% pure, the required impure mass is the pure mass divided by 0.95.
Final answer: 1.32 g
Q58Single correctClassification of Elements and Periodicity
The IUPAC name of an element with atomic number 119 is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1ununennium
Approach:
The systematic IUPAC name is built from the numerical roots of each digit of the atomic number followed by the suffix -ium.
Step 1:The atomic number 119 is split into digits 1, 1 and 9 with roots un, un and enn.
Step 2:Joining the roots and adding the suffix -ium gives the systematic name, with the printed solution stating the IUPAC name of element 119 as ununennium.
Final answer: ununennium
Q59Single correctThe p-Block Elements (Group 14)
Choose the correct statement:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Diamond is hybridised and graphite is hybridized.
Approach:
The bonding and structural features of diamond and graphite are compared to find the correct statement.
Step 1:Diamond forms a three dimensional rigid network, whereas graphite has a layered structure, so statement 1 is false.
Step 2:Both diamond and graphite are covalent allotropes of carbon, so the claim that graphite is ionic is false.
Step 3:Each carbon in diamond is hybridised with four single bonds, while in graphite each carbon is hybridised with delocalised electrons, making statement 3 correct.
Step 4:Graphite acts as a dry lubricant because of its slippery layers, but diamond is hard and does not, so statement 4 is false.
Final answer: Diamond is hybridised and graphite is hybridized.
Q60Single correctSurface Chemistry
Given below are two statements
Statement I:
In the coagulation of a negative sol, the flocculating power of the three given ions is in the order
Statement II:
In the coagulation of a positive sol, the flocculating power of the three given salts is in the order
In the light of the above statements, choose the most appropriate answer from the options given below
Statement I:
In the coagulation of a negative sol, the flocculating power of the three given ions is in the order
Statement II:
In the coagulation of a positive sol, the flocculating power of the three given salts is in the order
In the light of the above statements, choose the most appropriate answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
The Hardy-Schulze rule, which links coagulating power to the charge of the oppositely charged ion, is applied to each statement.
Step 1:A negative sol is coagulated by cations, and flocculating power increases with cation charge, giving the order ; Statement I is correct.
Step 2:A positive sol is coagulated by anions, so power increases with anion charge as , making the order ; the reversed order in Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect.
Q61Single correctThermodynamics
Which of the following p-V curve represents maximum work done?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A falling p-V curve labelled Isothermal that runs across the full width of the volume axis, from the smallest to the largest volume shown
Approach:
The magnitude of work in an expansion equals the area under the path on a p-V diagram, so the path sweeping the greatest area between its end volumes does the most work.
Step 1:For a gas expanding from one volume to another, the magnitude of the work equals the area beneath the path drawn on the p-V plane.
Step 2:A vertical line has no change in volume and therefore sweeps no area, so it does no work at all; a closed loop returns to its starting state and its net work is only the small area it encloses.
Step 3:Of the two expansions, one falls across the whole width of the volume axis while the other stops after a small increase in volume, so the first sweeps the larger area and does the greater work.
Final answer: A falling p-V curve labelled Isothermal that runs across the full width of the volume axis, from the smallest to the largest volume shown
Q62Single correctAmines
Given below are two statements
Statement I:
Primary aliphatic amines react with to give unstable diazonium salts.
Statement II:
Primary aromatic amines react with to form diazonium salts which are stable even above 300 K.
In the light of the above statements, choose the most appropriate answer from the options given below
Statement I:
Primary aliphatic amines react with to give unstable diazonium salts.
Statement II:
Primary aromatic amines react with to form diazonium salts which are stable even above 300 K.
In the light of the above statements, choose the most appropriate answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
The stability of diazonium salts formed from aliphatic and aromatic primary amines with nitrous acid is examined.
Step 1:Primary aliphatic amines react with nitrous acid to give aliphatic diazonium salts that are unstable and decompose readily, so Statement I is correct.
Step 2:Primary aromatic amines form aryldiazonium salts that are stabilised by resonance with the ring but only at low temperature, typically below 278 K; they decompose well below 300 K, so the claim of stability above 300 K is false.
Final answer: Statement I is correct but Statement II is incorrect.
Q63Single correctChemical Bonding and Molecular Structure
Which amongst the following is incorrect statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 ion is diamagnetic
Approach:
Each statement is tested against molecular orbital theory, focusing on bond orders, electron counts and magnetic behaviour.
Step 1:The carbon molecule has its highest-energy electrons in two degenerate bonding orbitals holding four electrons, so statement 2 is correct.
Step 2:The dihydrogen cation has a single electron in the bonding orbital, so statement 3 is correct.
Step 3:The dioxygenyl ion has 15 electrons and retains one unpaired electron in its orbital, so it is paramagnetic, not diamagnetic; statement 4 is the incorrect one.
Final answer: ion is diamagnetic
Q64Single correctAldehydes, Ketones and Carboxylic Acids
What is Y in the above reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The mechanism of carboxylic acid formation from a Grignard reagent and carbon dioxide is followed to identify the intermediate Y.
Step 1:The Grignard reagent adds across one carbon-oxygen bond of carbon dioxide, attaching the alkyl group to carbon and the magnesium-halide unit to oxygen.
Step 2:This intermediate magnesium carboxylate Y is then hydrolysed by acidic water to release the carboxylic acid.
Final answer:
Q65Single correctPolymers
Which statement regarding polymers is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Thermosetting polymers are reusable
Approach:
Each statement on polymer classes is checked, with attention to whether thermosetting polymers can be reused.
Step 1:Elastomers have coiled chains held by weak intermolecular forces that allow stretching, so statement 1 is correct.
Step 2:Fibres have strong intermolecular forces and close packing giving high tensile strength, so statement 2 is correct.
Step 3:Thermoplastic polymers soften on heating and harden on cooling and can be remoulded, so statement 3 is correct.
Step 4:Thermosetting polymers form a permanent cross-linked network on heating and cannot be remelted or reused, so the claim that they are reusable is false.
Final answer: Thermosetting polymers are reusable
Q66Single correctElectrochemistry
Given below are half cell reactions:
Will the permanganate ion, liberate from water in the presence of an acid?
Will the permanganate ion, liberate from water in the presence of an acid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Yes, because
Approach:
The overall cell potential is obtained by combining the two half-reactions so that permanganate is reduced and water is oxidised; a positive value indicates a feasible reaction.
Step 1:Permanganate acts as the oxidising agent and is reduced at the cathode with standard potential , while water is oxidised to oxygen at the anode with standard potential .
Step 2:Subtracting the anode potential from the cathode potential gives the overall cell potential.
Step 3:A positive cell potential means the reaction is spontaneous, so permanganate does liberate oxygen from water in acidic medium.
Final answer: Yes, because
Q67Single correctOrganic Chemistry - Some Basic Principles and Techniques
The Kjeldahl's method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Aniline,
Approach:
The applicability of the Kjeldahl method is judged by whether the nitrogen can be converted quantitatively to ammonium sulphate on digestion.
Step 1:The Kjeldahl method works when nitrogen is bonded to carbon or hydrogen and can be converted to ammonium sulphate during digestion with concentrated sulphuric acid.
Step 2:Nitrogen present in a nitro group, in a ring as in pyridine, or in an azo linkage is not quantitatively converted, so those structures fail. Aniline carries an group whose nitrogen is converted to ammonia, making it suitable.
Final answer: Aniline,
Q68Single correctBiomolecules
The incorrect statement regarding enzymes is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Enzymes are polysaccharides.
Approach:
Each property is checked against the known chemical nature and behaviour of enzymes.
Step 1:Enzymes are biological catalysts, so statement 1 is correct.
Step 2:Like chemical catalysts, enzymes lower the activation energy of biochemical reactions, so statement 2 is correct.
Step 3:Enzymes are globular proteins, not polysaccharides, so this statement is false.
Step 4:Enzymes are highly specific to a particular substrate and reaction, so statement 4 is correct.
Final answer: Enzymes are polysaccharides.
Q69Single correctCoordination Compounds
The IUPAC name of the complex- is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4diaquasilver(I) dicyanidoargentate(I)
Approach:
The complex contains separate cationic and anionic coordination spheres, each named by IUPAC rules with ligands in alphabetical order and the oxidation state of silver determined in each.
Step 1:In the cation two neutral aqua ligands give silver an oxidation state of , named diaquasilver(I).
Step 2:In the anion two cyanide ligands give silver an oxidation state of ; as the anionic sphere the metal takes the suffix -ate, named dicyanidoargentate(I).
Step 3:The cation is named before the anion to give the complete name.
Final answer: diaquasilver(I) dicyanidoargentate(I)
Q70Single correctChemistry in Everyday Life
Choose the correct answer from the options given below :
| List-I (Drug class) | List-II (Drug molecule) |
|---|---|
| (a). Antacids | (i). Salvarsan |
| (b). Antihistamines | (ii). Morphine |
| (c). Analgesics | (iii). Cimetidine |
| (d). Antimicrobials | (iv). Seldane |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
Approach:
Each drug class in List-I is matched to its representative molecule in List-II based on therapeutic action.
Step 1:Cimetidine is a histamine blocker used to reduce stomach acid, so antacids match (iii).
Step 2:Seldane (terfenadine) is an antihistamine, so antihistamines match (iv).
Step 3:Morphine is a narcotic analgesic, so analgesics match (ii).
Step 4:Salvarsan is an antimicrobial used against syphilis, so antimicrobials match (i).
Final answer: (a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
Q71Single correctThe p-Block Elements (Group 17 and 18)
Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The number of lone pairs on the central atom is counted for each species, since more lone pairs increase lone-pair to lone-pair repulsion.
Step 1:In chlorine has two lone pairs, in iodine has one lone pair, and in silicon has no lone pair.
Step 2:In xenon carries three lone pairs in its trigonal bipyramidal arrangement, the largest count, giving the maximum lone-pair to lone-pair repulsion.
Final answer:
Q72Single correctElectrochemistry
At 298 K, the standard electrode potentials of / Cu, / Zn, / Fe and / Ag are 0.34 V, V, V and 0.80 V, respectively.
On the basis of standard electrode potential, predict which of the following reaction cannot occur?
On the basis of standard electrode potential, predict which of the following reaction cannot occur?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A displacement reaction occurs only if the added metal is a stronger reducing agent, that is has a more negative standard potential, than the metal in solution; feasibility is confirmed by a positive overall cell potential.
Step 1:Zinc at and iron at both lie below copper at , so each displaces copper, making reactions 1 and 2 feasible. Zinc lies below iron, so it displaces iron in reaction 3, which is also feasible.
Step 2:Silver at is higher than copper at , so silver cannot displace copper from solution; the cell potential for reaction 4 is , which is negative.
Step 3:A negative cell potential means the reaction is non-spontaneous, so reaction 4 cannot occur.
Final answer:
Q73Single correctThe d- and f-Block Elements
Identify the incorrect statement from the following.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The shapes of , and orbitals are similar to each other; and and are similar to each other.
Approach:
Each statement on the size, shape and energy of d orbitals is examined, with focus on the shapes of the five d orbitals.
Step 1:Orbitals of higher principal quantum number are larger, so the orbitals differ in size from the orbitals; statement 1 is correct.
Step 2:Orbitals with the same azimuthal quantum number share the same shape across shells, so the orbitals match the orbitals in shape; statement 2 is correct.
Step 3:In a free atom the five orbitals are degenerate, equal in energy; statement 3 is correct.
Step 4:The three orbitals , and have the same four-lobed shape between axes, but lies along axes and has a distinct ring-and-lobe shape, so these two are not similar to each other; statement 4 is incorrect.
Final answer: The shapes of , and orbitals are similar to each other; and and are similar to each other.
Q74Single correctSolutions
In one molal solution that contains 0.5 mole of a solute, there is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2500 g of solvent
Approach:
Molality is defined as moles of solute per kilogram of solvent, so the mass of solvent is found from the given molality and moles of solute.
Step 1:Rearranging the molality definition gives the solvent mass as moles of solute divided by molality.
Step 2:Converting kilograms to grams gives the answer.
Final answer: 500 g of solvent
Q75Single correctThe p-Block Elements (Group 17)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): ICl is more reactive than .
Reason (R): I-Cl bond is weaker than I-I bond.
In the light of the above statements, choose the most appropriate answer from the options given below:
Assertion (A): ICl is more reactive than .
Reason (R): I-Cl bond is weaker than I-I bond.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both (A) and (R) are correct and (R) is the correct explanation of (A).
Approach:
The reactivity of the interhalogen ICl relative to iodine is linked to the relative strengths of the I-Cl and I-I bonds.
Step 1:Interhalogen compounds are generally more reactive than the halogens, except fluorine, so ICl is more reactive than iodine; the assertion is correct.
Step 2:The I-Cl bond between two different atoms is weaker than the I-I bond, so the reason is correct.
Step 3:A weaker I-Cl bond breaks more readily, which is the direct cause of the higher reactivity of ICl, so the reason correctly explains the assertion.
Final answer: Both (A) and (R) are correct and (R) is the correct explanation of (A).
Q76Single correctHydrocarbons / Aromaticity
Which compound amongst the following is not an aromatic compound?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cyclohepta-1,3,5-triene
Approach:
An aromatic compound must be cyclic, planar, fully conjugated and carry electrons in the ring. Each drawn species is tested against these requirements.
Step 1:The cyclopentadienyl anion is a planar five-membered ring with a fully conjugated system of 6 electrons, satisfying the rule with .
Step 2:The cycloheptatrienyl (tropylium) cation is a planar seven-membered ring carrying 6 electrons, and the cyclopropenyl cation is a planar three-membered ring carrying 2 electrons with .
Step 3:Cyclohepta-1,3,5-triene has three double bonds but also one carbon bearing two hydrogens. That saturated centre breaks the ring conjugation, so the compound is not aromatic.
Final answer: Cyclohepta-1,3,5-triene
Q77Single correctThe p-Block Elements (Group 16)
Given below are two statements
Statement I
The boiling points of the following hydrides of group 16 elements increases in the order –
Statement II
The boiling points of these hydrides increase with increase in molar mass.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I
The boiling points of the following hydrides of group 16 elements increases in the order –
Statement II
The boiling points of these hydrides increase with increase in molar mass.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are incorrect
Approach:
Each statement is tested against the measured boiling points of the group 16 hydrides.
Step 1:The boiling points are 373 K for water, 213 K for hydrogen sulphide, 232 K for hydrogen selenide and 271 K for hydrogen telluride, so the true order is .
Step 2:Statement I places water lowest in the series, which contradicts the measured order, so Statement I is incorrect.
Step 3:Statement II claims the boiling points of these hydrides rise with molar mass. Water is the lightest of the four yet boils highest, because its small, highly electronegative oxygen supports strong O-H.O hydrogen bonding. The molar-mass rule therefore fails for the series as a whole, so Statement II is also incorrect.
Final answer: Both Statement I and Statement II are incorrect
Q78Single correctThe s-Block Elements
Match List-I with List-II Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| a. Li | i. absorbent for carbon dioxide |
| b. Na | ii. electrochemical cells |
| c. KOH | iii. coolant in fast breeder reactors |
| d. Cs | iv. photoelectric cell |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)
Approach:
Each alkali metal entry is paired with its characteristic application based on standard properties.
Step 1:Lithium is used in electrochemical cells because of its high standard electrode potential and light weight, matching (ii).
Step 2:Liquid sodium serves as a coolant in fast breeder reactors due to its wide liquid range and high thermal conductivity, matching (iii).
Step 3:Potassium hydroxide absorbs carbon dioxide, matching (i), while caesium is used in photoelectric cells owing to its very low ionisation enthalpy, matching (iv).
Final answer: (a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)
Q79Single correctHaloalkanes and Haloarenes
Which of the following sequence of reactions is suitable to synthesize chlorobenzene?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Benzene, , anhydrous
Approach:
Chlorobenzene forms by electrophilic aromatic substitution of benzene with chlorine using a Lewis acid catalyst. Each route is checked for feasibility.
Step 1:Treatment of benzene with chlorine in the presence of anhydrous ferric chloride generates the chloronium electrophile and substitutes a ring hydrogen with chlorine.
Step 2:Direct reaction of benzene or aniline with hydrochloric acid does not introduce chlorine onto the aromatic ring under the stated conditions.
Step 3:The phenol diazonium route is not applicable because phenol does not form a stable diazonium salt, so only the first sequence gives chlorobenzene.
Final answer: Benzene, , anhydrous
Q80Single correctAldehydes, Ketones and Carboxylic Acids
Given below are two statements :
Statement I : The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions.
Statement II : The boiling points of aldehydes and ketones are lower than that the alcohols of similar molecular masses due to the absence of H-bonding.
In the light of the above statements, choose the most appropriate answer from the given below
Statement I : The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions.
Statement II : The boiling points of aldehydes and ketones are lower than that the alcohols of similar molecular masses due to the absence of H-bonding.
In the light of the above statements, choose the most appropriate answer from the given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
The relative boiling points of aldehydes, ketones, hydrocarbons, and alcohols are governed by their intermolecular forces. Each statement is compared with these forces.
Step 1:The polar carbonyl group produces dipole-dipole attractions that are absent in non-polar hydrocarbons, raising the boiling points of aldehydes and ketones above hydrocarbons of similar mass, so Statement I is correct.
Step 2:Alcohols form intermolecular hydrogen bonds, which are stronger than dipole-dipole forces, so alcohols boil higher than aldehydes and ketones of similar mass, confirming Statement II.
Step 3:Both statements correctly describe the trend in intermolecular forces.
Final answer: Both Statement I and Statement II are correct
Q81Single correctAldehydes, Ketones and Carboxylic Acids
Match List-I with List-II.
| List-I | List-II |
|---|---|
| a. Cyanohydrin | i. |
| b. Acetal | ii. |
| c. Schiff's base | iii. alcohol |
| d. Oxime | iv. HCN |
Choose the correct answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
Approach:
Each named product is matched with the reagent that adds across or condenses with the carbonyl group.
Step 1:Hydrogen cyanide adds to a carbonyl to give a cyanohydrin, so (a) pairs with (iv).
Step 2:Two equivalents of alcohol convert a carbonyl into an acetal, so (b) pairs with (iii).
Step 3:Primary amines give Schiff's bases and hydroxylamine gives oximes, so (c) pairs with (ii) and (d) pairs with (i).
Final answer: (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
Q82Single correctHaloalkanes and Haloarenes
The incorrect statement regarding chirality is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Enantiomers are superimposable mirror images on each other
Approach:
Each statement about chirality and substitution stereochemistry is tested against the definitions of enantiomers and the mechanisms of nucleophilic substitution.
Step 1:The pathway proceeds through a planar carbocation that is attacked from both faces, giving equal amounts of the two enantiomers, so statement 1 is correct.
Step 2:The pathway proceeds by backside attack, inverting the configuration at a chiral reactive centre, so statement 2 is correct, and a racemic mixture has zero net rotation, so statement 4 is correct.
Step 3:Enantiomers are non-superimposable mirror images, so the claim that they are superimposable is false.
Final answer: Enantiomers are superimposable mirror images on each other
Q83Single correctChemical Bonding and Molecular Structure
Match List-I with List-II.
| List-I | List-II |
|---|---|
| a. | i. Electron precise |
| b. | ii. Electron deficient |
| c. | iii. Electron rich |
| d. HF | iv. Ionic |
Choose the correct answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
Approach:
Each hydride is classified by comparing its available bonding electrons with the number of bonds it forms.
Step 1:Magnesium hydride is a saline hydride containing the hydride ion, making it ionic, so (a) pairs with (iv).
Step 2:Germane has exactly the number of electrons needed for its covalent bonds, making it electron precise, so (b) pairs with (i).
Step 3:Diborane lacks sufficient electrons for normal two-centre bonds, making it electron deficient, while hydrogen fluoride has lone pairs in excess, making it electron rich, so (c) pairs with (ii) and (d) pairs with (iii).
Final answer: (a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
Q84Single correctThe p-Block Elements (Group 13)
Which of the following statement is not correct about diborane?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both the Boron atoms are hybridised.
Approach:
The bonding picture of diborane is reviewed and each statement is compared with the accepted structure.
Step 1:Diborane has two bridging hydrogens involved in two three-centre two-electron bonds and four terminal B-H bonds that are ordinary two-centre two-electron bonds, so statements 1 and 2 are correct.
Step 2:The four terminal hydrogens and the two boron atoms lie in one plane while the two bridging hydrogens lie above and below it, so statement 3 is correct.
Step 3:Each boron is hybridised, not , so statement 4 is incorrect.
Final answer: Both the Boron atoms are hybridised.
Q85Single correctChemical Kinetics
The given graph is a representation of kinetics of a reaction.
The y and x axes for zero and first order reactions, respectively are
The y and x axes for zero and first order reactions, respectively are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3zero order (y = rate and x = concentration), first order (y = and x = concentration)
Approach:
The printed graph is a horizontal line at constant temperature T, meaning the plotted quantity is independent of the variable on the x-axis. Each axis assignment is tested for this behaviour.
Step 1:For a zero-order reaction the rate equals the rate constant and does not depend on concentration, giving a horizontal line when rate is plotted against concentration.
Step 2:For a first-order reaction the half-life is independent of the starting concentration, giving a horizontal line when half-life is plotted against concentration.
Step 3:Only the assignment in which both quantities stay constant matches the horizontal graph.
Final answer: zero order (y = rate and x = concentration), first order (y = and x = concentration)
Q86Single correctGeneral Principles and Processes of Isolation of Elements
Match List-I with List-II.
List-I (Ores): (a) Haematite (b) Magnetite (c) Calamine (d) Kaolinite
List-II (Composition): (i) (ii) (iii) (iv)
Choose the correct answer from the options given below:
List-I (Ores): (a) Haematite (b) Magnetite (c) Calamine (d) Kaolinite
List-II (Composition): (i) (ii) (iii) (iv)
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Approach:
Each ore is matched with its standard chemical composition. Haematite and magnetite are iron oxides, calamine is a zinc carbonate ore, and kaolinite is an aluminosilicate clay mineral. Pairing these gives a unique combination.
Step 1:Haematite is the common iron(III) oxide ore, so (a) pairs with (iii).
Step 2:Magnetite is the mixed iron oxide (ferrous-ferric), so (b) pairs with (i).
Step 3:Calamine is a zinc carbonate ore, so (c) pairs with (ii).
Step 4:Kaolinite is a hydrated aluminosilicate clay mineral, so (d) pairs with (iv).
Final answer: (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Q87Single correctStates of Matter / Thermodynamics
A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume gas is behaving ideally). The pressure inside the flask in bar is (Given R = 0.0831 L bar mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 44.9
Approach:
The ideal gas equation relates pressure to the amount of gas, temperature, and volume. The number of moles is found from the mass, then the pressure is computed.
Step 1:Dividing the mass of oxygen by its molar mass gives the amount of gas.
Step 2:The temperature is converted to kelvin.
Step 3:Substituting the values into the ideal gas equation gives the pressure.
Final answer: 4.9
Q88Single correctChemical Kinetics
For a first order reaction A Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in mi is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.9212
Approach:
The integrated first-order rate law connects the rate constant to the initial and final concentrations over the elapsed time.
Step 1:The concentration ratio between the start and end of the interval is found.
Step 2:The logarithm of the ratio equals 2.
Step 3:Substituting the time and logarithm gives the rate constant.
Final answer: 0.9212
Q89Single correctCoordination Compounds
The order of energy absorbed which is responsible for the color of complexes
(A)
(B) and
(C)
is
(A)
(B) and
(C)
is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The energy absorbed that produces colour equals the crystal field splitting energy, which grows with the field strength of the ligand set. Ethylenediamine is a stronger-field ligand than water, so the complex with the most en ligands has the largest splitting and absorbs the highest energy. Counting en donors in each complex orders the splitting and hence the energy absorbed.
Step 1:In the spectrochemical series ethylenediamine lies above water, so replacing water by en raises the octahedral splitting and the energy of the absorbed light.
Step 2:Complex (C) has three bidentate en ligands and no water, giving the strongest field and the largest splitting.
Step 3:Complex (A) has two en and two water donors, an intermediate field, while complex (B) has one en and four water donors, the weakest field of the three.
Step 4:Ranking the splitting from strongest to weakest field gives the order of absorbed energy as (C) greater than (A) greater than (B).
Final answer:
Q90Single correctEquilibrium
for the above reaction at 298 K, is found to be . If the concentration of at equilibrium is 0.040 M then concentration of in M is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The equilibrium constant expression links the ozone and oxygen concentrations. Rearranging it gives the ozone concentration.
Step 1:Rearranging the equilibrium expression isolates the square of the ozone concentration.
Step 2:Substituting the equilibrium values gives the square of the ozone concentration.
Step 3:Taking the square root gives the ozone concentration.
Final answer:
Q91Single correctElectrochemistry
Find the emf of the cell in which the following reaction takes place at 298 K
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.9615 V
Approach:
The Nernst equation gives the cell emf from the standard potential and the reaction quotient. The number of electrons transferred is two.
Step 1:The reaction quotient uses the nickel ion concentration over the square of the silver ion concentration.
Step 2:The logarithm of the reaction quotient equals 3, and two electrons are transferred.
Step 3:Substituting into the Nernst equation gives the emf.
Final answer: 0.9615 V
Q92Single correctAldehydes, Ketones and Carboxylic Acids
Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 25-Ethyl-4-methylhept-4-en-3-one
Approach:
A crossed aldol condensation between acetone and 2-pentanone gives, after dehydration, the enones formed from every available enolate-carbonyl pairing. The listed products are checked against those pairings.
Step 1:Acetone condensing with itself and losing water gives 4-methylpent-3-en-2-one.
Step 2:The acetone enolate adding to the carbonyl of 2-pentanone and losing water gives 4-methylhept-3-en-2-one, while the C-3 enolate of 2-pentanone adding to acetone gives 3-ethyl-4-methylpent-3-en-2-one.
Step 3:5-Ethyl-4-methylhept-4-en-3-one is an ethyl ketone, so its carbonyl fragment would have to come from pentan-3-one. Neither acetone nor 2-pentanone can supply that fragment, and every enolate available from the two reactants keeps the carbonyl as a methyl ketone.
Final answer: 5-Ethyl-4-methylhept-4-en-3-one
Q93Single correctAlcohols, Phenols and Ethers / Nomenclature
The correct IUPAC name of the compound is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11-bromo-5-chloro-4-methylhexan-3-ol
Approach:
The principal chain and the locants are chosen so that the hydroxyl group, which is the principal characteristic group, gets the lowest possible number, then substituents are listed alphabetically.
Step 1:The longest chain bearing the hydroxyl group is a six-carbon chain, making the parent a hexan-ol.
Step 2:Numbering to give the hydroxyl group the lowest locant places the OH at carbon 3, with bromine at carbon 1, the methyl branch at carbon 4, and chlorine at carbon 5.
Step 3:Listing the substituents alphabetically gives bromo before chloro before methyl, yielding the full name.
Final answer: 1-bromo-5-chloro-4-methylhexan-3-ol
Q94Single correctStructure of Atom
If radius of second Bohr orbit of the He ion is 105.8 pm, what is the radius of third Bohr orbit of L ion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1158.7 pm
Approach:
The Bohr radius of a hydrogen-like ion scales as the square of the principal quantum number divided by the nuclear charge. The He value fixes the proportionality constant.
Step 1:For He with n = 2 and Z = 2, the radius gives the base constant.
Step 2:For L with n = 3 and Z = 3, the radius factor becomes three.
Step 3:Substituting the base constant gives the required radius.
Final answer: 158.7 pm
Q95Single correctHydrocarbons / Aldehydes and Ketones
Compound X on reaction with followed by Zn/ gives formaldehyde and 2-methyl propanal as products. The compound X is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 13-Methylbut-1-ene
Approach:
Reductive ozonolysis cleaves the double bond and replaces it with two carbonyl groups. The alkene is reconstructed by joining the two carbonyl carbons of the products with a double bond.
Step 1:Formaldehyde supplies a terminal fragment, indicating a terminal double bond in X.
Step 2:2-methylpropanal supplies the fragment from the other carbonyl carbon.
Step 3:Joining the two fragments by a double bond gives , which is 3-methylbut-1-ene.
Final answer: 3-Methylbut-1-ene
Q96Single correctThe d- and f-Block Elements / Redox Reactions
In the neutral or faintly alkaline medium, KMn oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1+7 to +4
Approach:
The oxidation state of manganese is found in the reactant permanganate and in the product formed in neutral or faintly alkaline medium.
Step 1:In permanganate the manganese carries an oxidation state of plus seven.
Step 2:In neutral or faintly alkaline medium permanganate is reduced to manganese dioxide, where manganese is plus four.
Step 3:The oxidation state of manganese therefore changes from plus seven to plus four.
Final answer: +7 to +4
Q97Single correctEnvironmental Chemistry
The pollution due to oxides of sulphur gets enhanced due to the presence of:
(a) particulate matter
(b) ozone
(c) hydrocarbons
(d) hydrogen peroxide
Choose the most appropriate answer from the options given below:
(a) particulate matter
(b) ozone
(c) hydrocarbons
(d) hydrogen peroxide
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (b), (d) only
Approach:
Sulphur dioxide pollution intensifies when oxidising agents convert it to sulphur trioxide and sulphuric acid. Each listed species is judged for its oxidising role.
Step 1:Particulate matter offers catalytic surfaces that promote the oxidation of sulphur dioxide, so (a) contributes.
Step 2:Ozone and hydrogen peroxide are strong oxidants that convert sulphur dioxide into sulphur trioxide and sulphuric acid, so (b) and (d) contribute.
Step 3:Hydrocarbons do not directly enhance sulphur oxide pollution, so the contributing set is (a), (b) and (d).
Final answer: (a), (b), (d) only
Q98Single correctAlcohols, Phenols and Ethers
Given below are two statements:
Statement I:
In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnC known as Lucas Reagent.
Statement II:
Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I:
In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnC known as Lucas Reagent.
Statement II:
Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
The Lucas test distinguishes alcohols by the rate at which they form an insoluble chloride. Each statement is judged against the known reactivity order.
Step 1:The Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride that distinguishes primary, secondary and tertiary alcohols by their reactivity, so Statement I is correct.
Step 2:Tertiary alcohols are the most reactive and give immediate turbidity, while primary alcohols react slowest, so the claim that primary alcohols are most reactive is wrong.
Step 3:Statement I holds while Statement II reverses the reactivity order.
Final answer: Statement I is correct but Statement II is incorrect
Q99Single correctThe Solid State
Copper crystallises in fcc unit cell with cell edge length of cm. The density of copper is 8.92 g c. Calculate the atomic mass of copper.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 163.1 u
Approach:
The density of a cubic crystal relates the atomic mass to the number of atoms per unit cell, the cell volume, and Avogadro's number. Rearranging gives the atomic mass.
Step 1:For a face-centred cubic cell the number of atoms per unit cell is four.
Step 2:Rearranging the density relation isolates the atomic mass.
Step 3:Substituting the density, the cube of the edge length, Avogadro's number and Z gives the atomic mass.
Final answer: 63.1 u
Q100Single correctAmines / Aldehydes, Ketones and Carboxylic Acids
The product formed from the following reaction sequence is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Benzyl alcohol ()
Approach:
The starting benzonitrile is first reduced, then the resulting amine is diazotised and hydrolysed. Each step transforms the functional group toward the final product.
Step 1:Lithium aluminium hydride reduces the nitrile of benzonitrile to a primary amine, giving benzylamine.
Step 2:Treatment with sodium nitrite and hydrochloric acid converts the aliphatic primary amine into an unstable diazonium species that loses nitrogen.
Step 3:Aqueous workup traps the benzylic carbon as benzyl alcohol.
Final answer: Benzyl alcohol ()
Biology100 questions
Q101Single correctBiotechnology: Principles and Processes
Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason (R).
Assertion (A) :
Polymerase chain reaction is used in DNA amplification.
Reason (R) :
The ampicillin resistant gene is used as a selectable marker to check transformation
In the light of the above statements, choose the correct answer from the options given below :
Assertion (A) and the other is labelled as Reason (R).
Assertion (A) :
Polymerase chain reaction is used in DNA amplification.
Reason (R) :
The ampicillin resistant gene is used as a selectable marker to check transformation
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Approach:
Each statement is evaluated independently, then the relationship between them is examined.
Step 1:Polymerase chain reaction repeatedly copies a target DNA segment using a thermostable DNA polymerase, so it is a method of in vitro DNA amplification. Assertion (A) is correct.
Step 2:The ampicillin resistance gene on a vector serves as a selectable marker; transformed cells grow on ampicillin medium while non-transformed cells die, confirming transformation. Reason (R) is correct.
Step 3:Amplification of DNA by PCR and selection of transformants by an antibiotic resistance marker are independent processes, so (R) does not explain (A).
Final answer: Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Q102Single correctMolecular Basis of Inheritance
The process of translation of mRNA to proteins begins as soon as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The small subunit of ribosome encounters mRNA
Approach:
The initiation step of translation is identified from the sequence of ribosome assembly on mRNA.
Step 1:Translation initiation requires recognition of the mRNA. The small (smaller) ribosomal subunit first binds the mRNA at the start codon region.
Step 2:Once the small subunit is positioned with the initiator tRNA, the larger subunit joins to form the complete ribosome and elongation proceeds. Translation is therefore considered to begin when the small subunit encounters the mRNA.
Final answer: The small subunit of ribosome encounters mRNA
Q103Single correctPlant Growth and Development
The gaseous plant growth regulator is used in plants to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2promote root growth and roothair formation to increase the absorption surface
Approach:
The only gaseous plant growth regulator is identified and its established physiological role is recalled.
Step 1:The only gaseous plant growth regulator is ethylene.
Step 2:Ethylene promotes growth of lateral roots and root hairs, thereby increasing the absorptive surface of the root system. The other listed effects belong to gibberellins (malting), cytokinins (overcoming apical dominance) and synthetic auxins such as 2,4-D (killing dicot weeds).
Final answer: promote root growth and roothair formation to increase the absorption surface
Q104Single correctBiomolecules
Exoskeleton of arthropods is composed of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Chitin
Approach:
The structural polysaccharide forming the arthropod exoskeleton is recalled.
Step 1:Chitin is a nitrogen-containing homopolymer of N-acetylglucosamine units and forms the hard exoskeleton of arthropods.
Step 2:Cutin is a plant cuticle polymer, cellulose is a plant cell-wall polysaccharide, and glucosamine is only the monomeric derivative, not the structural polymer.
Final answer: Chitin
Q105Single correctTransport in Plants
Which of the following is not observed during apoplastic pathway ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The movement is aided by cytoplasmic streaming
Approach:
Each statement is tested against the defining features of the apoplastic route of water movement.
Step 1:The apoplast comprises cell walls and intercellular spaces, so water moves through them without crossing any cell membrane, making statements 1, 2 and 4 features of this pathway.
Step 2:Cytoplasmic streaming aids movement through the symplast, where water travels via the cytoplasm and plasmodesmata, not through the apoplast. The statement about cytoplasmic streaming is therefore not observed during the apoplastic pathway.
Final answer: The movement is aided by cytoplasmic streaming
Q106Single correctBiodiversity and Conservation
Which of the following is not a method of ex\ situ conservation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2National Parks
Approach:
Each listed practice is classified as ex situ or in situ conservation.
Step 1:Ex situ conservation protects organisms outside their natural habitat. In vitro fertilization, micropropagation and cryopreservation are laboratory-based ex situ methods.
Step 2:National parks protect species within their natural habitat and are therefore an example of in situ conservation, not ex situ.
Final answer: National Parks
Q107Single correctMineral Nutrition
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| (a). Manganese | (i). Activates the enzyme catalase |
| (b). Magnesium | (ii). Required for pollen germination |
| (c). Boron | (iii). Activates enzymes of respiration |
| (d). Iron | (iv). Functions in splitting of water during photosynthesis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Approach:
Each micronutrient is matched with its characteristic metabolic role.
Step 1:Manganese activates the water-splitting reaction of photosystem II during the light reactions, matching (iv).
Step 2:Magnesium activates several enzymes of respiration and photosynthesis, matching (iii).
Step 3:Boron is required for pollen germination and pollen tube growth, matching (ii).
Step 4:Iron is a constituent and activator of catalase, matching (i).
Final answer: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Q108Single correctBiotechnology: Principles and Processes
Which one of the following statement is not true regarding gel electrophoresis technique?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The presence of chromogenic substrate gives blue coloured DNA bands on the gel.
Approach:
Each statement is checked against the standard procedure of agarose gel electrophoresis.
Step 1:Recovery of separated DNA fragments from the gel is termed elution, and ethidium bromide is the dye used to stain DNA, so statements 1 and 2 are true.
Step 2:Ethidium bromide-stained DNA fluoresces bright orange under ultraviolet light, so statement 4 is true.
Step 3:DNA bands are visualized through ethidium bromide fluorescence, not through a chromogenic substrate producing blue bands. The statement about a chromogenic substrate giving blue bands is therefore false.
Final answer: The presence of chromogenic substrate gives blue coloured DNA bands on the gel.
Q109Single correctPhotosynthesis in Higher Plants
Which one of the following is not true regarding the release of energy during ATP synthesis through chemiosmosis? It involves:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Breakdown of electron gradient
Approach:
The driving force of chemiosmotic ATP synthesis is identified, and each statement is judged accordingly.
Step 1:Chemiosmosis depends on a proton gradient across the thylakoid membrane. Its breakdown as protons flow through ATP synthase into the stroma releases energy for ATP synthesis, and NADP reduction occurs on the stroma side, making statements 1, 3 and 4 true.
Step 2:There is no electron gradient that breaks down during chemiosmosis; the gradient involved is a proton (electrochemical) gradient. The statement about breakdown of an electron gradient is therefore not true.
Final answer: Breakdown of electron gradient
Q110Single correctMolecular Basis of Inheritance
DNA polymorphism forms the basis of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both genetic mapping and DNA finger printing
Approach:
The applications that rely on inherited variation in DNA sequence are identified.
Step 1:DNA polymorphism refers to inheritable variation at the DNA sequence level within a population. Such variation underlies DNA fingerprinting for individual identification.
Step 2:Polymorphic markers also provide reference points for constructing genetic maps. Therefore DNA polymorphism forms the basis of both genetic mapping and DNA fingerprinting, while translation is unrelated.
Final answer: Both genetic mapping and DNA finger printing
Q111Single correctBiodiversity and Conservation
Habitat loss and fragmentation, over exploitation, alien species invasion and co-extinction are causes for:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Biodiversity loss
Approach:
The listed factors are recognized as the recognized drivers of a specific ecological consequence.
Step 1:Habitat loss and fragmentation, over-exploitation, alien species invasion and co-extinction are the four major causes of biodiversity loss, often called the evil quartet.
Step 2:These factors reduce the variety of species and are not the causes of population explosion, competition or natality.
Final answer: Biodiversity loss
Q112Single correctEnvironmental Issues
The device which can remove particulate matter present in the exhaust from a thermal power plant is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Electrostatic Precipitator
Approach:
The device used for removing particulates from flue gases is identified.
Step 1:An electrostatic precipitator charges and collects particulate matter from the exhaust of thermal power plants, removing more than 99 percent of particulates.
Step 2:A sewage treatment plant treats wastewater, an incinerator burns solid waste, and a catalytic converter reduces gaseous pollutants from vehicles, so these do not serve the stated purpose.
Final answer: Electrostatic Precipitator
Q113Single correctPlant Growth and Development
Which one of the following plants does not show plasticity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Maize
Approach:
Plants known to show heterophylly or leaf-form plasticity are recalled, and the exception is selected.
Step 1:Plasticity is the ability of a plant to alter its form in response to environment or developmental stage. Cotton, coriander and buttercup show different leaf shapes (heterophylly) and are standard examples of plasticity.
Step 2:Maize is not listed among the standard examples of leaf plasticity and is therefore the plant that does not show plasticity in this context.
Final answer: Maize
Q114Single correctOrganisms and Populations
Which one of the following statements cannot be connected to Predation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both the interacting species are negatively impacted
Approach:
The defining nature of predation as a (+, −) interaction is applied to each statement.
Step 1:Predation maintains species diversity, can drive prey to extinction, and helps maintain ecological balance and energy transfer, so statements 1, 2 and 4 are linked to predation.
Step 2:In predation the predator benefits while the prey is harmed, a (+, −) interaction. The claim that both species are negatively impacted describes competition, not predation, so it cannot be connected to predation.
Final answer: Both the interacting species are negatively impacted
Q115Single correctRespiration in Plants
What amount of energy is released from glucose during lactic acid fermentation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Less than 7%
Approach:
The fraction of glucose energy released during fermentation is recalled from the energetics of anaerobic respiration.
Step 1:During fermentation, glucose is only partially oxidized and less than seven percent of the energy stored in glucose is released, with most energy retained in the lactic acid or ethanol produced.
Step 2:The larger percentages listed overstate the small yield of anaerobic fermentation.
Final answer: Less than 7%
Q116Single correctPrinciples of Inheritance and Variation
Given below are two statements :
Statement I :
Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance.
Statement II :
Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height.
In the light of the above statements, choose the correct answer from the options given below :
Statement I :
Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance.
Statement II :
Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Both statements about Mendel's work are checked against the recorded set of seven contrasting characters.
Step 1:Mendel selected seven pairs of contrasting traits in the garden pea and from these experiments proposed the laws of inheritance, so Statement I is correct.
Step 2:The seven characters were seed shape, seed colour, flower colour, pod shape, pod colour, flower position and stem height, which matches Statement II, making it correct as well.
Final answer: Both Statement I and Statement II are correct
Q117Single correctEcosystem
Given below are two statements:
Statement I: Decomposition is a process in which the detritus is degraded into simpler substances by microbes.
Statement II: Decomposition is faster if the detritus is rich in lignin and chitin.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Decomposition is a process in which the detritus is degraded into simpler substances by microbes.
Statement II: Decomposition is faster if the detritus is rich in lignin and chitin.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement on decomposition is evaluated against the effect of detritus composition on decomposition rate.
Step 1:Decomposition is the process by which decomposers break detritus into simpler inorganic substances, so Statement I is correct.
Step 2:Detritus rich in lignin and chitin decomposes slowly, whereas detritus rich in nitrogen and water-soluble substances decomposes faster. Statement II is therefore incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q118Single correctMolecular Basis of Inheritance
Read the following statements and choose the set of correct statements :
(a) Euchromatin is loosely packed chromatin
(b) Heterochromatin is transcriptionally active
(c) Histone octomer is wrapped by negatively charged DNA in nucleosome
(d) Histones are rich in lysine and arginine
(e) A typical nucleosome contains 400 bp of DNA helix
Choose the correct answer from the options given below :
(a) Euchromatin is loosely packed chromatin
(b) Heterochromatin is transcriptionally active
(c) Histone octomer is wrapped by negatively charged DNA in nucleosome
(d) Histones are rich in lysine and arginine
(e) A typical nucleosome contains 400 bp of DNA helix
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (c), (d) Only
Approach:
Each statement on chromatin and nucleosome structure is judged correct or incorrect.
Step 1:Euchromatin is loosely packed and transcriptionally active, so (a) is correct while (b) is incorrect because heterochromatin is condensed and inactive.
Step 2:In a nucleosome the negatively charged DNA wraps around the positively charged histone octamer, and histones are rich in the basic residues lysine and arginine, so (c) and (d) are correct.
Step 3:A typical nucleosome contains about 200 base pairs of DNA, not 400, so (e) is incorrect. The correct set is (a), (c) and (d).
Final answer: (a), (c), (d) Only
Q119Single correctMorphology of Flowering Plants
Which one of the following plants shows vexillary aestivation and diadelphous stamens?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pisum sativum
Approach:
The plant having both vexillary aestivation and diadelphous stamens is identified from its family characters.
Step 1:Vexillary (papilionaceous) aestivation and diadelphous stamens are characteristic of the family Fabaceae (Papilionoideae).
Step 2:Pisum sativum, the garden pea, belongs to this family, whereas Colchicum and Allium are Liliaceae and Solanum is Solanaceae.
Final answer: Pisum sativum
Q120Single correctAnatomy of Flowering Plants
In all trees the greater part of secondary xylem is dark brown and resistant to insect attack due to :
(a) secretion of secondary metabolites and their deposition in the lumen of vessels.
(b) deposition of organic compounds like tannins and resins in the central layers of stem.
(c) deposition of suberin and aromatic substances in the outer layer of stem.
(d) deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.
(e) presence of parenchyma cells, functionally active xylem elements and essential oils.
Choose the correct answer from the options given below :
(a) secretion of secondary metabolites and their deposition in the lumen of vessels.
(b) deposition of organic compounds like tannins and resins in the central layers of stem.
(c) deposition of suberin and aromatic substances in the outer layer of stem.
(d) deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.
(e) presence of parenchyma cells, functionally active xylem elements and essential oils.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a) and (b) Only
Approach:
The cause of dark, durable heartwood is identified from the process of heartwood formation.
Step 1:In an old tree the central secondary xylem is converted into heartwood. Secondary metabolites are secreted and deposited in the lumen of its vessels and tracheids, which plugs them, so statement (a) holds.
Step 2:The deposited material is tannins, resins, gums, oils and aromatic substances, and it accumulates in the central layers of the stem, so statement (b) holds.
Step 3:Those deposits darken the wood and make it hard and resistant to insect and microbial attack. Deposition in the outer or peripheral layers, or in still-active xylem, describes sapwood rather than heartwood, so statements (c), (d) and (e) do not apply.
Final answer: (a) and (b) Only
Q121Single correctAnatomy of Flowering Plants
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 :
(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 :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a), (b), (c) and (d) Only
Approach:
Each of the five statements about vascular bundle organisation is tested independently.
Step 1:In roots the xylem and phloem lie on different radii in an alternate (radial) arrangement, so (a) is correct, and a conjoint closed bundle has no cambium between its xylem and phloem, so (b) is correct.
Step 2:An open vascular bundle is defined by the presence of cambium between xylem and phloem, so (c) is correct, and the dicot stem bundle has its protoxylem towards the centre, an endarch condition, so (d) is correct.
Step 3:A monocot root is typically polyarch, with more than six xylem bundles, so (e) is correct as well. All five statements are individually valid, so no single set of four can be singled out as the only correct one.
Final answer: (a), (b), (c) and (d) Only
Q122Single correctCell Cycle and Cell Division
Which one of the following never occurs during mitotic cell division?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pairing of homologous chromosomes
Approach:
Events of mitosis are compared with those unique to meiosis to find the one absent from mitosis.
Step 1:During mitosis spindle fibres attach to kinetochores, centrioles move to opposite poles, and chromatids coil and condense, so statements 1, 2 and 4 occur in mitosis.
Step 2:Pairing (synapsis) of homologous chromosomes is restricted to prophase I of meiosis and never occurs during mitosis.
Final answer: Pairing of homologous chromosomes
Q123Single correctPlant Growth and Development
Production of Cucumber has increased manifold in recent years. Application of which of the following phytohormones has resulted in this increased yield as the hormone is known to produce female flowers in the plants :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ethylene
Approach:
The phytohormone that promotes female flower formation in cucurbits is identified.
Step 1:Ethylene promotes the formation of female flowers in cucurbits such as cucumber, increasing the number of fruits and hence the yield.
Step 2:Abscisic acid promotes dormancy, gibberellin tends to favour male flowers, and cytokinin is not associated with this sex expression.
Final answer: Ethylene
Q124Single correctMorphology of Flowering Plants
The flowers are Zygomorphic in:
(a) Mustard
(b) Gulmohar
(c) Cassia
(d) Datura
(e) Chilly
Choose the correct answer from the options given below:
(a) Mustard
(b) Gulmohar
(c) Cassia
(d) Datura
(e) Chilly
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(b), (c) Only
Approach:
Each plant is classified by floral symmetry to select those with zygomorphic flowers.
Step 1:Gulmohar and Cassia bear zygomorphic (bilaterally symmetrical) flowers, so (b) and (c) qualify.
Step 2:Mustard, Datura and Chilly bear actinomorphic (radially symmetrical) flowers, so (a), (d) and (e) are excluded.
Final answer: (b), (c) Only
Q125Single correctMorphology of Flowering Plants
Identify the correct set of statements :
(a) The leaflets are modified into pointed hard thorns in and
(b) Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin
(c) Stem is flattened and fleshy in and modified to perform the function of leaves
(d) shows vertically upward growing roots that help to get oxygen for respiration
(e) Subaerially growing stems in grasses and strawberry help in vegetative propagation
Choose the correct answer from the options given below :
(a) The leaflets are modified into pointed hard thorns in and
(b) Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin
(c) Stem is flattened and fleshy in and modified to perform the function of leaves
(d) shows vertically upward growing roots that help to get oxygen for respiration
(e) Subaerially growing stems in grasses and strawberry help in vegetative propagation
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(b), (c), (d) and (e) Only
Approach:
Each statement on plant part modifications is verified against known examples.
Step 1:Axillary buds form coiled tendrils in cucumber and pumpkin, the stem is flattened and fleshy (phylloclade) in Opuntia, Rhizophora shows upward pneumatophores for respiration, and grasses and strawberry have subaerial runners for vegetative propagation, so (b), (c), (d) and (e) are correct.
Step 2:In Citrus and Bougainvillea the axillary buds, not the leaflets, are modified into thorns, so statement (a) is incorrect.
Final answer: (b), (c), (d) and (e) Only
Q126Single correctBiological Classification
Which of the following is matched?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 – Mannitol
Approach:
Each alga is paired with a characteristic stored food or pigment, and the mismatched pair is identified.
Step 1:Brown algae such as Ectocarpus contain the pigment fucoxanthin, so this pairing is correct.
Step 2:Ulothrix is a green alga (Chlorophyceae) that stores food as starch; mannitol is the reserve food of brown algae, so this pairing is wrong.
Step 3:Red algae such as Porphyra store floridean starch, and Volvox, a green alga, stores starch, so both pairings are correct.
Final answer: – Mannitol
Q127Single correctMineral Nutrition
Which one of the following produces nitrogen fixing nodules on the roots of ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The symbiont that forms root nodules on the non-leguminous tree Alnus is identified.
Step 1:Frankia is an actinomycete that induces nitrogen fixing root nodules in non-leguminous plants such as Alnus and Casuarina.
Step 2:Rhizobium nodulates legumes, while Rhodospirillum and Beijerinckia are free-living nitrogen fixers and do not form nodules.
Final answer:
Q128Single correctSexual Reproduction in Flowering Plants
Identify the statement related to Pollination :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Moths and butterflies are the most dominant pollinating agents among insects
Approach:
Each statement on pollination is tested against established facts, and the false statement is selected.
Step 1:Water pollination is rare and limited to a few aquatic plants, and among abiotic agents wind is more common than water, so the first two statements are correct.
Step 2:Some flowers emit foul odours to attract flies and beetles for pollination, confirming the third statement.
Step 3:Bees, not moths and butterflies, are the most dominant insect pollinators, so the fourth statement is incorrect.
Final answer: Moths and butterflies are the most dominant pollinating agents among insects
Q129Single correctSexual Reproduction in Flowering Plants
Given below are two statements :
Statement I :
Cleistogamous flowers are invariably autogamous
Statement II :
Cleistogamy is disadvantageous as there is no chance for cross pollination
In the light of the above statements, choose the correct answer from the options given below :
Statement I :
Cleistogamous flowers are invariably autogamous
Statement II :
Cleistogamy is disadvantageous as there is no chance for cross pollination
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Both statements about cleistogamous flowers are evaluated for correctness.
Step 1:Cleistogamous flowers do not open, so pollen is released onto the stigma within the closed bud, making them invariably autogamous; Statement I is correct.
Step 2:Because the flowers never open, cross pollination is impossible, which is a disadvantage of cleistogamy; Statement II is correct.
Final answer: Both Statement I and Statement II are correct
Q130Single correctBiological Classification
Hydrocolloid carrageen is obtained from:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Rhodophyceae only
Approach:
The algal class yielding the hydrocolloid carrageen is identified.
Step 1:Carrageen is a commercial hydrocolloid obtained from red algae belonging to Rhodophyceae.
Step 2:Algin, not carrageen, is the hydrocolloid from brown algae of Phaeophyceae.
Final answer: Rhodophyceae only
Q131Single correctRespiration in Plants
What is the net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Two
Approach:
The net ATP balance of glycolysis is computed from ATP consumed and ATP produced.
Step 1:Glycolysis consumes two ATP in the preparatory phase and produces four ATP in the payoff phase.
Step 2:The conversion of one glucose to two pyruvate therefore yields a net of two ATP.
Final answer: Two
Q132Single correctSexual Reproduction in Flowering Plants
The appearance of recombination nodules on homologous chromosomes during meiosis characterizes :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Sites at which crossing over occurs
Approach:
The role of recombination nodules during meiotic prophase is recalled.
Step 1:Recombination nodules appear along the synaptonemal complex at pachytene and mark the points where crossing over takes place between homologous chromosomes.
Step 2:While the synaptonemal complex and bivalent form during pachytene, the nodules specifically denote where recombination occurs.
Final answer: Sites at which crossing over occurs
Q133Single correctPhotosynthesis in Higher Plants
Given below are two statements :
Statement I :
The primary acceptor in plants is phosphoenolpyruvate and is found in the mesophyll cells.
Statement II :
Mesophyll cells of plants lack RuBisCo enzyme. In the light of the above statements, choose the correct answer from the options given below:
Statement I :
The primary acceptor in plants is phosphoenolpyruvate and is found in the mesophyll cells.
Statement II :
Mesophyll cells of plants lack RuBisCo enzyme. In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Both statements about the C4 pathway and the distribution of its enzymes are checked.
Step 1:In C4 plants the primary CO2 acceptor is phosphoenolpyruvate, located in the mesophyll cells where PEP carboxylase operates; Statement I is correct.
Step 2:RuBisCo is confined to the bundle sheath cells, so the mesophyll cells of C4 plants lack RuBisCo; Statement II is correct.
Final answer: Both Statement I and Statement II are correct
Q134Single correctTransport in Plants
"Girdling Experiment" was performed by Plant Physiologists to identify the plant tissue through which:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2food is transported
Approach:
The purpose of the girdling experiment in demonstrating tissue function is recalled.
Step 1:Girdling removes a ring of bark including the phloem while leaving the xylem intact.
Step 2:Tissue above the ring swells because downward food transport is blocked, showing that organic food is translocated through the phloem.
Final answer: food is transported
Q135Single correctPrinciples of Inheritance and Variation
XO type of sex determination can be found in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Grasshoppers
Approach:
The organism exhibiting the XO mechanism of sex determination is identified.
Step 1:In grasshoppers, males are XO with a single X and no Y, while females are XX, illustrating the XO type of sex determination.
Step 2:Drosophila and monkeys follow XY type, and birds follow ZW type, so they are eliminated.
Final answer: Grasshoppers
Q136Single correctTransport in Plants
Addition of more solutes in a given solution will :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2lower its water potential
Approach:
The effect of added solute on the water potential of a solution is determined.
Step 1:Dissolving solute introduces a negative solute potential that adds to the water potential of the solution.
Step 2:As a result the water potential of the solution falls below that of pure water.
Final answer: lower its water potential
Q137Single correctMolecular Basis of Inheritance
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sequence annotation
Approach:
The genome-sequencing strategy that sequences the whole genome first and assigns functions afterward is identified.
Step 1:In the sequence annotation approach the entire genome is sequenced blindly and functions are later assigned to the segments.
Step 2:The expressed sequence tags approach instead focuses only on the expressed genes, so it is eliminated.
Final answer: Sequence annotation
Q138Single correctPrinciples of Inheritance and Variation
Which of the following occurs due to the presence of autosome linked dominant trait ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Myotonic dystrophy
Approach:
The inheritance pattern of each listed disorder is recalled to find the autosomal dominant one.
Step 1:Myotonic dystrophy is inherited as an autosomal dominant condition.
Step 2:Sickle cell anaemia and thalassemia are autosomal recessive, while haemophilia is X-linked recessive, so they are eliminated.
Final answer: Myotonic dystrophy
Q139Single correctPrinciples of Inheritance and Variation
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A) : Mendel's law of Independent assortment does not hold good for the genes that are located closely on the same chromosome.
Reason (R) : Closely located genes assort independently.
In the light of the above statements, choose the correct answer from the options given below:
Assertion (A) : Mendel's law of Independent assortment does not hold good for the genes that are located closely on the same chromosome.
Reason (R) : Closely located genes assort independently.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A) is correct but (R) is not correct
Approach:
The Assertion and Reason about independent assortment of linked genes are evaluated separately.
Step 1:Genes located closely on the same chromosome are linked and tend to be inherited together, so the law of independent assortment fails for them; the Assertion is correct.
Step 2:Closely located genes do not assort independently but stay linked, so the Reason is incorrect.
Final answer: (A) is correct but (R) is not correct
Q140Single correctSexual Reproduction in Flowering Plants
Which part of the fruit, labelled in the given figure makes it a false fruit?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C → Thalamus
Approach:
The labelled part of the fruit responsible for false fruit formation is identified from the diagram.
Step 1:A false fruit develops when a floral part other than the ovary, typically the thalamus, contributes to fruit formation.
Step 2:In the figure the part labelled C corresponds to the thalamus, so its participation makes the structure a false fruit.
Final answer: C → Thalamus
Q141Single correctBiomolecules
Read the following statements on lipids and find out correct set of statements:
(a) Lecithin found in the plasma membrane is a glycolipid
(b) Saturated fatty acids possess one or more bonds
(c) Gingely oil has lower melting point, hence remains as oil in winter
(d) Lipids are generally insoluble in water but soluble in some organic solvents
(e) When fatty acid is esterified with glycerol, monoglycerides are formed
Choose the correct answer from the option given below:
(a) Lecithin found in the plasma membrane is a glycolipid
(b) Saturated fatty acids possess one or more bonds
(c) Gingely oil has lower melting point, hence remains as oil in winter
(d) Lipids are generally insoluble in water but soluble in some organic solvents
(e) When fatty acid is esterified with glycerol, monoglycerides are formed
Choose the correct answer from the option given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(c), (d) and (e) only
Approach:
Each statement on lipids is tested and the set of correct statements is assembled.
Step 1:Lecithin is a phospholipid, not a glycolipid, and saturated fatty acids have no double bonds, so statements (a) and (b) are wrong.
Step 2:Gingely oil has a low melting point and stays liquid in winter, lipids are insoluble in water but soluble in organic solvents, and esterification of one fatty acid with glycerol gives a monoglyceride, so (c), (d) and (e) are correct.
Final answer: (c), (d) and (e) only
Q142Single correctBiotechnology - Principles and Processes
Transposons can be used during which one of the following ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gene Silencing
Approach:
Transposons are mobile genetic elements that move within a genome and can disrupt or inactivate genes at their site of insertion, a property exploited in functional genomics.
Step 1:Transposons are DNA segments capable of changing their position within the genome through transposition, inserting into and interrupting host gene sequences.
Step 2:Insertion of a transposon into a coding or regulatory sequence prevents normal expression of that gene, a technique used to switch off or knock out target genes.
Step 3:This transposon-mediated inactivation is applied in gene silencing, whereas Polymerase Chain Reaction, autoradiography and gene sequencing rely on primers, radioactive labelling and base-reading chemistry respectively rather than on transposable elements.
Final answer: Gene Silencing
Q143Single correctOrganisms and Populations
While explaining interspecific interaction of population, (+) sign is assigned for beneficial interaction, () sign is assigned for detrimental interaction and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and () for another species involved in the interaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Predation
Approach:
The interaction giving a benefit to one species and harm to the other is identified from the sign convention.
Step 1:Predation benefits the predator (+) and harms the prey (−), matching the required sign pattern.
Step 2:Amensalism is (−/0), commensalism is (+/0) and competition is (−/−), so they do not fit the pattern.
Final answer: Predation
Q144Single correctBiotechnology - Principles and Processes
In the following palindromic base sequences of DNA, which one can be cut easily by particular restriction enzyme?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The sequence forming a true palindrome recognized by restriction enzymes is identified.
Step 1:A restriction site reads the same on both strands in the 5' to 3' direction; for GAATTC the complementary strand read 5' to 3' is also GAATTC, making it a palindrome.
Step 2:The other sequences do not read identically on both strands in the 5' to 3' direction, so they are not palindromic restriction sites.
Final answer:
Q145Single correctEcosystem
Which one of the following will accelerate phosphorus cycle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Weathering of rocks
Approach:
The process that releases phosphorus into the cycle is identified.
Step 1:The natural reservoir of phosphorus is rock, and weathering of these rocks releases phosphates into the soil and water, driving the phosphorus cycle.
Step 2:The phosphorus cycle has no significant gaseous phase, so atmospheric processes do not accelerate it.
Final answer: Weathering of rocks
Q146Single correctEnvironmental Issues
The entire fleet of buses in Delhi were converted to CNG from diesel. In reference to this, which one of the following statements is false?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The same diesel engine is used in CNG buses making the cost of conversion low
Approach:
Each statement on CNG conversion of buses is checked and the false one is selected.
Step 1:CNG burns more efficiently, is cheaper than diesel and cannot be easily adulterated, so those statements are true.
Step 2:Converting a vehicle to CNG requires fitting CNG-compatible equipment rather than reusing the same diesel engine, so this statement is false.
Final answer: The same diesel engine is used in CNG buses making the cost of conversion low
Q147Single correctPlant Kingdom
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| (a). | (i). Dominant diploid sporophyte vascular plant, with highly reduced male or female gametophyte |
| (b). Fern | (ii). Dominant haploid free-living gametophyte |
| (c). | (iii). Dominant diploid sporophyte alternating with reduced gametophyte called prothallus |
| (d). | (iv). Dominant haploid leafy gametophyte alternating with partially dependent multicellular sporophyte |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each plant is matched to its characteristic life cycle and dominant generation.
Step 1:Spirogyra is a green alga with a dominant haploid free-living gametophyte, matching (ii).
Step 2:The fern has a dominant diploid sporophyte alternating with a reduced gametophyte called prothallus, matching (iii).
Step 3:Funaria, a moss, has a dominant haploid leafy gametophyte with a partially dependent sporophyte, matching (iv), while Cycas, a gymnosperm, has a dominant diploid sporophyte with highly reduced gametophytes, matching (i).
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q148Single correctCell Cycle and Cell Division
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| (a). Metacentric chromosome | (i). Centromere situated close to the end forming one extremely short and one very long arms |
| (b). Acrocentric chromosome | (ii). Centromere at the terminal end |
| (c). Submetacentric | (iii). Centromere in the middle forming two equal arms of chromosomes |
| (d). Telocentric chromosome | (iv). Centromere slightly away from the middle forming one shorter arm and one longer arm |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Each chromosome type is matched to its centromere position and resulting arm lengths.
Step 1:A metacentric chromosome has the centromere in the middle forming two equal arms, matching (iii).
Step 2:An acrocentric chromosome has the centromere near the end forming one very short and one long arm, matching (i).
Step 3:A submetacentric chromosome has the centromere slightly off centre forming one shorter and one longer arm, matching (iv), and a telocentric chromosome has the centromere at the terminal end, matching (ii).
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q149Single correctAnatomy of Flowering Plants
The anatomy of springwood shows some peculiar features. Identify the correct set of statements about springwood.
(a) It is also called as the earlywood
(b) In spring season cambium produces xylem elements with narrow vessels
(c) It is lighter in colour
(d) The springwood along with autumnwood shows alternate concentric rings forming annual rings
(e) It has lower density
Choose the correct answer from the options given below :
(a) It is also called as the earlywood
(b) In spring season cambium produces xylem elements with narrow vessels
(c) It is lighter in colour
(d) The springwood along with autumnwood shows alternate concentric rings forming annual rings
(e) It has lower density
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (c), (d) and (e) Only
Approach:
Each statement on springwood is verified and the correct set is assembled.
Step 1:Springwood, also called earlywood, is lighter in colour and has lower density, and together with autumnwood it forms the alternating concentric annual rings, so (a), (c), (d) and (e) are correct.
Step 2:In spring the cambium is more active and produces xylem with wider, larger vessels, not narrow vessels, so statement (b) is incorrect.
Final answer: (a), (c), (d) and (e) Only
Q150Single correctPhotosynthesis in Higher Plants
What is the role of large bundle sheath cells found around the vascular bundles in plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2To increase the number of chloroplast for the operation of Calvin cycle
Approach:
The function of the large bundle sheath cells in C4 leaf anatomy is recalled.
Step 1:The bundle sheath cells of C4 plants are large and contain numerous chloroplasts where the Calvin cycle operates with high concentrations of CO2.
Step 2:These cells thus increase the number of chloroplasts available for the Calvin cycle, while photorespiration is suppressed in C4 plants.
Final answer: To increase the number of chloroplast for the operation of Calvin cycle
Q151Single correctDigestion and Absorption
Given below are two statements :
Statement I : Fatty acids and glycerols cannot be absorbed into the blood.
Statement II : Specialized lymphatic capillaries called lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Fatty acids and glycerols cannot be absorbed into the blood.
Statement II : Specialized lymphatic capillaries called lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement is tested against the described route by which the products of fat digestion reach the circulation.
Step 1:Fatty acids and glycerol are water insoluble, so they cannot pass directly into the blood capillaries of the villus. They are first packed into micelles, taken into the intestinal mucosa and re-formed there as protein-coated fat globules, the chylomicrons. Statement I is therefore correct.
Step 2:The lacteals are themselves the lymph vessels of the villus. The chylomicrons are transported into the lacteals, the lymph then drains through the lymphatic system and is finally released into the blood stream.
Step 3:Statement II instead describes the lacteals as capillaries that carry chylomicrons onward into lymphatic vessels, which places an extra step in the pathway and treats the lacteals as something other than the lymph vessels they are. As worded, Statement II is not correct.
Final answer: Statement I is correct but Statement II is incorrect
Q152Single correctHuman Reproduction
Given below are two statements:
Statement I :
The release of sperms into the seminiferous tubules is called spermiation.
Statement II :
Spermiogenesis is the process of formation of sperms from spermatogonia.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I :
The release of sperms into the seminiferous tubules is called spermiation.
Statement II :
Spermiogenesis is the process of formation of sperms from spermatogonia.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are incorrect
Approach:
Each statement is checked against the defined terms for the formation and the release of spermatozoa.
Step 1:Spermiogenesis is the transformation of spermatids into spermatozoa. Formation of sperms all the way from spermatogonia is spermatogenesis, so Statement II is incorrect.
Step 2:Spermiation is the release of the mature spermatozoa FROM the seminiferous tubules, after their heads have been embedded in the Sertoli cells. Statement I describes it as a release of sperms INTO the seminiferous tubules, which reverses the direction of the process.
Step 3:Both statements therefore misstate their terms.
Final answer: Both Statement I and Statement II are incorrect
Q153Single correctBreathing and Exchange of Gases
Which of the following is not the function of conducting part of respiratory system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Provides surface for diffusion of and
Approach:
The conducting part transports air, while the exchange part handles gas diffusion.
Step 1:The conducting part, from nostrils to terminal bronchioles, cleans the incoming air of foreign particles, humidifies it and warms it to body temperature.
Step 2:Gaseous exchange of oxygen and carbon dioxide occurs across the alveoli of the exchange part, not the conducting part.
Final answer: Provides surface for diffusion of and
Q154Single correctMicrobes in Human Welfare
Identify the microorganism which is responsible for the production of an immunosuppressive molecule cyclosporin A :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Each microbe is matched with its characteristic bioactive product.
Step 1:Cyclosporin A, used as an immunosuppressive agent in organ transplant patients, is produced by the fungus Trichoderma polysporum.
Step 2:Clostridium butylicum is an industrial butyric acid producer, Aspergillus niger produces citric acid, and the remaining name is not associated with cyclosporin A.
Final answer:
Q155Single correctBody Fluids and Circulation
Under normal physiological conditions in human being every 100 ml of oxygenated blood can deliver ____ ml of to the tissues.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 25 ml
Approach:
The fixed delivery value of oxygen per unit volume of blood is recalled.
Step 1:Oxygenated blood carries about 20 ml of oxygen per 100 ml, and under normal conditions about 5 ml of this is unloaded to the tissues from every 100 ml of blood.
Final answer: 5 ml
Q156Single correctStructural Organisation in Animals
Tegmina in cockroach, arises from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mesothorax
Approach:
The wing type of each thoracic segment of the cockroach is identified.
Step 1:The first pair of wings, the leathery opaque tegmina, are borne on the mesothorax (second thoracic segment), while the membranous hind wings arise from the metathorax.
Final answer: Mesothorax
Q157Single correctBiodiversity and Conservation
conservation refers to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Protect and conserve the whole ecosystem
Approach:
The scope of in-situ conservation is established from what it protects.
Step 1:In-situ conservation protects species within their natural habitats, which conserves the entire ecosystem along with all its biodiversity rather than only selected species.
Final answer: Protect and conserve the whole ecosystem
Q158Single correctEcosystem
Detritivores breakdown detritus into smaller particles. This process is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Fragmentation
Approach:
The breaking of detritus into smaller particles is mapped to its named decomposition step.
Step 1:Decomposition proceeds through fragmentation, leaching, catabolism, humification and mineralisation. The breakdown of detritus into smaller particles by detritivores is the fragmentation step.
Final answer: Fragmentation
Q159Single correctBiomolecules
A dehydration reaction links two glucose molecules to product maltose. If the formula for glucose is then what is the formula for maltose?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A dehydration (condensation) reaction joins two glucose units with loss of one water molecule.
Step 1:Adding two glucose formulae gives .
Step 2:Removing one water molecule during the glycosidic bond formation subtracts two hydrogen and one oxygen atom.
Final answer:
Q160Single correctReproduction in Organisms
Identify the asexual reproductive structure associated with :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Conidia
Approach:
Each asexual structure is matched to its representative organism.
Step 1:Penicillium reproduces asexually by non-motile conidia produced exogenously on conidiophores. Zoospores occur in algae and lower fungi, gemmules in sponges, and buds in yeast and Hydra.
Final answer: Conidia
Q161Single correctCell Cycle and Cell Division
Select the incorrect statement with reference to mitosis:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Spindle fibres attach to centromere of chromosomes
Approach:
Each statement about mitotic phases is tested for accuracy.
Step 1:Spindle fibres attach to the kinetochores, the disc-shaped protein structures on the centromere, not directly to the centromere itself, so this statement is incorrect.
Step 2:Chromosomes do align at the metaphase plate, they decondense at telophase, and the centromere does split at anaphase; those three statements are correct.
Final answer: Spindle fibres attach to centromere of chromosomes
Q162Single correctCell : The Unit of Life
Which of the following statements with respect to Endoplasmic Reticulum is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3In prokaryotes only RER are present
Approach:
Each property of the endoplasmic reticulum is verified.
Step 1:Prokaryotic cells lack membrane bound organelles and therefore have no endoplasmic reticulum of any type, so the claim that prokaryotes possess RER is incorrect.
Step 2:Rough ER bears ribosomes, smooth ER lacks ribosomes, and smooth ER is the site of lipid synthesis; these statements are correct.
Final answer: In prokaryotes only RER are present
Q163Single correctAnimal Kingdom
In the taxonomic categories which hierarchical arrangement in ascending order is correct in case of animals?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Kingdom, Phylum, Class, Order, Family, Genus, Species
Approach:
The descending taxonomic hierarchy is recalled and matched to the listed order.
Step 1:The taxonomic hierarchy from highest to lowest category in animals is Kingdom, Phylum, Class, Order, Family, Genus, Species.
Final answer: Kingdom, Phylum, Class, Order, Family, Genus, Species
Q164Single correctDigestion and Absorption
In which of the following animals, digestive tract has additional chambers like crop and gizzard?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Crop and gizzard are characteristic of the avian digestive tract, so a group of birds is required.
Step 1:Pavo (peafowl), Psittacula (parakeet) and Corvus (crow) are all birds, and birds possess a crop for food storage and a muscular gizzard for grinding.
Step 2:The other groups include Chameleon, Bufo, Bangarus, Catla and Crocodilus, which are not birds, so those sets are excluded.
Final answer:
Q165Single correctBiological Classification
Given below are two statements :
Statement I : Mycoplasma can pass through less than 1 micron filter size.
Statement II : Mycoplasma are bacteria with cell wall.
In the light of the above statements, choose the most appropriate answer from the options given below
Statement I : Mycoplasma can pass through less than 1 micron filter size.
Statement II : Mycoplasma are bacteria with cell wall.
In the light of the above statements, choose the most appropriate answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement is checked against the known features of Mycoplasma.
Step 1:Mycoplasmas are the smallest living cells, and their tiny size allows them to pass through filters of less than 1 micron pore size, so Statement I holds.
Step 2:Mycoplasmas completely lack a cell wall, so describing them as bacteria with a cell wall is wrong and Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q166Single correctStructural Organisation in Animals
Which of the following is not a connective tissue?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Neuroglia
Approach:
Each tissue is classified into its tissue type.
Step 1:Blood is a fluid connective tissue, adipose tissue is a loose connective tissue, and cartilage is a specialised connective tissue.
Step 2:Neuroglia are supporting cells of the nervous tissue, so they belong to nervous tissue and not connective tissue.
Final answer: Neuroglia
Q167Single correctExcretory Products and their Elimination
Nitrogenous waste is excreted in the form of pellet or paste by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Excretion as a pellet or paste indicates uricotelism, characteristic of birds and reptiles.
Step 1:Uricotelic animals excrete uric acid as a semisolid pellet or paste to conserve water. Among the listed forms, Pavo (peafowl) is a bird and is uricotelic.
Step 2:Ornithorhynchus is a mammal, Salamandra an amphibian and Hippocampus a fish, and these are not uricotelic, so they are excluded.
Final answer:
Q168Single correctAnimal Kingdom
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : All vertebrates are chordates but all chordates are not vertebrates.
Reason (R) : Notochord is replaced by vertebral column in the adult vertebrates.
In the light of the above statements, choose the most appropriate answer from the option given below :
Assertion (A) : All vertebrates are chordates but all chordates are not vertebrates.
Reason (R) : Notochord is replaced by vertebral column in the adult vertebrates.
In the light of the above statements, choose the most appropriate answer from the option given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both (A) and (R) are correct and (R) is the correct explanation of (A)
Approach:
The truth of the assertion and reason is evaluated, then their explanatory link.
Step 1:All vertebrates possess the basic chordate features, but many chordates such as urochordates and cephalochordates never develop a vertebral column, so the assertion is correct.
Step 2:In vertebrates the embryonic notochord is replaced by a bony or cartilaginous vertebral column in the adult, which is exactly the feature that distinguishes vertebrates from the other chordates, so the reason is correct and explains the assertion.
Final answer: Both (A) and (R) are correct and (R) is the correct explanation of (A)
Q169Single correctLocomotion and Movement
Which of the following is a correct match for disease and its symptoms?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Arthritis – Inflammed joints
Approach:
Each disorder is paired with its correct cause and symptom.
Step 1:Arthritis is inflammation of the joints, so this pair is correctly matched.
Step 2:Tetany results from low calcium causing wild muscle contractions, myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction, and muscular dystrophy is a genetic disorder of progressive muscle degeneration. The descriptions given for tetany, myasthenia gravis and muscular dystrophy are interchanged and incorrect.
Final answer: Arthritis – Inflammed joints
Q170Single correctLocomotion and Movement
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Osteoporosis is characterised by decreased bone mass and increased chance of fractures.
Reason (R): Common cause of osteoporosis is increased levels of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below.
Assertion (A): Osteoporosis is characterised by decreased bone mass and increased chance of fractures.
Reason (R): Common cause of osteoporosis is increased levels of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A) is correct but (R) is not correct
Approach:
The assertion and reason are evaluated separately for correctness.
Step 1:Osteoporosis is an age related disorder marked by decreased bone mass and mineral content with an increased risk of fractures, so the assertion is correct.
Step 2:Osteoporosis is commonly caused by a decreased level of estrogen, especially after menopause, not an increased level, so the reason is incorrect.
Final answer: (A) is correct but (R) is not correct
Q171Single correctMolecular Basis of Inheritance
In an E.\ coli strain i gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4RNA polymerase will bind the promoter region
Approach:
The gene of the lac operon encodes the repressor. Lactose, converted to allolactose, is the inducer that normally binds the repressor and releases it from the operator. A repressor that cannot bind the inducer changes what happens when lactose is supplied.
Step 1:A repressor unable to bind the inducer stays in its active, operator-binding conformation even in the presence of lactose, behaving as a super-repressor.
Step 2:RNA polymerase recognises and binds the promoter of the operon regardless of the repressor, because the repressor occupies the adjacent operator and not the promoter itself.
Step 3:With the operator occupied, the bound polymerase cannot move on into the structural genes, so no mRNA and hence no enzyme is produced from , and .
Final answer: RNA polymerase will bind the promoter region
Q172Single correctMolecular Basis of Inheritance
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 bp
Approach:
The number of base pairs is the total length divided by the rise per base pair.
Step 1:The distance between two adjacent base pairs in B-DNA is about 0.34 nm, which equals m.
Step 2:Dividing the total length 1.1 m by the rise per base pair gives the number of base pairs.
Final answer: bp
Q173Single correctHuman Reproduction
Which of the following statements are true for spermatogenesis but do not hold true for Oogenesis?
(a) It results in the formation of haploid gametes
(b) Differentiation of gamete occurs after the completion of meiosis
(c) Meiosis occurs continuously in a mitotically dividing stem cell population
(d) It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary
(e) It is initiated at puberty
Choose the most appropriate answer from the options given below:
(a) It results in the formation of haploid gametes
(b) Differentiation of gamete occurs after the completion of meiosis
(c) Meiosis occurs continuously in a mitotically dividing stem cell population
(d) It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary
(e) It is initiated at puberty
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(b), (c) and (e) only
Approach:
Each statement is tested for being true of spermatogenesis yet false for oogenesis.
Step 1:Both processes form haploid gametes and both are controlled by LH and FSH from the anterior pituitary, so statements (a) and (d) are common to both and do not distinguish them.
Step 2:In spermatogenesis, differentiation into spermatozoa occurs after meiosis is complete, the spermatogonia divide mitotically and continuously feed meiosis, and the process begins at puberty. In oogenesis, meiosis is arrested before birth and is not continuous, and oogonia formation occurs in the fetus, so (b), (c) and (e) are true only for spermatogenesis.
Final answer: (b), (c) and (e) only
Q174Single correctLocomotion and Movement
Which of the following is present between the adjacent bones of the vertebral column?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cartilage
Approach:
The structure joining successive vertebrae is identified.
Step 1:Adjacent vertebrae are connected by cartilaginous joints, where fibrocartilage intervertebral discs lie between the vertebral bodies and allow limited movement.
Step 2:Intercalated discs occur in cardiac muscle, so this pairing is unrelated to the vertebral column.
Final answer: Cartilage
Q175Single correctCell Cycle and Cell Division
Regarding Meiosis, which of the statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2DNA replication occurs in S phase of Meiosis-II
Approach:
Each statement about meiotic events is checked for accuracy.
Step 1:DNA replication occurs only once, during the S phase before Meiosis-I, and there is no DNA replication before Meiosis-II, so this statement is incorrect.
Step 2:Meiosis comprises two stages, homologous pairing and recombination occur in Meiosis-I, and four haploid cells result at the end of Meiosis-II; these statements are correct.
Final answer: DNA replication occurs in S phase of Meiosis-II
Q176Single correctHuman Health and Disease
Given below are two statements:
Statement I:
Autoimmune disorder is a condition where body defense mechanism recognizes its own cells as foreign bodies.
Statement II:
Rheumatoid arthritis is a condition where body does not attack self cells.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I:
Autoimmune disorder is a condition where body defense mechanism recognizes its own cells as foreign bodies.
Statement II:
Rheumatoid arthritis is a condition where body does not attack self cells.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement is judged against the definition of an autoimmune disorder and the immunological basis of rheumatoid arthritis.
Step 1:An autoimmune disorder arises when the immune system fails to distinguish self molecules from non-self and treats the body's own cells as foreign, mounting an immune response against them. Statement I states this correctly.
Step 2:Rheumatoid arthritis is itself an autoimmune disease in which the immune system attacks the synovial joints, that is, it does attack self cells. Statement II claims the body does not attack self cells, which contradicts this.
Step 3:Combining both judgments, Statement I is correct while Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q177Single correctEvolution
Natural selection where more individuals acquire specific character value other than the mean character value, leads to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Directional change
Approach:
The three classic forms of natural selection are distinguished by how the population mean shifts relative to the original mean value.
Step 1:In stabilising selection intermediate individuals are favoured and the mean value is retained, so the population does not move toward a new character value.
Step 2:In disruptive selection both extreme phenotypes are favoured and the population splits into two peaks, rather than more individuals acquiring one specific value.
Step 3:In directional selection one extreme is favoured, so more individuals acquire a specific character value that differs from the original mean, shifting the mean in that direction.
Final answer: Directional change
Q178Single correctBody Fluids and Circulation
Given below are two statements :
Statement I : The coagulum is formed of network of threads called thrombins.
Statement II : Spleen is the graveyard of erythrocytes.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : The coagulum is formed of network of threads called thrombins.
Statement II : Spleen is the graveyard of erythrocytes.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are incorrect
Approach:
Each statement is tested on its own wording.
Step 1:During coagulation, thrombin converts soluble fibrinogen into insoluble fibrin. It is the fibrin threads, not thrombin, that form the network of the clot, so Statement I is incorrect.
Step 2:The spleen is a large lymphoid organ whose defined roles are to filter blood-borne micro-organisms and to act as a reservoir of erythrocytes. Worn-out red cells are broken down by macrophages of the spleen, liver and bone marrow together, so describing the spleen alone as the graveyard of erythrocytes is not an accurate statement of its function.
Final answer: Both Statement I and Statement II are incorrect
Q179Single correctMicrobes in Human Welfare
Breeding crops with higher levels of vitamins and minerals or higher proteins and healthier fats is called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Bio-fortification
Approach:
The defining feature of breeding crops for enhanced nutritional content is matched against each listed term.
Step 1:Biofortification is the breeding of crops with higher levels of vitamins, minerals, proteins and healthier fats to improve public health, which matches the statement exactly.
Step 2:Biomagnification and bioaccumulation describe increasing concentration of toxicants through a food chain or in an organism, and bioremediation describes the use of organisms to clean pollutants, none of which concern improving crop nutrition.
Final answer: Bio-fortification
Q180Single correctBiotechnology and its Applications
In gene therapy of Adenosine Deaminase (ADA) deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Genetically engineered lymphocytes are not immortal cells.
Approach:
The reason for repeated infusions in ADA-deficiency gene therapy is identified from the life span of the treated cells.
Step 1:In gene therapy for ADA deficiency, lymphocytes are taken from the patient, a functional ADA gene is introduced using a retroviral vector, and the cells are returned to the patient.
Step 2:Because these genetically engineered lymphocytes are not immortal, they have a limited life span and die over time, so the procedure must be repeated periodically. A permanent cure would require introducing the gene into bone marrow cells at early embryonic stages.
Final answer: Genetically engineered lymphocytes are not immortal cells.
Q181Single correctHuman Reproduction
At which stage of life the oogenesis process is initiated?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Embryonic development stage
Approach:
The timing of the onset of oogenesis in the human female is recalled from the developmental sequence of the ovary.
Step 1:Oogenesis is initiated during the embryonic development stage, when the gamete mother cells or oogonia are formed in the foetal ovary.
Step 2:These oogonia enter meiosis and are arrested as primary oocytes; meiosis only resumes around puberty and at fertilisation, so the initiation itself is embryonic.
Final answer: Embryonic development stage
Q182Single correctReproductive Health
Lippe's loop is a type of contraceptive used as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Non-Medicated IUD
Approach:
Lippe's loop is classified within the categories of intrauterine devices and barrier methods.
Step 1:Intrauterine devices are grouped into non-medicated devices, copper releasing devices and hormone releasing devices.
Step 2:Lippe's loop is a plain plastic device that carries no copper or hormone, placing it among the non-medicated intrauterine devices.
Final answer: Non-Medicated IUD
Q183Single correctDigestion and Absorption
Which of the following functions is not performed by secretions from salivary glands?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Digestion of disaccharides
Approach:
The known components of saliva and their actions are compared with each listed function to find the one not performed.
Step 1:Saliva contains the enzyme salivary amylase, which hydrolyses starch, a complex carbohydrate, into maltose, so digestion of complex carbohydrates is performed.
Step 2:Saliva contains lysozyme, which controls bacterial growth in the mouth, and mucus, which lubricates the oral cavity, so those functions are performed.
Step 3:Saliva lacks the enzymes that split disaccharides such as maltose, sucrose and lactose; these are digested by intestinal enzymes, not by salivary secretions.
Final answer: Digestion of disaccharides
Q184Single correctOrganisms and Populations
If '8' Drosophila in a laboratory population of '80' died during a week, the death rate in the population is _______ individuals per Drosophila per week.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The per capita death rate equals the number of deaths divided by the total population over the time interval.
Step 1:Eight individuals die out of a population of eighty in one week, so the deaths and the population size are taken as given.
Step 2:Dividing the number of deaths by the total population gives the death rate per individual per week.
Final answer:
Q185Single correctBiotechnology - Principles and Processes
Given below are two statements:
Statement I:
Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence.
Statement II:
Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I:
Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence.
Statement II:
Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
The recognition and cutting behaviour of restriction endonucleases is assessed against each statement.
Step 1:Restriction endonucleases inspect the DNA and bind to specific recognition sequences that read the same on both strands in the same orientation, that is, palindromic sequences. Statement I is correct.
Step 2:These enzymes cut each strand at a point a little away from the centre of the palindrome, between the same two bases on opposite strands, producing single-stranded overhangs called sticky ends. Statement II is correct.
Step 3:Both statements describe restriction endonuclease action accurately.
Final answer: Both Statement I and Statement II are correct
Q186Single correctBiological Classification
Which of the following is a correct statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cyanobacteria are a group of autotrophic organisms classified under kingdom Monera.
Approach:
Each statement is tested against the established classification and nutrition of these organisms.
Step 1:Cyanobacteria are photosynthetic prokaryotes placed in kingdom Monera, so describing them as autotrophic organisms of Monera is correct.
Step 2:Bacteria show diverse nutrition including autotrophic forms, so they are not exclusively heterotrophic; slime moulds are protists, not Monera; and mycoplasma are the smallest cells that lack a cell wall.
Final answer: Cyanobacteria are a group of autotrophic organisms classified under kingdom Monera.
Q187Single correctBiotechnology and its Applications
Statements related to human Insulin are given below.
Which statement(s) is/are correct about genetically engineered Insulin?
(a) Pro-hormone insulin contain extra stretch of C-peptide
(b) A-peptide and B-peptide chains of insulin were produced separately in , extracted and combined by creating disulphide bond between them.
(c) Insulin used for treating Diabetes was extracted from Cattles and Pigs.
(d) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone.
(e) Some patients develop allergic reactions to the foreign insulin.
Choose the most appropriate answer from the options given below:
Which statement(s) is/are correct about genetically engineered Insulin?
(a) Pro-hormone insulin contain extra stretch of C-peptide
(b) A-peptide and B-peptide chains of insulin were produced separately in , extracted and combined by creating disulphide bond between them.
(c) Insulin used for treating Diabetes was extracted from Cattles and Pigs.
(d) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone.
(e) Some patients develop allergic reactions to the foreign insulin.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(b) only
Approach:
The question asks which statements are correct specifically about genetically engineered insulin, so each statement is tested against that restriction.
Step 1:Proinsulin contains an additional C-peptide stretch that is removed during maturation, so statement (a) is biologically true and statement (d), that proinsulin must be processed into the mature hormone, is also true.
Step 2:For genetically engineered insulin, the A and B chains were produced separately in E. coli, extracted and then joined by disulphide bonds, so statement (b) describes the engineering correctly.
Step 3:Statement (c) describes insulin from cattle and pigs, which is the older animal source and not the genetically engineered product, and statement (e) refers to allergic reactions to such foreign insulin, so both lie outside the description of engineered insulin.
Final answer: (b) only
Q188Single correctEnvironmental Issues
Given below are two statements:
Statements I : In a scrubber the exhaust from the thermal plant is passed through the electric wires to charge the dust particles.
Statement II : Particulate matter (PM 2.5) cannot be removed by scrubber but can be removed by an electrostatic precipitator.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statements I : In a scrubber the exhaust from the thermal plant is passed through the electric wires to charge the dust particles.
Statement II : Particulate matter (PM 2.5) cannot be removed by scrubber but can be removed by an electrostatic precipitator.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct
Approach:
The working of a scrubber and an electrostatic precipitator is matched against each statement.
Step 1:Passing the exhaust through electrically charged wires to charge dust particles is the action of an electrostatic precipitator, not a scrubber. A scrubber removes gases such as sulphur dioxide by passing the exhaust through a spray of water or lime. Statement I wrongly attributes the precipitator action to the scrubber, so it is incorrect.
Step 2:Fine particulate matter such as PM 2.5 is collected by an electrostatic precipitator through charged plates and is not removed by a scrubber, which mainly handles gases. Statement II is correct.
Step 3:Therefore Statement I is incorrect while Statement II is correct.
Final answer: Statement I is incorrect but Statement II is correct
Q189Single correctPrinciples of Inheritance and Variation
The recombination frequency between the genes a & c is 5%, b & c is 15%, b & d is 9%, a & b is 20%, c & d is 24% and a & d is 29%. What will be the sequence of these genes on a linear chromosome?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4a, c, b, d
Approach:
Recombination frequencies are treated as map distances, and the gene order is built so that distances between adjacent and end genes are additive.
Step 1:The pairs around a give a to c as 5, a to b as 20 and a to b passing through c, since a to c plus c to b equals five plus fifteen which is twenty. This places c between a and b in the order a, c, b.
Step 2:For d, the distance a to d is 29 and c to d is 24; adding the c to b and b to d segments must reproduce these, and placing d beyond b gives a to b as twenty plus b to d, so b to d is nine, making a to d twenty-nine, consistent with the data.
Step 3:Assembling the segments in order gives the linear sequence a, c, b, d.
Final answer: a, c, b, d
Q190Single correctBiomolecules
Choose the correct answer from the options given below:
| List-I (Biological Molecules) | List-II (Biological functions) |
|---|---|
| (a). Glycogen | (i). Hormone |
| (b). Globulin | (ii). Biocatalyst |
| (c). Steroids | (iii). Antibody |
| (d). Thrombin | (iv). Storage product |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
Approach:
Each biological molecule in List-I is assigned its characteristic function from List-II.
Step 1:Glycogen is the storage polysaccharide of animals, matching the storage product, so (a) pairs with (iv).
Step 2:Globulins include the immunoglobulins that act as antibodies, so (b) pairs with (iii); steroids such as the sex hormones act as hormones, so (c) pairs with (i).
Step 3:Thrombin is an enzyme of the clotting cascade and therefore a biocatalyst, so (d) pairs with (ii).
Final answer: (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
Q191Single correctReproductive Health
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| (a). Diaphragms | (i). Inhibit ovulation and Implantation |
| (b). Contraceptive Pills | (ii). Increase phagocytosis of sperm within Uterus |
| (c). Intra Uterine Devices | (iii). Absence of Menstrual cycle and ovulation following parturition |
| (d). Lactational Amenorrhea | (iv). They cover the cervix blocking the entry of sperms |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
Approach:
Each contraceptive method in List-I is matched with its mechanism of action in List-II.
Step 1:Diaphragms are barriers placed to cover the cervix and block the entry of sperm, so (a) pairs with (iv); contraceptive pills act mainly by inhibiting ovulation and implantation, so (b) pairs with (i).
Step 2:Intra uterine devices increase phagocytosis of sperm within the uterus, so (c) pairs with (ii).
Step 3:Lactational amenorrhea is the absence of menstrual cycle and ovulation following parturition during intense breast feeding, so (d) pairs with (iii).
Final answer: (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
Q192Single correctChemical Coordination and Integration
Which of the following are not the effects of Parathyroid hormone?
(a) Stimulates the process of bone resorption
(b) Decreases C level in blood
(c) Reabsorption of C by renal tubules
(d) Decreases the absorption of C from digested food
(e) Increases metabolism of carbohydrates
Choose the most appropriate answer from the options given below:
(a) Stimulates the process of bone resorption
(b) Decreases C level in blood
(c) Reabsorption of C by renal tubules
(d) Decreases the absorption of C from digested food
(e) Increases metabolism of carbohydrates
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(b), (d) and (e) only
Approach:
The hypercalcaemic actions of parathyroid hormone are identified, and statements describing the opposite effect are marked as not its effects.
Step 1:Parathyroid hormone raises blood calcium by stimulating bone resorption and by promoting reabsorption of calcium from the renal tubules, so statements (a) and (c) are genuine effects.
Step 2:It increases, not decreases, blood calcium and increases calcium absorption from digested food, so statements (b) and (d), which state decreases, are not its effects; it has no role in carbohydrate metabolism, so statement (e) is also not its effect.
Step 3:Collecting the statements that are not effects gives (b), (d) and (e).
Final answer: (b), (d) and (e) only
Q193Single correctHuman Health and Disease
Select the incorrect statement with respect to acquired immunity.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Acquired immunity is non-specific type of defense present at the time of birth.
Approach:
Each statement is checked against the defining features of acquired immunity, looking for the false one.
Step 1:Acquired immunity is pathogen specific and develops after exposure; the first encounter gives a primary response and later encounters of the same pathogen give a stronger secondary or anamnestic response based on memory, so statements 1, 2 and 3 are correct.
Step 2:Acquired immunity is specific and develops over life, whereas the non-specific defense present at birth is innate immunity. Statement 4 therefore describes innate immunity and is the incorrect statement about acquired immunity.
Final answer: Acquired immunity is non-specific type of defense present at the time of birth.
Q194Single correctMolecular Basis of Inheritance
Ten E.coli cells with N - dsDNA are incubated in medium containing N nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from N?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 360 cells
Approach:
The number of cell generations in 60 minutes is found, then the semiconservative pattern is applied to count cells whose DNA has no heavy nitrogen strand.
Step 1:The generation time of E. coli is taken as 20 minutes, so in 60 minutes three rounds of replication occur and each starting cell gives eight cells.
Step 2:Because DNA replicates semiconservatively, the two original heavy strands are conserved and end up in two daughter cells per starting cell, while the remaining six daughter cells contain only newly made light DNA free from heavy nitrogen.
Step 3:For ten starting cells the number of cells with DNA totally free from heavy nitrogen is ten multiplied by six.
Final answer: 60 cells
Q195Single correctPrinciples of Inheritance and Variation
If a colour blind female marries a man whose mother was also colour blind, what are the chances of her progeny having colour blindness?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Genotypes are assigned for X-linked recessive colour blindness, then the cross is worked out to find the fraction of affected progeny.
Step 1:A colour blind female carries two affected X chromosomes. Her husband's mother was colour blind, so the son must have received an affected X from his mother and is himself colour blind.
Step 2:Every daughter receives one affected X from each parent and is colour blind, and every son receives the affected X from the mother and is colour blind, so all offspring are affected.
Step 3:Since every child is colour blind, the chance of affected progeny is the whole.
Final answer:
Q196Single correctBiotechnology - Principles and Processes
Which of the following is not a desirable feature of a cloning vector?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Presence of two or more recognition sites
Approach:
The required properties of a good cloning vector are recalled, and the listed feature that conflicts with them is identified.
Step 1:A useful cloning vector must have an origin of replication so it can multiply, a selectable marker gene to identify transformants, and ideally a single recognition site for a given restriction enzyme so the foreign DNA inserts at one defined position.
Step 2:If a vector has two or more recognition sites for the same enzyme it would be cut into several fragments and could not retain the insert at a single defined site, so multiple recognition sites are not desirable.
Final answer: Presence of two or more recognition sites
Q197Single correctStructural Organisation in Animals
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| (a). Bronchioles | (i). Dense Regular Connective Tissue |
| (b). Goblet Cell | (ii). Loose Connective Tissue |
| (c). Tendons | (iii). Glandular Tissue |
| (d). Adipose Tissue | (iv). Ciliated Epithelium |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
Approach:
Each structure in List-I is assigned the tissue type in List-II to which it belongs.
Step 1:Bronchioles are lined by ciliated epithelium that sweeps out mucus, so (a) pairs with (iv); goblet cells are unicellular mucus-secreting glands, that is glandular tissue, so (b) pairs with (iii).
Step 2:Tendons connect muscle to bone and are dense regular connective tissue, so (c) pairs with (i); adipose tissue is a loose connective tissue storing fat, so (d) pairs with (ii).
Final answer: (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
Q198Single correctBody Fluids and Circulation
Which one of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Blood moves freely from atrium to the ventricle during joint diastole.
Approach:
Each statement is checked against the events of the cardiac cycle and the role of the conducting nodes.
Step 1:The action potential that begins each heartbeat is generated by the sino-atrial node, not the atrio-ventricular node, so statement 1 is wrong; the atrioventricular valves open because of pressure from the contracting atria pushing blood into the ventricles, but the tricuspid and bicuspid valves are themselves the atrioventricular valves and their opening is not driven simultaneously as described, making statement 2 incorrect.
Step 2:During joint diastole all chambers are relaxed, the atrioventricular valves are open, and blood flows freely from the atria into the ventricles, so statement 3 is correct.
Step 3:The semilunar valves close when ventricular pressure falls below that in the aorta and pulmonary artery, that is during a pressure drop, not when ventricular pressure increases, so statement 4 is wrong.
Final answer: Blood moves freely from atrium to the ventricle during joint diastole.
Q199Single correctNeural Control and Coordination
Select the incorrect statement regarding synapses :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Impulse transmission across a chemical synapse is always faster than that across an electrical synapse.
Approach:
The properties of electrical and chemical synapses are compared with each statement to find the false one.
Step 1:At an electrical synapse the pre and postsynaptic membranes lie very close together so that current passes directly from one neuron to the other, so statements 1 and 2 are correct, and chemical synapses do transmit signals using neurotransmitters, so statement 3 is correct.
Step 2:Transmission across an electrical synapse is faster than across a chemical synapse, because the chemical synapse involves the release, diffusion and binding of neurotransmitters, which introduces a delay. The claim that the chemical synapse is always faster is therefore wrong.
Final answer: Impulse transmission across a chemical synapse is always faster than that across an electrical synapse.
Q200Single correctEvolution
Which of the following statements is not true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Flippers of penguins and dolphins are a pair of homologous organs
Approach:
Each statement is checked against the definitions of homology, analogy and convergent evolution to find the untrue one.
Step 1:Analogous structures perform similar functions but have different origins and arise through convergent evolution, and sweet potato (a modified root) and potato (a modified stem) are a classic example of analogy, so statements 1 and 2 are true; homology, shared structural design from a common ancestor, indicates common ancestry, so statement 3 is true.
Step 2:The flippers of penguins and dolphins look alike and serve the same swimming function but evolved independently from different ancestral structures, making them analogous rather than homologous. Calling them homologous is untrue.
Final answer: Flippers of penguins and dolphins are a pair of homologous organs
Frequently Asked Questions
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