Back to NEET PYQs







In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are













NEET 2024 May 05 Question Paper with Solutions
All 200 questions from the NEET 2024 (May 05) 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 2024Chemistry PYQs 2024Biology PYQs 2024
- Questions
- 200
- Physics
- 50
- Chemistry
- 50
- Biology
- 100
Physics50 questions
Q1Single correctSemiconductor Electronics
A logic circuit provides the output as per the following truth table :
A | B | Y
0 | 0 | 1
0 | 1 | 0
1 | 0 | 1
1 | 1 | 0
The expression for the output is :
A | B | Y
0 | 0 | 1
0 | 1 | 0
1 | 0 | 1
1 | 1 | 0
The expression for the output is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The output is determined by comparing each row of the truth table with the candidate Boolean expressions.
Step 1:The truth table gives whenever and whenever , independent of the value of .
Step 2:The expression matching this behaviour is the complement of .
Final answer:
Q2Single correctSystems of Particles and Rotational Motion
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is in the direction shown, which one of the following options is correct ( and are any highest and lowest points on the wheel, respectively)?

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Point moves faster than point
Approach:
In pure rolling the velocity of any point on the wheel is the sum of the translational velocity of the centre and the rotational velocity about the centre.
Step 1:For pure rolling the contact point at the bottom is instantaneously at rest, so point has zero velocity.
Step 2:The topmost point has velocity equal to twice the centre velocity.
Step 3:Comparison of the two velocities shows the topmost point moves faster than the lowest point.
Final answer: Point moves faster than point
Q3Single correctElectrostatic Potential and Capacitance
In the following circuit, the equivalent capacitance between terminal and terminal is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The given network forms a balanced Wheatstone bridge of capacitors, so the bridge capacitor carries no charge and can be removed.
Step 1:The four arm capacitors of each satisfy the balance condition of a Wheatstone bridge, so the central capacitor carries no charge and is removed.
Step 2:Each branch of the bridge has two capacitors in series.
Step 3:The two branches are in parallel between and .
Final answer:
Q4Single correctWork, Energy and Power
At any instant of time t, the displacement of any particle is given by (SI unit) under the influence of force of . The value of instantaneous power is (in SI unit):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Instantaneous power is the product of the applied force and the instantaneous velocity, where velocity is the time derivative of displacement.
Step 1:Differentiating the displacement with respect to time gives the velocity.
Step 2:Multiplying the force by the velocity gives the instantaneous power.
Final answer:
Q5Single correctElectric Charges and Fields
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential (V) at any axial point, at m distance (r) from the centre of the dipole of dipole moment vector of magnitude, C m, is V.
(Take SI units)
Reason R: , where r is the distance of any axial point, situated at m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
Assertion A: The potential (V) at any axial point, at m distance (r) from the centre of the dipole of dipole moment vector of magnitude, C m, is V.
(Take SI units)
Reason R: , where r is the distance of any axial point, situated at m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false.
Approach:
The potential due to a dipole at a general point is evaluated and then specialised to the axial point to test the assertion, while the reason expression is checked for correctness.
Step 1:The potential at any point at distance from the centre of a dipole is given by the standard expression.
Step 2:On the axial point in the direction of the dipole moment, the angle is zero, giving the positive potential.
Step 3:At the axial point on the opposite side the angle is , giving the negative potential.
Step 4:The assertion value is correct, but the reason expression for the axial potential carries an incorrect factor of and is therefore false.
Final answer: A is true but R is false.
Q6Single correctWork, Energy and Power
Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity while body B is at rest before collision. The velocity of the system after collision is . The ratio is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Conservation of linear momentum is applied to the completely inelastic collision, in which both bodies move together afterwards.
Step 1:Initial momentum equals final momentum for the completely inelastic collision, with both equal masses moving together after impact.
Step 2:Cancelling the mass and rearranging gives the velocity ratio.
Final answer:
Q7Single correctMagnetism and Matter
In a uniform magnetic field of T, a magnetic needle performs complete oscillations in seconds as shown. The moment of inertia of the needle is . If the magnitude of magnetic moment of the needle is , then the value of 'x' is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The period of oscillation of a magnetic needle in a uniform field relates the moment of inertia, magnetic moment and field strength, and is rearranged for the magnetic moment.
Step 1:The period is the total time divided by the number of oscillations.
Step 2:Squaring the period equation and isolating the magnetic moment gives the working expression.
Step 3:Substituting the values gives the magnetic moment.
Step 4:Comparing with the given form identifies the value of .
Final answer:
Q8Single correctWave Optics
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The reflected light will be completely polarised but the refracted light will be partially polarised.
Approach:
Brewster's law describes the polarisation states of reflected and refracted light when unpolarised light is incident at the polarising angle.
Step 1:At Brewster's angle the reflected ray and refracted ray are perpendicular to each other.
Step 2:By Brewster's law the reflected ray is completely polarised, while the refracted ray remains only partially polarised.
Final answer: The reflected light will be completely polarised but the refracted light will be partially polarised.
Q9Single correctSemiconductor Electronics
Consider the following statements A and B and identify the correct answer:
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in (A), is due to majority charge carriers.
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in (A), is due to majority charge carriers.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A is correct but B is incorrect
Approach:
Each statement is assessed against the known characteristics of a solar cell and a reverse biased junction diode.
Step 1:The solar cell operates with a positive voltage and a negative (reverse) current, so its I-V characteristic lies in the fourth quadrant, making statement A correct.
Step 2:In a reverse biased pn junction diode the small reverse current measured in microamperes is due to minority charge carriers, not majority carriers, so statement B is incorrect.
Final answer: A is correct but B is incorrect
Q10Single correctDual Nature of Radiation and Matter
The graph which shows the variation of and its kinetic energy, E is (where is de Broglie wavelength of a free particle):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Graph of versus E : a straight line through the origin.
Approach:
The de Broglie wavelength is expressed in terms of kinetic energy, and the relation between the inverse square of the wavelength and energy is examined.
Step 1:Writing the de Broglie wavelength in terms of kinetic energy and squaring both sides gives the inverse square of wavelength.
Step 2:Taking the reciprocal shows that the inverse square of wavelength is directly proportional to the kinetic energy.
Step 3:A direct proportionality produces a straight line through the origin with constant slope.
Final answer: Graph of versus E : a straight line through the origin.
Q11Single correctCurrent Electricity
The terminal voltage of the battery, whose emf is V and internal resistance , when connected through an external resistance of as shown in the figure is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 V
Approach:
The circuit current is found from the emf and total resistance, and the terminal voltage is the emf less the drop across the internal resistance.
Step 1:The current in the circuit is the emf divided by the sum of external and internal resistances.
Step 2:The terminal voltage is the emf minus the voltage drop across the internal resistance.
Final answer: V
Q12Single correctUnits and Measurements
The quantities which have the same dimensions as those of solid angle are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1strain and angle
Approach:
The dimensions of solid angle, strain and plane angle are compared to find which quantities are dimensionless like solid angle.
Step 1:Solid angle is area divided by the square of distance, which is dimensionless.
Step 2:Strain is a ratio of lengths and is also dimensionless.
Step 3:Plane angle measured in radians is the ratio of arc to radius and is also dimensionless.
Final answer: strain and angle
Q13Single correctElectromagnetic Induction
In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
The induced current opposes the change in magnetic flux through each solenoid (Lenz's law). The pole that forms on the face of each solenoid nearest the magnet fixes the sense of the induced current, and hence the terminal-to-terminal direction.
Step 1:The bar magnet carries its north pole facing solenoid-1 and its south pole facing solenoid-2, and it moves to the right (towards solenoid-2).
Step 2:The north pole moves away from solenoid-1, so the flux linking solenoid-1 decreases. To oppose this decrease, the near face of solenoid-1 (end B) develops a south pole that attracts the receding north pole.
Step 3:The south pole approaches solenoid-2, so the flux linking solenoid-2 increases. To oppose this increase, the near face of solenoid-2 (end C) develops a south pole that repels the approaching south pole.
Step 4:Tracing the winding sense that produces a south pole at end B and a south pole at end C fixes the terminal directions of the induced currents as A to B in solenoid-1 and D to C in solenoid-2.
Final answer: and
Q14Single correctSystems of Particles and Rotational Motion
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is g c. The length of the g rod is nearly :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 cm
Approach:
The moment of inertia of a rod about its centre is used to solve for the length given the mass and moment of inertia.
Step 1:Substituting the given moment of inertia and mass into the rod formula gives an equation for the length.
Step 2:Simplifying yields the square of the length.
Step 3:Taking the square root gives the length.
Final answer: cm
Q15Single correctOscillations
If m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m, s
Approach:
The amplitude and angular frequency are read directly from the standard SHM displacement equation, and the period is obtained from the angular frequency.
Step 1:Comparing with the standard SHM form, the amplitude is the coefficient of the sine, giving m.
Step 2:The angular frequency is the coefficient of inside the sine.
Step 3:Solving for the time period gives the period of motion.
Final answer: m, s
Q16Single correctGravitation
The mass of a planet is that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The surface gravity is expressed in terms of mass and radius, and the ratios of the planet's mass and radius to those of the earth are substituted.
Step 1:The planet's mass is one tenth of the earth's mass and its radius is half of the earth's radius.
Step 2:Simplifying the ratio of the new gravity to the earth's gravity gives a factor of .
Step 3:Evaluating the product gives the acceleration due to gravity on the planet.
Final answer:
Q17Single correctLaws of Motion
A horizontal force N is applied to a block as shown in figure. The mass of blocks and are kg and kg respectively. The blocks slide over a frictionless surface. The force exerted by block on block is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 N
Approach:
The common acceleration of the connected blocks is found first, then the contact force on the second block is obtained from its own mass and acceleration.
Step 1:The applied force accelerates both blocks together, giving the common acceleration.
Step 2:The force exerted by block on block provides the acceleration of block alone.
Final answer: N
Q18Single correctThermodynamics
A thermodynamic system is taken through the cycle . The work done by the gas along the path is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Zero
Approach:
The nature of the path on the pressure-volume diagram is identified to determine the work done.
Step 1:Along the path the volume remains constant, so the process is isochoric.
Step 2:Since the volume does not change, the work done by the gas along is zero.
Final answer: Zero
Q19Single correctAtoms
Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below.
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement about atomic neutrality and atomic stability is evaluated against established atomic physics.
Step 1:Atoms are electrically neutral because they contain equal numbers of positive and negative charges, so the first statement is true.
Step 2:Atoms of every element are not stable and do not all emit a characteristic spectrum, so the second statement is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q20Single correctNuclei
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each emission changes the mass number and atomic number according to the type of particle emitted, and the changes are tracked through the decay chain.
Step 1:Alpha emission reduces the mass number by four and the atomic number by two, giving .
Step 2:Positron emission keeps the mass number and lowers the atomic number by one, giving .
Step 3:Beta-minus emission keeps the mass number and raises the atomic number by one, giving .
Step 4:Electron emission keeps the mass number and raises the atomic number by one, giving .
Final answer:
Q21Single correctMotion in a Plane
A particle moving with uniform speed in a circular path maintains:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Varying velocity and varying acceleration
Approach:
The vector nature of velocity and acceleration in uniform circular motion is analysed despite the constant speed.
Step 1:Although the speed is constant, the direction of velocity changes continuously around the circle, so the velocity vector varies.
Step 2:The centripetal acceleration is directed toward the centre, and its direction changes continuously, so the acceleration also varies.
Final answer: Varying velocity and varying acceleration
Q22Single correctRay Optics and Optical Instruments
A light ray enters through a right angled prism at point P with the angle of incidence as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The geometry of the right angled prism relates the refraction angle at the entry face to the prism angle, and Snell's law at the incidence surface determines the refractive index.
Step 1:With the ray travelling parallel to the base and the right angle at the apex, the refraction angle at the first face satisfies the prism relation.
Step 2:The critical angle relation at the second face gives the sine and cosine of the angle in terms of the refractive index.
Step 3:Applying Snell's law at the incidence surface and substituting the angle gives an equation for the refractive index.
Step 4:Squaring and solving gives the refractive index of the prism.
Final answer:
Q23Single correctUnits and Measurements
In a vernier callipers, divisions of vernier scale coincide with divisions of main scale. If MSD represents mm, the vernier constant (in cm) is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The value of one vernier scale division is found from the coincidence condition, and the vernier constant is the difference between one main scale division and one vernier scale division.
Step 1:The coincidence condition gives one vernier scale division in terms of one main scale division.
Step 2:The vernier constant is the difference of one main scale division and one vernier scale division.
Step 3:Substituting one main scale division equal to mm, that is cm, gives the vernier constant.
Final answer:
Q24Single correctWave Optics
If the monochromatic source in Young's double slit experiment is replaced by white light, then:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3There will be a central bright white fringe surrounded by a few coloured fringes
Approach:
The central fringe lies where the path difference is zero for every wavelength, so all colours superpose there. Away from the centre the fringe spacing depends on wavelength, so the components separate into coloured fringes.
Step 1:At the central point on the screen the geometric path difference is zero independent of wavelength, so every constituent colour of white light forms a bright maximum at that point.
Step 2:The fringe width depends on wavelength, so each colour produces fringes of a different spacing on either side of the centre.
Step 3:Because the colours no longer coincide away from the centre, the fringes adjacent to the central maximum show coloured edges, and after a few orders the overlap washes out the pattern.
Final answer: There will be a central bright white fringe surrounded by a few coloured fringes
Q25Single correctDual Nature of Radiation and Matter
If c is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is .
B. The velocity of a photon is c.
C. The momentum of a photon, .
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
A. The energy of a photon is .
B. The velocity of a photon is c.
C. The momentum of a photon, .
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B, C and D only
Approach:
Each statement about photon energy, speed, momentum, conservation and charge is checked against the properties of photons.
Step 1:The energy of a photon is the product of Planck's constant and frequency, so statement A is correct.
Step 2:A photon travels at the velocity of light in free space, so statement B is correct.
Step 3:The momentum of a photon equals Planck's constant divided by wavelength, which can be written in terms of frequency and the speed of light, so statement C is correct.
Step 4:In a photon-electron collision both total energy and total momentum are conserved, so statement D is correct, while a photon carries no charge, so statement E is incorrect.
Final answer: A, B, C and D only
Q26Single correctSemiconductor Electronics
The output () of the given logic gate is similar to the output of an/a

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4AND gate
Approach:
The combinational circuit is simplified using De Morgan's theorems to identify the equivalent single gate.
Step 1:The first NAND with both inputs tied to A gives the complement of A.
Step 2:The second NAND with both inputs tied to B gives the complement of B.
Step 3:The final NOR combines the two intermediate outputs, and De Morgan's theorem reduces the expression.
Step 4:The expression corresponds to the logical AND operation.
Final answer: AND gate
Q27Single correctAtoms
Match List I with List II. Choose the correct answer from the options given below:
| List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelengths (nm)) |
|---|---|
| A. to | I. 410.2 |
| B. to | II. 434.1 |
| C. to | III. 656.3 |
| D. to | IV. 486.1 |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-II, D-I
Approach:
Transitions of the Balmer series ending at are ordered by photon energy, and energy is related inversely to wavelength.
Step 1:Wavelength is inversely proportional to the energy difference of a transition.
Step 2:Among the Balmer transitions ending at , the energy gap increases as the upper level rises.
Step 3:Inverting the energy order gives the wavelength order.
Step 4:Matching each transition with its wavelength produces the final pairing.
Final answer: A-III, B-IV, C-II, D-I
Q28Single correctMechanical Properties of Fluids
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N , then the excess force required to take it away from the surface is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The excess force needed equals surface tension acting along the circumference of the disc.
Step 1:The water contacts the disc along its full circumference, so the force is surface tension times the perimeter.
Step 2:Substituting the given values yields the force.
Step 3:Expressing the result in millinewtons.
Final answer:
Q29Single correctAlternating Current
In an ideal transformer, the turns ratio is . The ratio is equal to (the symbols carry their usual meaning) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
In an ideal transformer the voltage ratio equals the turns ratio of the corresponding windings.
Step 1:The secondary-to-primary voltage ratio equals the secondary-to-primary turns ratio.
Step 2:Substituting the given turns ratio with primary half of secondary.
Step 3:The required ratio follows directly.
Final answer:
Q30Single correctMechanical Properties of Solids
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are N and N , is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Maximum elongation occurs when the stress equals the elastic limit, and strain follows from Young's modulus.
Step 1:At maximum elongation the stress reaches the elastic limit.
Step 2:Substituting the elastic limit, length and Young's modulus gives the elongation.
Step 3:Expressing the elongation in millimetres.
Final answer:
Q31Single correctElectric Charges and Fields
A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is:
(Take SI units)
cm,
(Take SI units)
cm,

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Zero
Approach:
The potential inside a uniformly charged spherical shell is constant, so any two interior points share the same potential.
Step 1:Both points C and P lie inside the shell, where the potential is uniform.
Step 2:The potential difference between two points at equal potential is zero.
Final answer: Zero
Q32Single correctCurrent Electricity
A wire of length 'l' and resistance 100 is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Each part has one tenth of the total resistance; the series and parallel sub-combinations are evaluated and then added.
Step 1:Dividing the wire into ten equal parts gives each part a resistance one tenth of the whole.
Step 2:Five such parts in series add directly.
Step 3:Five equal parts in parallel reduce the resistance by a factor of five.
Step 4:The two combinations in series add together.
Final answer:
Q33Single correctMoving Charges and Magnetism
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as SI units):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The magnetic field at the centre of a multi-turn circular coil is computed from the standard expression.
Step 1:The field at the centre of N circular turns is given by the standard formula.
Step 2:Substituting permeability, number of turns, current and radius.
Step 3:Expressing the result in millitesla.
Final answer:
Q34Single correctMagnetism and Matter
Match List-I with List-II. Choose the correct answer from the options given below
| List-I (Material) | List-II (Susceptibility ()) |
|---|---|
| A. Diamagnetic | I. |
| B. Ferromagnetic | II. |
| C. Paramagnetic | III. |
| D. Non-magnetic | IV. (a small positive number) |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each magnetic class is identified by the sign and magnitude of its susceptibility.
Step 1:Diamagnetic materials have small negative susceptibility bounded below by minus one.
Step 2:Ferromagnetic materials have very large positive susceptibility.
Step 3:Paramagnetic materials have a small positive susceptibility.
Step 4:A non-magnetic material has zero susceptibility.
Final answer: A-II, B-III, C-IV, D-I
Q35Single correctSystems of Particles and Rotational Motion
A bob is whirled in a horizontal plane by means of a string with an initial speed of rpm. The tension in the string is T. If speed becomes while keeping the same radius, the tension in the string becomes:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a horizontal circle of fixed radius the string tension provides the centripetal force, which scales with the square of angular speed.
Step 1:At the initial angular speed the tension supplies the centripetal force.
Step 2:Doubling the angular speed at the same radius increases the required force.
Step 3:Comparing with the initial tension gives the new value.
Final answer:
Q36Single correctElectromagnetic Waves
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If is the current in the circuit, then in the gap between the plates:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Displacement current of magnitude equal to flows in the same direction as
Approach:
The continuity of current is maintained across the capacitor gap by Maxwell's displacement current.
Step 1:For a loop enclosing the conducting wire the conduction current dominates and displacement current vanishes.
Step 2:For a loop crossing the gap there is no conduction current, only displacement current.
Step 3:Charge conservation forces the displacement current to equal and continue the conduction current in the same direction.
Final answer: Displacement current of magnitude equal to flows in the same direction as
Q37Single correctAlternating Current
A 10 F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ():

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The capacitive reactance is found, the rms current follows from Ohm's law, and the peak current is the rms value times root two.
Step 1:The capacitive reactance is computed from the frequency and capacitance.
Step 2:The rms current is the rms voltage divided by the reactance.
Step 3:The peak current is root two times the rms current.
Final answer:
Q38Single correctOscillations
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is times its original time period. Then the value of x is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The pendulum period is independent of mass and depends on the square root of length, so the new period is compared to the old.
Step 1:The period is independent of the bob mass, so the mass change has no effect.
Step 2:Halving the length gives the new period in terms of the original length.
Step 3:Equating the new period to determines x.
Final answer:
Q39Single correctThermodynamics
The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures , and compared with those of Charles' law represented as dotted lines. Then the correct relation is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
On a temperature versus volume plot the slope at fixed volume is proportional to pressure, so steeper lines represent higher pressure.
Step 1:For an ideal gas the ratio of temperature to volume is proportional to pressure.
Step 2:At the same temperature a curve reaching a higher volume corresponds to lower pressure.
Step 3:The steepest line on the T-V graph therefore represents the largest pressure.
Final answer:
Q40Single correctGravitation
The minimum energy required to launch a satellite of mass from the surface of earth of mass and radius in a circular orbit at an altitude of from the surface of the earth is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Energy conservation equates the energy supplied plus the surface energy to the total orbital energy at the final radius.
Step 1:The orbital radius is three times the earth radius since the altitude is two radii.
Step 2:Energy conservation relates the surface energy and supplied energy to the final mechanical energy.
Step 3:Solving for the supplied energy combines the potential and orbital kinetic terms.
Final answer:
Q41Single correctElectromagnetic Waves
The property which is not of an electromagnetic wave travelling in free space is that:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4They originate from charges moving with uniform speed
Approach:
Each stated property is checked against the known characteristics of electromagnetic waves to find the incorrect one.
Step 1:Electromagnetic waves are transverse, with equal electric and magnetic energy densities, and travel at the speed of light, so the first three statements are correct.
Step 2:Electromagnetic waves originate from accelerating charges, not from charges moving with uniform speed, so the fourth statement is the incorrect one.
Final answer: They originate from charges moving with uniform speed
Q42Single correctCurrent Electricity
Choose the correct circuit which can achieve the bridge balance.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Circuit (1)
Approach:
Bridge balance requires the ratio of resistances in the two arms to be equal so that no current flows through the galvanometer branch.
Step 1:A Wheatstone bridge is balanced when the resistance ratios in the two adjacent arms are equal.
Step 2:Among the four arrangements, only the configuration with arm resistances satisfying the equal-ratio condition produces zero galvanometer current.
Final answer: Circuit (1)
Q43Single correctPhysical World and Measurement
A force defined by acts on a particle at a given time t. The factor which is dimensionless, if and are constants, is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By the principle of homogeneity each additive term has the dimensions of force, fixing the dimensions of the constants.
Step 1:Both terms must have the dimension of force, giving the dimensions of the constants.
Step 2:Forming the ratio of to cancels all dimensions.
Final answer:
Q44Single correctCurrent Electricity
Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The heater resistances are found from their ratings at a common voltage, and the total power in each connection is computed at the fixed source voltage.
Step 1:Power at a fixed voltage is inversely proportional to resistance, so the 1 kW heater has twice the resistance of the 2 kW heater.
Step 2:In series the resistances add, giving the series power.
Step 3:In parallel the combined resistance is found, giving the parallel power.
Step 4:Taking the ratio of series to parallel power eliminates the common factors.
Final answer:
Q45Single correctMagnetism and Matter
An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle with each other. The magnetic moment of this new magnet is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Each half retains its pole strength and length, and the net moment is the vector sum of the two half-magnet moments separated by the bend angle.
Step 1:Each arm has half the length and the same pole strength, so each half carries moment .
Step 2:The bend places the two half-moments at , and the resultant is twice one component times the sine of half the angle.
Step 3:Evaluating the sine completes the calculation.
Final answer:
Q46Single correctMechanical Properties of Solids
A metallic bar of Young's modulus, N and coefficient of linear thermal expansion , length 1 m and area of cross-section is heated from to without expansion or bending. The compressive force developed in it is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Prevented thermal expansion produces a thermal strain, and the resulting stress times the area gives the compressive force.
Step 1:The longitudinal thermal strain equals the product of the expansion coefficient and the temperature rise.
Step 2:The compressive stress is the strain times Young's modulus.
Step 3:The compressive force is the stress times the cross-sectional area.
Final answer:
Q47Single correctMotion in a Straight Line
The velocity () - time () plot of the motion of a body is shown below. The acceleration () - time () graph that best suits this motion is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Graph (3)
Approach:
Acceleration is the slope of the velocity-time graph, so each segment of the v-t plot is converted to the corresponding acceleration value.
Step 1:Initially the velocity is zero and the slope is zero, so the acceleration is zero at the start.
Step 2:When the velocity rises linearly the slope is constant and positive, giving a constant positive acceleration.
Step 3:When the velocity falls linearly the slope is constant and negative, giving a constant negative acceleration.
Step 4:The acceleration profile of zero, then positive constant, then negative constant matches graph (3).
Final answer: Graph (3)
Q48Single correctElectrostatic Potential and Capacitance
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, C and E only
Approach:
With the battery connected the voltage stays fixed, and reducing the plate separation raises the capacitance, which determines the changes in charge and energy.
Step 1:Decreasing the separation increases the capacitance at constant voltage.
Step 2:At fixed voltage the larger capacitance stores more charge.
Step 3:The stored energy at constant voltage grows with capacitance.
Step 4:The product of charge and voltage equals twice the energy and therefore increases.
Final answer: A, C and E only
Q49Single correctRay Optics and Optical Instruments
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a telescope viewing a distant object the magnifying power is the ratio of the objective to eyepiece focal lengths.
Step 1:For a distant object the magnifying power equals the objective focal length divided by the eyepiece focal length.
Step 2:Substituting the two focal lengths gives the magnification.
Final answer:
Q50Single correctMoving Charges and Magnetism
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and C only
Approach:
Each case is examined for whether a magnetic interaction or an induced eddy-current force requires an external force to hold or move the sheet.
Step 1:A magnetic sheet is attracted by the pole, so an external force is needed to hold it in place.
Step 2:A non-magnetic stationary sheet experiences no magnetic force, so no holding force is required.
Step 3:Moving a conducting sheet changes the flux through it and induces eddy currents that oppose the motion, so a force is required.
Step 4:A non-conducting non-polar sheet supports no induced currents and does not interact with the field, so no force is needed.
Final answer: A and C only
Chemistry49 questions
Q51Single correctClassification of Elements and Periodicity in Properties
Arrange the following elements in increasing order of electronegativity:
N, O, F, C, Si
Choose the correct answer from the options given below:
N, O, F, C, Si
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Electronegativity increases across a period from left to right and decreases down a group, so the relative order is fixed by the position of each element in the periodic table.
Step 1:Silicon lies in period 3, group 14, far to the left and one period below carbon, giving it the lowest electronegativity of the set.
Step 2:Within period 2, electronegativity rises along the sequence carbon, nitrogen, oxygen, fluorine as nuclear charge increases and atomic size decreases.
Step 3:Combining the position of silicon with the period-2 trend gives the complete increasing order.
Final answer:
Q52Single correctHydrocarbons
Identify the correct reagents that would bring about the following transformation.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(i)
(ii)
(iii) PCC
(ii)
(iii) PCC
Approach:
Converting a terminal alkene to the corresponding aldehyde requires anti-Markovnikov addition of water to place the hydroxyl on the terminal carbon, followed by a mild oxidation that stops at the aldehyde stage.
Step 1:Hydroboration of the terminal alkene with diborane adds boron to the less substituted terminal carbon.
Step 2:Oxidation of the organoborane with hydrogen peroxide in base replaces boron with a hydroxyl group, giving the anti-Markovnikov primary alcohol Ph-C-C-C-OH.
Step 3:Pyridinium chlorochromate oxidises the primary alcohol to the aldehyde without over-oxidation to the carboxylic acid.
Final answer: (i)
(ii)
(iii) PCC
(ii)
(iii) PCC
Q53Single correctHaloalkanes and Haloarenes
The compound that will undergo 1 reaction with the fastest rate is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The benzylic bromide a benzene ring bearing CH(C)Br (drawn structure, option 4)
Approach:
The rate of an 1 reaction depends on the stability of the carbocation formed after the leaving group departs; the substrate giving the most stable cation reacts fastest.
Step 1:Ionisation of 1-phenylethyl bromide gives a benzylic carbocation that is both resonance-stabilised by the ring and stabilised by the adjacent methyl group, making it a secondary benzylic cation.
Step 2:Benzyl bromide gives a primary benzylic carbocation stabilised only by resonance, which is less stable than the secondary benzylic cation.
Step 3:The cyclohexyl and cyclohexylmethyl bromides give a secondary and a primary alkyl cation respectively, neither of which enjoys aromatic resonance stabilisation.
Final answer: The benzylic bromide a benzene ring bearing CH(C)Br (drawn structure, option 4)
Q54Single correctEquilibrium
For the reaction , . At a given time, the composition of reaction mixture is:
M.
Then, which of the following is correct?
M.
Then, which of the following is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Reaction has a tendency to go in backward direction.
Approach:
The reaction quotient is evaluated from the given instantaneous concentrations and compared with the equilibrium constant to determine the direction of net change.
Step 1:Substituting the instantaneous concentrations into the reaction-quotient expression gives the current value.
Step 2:Comparing the quotient with the equilibrium constant shows the quotient far exceeds the constant.
Step 3:When the quotient is greater than the equilibrium constant the system shifts to consume products and form reactants, that is the backward direction.
Final answer: Reaction has a tendency to go in backward direction.
Q55Single correctSome Basic Concepts of Chemistry
The highest number of helium atoms is in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 14 mol of helium
Approach:
Each amount is converted to a number of helium atoms using the mole, the atomic mass unit, the molar mass and the molar volume, then the largest count is selected.
Step 1:Four moles of helium contain four times the Avogadro number of atoms.
Step 2:Four atomic mass units of helium correspond to the mass of one atom, since the atomic mass of helium is 4 u.
Step 3:Four grams of helium equal one mole, which is the Avogadro number of atoms.
Step 4:At STP the molar volume is 22.710982 L, so 2.271098 L corresponds to 0.1 mole and hence one tenth of the Avogadro number of atoms.
Step 5:Comparing the four counts shows four moles gives the largest number of atoms.
Final answer: 4 mol of helium
Q56Single correctThermodynamics
Choose the correct answer from the options given below:
| List-I (Process) | List-II (Conditions) |
|---|---|
| A. Isothermal process | I. No heat exchange |
| B. Isochoric process | II. Carried out at constant temperature |
| C. Isobaric process | III. Carried out at constant volume |
| D. Adiabatic process | IV. Carried out at constant pressure |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Each named thermodynamic process is paired with the variable held constant or the heat condition that defines it.
Step 1:An isothermal process keeps temperature constant throughout the process.
Step 2:An isochoric process keeps volume constant throughout the process.
Step 3:An isobaric process keeps pressure constant throughout the process.
Step 4:An adiabatic process allows no exchange of heat between the system and the surroundings.
Final answer: A-II, B-III, C-IV, D-I
Q57Single correctStructure of Atom
The energy of an electron in the ground state () for He ion is J, then that for an electron in state for B ion in J is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The Bohr energy of a hydrogen-like ion depends on the square of the nuclear charge divided by the square of the principal quantum number, so the unknown constant fixes the energy of the second species.
Step 1:For the helium ion with charge two in the ground state the energy equals minus four times the Rydberg constant, and this is set equal to minus x.
Step 2:For the beryllium ion with charge four in the second level the energy is minus the Rydberg constant times sixteen over four.
Final answer:
Q58Single correctChemical Bonding and Molecular Structure
Choose the correct answer from the options given below:
| List I (Molecule) | List II (Number and types of bonds(s) between two carbon atoms) |
|---|---|
| A. ethane | I. one -bond and two -bonds |
| B. ethene | II. two -bonds |
| C. carbon molecule, | III. one -bond |
| D. ethyne | IV. one -bond and one -bond |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-II, D-I
Approach:
The carbon-carbon bond order of each molecule fixes the number and type of sigma and pi bonds between the two carbon atoms.
Step 1:Ethane has a single carbon-carbon bond, which is one sigma bond.
Step 2:Ethene has a double bond, comprising one sigma bond and one pi bond.
Step 3:The diatomic carbon molecule has a bond consisting of two pi bonds between the carbon atoms according to molecular orbital filling.
Step 4:Ethyne has a triple bond, comprising one sigma bond and two pi bonds.
Final answer: A-III, B-IV, C-II, D-I
Q59Single correctChemical Bonding and Molecular Structure
Choose the correct answer from the options given below:
| List I (Compound) | List II (Shape/geometry) |
|---|---|
| A. | I. Trigonal Pyramidal |
| B. | II. Square Planar |
| C. | III. Octahedral |
| D. | IV. Square Pyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-I, B-IV, C-II, D-III
Approach:
The hybridisation and the number of lone pairs on the central atom determine the molecular shape of each species.
Step 1:Ammonia has sp3 hybridisation with one lone pair, giving a trigonal pyramidal shape.
Step 2:Bromine pentafluoride has sp3d2 hybridisation with one lone pair, giving a square pyramidal shape.
Step 3:Xenon tetrafluoride has sp3d2 hybridisation with two lone pairs, giving a square planar shape.
Step 4:Sulphur hexafluoride has sp3d2 hybridisation with no lone pair, giving an octahedral shape.
Final answer: A-I, B-IV, C-II, D-III
Q60Single correctThe d- and f-Block Elements
The E value for the M/M couple is more positive than that of C/C or F/F due to change of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 to configuration
Approach:
The electrode potential reflects the stability change accompanying reduction; the manganese couple is unusual because reduction yields a half-filled d subshell.
Step 1:Reduction of the manganese ion changes the d-electron count from four to five, producing the extra-stable half-filled configuration.
Step 2:The greater stability of the half-filled d5 ion released on reduction makes reduction strongly favourable, raising the standard potential above that of the chromium and iron couples.
Final answer: to configuration
Q61Single correctEquilibrium
In which of the following equilibria, K and K are NOT equal?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The two equilibrium constants differ only when the change in the number of moles of gas between products and reactants is non-zero.
Step 1:The two constants are equal only when the change in gaseous moles is zero.
Step 2:For the dissociation of phosphorus pentachloride the gaseous moles increase from one to two.
Step 3:For the remaining three equilibria the gaseous moles are equal on both sides, giving a zero change.
Final answer:
Q62Single correctThe p-Block Elements
Among Group 16 elements, which one does NOT show oxidation state?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Po
Approach:
The tendency to show the minus two state decreases down group 16, and the heaviest member loses this state owing to its metallic character.
Step 1:Oxygen shows oxidation states of minus two, minus one, plus one and plus two.
Step 2:Selenium shows oxidation states of minus two, plus two, plus four and plus six.
Step 3:Tellurium shows oxidation states of minus two, plus two, plus four and plus six.
Step 4:Polonium, the most metallic group 16 element, shows only the plus two and plus four states and not the minus two state.
Final answer: Po
Q63Single correctThe d- and f-Block Elements
'Spin only' magnetic moment is same for which of the following ions?
A. T
B. C
C. M
D. F
E. S
Choose the most appropriate answer from the options given below:
A. T
B. C
C. M
D. F
E. S
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B and D only
Approach:
The spin-only magnetic moment depends only on the number of unpaired electrons, so ions with equal unpaired electron counts share the same moment.
Step 1:The titanium(III) ion is 3d1 and carries one unpaired electron.
Step 2:The chromium(II) ion is 3d4 and carries four unpaired electrons.
Step 3:The manganese(II) ion is 3d5 with five unpaired electrons, the iron(II) ion is 3d6 with four unpaired electrons, and the scandium(III) ion is 3d0 with none.
Step 4:Chromium(II) and iron(II) both carry four unpaired electrons, so both give the same spin-only moment.
Final answer: B and D only
Q64Single correctOrganic Chemistry - Some Basic Principles and Techniques
A compound with a molecular formula has two tertiary carbons. Its IUPAC name is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 32,3-dimethylbutane
Approach:
A tertiary carbon is bonded to three other carbon atoms; each isomer of the formula is examined for the number of such carbons.
Step 1:n-Hexane is an unbranched chain whose carbons bond to at most two other carbons, so it has no tertiary carbon.
Step 2:2-Methylpentane has a single branch point and therefore only one tertiary carbon.
Step 3:2,3-Dimethylbutane has two adjacent branch points, each bonded to three carbons, giving two tertiary carbons.
Step 4:2,2-Dimethylbutane has a quaternary carbon and no tertiary carbon.
Final answer: 2,3-dimethylbutane
Q65Single correctOrganic Chemistry - Some Basic Principles and Techniques
Given below are two statements:
Statement I : The boiling point of three isomeric pentanes follows the order
n-pentane > isopentane > neopentane
Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I : The boiling point of three isomeric pentanes follows the order
n-pentane > isopentane > neopentane
Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Both the stated boiling-point order and the explanation based on branching and surface area are tested against the behaviour of the isomeric pentanes.
Step 1:The boiling points of the three pentane isomers fall in the order quoted, with the straight chain highest and the most branched lowest.
Step 2:Increased branching makes a molecule more compact and sphere-like, reducing the surface area available for intermolecular contact.
Step 3:Weaker intermolecular forces between the compact molecules lower the boiling point, so the second statement correctly explains the first.
Final answer: Both Statement I and Statement II are correct
Q66Single correctChemical Kinetics
Which plot of ln k vs is consistent with Arrhenius equation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A straight line of negative slope with a positive intercept (drawn graph, option 4)
Approach:
Taking the natural logarithm of the Arrhenius equation produces a linear relation between ln k and the reciprocal of temperature, whose slope and intercept fix the correct plot.
Step 1:Taking the logarithm of the Arrhenius equation gives a linear equation in the reciprocal of temperature.
Step 2:The slope of the line equals minus the activation energy divided by the gas constant, which is negative, and the intercept equals the logarithm of the pre-exponential factor.
Step 3:A plot of ln k against the reciprocal of temperature is therefore a straight line falling from left to right.
Final answer: A straight line of negative slope with a positive intercept (drawn graph, option 4)
Q67Single correctSolutions
The Henry's law constant (K) values of three gases (A, B, C) in water are 145, and 35 kbar, respectively. The solubility of these gases in water follow the order:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B > C > A
Approach:
The solubility of a gas at a fixed pressure is inversely proportional to its Henry's law constant, so the gas with the smallest constant is the most soluble.
Step 1:A higher Henry's law constant at a given partial pressure corresponds to a lower solubility of the gas in the liquid.
Step 2:Ordering the three constants from smallest to largest places gas B lowest, gas C next and gas A highest.
Step 3:Inverting this gives the solubility order with gas B most soluble and gas A least soluble.
Final answer: B > C > A
Q68Single correctAldehydes, Ketones and Carboxylic Acids
Fehling's solution 'A' is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1aqueous copper sulphate
Approach:
Fehling's reagent is prepared from two separate solutions, and the identity of solution A follows from its standard composition.
Step 1:Fehling's solution A is an aqueous solution of copper sulphate.
Step 2:Fehling's solution B is an alkaline solution of sodium potassium tartrate, known as Rochelle salt, which is mixed with A before use.
Final answer: aqueous copper sulphate
Q69Single correctThe p-Block Elements
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order
O > Te > Se > S.
Statement II: On the basis of molecular mass, O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: The boiling point of hydrides of Group 16 elements follow the order
O > Te > Se > S.
Statement II: On the basis of molecular mass, O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in O, it has higher boiling point.
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 true
Approach:
Both the quoted boiling-point order and the hydrogen-bonding explanation for the water anomaly are checked against the known behaviour of group 16 hydrides.
Step 1:The boiling points of the group 16 hydrides follow the order with water highest, then the heavier hydrides falling as molar mass decreases.
Step 2:On molar mass grounds alone water would have the lowest boiling point, but extensive intermolecular hydrogen bonding raises it above the others.
Step 3:Both statements describe the correct order and the correct reason, so both are true.
Final answer: Both Statement I and Statement II are true
Q70Single correctCoordination Compounds
Given below are two statements :
Statement I: Both and complexes are octahedral but differ in their magnetic behaviour.
Statement II: is diamagnetic whereas is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Both and complexes are octahedral but differ in their magnetic behaviour.
Statement II: is diamagnetic whereas is paramagnetic.
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 true
Approach:
The cobalt(III) centre in both complexes is octahedral, and the strength of each ligand decides whether the electrons pair up, fixing the magnetic behaviour.
Step 1:In both complexes the cobalt ion is in the plus three oxidation state with a d6 configuration and a six-coordinate octahedral arrangement.
Step 2:Ammonia is a strong-field ligand that pairs the d electrons, giving inner-orbital d2sp3 hybridisation and a diamagnetic complex.
Step 3:Fluoride is a weak-field ligand that leaves four unpaired electrons, giving outer-orbital sp3d2 hybridisation and a paramagnetic complex.
Final answer: Both Statement I and Statement II are true
Q71Single correctOrganic Chemistry - Some Basic Principles and Techniques
The most stable carbocation among the following is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The tertiary carbocation on the methyl-bearing ring carbon of cyclohexane (drawn structure, option 4)
Approach:
The stability of a carbocation is governed by hyperconjugation, which increases with the number of alpha hydrogen atoms available for delocalisation into the empty p orbital.
Step 1:The number of alpha hydrogens on each carbocation determines the extent of hyperconjugative stabilisation.
Step 2:Counting the alpha hydrogens gives seven for option four, five for option two, three for option one and one for option three.
Step 3:The carbocation with seven alpha hydrogens enjoys the greatest hyperconjugation and is therefore the most stable.
Final answer: The tertiary carbocation on the methyl-bearing ring carbon of cyclohexane (drawn structure, option 4)
Q72Single correctOrganic Chemistry - Some Basic Principles and Techniques
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Sublimation
Approach:
Each technique is matched to its underlying principle, and the one based on a direct solid-to-vapour change is identified.
Step 1:Crystallization relies on the difference in solubility of the compound and its impurities in a suitable solvent.
Step 2:Sublimation purifies a solid that passes directly from solid to vapour on heating without forming a liquid.
Step 3:Distillation separates volatile liquids and chromatography separates components using stationary and mobile phases, neither matching the stated principle.
Final answer: Sublimation
Q73Single correctCoordination Compounds
Choose the correct answer from the options given below:
| List I (Complex) | List II (Type of isomerism) |
|---|---|
| A. | I. Solvate isomerism |
| B. | II. Linkage isomerism |
| C. | III. Ionization isomerism |
| D. | IV. Coordination isomerism |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each complex is assigned the type of isomerism it can display, based on the nature of its ligands, counter ions and metal centres.
Step 1:The nitrito complex can bind through either the nitrogen or the oxygen of the ambidentate group, showing linkage isomerism.
Step 2:The sulphate-bromide complex can exchange the coordinated and ionic groups, showing ionization isomerism.
Step 3:The complex with both cobalt and chromium centres can interchange ligands between the two coordination spheres, showing coordination isomerism.
Step 4:The hexaaqua complex can vary the number of water molecules inside and outside the coordination sphere, showing solvate isomerism.
Final answer: A-II, B-III, C-IV, D-I
Q74Single correctBiomolecules
The reagents with which glucose does not react to give the corresponding tests/products are
A. Tollen's reagent
B. Schiff's reagent
C. HCN
D. NOH
E. NaHS
Choose the correct options from the given below:
A. Tollen's reagent
B. Schiff's reagent
C. HCN
D. NOH
E. NaHS
Choose the correct options from the given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B and E
Approach:
Glucose exists predominantly as a cyclic hemiacetal, so its potential aldehyde group is masked. Reactions that need a free aldehyde in solution fail, while reactions that the open-chain form can undergo succeed.
Step 1:Glucose reduces Tollen's reagent, forms a cyanohydrin with HCN, and forms an oxime with NH2OH, so A, C and D give positive results.
Step 2:Despite carrying a carbonyl-like centre, glucose fails the Schiff's test because the free aldehyde is locked in the cyclic hemiacetal form.
Step 3:Glucose does not form the hydrogen sulphite (bisulphite) addition product with NaHSO3 for the same reason, the carbonyl being unavailable.
Final answer: B and E
Q75Single correctAldehydes, Ketones and Carboxylic Acids
Choose the correct answer from the options given below:
| List I (Reaction) | List II (Reagents/Condition) |
|---|---|
A. ![]() | I. ![]() |
B. ![]() | II. |
C. ![]() | III. KMn/KOH, |
D. ![]() | IV. (i) (ii) Zn-O |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-I, C-II, D-III
Approach:
Each transformation is paired with the reagent that effects it, based on the functional-group change required.
Step 1:Reaction A cleaves the carbon-carbon double bond to two carbonyl compounds, which is reductive ozonolysis using ozone followed by zinc and water.
Step 2:Reaction B introduces an alkyl group onto the aromatic ring, a Friedel-Crafts type process using the alkyl chloride with anhydrous aluminium chloride.
Step 3:Reaction C oxidises a secondary alcohol to a ketone using chromium trioxide.
Step 4:Reaction D oxidises an aromatic side chain to a carboxylic acid using hot alkaline potassium permanganate.
Final answer: A-IV, B-I, C-II, D-III
Q76Single correctChemical Kinetics
Activation energy of any chemical reaction can be calculated if one knows the value of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4rate constant at two different temperatures
Approach:
The activation energy is obtained from the Arrhenius equation in its two-temperature form, which relates the rate constants at two temperatures to the energy of activation.
Step 1:The Arrhenius equation expressed for two temperatures relates the ratio of rate constants to the activation energy.
Step 2:Rearranging for the activation energy shows that knowledge of the rate constants at two temperatures determines its value.
Final answer: rate constant at two different temperatures
Q77Single correctElectrochemistry
Match List I with List II. Choose the correct answer from the options given below:
| List I (Conversion) | List II (Number of Faraday required) |
|---|---|
| A. 1 mol of O to | I. 3F |
| B. 1 mol of MnO to M | II. 2F |
| C. 1.5 mol of Ca from molten CaC | III. 1F |
| D. 1 mol of FeO to F | IV. 5F |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-I, D-III
Approach:
The charge required is found from the change in oxidation state per ion multiplied by the number of moles, with one mole of electrons equal to one Faraday.
Step 1:Oxidation of water to oxygen releases four electrons per O2 molecule, so two moles of H2O supply 4F; one mole of H2O therefore requires 2F.
Step 2:Permanganate is reduced from the +7 to the +2 state, a gain of five electrons, so one mole of MnO4- requires 5F.
Step 3:Calcium is deposited as Ca2+ gaining two electrons, so one mole needs 2F and 1.5 mol needs 3F.
Step 4:Conversion of FeO to Fe2O3 oxidises iron from +2 to +3, a loss of one electron per iron, so one mole of FeO requires 1F.
Final answer: A-II, B-IV, C-I, D-III
Q78Single correctHydrogen Bonding / General Organic Chemistry
Intramolecular hydrogen bonding is present in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1ortho-Nitrophenol N and OH on adjacent ring carbons (drawn structure, option 1)
Approach:
Intramolecular hydrogen bonding requires a donor and acceptor positioned close enough within the same molecule, which occurs only in the ortho isomer of nitrophenol.
Step 1:In ortho-nitrophenol the hydroxyl hydrogen lies adjacent to an oxygen of the nitro group, allowing a hydrogen bond within the same molecule.
Step 2:In the meta and para isomers the two groups are too far apart, so only intermolecular hydrogen bonding is possible; hydrogen fluoride forms only intermolecular hydrogen bonds.
Final answer: ortho-Nitrophenol N and OH on adjacent ring carbons (drawn structure, option 1)
Q79Single correctAmines
Given below are two statements :
Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both statement I and Statement II are true
Approach:
Both statements are assessed against the known behaviour of aniline towards Lewis-acid catalysed alkylation and towards the Gabriel phthalimide route.
Step 1:Aniline does not undergo Friedel-Crafts alkylation because its lone pair forms a salt with the Lewis acid catalyst aluminium chloride, leaving the ring strongly deactivated.
Step 2:Aromatic primary amines cannot be made by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion derived from phthalimide.
Final answer: Both statement I and Statement II are true
Q80Single correctAlcohols, Phenols and Ethers
Which one of the following alcohols reacts instantaneously with Lucas reagent?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The tertiary alcohol a carbon bearing three C groups and OH (drawn structure, option 4)
Approach:
Reactivity towards Lucas reagent follows the order tertiary greater than secondary greater than primary, since the reaction proceeds through a carbocation whose stability governs the rate.
Step 1:Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride that converts alcohols to alkyl chlorides through carbocation intermediates.
Step 2:Among the choices, only is a tertiary alcohol, forming the most stable tertiary carbocation and producing immediate turbidity.
Final answer: The tertiary alcohol a carbon bearing three C groups and OH (drawn structure, option 4)
Q81Single correctClassification of Elements and Periodicity
Arrange the following elements in increasing order of first ionization enthalpy:
Li, Be, B, C, N
Choose the correct answer from the options given below:
Li, Be, B, C, N
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Li < B < Be < C < N
Approach:
Ionization enthalpy generally rises across a period, but boron is lower than beryllium and the order is settled using the reported values for the second-period elements.
Step 1:Beryllium has a fully filled 2s subshell, giving it a higher ionization enthalpy than boron, whose electron is removed from the higher-energy 2p orbital.
Step 2:The first ionization enthalpy values in kJ per mol are Li 520, B 801, Be 899, C 1086 and N 1402, which place the elements in increasing order.
Final answer: Li < B < Be < C < N
Q82Single correctRedox Reactions
Which reaction is NOT a redox reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A reaction is non-redox when no element changes its oxidation state; each option is checked for oxidation-number changes.
Step 1:In the zinc-copper sulphate, potassium chlorate-iodine and hydrogen-chlorine reactions the oxidation states of the elements change, so these are redox reactions.
Step 2:In the reaction of barium chloride with sodium sulphate, barium, chlorine, sodium and the sulphate group all keep their oxidation states; it is a double-displacement precipitation reaction.
Final answer:
Q83Single correctStructure of Atom
Choose the correct answer from the options given below :
| List I (Quantum Number) | List II (Information provided) |
|---|---|
| A. | I. Shape of orbital |
| B. | II. Size of orbital |
| C. | III. Orientation of orbital |
| D. | IV. Orientation of spin of electron |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Each quantum number is matched to the property it specifies for an electron in an atom.
Step 1:The spin quantum number gives the orientation of spin of the electron, and the magnetic quantum number gives the orientation of the orbital in space.
Step 2:The azimuthal quantum number determines the shape of the orbital, and the principal quantum number determines its size.
Final answer: A-IV, B-III, C-I, D-II
Q84Single correctThermodynamics
In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C.
D.
Choose the correct answer from the options given below:
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C.
D.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, C and D
Approach:
Entropy increases when disorder increases, which happens on forming gases or more moles of gaseous particles, and decreases on cooling a solid.
Step 1:Evaporation of a liquid to vapour increases molecular disorder, so entropy increases in process A.
Step 2:Lowering the temperature of a crystalline solid from 130 K toward 0 K decreases disorder, so entropy decreases in process B.
Step 3:Decomposition of sodium bicarbonate produces gaseous products from a solid, increasing the number of gaseous molecules; dissociation of chlorine forms two moles of gas from one, also increasing disorder.
Final answer: A, C and D
Q85Single correctSome Basic Concepts of Chemistry
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2250 mg
Approach:
The mass of HCl available is found from its molarity and volume, then stoichiometry of the one-to-one neutralization gives the mass of sodium hydroxide consumed and the amount left over.
Step 1:The mass of HCl present is obtained from molarity 0.75, molar mass 36.5 and volume 25 mL.
Step 2:Neutralization follows , so 36.5 g HCl reacts with 40 g NaOH; the NaOH consumed by 0.684 g HCl is computed proportionally.
Step 3:Subtracting the reacted mass from the initial 1 g gives the unreacted sodium hydroxide.
Final answer: 250 mg
Q86Single correctHaloalkanes and Haloarenes
The products A and B obtained in the following reactions, respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 and
Approach:
Both reactions convert alcohols to alkyl chlorides; the phosphorus-containing by-products are identified by completing each balanced equation.
Step 1:Reaction of three moles of alcohol with phosphorus trichloride yields alkyl chloride and phosphorous acid as the by-product A.
Step 2:Reaction of alcohol with phosphorus pentachloride yields alkyl chloride, hydrogen chloride and phosphorus oxychloride as the by-product B.
Final answer: and
Q87Single correctHaloalkanes and Haloarenes
Major products A and B formed in the following reaction sequence, are

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A = 1-bromo-2-methylcyclohexane, B = 1-methylcyclohex-1-ene (drawn structures, option 1)
Approach:
Phosphorus tribromide replaces the hydroxyl group by bromine, and alcoholic potassium hydroxide then carries out base-induced elimination to give the alkene.
Step 1:Phosphorus tribromide converts the secondary hydroxyl group into a bromide, giving 1-bromo-2-methylcyclohexane as product A.
Step 2:Alcoholic potassium hydroxide on heating causes dehydrohalogenation. Removing the proton from the methyl-bearing carbon builds the more substituted, and hence more stable, trisubstituted double bond, so the Zaitsev product 1-methylcyclohex-1-ene is the major alkene B.
Final answer: A = 1-bromo-2-methylcyclohexane, B = 1-methylcyclohex-1-ene (drawn structures, option 1)
Q88Single correctAmines / Nitriles
Identify the major product C formed in the following reaction sequence:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1propylamine
Approach:
The cyanide displaces iodide, partial hydrolysis gives an amide, and the amide undergoes Hoffmann bromamide degradation, which shortens the carbon chain by one carbon.
Step 1:Substitution of iodide by cyanide gives butanenitrile as A through an reaction.
Step 2:Partial hydrolysis of the nitrile converts it to butanamide as B.
Step 3:Hoffmann bromamide degradation of the amide with bromine and sodium hydroxide removes one carbon and produces the primary amine propylamine as C.
Final answer: propylamine
Q89Single correctRedox / d-Block Elements
During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of ion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4dilute sulphuric acid
Approach:
The acid added must suppress hydrolysis of the ferrous ion without introducing a foreign anion or oxidizing the iron, which singles out dilute sulphuric acid.
Step 1:Mohr's salt contains the sulphate anion, so adding dilute sulphuric acid keeps the solution free of extra ions while supplying acidity.
Step 2:The added acidity shifts the hydrolysis equilibrium back, preventing precipitation of basic ferrous compounds, while a non-oxidizing dilute acid leaves the iron in the state.
Final answer: dilute sulphuric acid
Q90Single correctEquilibrium
Consider the following reaction in a sealed vessel at equilibrium with concentrations of M, M and NO M.
If 0.1 mol of is taken in a closed vessel, what will be degree of dissociation () of at equilibrium?
If 0.1 mol of is taken in a closed vessel, what will be degree of dissociation () of at equilibrium?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 40.717
Approach:
The equilibrium constant is evaluated from the given concentrations, then applied to the dissociation of 0.1 M NO using an ICE treatment to solve for the degree of dissociation.
Step 1:Substituting the given equilibrium concentrations gives the equilibrium constant.
Step 2:For 0.1 M NO dissociating, the equilibrium amounts are for NO and each for nitrogen and oxygen.
Step 3:Setting this equal to 1.607 and taking the square root gives a linear equation in .
Step 4:Solving the linear equation yields the degree of dissociation.
Final answer: 0.717
Q91Single correctThermodynamics
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is
(Given R = 2.0 cal )
(Given R = 2.0 cal )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2-413.14 calories
Approach:
The work for reversible isothermal expansion of an ideal gas is found from the standard logarithmic relation using the ratio of initial to final pressure.
Step 1:Substituting one mole, R equal to 2 cal per kelvin per mole, temperature 298 K and the pressure ratio 20 over 10 into the work expression.
Step 2:Evaluating with the logarithm of two equal to 0.3 gives the work done by the system on expansion.
Final answer: -413.14 calories
Q92Single correctElectrochemistry
Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g , 1 F = 96487 C)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.315 g
Approach:
The mass deposited is found from Faraday's first law using the molar mass of copper, the two electrons per copper ion, and the total charge passed.
Step 1:Copper is deposited by the reduction , so two electrons are required per copper atom.
Step 2:Substituting molar mass 63, current 9.6487 A, time 100 s, two electrons and the faraday constant gives the deposited mass.
Final answer: 0.315 g
Q93Single correctChemical Kinetics
The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation.
(Given R = 8.314 J , log4 = 0.6021)
(Given R = 8.314 J , log4 = 0.6021)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 138.04 kJ/mol
Approach:
The activation energy is obtained from the two-temperature Arrhenius equation using the four-fold rate ratio and the two absolute temperatures.
Step 1:Converting the temperatures to kelvin gives 300 K and 330 K, and the rate constant ratio is four.
Step 2:The bracket evaluates to 30/(300 x 330), so the activation energy follows directly.
Final answer: 38.04 kJ/mol
Q94Single correctSolutions
The plot of osmotic pressure () vs concentration (mol ) for a solution gives a straight line with slope 25.73 L bar . The temperature at which the osmotic pressure measurement is done is
(Use R = 0.083 L bar )
(Use R = 0.083 L bar )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 137°C
Approach:
Osmotic pressure equals concentration times RT, so the slope of the pressure-concentration line equals RT, from which the temperature is obtained.
Step 1:Since osmotic pressure is the product of concentration and RT, the slope of the straight line equals the product RT.
Step 2:Solving for the absolute temperature and converting to the Celsius scale gives the measurement temperature.
Final answer: 37°C
Q95Single correctQualitative Analysis (Inorganic)
Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI.
A.
B.
C.
D.
E.
Choose the correct answer from the options given below:
A.
B.
C.
D.
E.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B, A, D, C, E
Approach:
Each cation is assigned to its analytical group, and the cations are ordered by increasing group number.
Step 1:Copper belongs to Group II, aluminium to Group III, cobalt to Group IV, barium to Group V and magnesium to Group VI.
Step 2:Arranging these in increasing group number gives the sequence of cations.
Final answer: B, A, D, C, E
Q96Single correctCoordination Compounds
Given below are two statements :
Statement I : is a homoleptic complex whereas is a heteroleptic complex.
Statement II : Complex has only one kind of ligands but has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : is a homoleptic complex whereas is a heteroleptic complex.
Statement II : Complex has only one kind of ligands but has more than one kind of ligands.
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 true
Approach:
Each statement is tested against the definitions of homoleptic and heteroleptic complexes and against the ligand composition of the two cobalt species.
Step 1:A homoleptic complex bears only one kind of ligand. The hexaamminecobalt(III) ion carries six ammonia ligands and is therefore homoleptic, while the tetraamminedichloridocobalt(III) ion carries both ammonia and chlorido ligands and is therefore heteroleptic. Statement I is true.
Step 2:Statement II asserts exactly that ligand composition: one kind of ligand in the hexaammine ion and more than one kind in the tetraamminedichlorido ion. That is the reason the first complex is homoleptic and the second heteroleptic, so Statement II is also true and it explains Statement I.
Final answer: Both Statement I and Statement II are true
Q97Single correctd- and f-Block Elements
The pair of lanthanoid ions which are diamagnetic is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
An ion is diamagnetic only when it has no unpaired electrons; the f-electron count of each lanthanoid ion is examined.
Step 1:Cerium in the state has the configuration and ytterbium in the state has , both with no unpaired electrons.
Step 2:With no unpaired electrons the spin-only moment is zero, so both ions are diamagnetic, while the ions in the other pairs carry unpaired f-electrons and are paramagnetic.
Final answer: and
Q99Single correctChemical Bonding and Molecular Structure
Identify the correct answer.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Three canonical forms can be drawn for ion
Approach:
Each statement is checked against the resonance, geometry and dipole properties of the species named.
Step 1:Ozone has only two resonating structures, not three, so statement one is incorrect; boron trifluoride is trigonal planar and symmetric, giving zero dipole moment, so statement two is incorrect.
Step 2:In nitrogen trifluoride the bond dipoles oppose the lone-pair moment, making its dipole smaller than that of ammonia, so statement three is incorrect.
Step 3:The carbonate ion has the negative charge delocalized over three equivalent oxygen atoms, giving three canonical resonance forms.
Final answer: Three canonical forms can be drawn for ion
Q100Single correctSome Basic Concepts of Chemistry
A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is :
(Given atomic masses of A = 64; B = 40; C = 32 u)
(Given atomic masses of A = 64; B = 40; C = 32 u)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The mole ratio of the three elements is found by dividing each percentage by its atomic mass and then scaling to the smallest value.
Step 1:The percentage of C is the remainder, equal to 48%; dividing each percentage by its atomic mass gives the relative moles of A, B and C.
Step 2:Dividing each relative mole value by the smallest gives the simplest whole-number ratio.
Final answer:
Biology100 questions
Q101Single correctBiological Classification
Match List I with List II Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. | I. Mushroom |
| B. | II. Smut fungus |
| C. | III. Bread mould |
| D. | IV. Rust fungus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-II, C-IV, D-I
Approach:
Each fungal genus is matched to its common name based on its characteristic structure and disease association.
Step 1:Rhizopus is a saprophytic mould that grows on stale bread, giving it the common name bread mould.
Step 2:Ustilago causes smut disease in cereals and is referred to as the smut fungus.
Step 3:Puccinia is the causal organism of rust diseases such as wheat rust and is called the rust fungus.
Step 4:Agaricus is the edible basidiomycete commonly known as mushroom.
Final answer: A-III, B-II, C-IV, D-I
Q102Single correctMorphology of Flowering Plants
Identify the part of the seed from the given figure which is destined to form root when the seed germinates.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C
Approach:
The part of the embryo axis that develops into the root system on germination is the radicle, which is identified from its position in the labelled seed diagram.
Step 1:In a seed, the embryonal axis carries the plumule at one end and the radicle at the other; the radicle gives rise to the root.
Step 2:In the labelled diagram, the lowermost embryonal axis tip marked C corresponds to the radicle.
Final answer: C
Q103Single correctBiomolecules
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Phospholipids
Approach:
Lecithin is classified by examining its chemical make-up, which contains a phosphorus-bearing group attached to a glycerol backbone.
Step 1:Lecithin contains phosphorus and a phosphorylated organic group, a hallmark of phospholipids found in cell membranes.
Step 2:Glycerides are a separate class of lipids lacking the phosphate group, so option (3) does not apply.
Step 3:Amino acids and carbohydrates form distinct classes of biomolecules unrelated to lecithin.
Final answer: Phospholipids
Q104Single correctPrinciples of Inheritance and Variation
Match List I with List II Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Two or more alternative forms of a gene | I. Back cross |
| B. Cross of progeny with homozygous recessive parent | II. Ploidy |
| C. Cross of progeny with any of the parents | III. Allele |
| D. Number of chromosome sets in plant | IV. Test cross |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Each genetic definition in List I is paired with the standard term it describes.
Step 1:Two or more alternative forms of a gene are called alleles.
Step 2:A cross of the F1 progeny with the homozygous recessive parent is a test cross.
Step 3:A cross of the F1 progeny with any of the parents is a back cross.
Step 4:The number of chromosome sets in a plant is termed ploidy.
Final answer: A-III, B-IV, C-I, D-II
Q105Single correctMolecular Basis of Inheritance
A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Promoter, Structural gene, Terminator
Approach:
A transcription unit is defined by its three principal regions located relative to the upstream and downstream ends of the coding strand.
Step 1:The promoter lies towards the 5'-end (upstream) of the structural gene relative to the polarity of the coding strand.
Step 2:The structural gene is the central region that is transcribed into RNA.
Step 3:The terminator is located towards the 3'-end (downstream) of the coding strand and defines the end of transcription.
Final answer: Promoter, Structural gene, Terminator
Q106Single correctPlant Growth and Development
Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3does not affect mature monocotyledonous plants.
Approach:
The selective herbicidal action of auxin on dicot weeds while sparing grasses is explained by the differing sensitivity of monocots and dicots to auxin.
Step 1:Auxin acts as a weedicide against broad-leaved dicot weeds but does not affect mature monocot plants.
Step 2:Grasses are monocots and show limited translocation of auxin, so the lawn grass is spared while dicot weeds undergo rapid uncontrolled growth and die.
Final answer: does not affect mature monocotyledonous plants.
Q107Single correctBiodiversity and Conservation
These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, B and D only
Approach:
The recognised causes of biodiversity loss, often called the evil quartet, are identified from the listed statements.
Step 1:Habitat loss and fragmentation, over exploitation, alien species invasions, and co-extinctions are the major causes of biodiversity loss.
Step 2:Mutation and migration are not counted among the major causes of biodiversity loss, so statements C and E are excluded.
Final answer: A, B and D only
Q108Single correctMicrobes in Human Welfare
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. | I. Ethanol |
| B. | II. Streptokinase |
| C. | III. Butyric acid |
| D. | IV. Cyclosporin-A |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Each microbe is paired with the specific commercial product it yields, following the industrial-microbiology pairings given in the chapter on microbes in human welfare.
Step 1:Clostridium butylicum is an anaerobic bacterium used in the fermentative production of butyric acid, pairing A with III.
Step 2:Saccharomyces cerevisiae, brewer's and baker's yeast, ferments sugars to ethanol, pairing B with I.
Step 3:Trichoderma polysporum produces cyclosporin-A, the immunosuppressant used in organ transplantation, pairing C with IV.
Step 4:Streptococcus sp. yields streptokinase, the clot-dissolving enzyme, pairing D with II.
Final answer: A-III, B-I, C-IV, D-II
Q109Single correctCell Cycle and Cell Division
Spindle fibers attach to kinetochores of chromosomes during
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Metaphase
Approach:
The phase of cell division in which spindle fibres make contact with the kinetochores is identified from the events of mitosis.
Step 1:During metaphase the spindle fibres attach to the kinetochores located on the centromeres of the chromosomes.
Step 2:The chromosomes then align at the metaphase plate held by these spindle attachments.
Final answer: Metaphase
Q110Single correctMorphology of Flowering Plants
Which of the following is an example of actinomorphic flower?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The flower with radial symmetry, able to be divided into equal halves along any plane through the centre, is the actinomorphic example.
Step 1:Datura bears radially symmetrical flowers and is therefore actinomorphic.
Step 2:Cassia, Pisum, and Sesbania bear zygomorphic flowers with bilateral symmetry.
Final answer:
Q111Single correctMineral Nutrition
The cofactor of the enzyme carboxypeptidase is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Zinc
Approach:
The mineral element serving as the metal cofactor of carboxypeptidase is identified from the role of micronutrients in enzyme activity.
Step 1:Zinc acts as the cofactor of the enzyme carboxypeptidase.
Step 2:Niacin and flavin are coenzyme precursors (NAD and FAD/NADP) rather than the cofactor here, and haem is the prosthetic group of peroxidase and catalase.
Final answer: Zinc
Q112Single correctCell - The Unit of Life
Match List I with List II Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Nucleolus | I. Site of formation of glycolipid |
| B. Centriole | II. Organization like the cartwheel |
| C. Leucoplasts | III. Site for active ribosomal RNA synthesis |
| D. Golgi apparatus | IV. For storing nutrients |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-II, C-IV, D-I
Approach:
Each cell organelle is matched to its characteristic function or structural feature.
Step 1:The nucleolus is the site for active ribosomal RNA synthesis.
Step 2:Centrioles in a centrosome have a cartwheel-like organisation.
Step 3:Leucoplasts are colourless plastids that store nutrients.
Step 4:The Golgi apparatus is important for the formation of glycoproteins and glycolipids.
Final answer: A-III, B-II, C-IV, D-I
Q113Single correctPlant Growth and Development
The capacity to generate a whole plant from any cell of the plant is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Totipotency
Approach:
The term describing the inherent ability of a single plant cell to regenerate an entire plant is identified.
Step 1:Totipotency is defined as the capacity to generate a whole plant from any cell of the plant.
Step 2:Micropropagation, differentiation, and somatic hybridization describe other tissue-culture processes and do not define this capacity.
Final answer: Totipotency
Q114Single correctBiotechnology - Principles and Processes
II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 26 bp
Approach:
The length of the recognition sequence of the first restriction endonuclease Hind II is recalled from the history of restriction enzymes.
Step 1:Hind II functioning depends on a specific DNA nucleotide sequence whose function was isolated as the recognition sequence.
Step 2:Hind II always cuts DNA molecules at a particular point within a set sequence of six base pairs.
Final answer: 6 bp
Q115Single correctSexual Reproduction in Flowering Plants
Identify the set of correct statements:
A. The flowers of are colourful and produce nectar.
B. The flowers of water lily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below:
A. The flowers of are colourful and produce nectar.
B. The flowers of water lily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B, C, D and E only
Approach:
Each statement on water and insect pollination is assessed against the known features of aquatic flowers.
Step 1:Flowers of Vallisneria are not colourful and do not produce nectar, so statement A is incorrect.
Step 2:Water lily is pollinated by insects or wind rather than by water, making statement B correct.
Step 3:In most water-pollinated species the pollen grains are protected from wetting by a mucilaginous covering, confirming statement C.
Step 4:Pollen grains of some hydrophytes are long and ribbon like, and in some hydrophytes the pollen is carried passively inside water, confirming statements D and E.
Final answer: B, C, D and E only
Q116Single correctAnatomy of Flowering Plants
Given below are two statements:
Statement I : Parenchyma is living but collenchyma is dead tissue.
Statement II : Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
Statement I : Parenchyma is living but collenchyma is dead tissue.
Statement II : Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is false but Statement II is true
Approach:
Each statement is evaluated against the nature of plant tissues and the vascular features of gymnosperms and angiosperms.
Step 1:Collenchyma is a living tissue, so the claim that it is dead makes Statement I false.
Step 2:Gymnosperms lack xylem vessels while the presence of xylem vessels is characteristic of angiosperms, making Statement II true.
Final answer: Statement I is false but Statement II is true
Q117Single correctPlant Growth and Development
Formation of interfascicular cambium from fully developed parenchyma cells is an example for
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Dedifferentiation
Approach:
The developmental process by which mature differentiated cells regain the capacity to divide is identified.
Step 1:The regaining of meristematic activity by fully differentiated parenchyma cells, as in formation of interfascicular cambium, is termed dedifferentiation.
Final answer: Dedifferentiation
Q118Single correctRespiration in Plants
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Competitive inhibition
Approach:
The type of enzyme inhibition shown by malonate on succinic dehydrogenase is determined from the structural relationship between malonate and the natural substrate.
Step 1:Malonate shows close structural similarity with the substrate succinate and competes with it for the substrate-binding site of succinic dehydrogenase.
Step 2:This competition for the active site defines competitive inhibition, so the other options of enzyme activation, feedback inhibition, and cofactor inhibition do not apply.
Final answer: Competitive inhibition
Q119Single correctMorphology of Flowering Plants
Identify the type of flowers based on the position of calyx, corolla and androecium with respect to the ovary from the given figures (a) and (b)

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) Perigynous; (b) Perigynous
Approach:
The flower type is determined by the level at which the calyx, corolla, and androecium are attached relative to the ovary in each diagram.
Step 1:When the gynoecium is situated in the centre and the other floral parts are located on the rim of the thalamus almost at the same level, the flower is perigynous.
Step 2:Both diagrams (a) and (b) display this perigynous arrangement of floral parts around a half-inferior ovary.
Final answer: (a) Perigynous; (b) Perigynous
Q120Single correctPrinciples of Inheritance and Variation
In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2bb
Approach:
The genotype of a dominant-phenotype plant is determined by crossing it with the homozygous recessive type, a test cross.
Step 1:To determine the genotype of a black seed colour at F2, the black seed plant is crossed with the white seed type, which is the homozygous recessive bb.
Step 2:The ratio of offspring phenotypes from this test cross reveals whether the black plant is homozygous BB or heterozygous Bb.
Final answer: bb
Q121Single correctBiodiversity and Conservation
Tropical regions show greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below:
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, C, D and E only
Approach:
Each proposed reason for high tropical species richness is checked against the recognised explanations.
Step 1:Long undisturbed time for diversification, greater solar energy, niche specialization in constant environments, and the constant predictable nature of the tropics all promote species richness, validating statements A, C, D and E.
Step 2:Tropical environments are less seasonal and relatively more constant, so statement B is incorrect.
Final answer: A, C, D and E only
Q122Single correctOrganisms and Populations
The equation of Verhulst-Pearl logistic growth is .
From this equation, K indicates:
From this equation, K indicates:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Carrying capacity
Approach:
The symbol K in the Verhulst-Pearl logistic growth equation is interpreted from the meaning of its terms.
Step 1:In the logistic growth equation, r is the intrinsic rate of natural increase and N is the population size, while K represents the maximum population the environment can sustain.
Final answer: Carrying capacity
Q123Single correctAnatomy of Flowering Plants
In the given figure, which component has thin outer walls and highly thickened inner walls?

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C
Approach:
The structure with thin outer and thick inner walls in the stomatal apparatus is identified from the labelled figure.
Step 1:Guard cells of the stomata have a thin outer wall and a highly thickened inner wall that controls opening and closing.
Step 2:In the labelled diagram the guard cell corresponds to the component marked C.
Final answer: C
Q124Single correctCell Cycle and Cell Division
Given below are two statements:
Statement I : Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements, choose the correct answer from the options given below:
Statement I : Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are true
Approach:
Each statement on the sub-stages of prophase I of meiosis is evaluated against their defining events.
Step 1:During the leptotene stage the chromosomes become gradually visible under the light microscope, validating Statement I.
Step 2:The beginning of diplotene is recognised by the dissolution of the synaptonemal complex and the tendency of recombined homologous chromosomes of a bivalent to separate from each other except at the site of crossover, validating Statement II.
Final answer: Both Statement I and Statement II are true
Q125Single correctBiodiversity and Conservation
List of endangered species was released by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4IUCN
Approach:
The organisation responsible for publishing the list of endangered species is identified.
Step 1:The International Union for Conservation of Nature, IUCN, releases the Red List of endangered species.
Final answer: IUCN
Q126Single correctPhotosynthesis in Higher Plants
Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C.
D. ATP
E. NADPH
Choose the correct answer from the options given below:
A. Light
B. Chlorophyll
C.
D. ATP
E. NADPH
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C, D and E only
Approach:
The biosynthetic (dark) phase of photosynthesis fixes carbon dioxide using the assimilatory power produced in the light reactions.
Step 1:The dark reaction proceeds in the stroma and does not directly require light or chlorophyll.
Step 2:The Calvin cycle requires carbon dioxide as the substrate together with ATP and NADPH supplied by the light reactions.
Final answer: C, D and E only
Q127Single correctOrganisms and Populations
The type of conservation in which the threatened species are taken out from their natural habitat and placed in special setting where they can be protected and given special care is called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Biodiversity conservation
Approach:
Removing a threatened species from its natural habitat and giving it protective care in a special setting defines off-site conservation.
Step 1:Conservation carried out away from the natural habitat, in places such as zoological parks, botanical gardens and wildlife safari parks, is ex-situ or off-site conservation, and the stem describes exactly that.
Step 2:Ex-situ conservation is one of the two approaches, alongside in-situ conservation, that together make up biodiversity conservation. Of the terms listed, biodiversity conservation is the one that covers it; in-situ conservation is its opposite, semi-conservative refers to DNA replication and sustainable development is a resource-use principle, not a conservation method.
Final answer: Biodiversity conservation
Q128Single correctMolecular Basis of Inheritance
What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B and C only
Approach:
A piece of DNA lacking an origin of replication cannot multiply on its own inside a host cell.
Step 1:Without an origin of replication the alien DNA fails to multiply independently in the progeny cells, ruling out statements A, D and E.
Step 2:Such a fragment may instead get integrated into the recipient genome, after which it multiplies and is inherited along with the host DNA.
Final answer: B and C only
Q129Single correctBiotechnology: Principles and Processes
The lactose present in the growth medium of bacteria is transported to the cell by the action of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Permease
Approach:
The lac operon encodes three structural genes whose products handle lactose metabolism.
Step 1:The gene y of the lac operon codes for the enzyme permease.
Step 2:Permease increases the permeability of the cell to beta-galactosides, thereby transporting lactose from the medium into the cell.
Final answer: Permease
Q130Single correctPrinciples of Inheritance and Variation
Which one of the following can be explained on the basis of Mendel's Law of Dominance?
A. Out of one pair of factors one is dominant and the other is recessive.
B. Alleles do not show any expression and both the characters appear as such in generation.
C. Factors occur in pairs in normal diploid plants.
D. The discrete unit controlling a particular character is called factor.
E. The expression of only one of the parental characters is found in a monohybrid cross.
Choose the correct answer from the options given below:
A. Out of one pair of factors one is dominant and the other is recessive.
B. Alleles do not show any expression and both the characters appear as such in generation.
C. Factors occur in pairs in normal diploid plants.
D. The discrete unit controlling a particular character is called factor.
E. The expression of only one of the parental characters is found in a monohybrid cross.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, C, D and E only
Approach:
Mendel's Law of Dominance explains the existence of factors, their pairing and the dominance of one allele over another.
Step 1:Characters are controlled by discrete units called factors, which occur in pairs in diploids, with one factor dominant and the other recessive, so only one parental character is expressed in a monohybrid cross.
Step 2:Statement B claims alleles show no expression and characters reappear unchanged in the second filial generation, which contradicts dominance and the law of segregation.
Final answer: A, C, D and E only
Q131Single correctBiological Classification
Which one of the following is a criterion for classification of fungi?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mode of nutrition
Approach:
Classification of fungi into classes rests on structural and reproductive criteria rather than nutritional mode.
Step 1:The morphology of the mycelium, the mode of spore formation and the nature of the fruiting body form the basis for dividing the kingdom Fungi into classes.
Step 2:All fungi are heterotrophic, so mode of nutrition does not distinguish the fungal classes.
Final answer: Mode of nutrition
Q132Single correctBiotechnology and its Applications
Given below are two statements:
Statement I : Bt toxins are insect group specific and coded by a gene cry IAc.
Statement II : Bt toxin exists as inactive protoxin in B.\ thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
Statement I : Bt toxins are insect group specific and coded by a gene cry IAc.
Statement II : Bt toxin exists as inactive protoxin in B.\ thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
Bt toxin specificity and its activation inside the insect gut determine the truth of each statement.
Step 1:Bt toxins are insect-group specific, and specific Bt toxin genes such as cry IAc were isolated from Bacillus thuringiensis and incorporated into crop plants.
Step 2:The protoxin is inactive in the bacterium and is converted to the active toxin in the insect gut owing to its alkaline pH, not acidic pH.
Final answer: Statement I is true but Statement II is false
Q133Single correctAnatomy of Flowering Plants
Bulliform cells are responsible for
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Inward curling of leaves in monocots.
Approach:
Bulliform cells are modified adaxial epidermal cells in grasses that control leaf rolling under water stress.
Step 1:In grasses certain adaxial epidermal cells along the veins enlarge into large, empty, colourless bulliform cells.
Step 2:When turgid, the leaf surface is exposed; when they become flaccid due to water stress, the leaves curl inwards to reduce water loss.
Final answer: Inward curling of leaves in monocots.
Q134Single correctPrinciples of Inheritance and Variation
A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Red flowered as well as pink flowered plants
Approach:
Flower colour in snapdragon shows incomplete dominance, with the heterozygote producing the pink intermediate.
Step 1:Red flower has genotype RR and pink flower has genotype Rr in snapdragon.
Step 2:The cross yields offspring in the ratio one RR to one Rr.
Final answer: Red flowered as well as pink flowered plants
Q135Single correctPhotosynthesis in Higher Plants
How many molecules of ATP and NADPH are required for every molecule of fixed in the Calvin cycle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 43 molecules of ATP and 2 molecules of NADPH
Approach:
The assimilatory power needed per carbon dioxide fixed follows from the Calvin cycle stoichiometry.
Step 1:Fixation of one molecule of carbon dioxide in the Calvin cycle consumes three molecules of ATP and two molecules of NADPH.
Final answer: 3 molecules of ATP and 2 molecules of NADPH
Q136Single correctOrganisms and Populations
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A.. Robert May | I.. Species-Area relationship |
| B.. Alexander von Humboldt | II.. Long term ecosystem experiment using out door plots |
| C.. Paul Ehrlich | III.. Global species diversity at about 7 million |
| D.. David Tilman | IV.. Rivet popper hypothesis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-I, C-IV, D-II
Approach:
Each ecologist is matched with the concept or experiment associated with their work.
Step 1:Robert May placed the global species diversity at about 7 million.
Step 2:Alexander von Humboldt described the species-area relationship.
Step 3:Paul Ehrlich used the analogy of the rivet popper hypothesis to explain the role of species in an ecosystem.
Step 4:David Tilman performed long term ecosystem experiments using outdoor plots.
Final answer: A-III, B-I, C-IV, D-II
Q137Single correctSexual Reproduction in Flowering Plants
Identify the correct description about the given figure:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Wind pollinated plant inflorescence showing flowers with well exposed stamens.
Approach:
The floral features in the figure indicate the agent of pollination.
Step 1:The diagram depicts a wind pollinated plant bearing a compact inflorescence with well exposed stamens.
Step 2:Because the stamens are exposed to disperse pollen by wind, complete autogamy does not occur.
Final answer: Wind pollinated plant inflorescence showing flowers with well exposed stamens.
Q138Single correctMolecular Basis of Inheritance
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A.. Frederick Griffith | I.. Genetic code |
| B.. Francois Jacob & Jacque Monod | II.. Semi-conservative mode of DNA replication |
| C.. Har Gobind Khorana | III.. Transformation |
| D.. Meselson & Stahl | IV.. operon |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Each scientist or pair is matched to their landmark discovery in molecular genetics.
Step 1:Frederick Griffith's experiment with miraculous transformation in bacteria demonstrated transformation.
Step 2:The elucidation of the Lac operon resulted from the close association between Francois Jacob and Jacques Monod.
Step 3:Har Gobind Khorana contributed to defining the combination of bases in the genetic code.
Step 4:Meselson and Stahl gave the semi-conservative mode of DNA replication.
Final answer: A-III, B-IV, C-I, D-II
Q139Single correctPlant Growth and Development
Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gibberellin
Approach:
The growth regulator that promotes internodal elongation enhances sugar storage in sugarcane stems.
Step 1:Sugarcanes store carbohydrate as sugar in their stems.
Step 2:Spraying sugarcane with gibberellins increases the length of the stem, thereby increasing the yield.
Final answer: Gibberellin
Q140Single correctSexual Reproduction in Flowering Plants
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Rose | I.. Twisted aestivation |
| B.. Pea | II.. Perigynous flower |
| C.. Cotton | III.. Drupe |
| D.. Mango | IV.. Marginal placentation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-I, D-III
Approach:
Each plant is matched with its characteristic floral or fruit feature.
Step 1:In rose the calyx, petals and stamens are borne on the rim of the thalamus, giving a perigynous flower.
Step 2:In pea the placenta develops along the ventral suture of the ovary, giving marginal placentation.
Step 3:In cotton the petals overlap so that one margin is covered while the other margin covers the next, giving twisted aestivation.
Step 4:In mango the fruit is a drupe with the stony endocarp enclosing the seed.
Final answer: A-II, B-IV, C-I, D-III
Q141Single correctBiological Classification
Read the following statements and choose the set of correct statements:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
Choose the correct answer from the options given below:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, C, D and E only
Approach:
The features of brown algae determine which statements are correct.
Step 1:In Phaeophyceae asexual reproduction is by biflagellate zoospores, stored food is mannitol or laminarin, the pigments are chlorophyll a and c with carotenoids and xanthophyll, and the cellulosic wall bears an algin coating.
Step 2:Sexual reproduction in members of Phaeophyceae may be isogamous, anisogamous or oogamous, so the claim of oogamy only is incorrect.
Final answer: A, C, D and E only
Q142Single correctRespiration in Plants
Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Succinyl-CoA Succinic acid
Approach:
Oxidation steps in the citric acid cycle involve loss of electrons and an increase in oxidation state of the substrate.
Step 1:Conversion of malic acid to oxaloacetic acid, succinic acid to malic acid, and isocitrate to alpha-ketoglutaric acid each involve removal of electrons and are oxidations.
Step 2:The conversion of succinyl-CoA to succinic acid is a substrate-level phosphorylation that does not oxidise the substrate.
Final answer: Succinyl-CoA Succinic acid
Q143Single correctPhotosynthesis in Higher Plants
Given below are two statements:
Statement I: In plants, some binds to RuBisCO, hence fixation is decreased.
Statement II: In plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: In plants, some binds to RuBisCO, hence fixation is decreased.
Statement II: In plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
Binding of oxygen to RuBisCO and the distribution of photorespiration determine the truth of each statement.
Step 1:In C3 plants RuBisCO also binds oxygen, and every oxygenation event diverts the enzyme from carboxylation, so carbon dioxide fixation falls.
Step 2:In C4 plants the mesophyll cells contain PEP carboxylase and no RuBisCO at all, so they show no photorespiration rather than 'very little'. The bundle sheath cells do hold RuBisCO, and although the C4 pump keeps the carbon dioxide concentration around it high, that is where any photorespiration would occur. Statement II therefore assigns the activity to the wrong cell type.
Final answer: Statement I is true but Statement II is false
Q144Single correctBiomolecules
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| A.. GLUT-4 | I.. Hormone |
| B.. Insulin | II.. Enzyme |
| C.. Trypsin | III.. Intercellular ground substance |
| D.. Collagen | IV.. Enables glucose transport into cells |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-I, C-II, D-III
Approach:
Each protein is matched with its biological role.
Step 1:GLUT-4 enables glucose transport into cells.
Step 2:Insulin is a hormone.
Step 3:Trypsin is an enzyme.
Step 4:Collagen forms the intercellular ground substance.
Final answer: A-IV, B-I, C-II, D-III
Q145Single correctMorphology of Flowering Plants
Choose the correct answer from the options given below:
| List I | List II (Example) |
|---|---|
| A.. Monoadelphous | I.. Citrus |
| B.. Diadelphous | II.. Pea |
| C.. Polyadelphous | III.. Lily |
| D.. Epiphyllous | IV.. China-rose |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-II, C-I, D-III
Approach:
Each type of cohesion of stamens is matched with its representative plant.
Step 1:In China-rose the stamens unite into one bundle, giving a monoadelphous androecium.
Step 2:In pea the stamens are arranged in two bundles, giving a diadelphous androecium.
Step 3:In citrus the stamens form many bundles, giving a polyadelphous androecium.
Step 4:In lily the stamens are attached to the petals, giving an epiphyllous androecium.
Final answer: A-IV, B-II, C-I, D-III
Q146Single correctMolecular Basis of Inheritance
The DNA present in chloroplast is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Circular, double stranded
Approach:
The organisation of chloroplast DNA resembles that of prokaryotic genomes.
Step 1:The DNA present in the chloroplast is circular and double stranded.
Final answer: Circular, double stranded
Q147Single correctEcosystem
In an ecosystem where the Net Primary Productivity (NPP) of first trophic level is , what would be the GPP (Gross Primary Productivity) of the third trophic level in the same ecosystem?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Only about ten percent of energy passes from one trophic level to the next, equating productivity across successive levels.
Step 1:The net primary productivity at the first trophic level becomes the gross primary productivity for the second trophic level.
Step 2:Applying the ten percent law, the net primary productivity at the second trophic level is one tenth of its gross value, which then serves as the gross primary productivity of the third trophic level.
Final answer:
Q148Single correctPlant Growth and Development
Which of the following are fused in somatic hybridization involving two varieties of plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Protoplasts
Approach:
Somatic hybridization combines whole cells stripped of their walls from two plant varieties.
Step 1:Protoplasts of two different plant varieties are fused in somatic hybridization.
Final answer: Protoplasts
Q149Single correctRespiration in Plants
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A.. Citric acid cycle | I.. Cytoplasm |
| B.. Glycolysis | II.. Mitochondrial matrix |
| C.. Electron transport system | III.. Intermembrane space of mitochondria |
| D.. Proton gradient | IV.. Inner mitochondrial membrane |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-I, C-IV, D-III
Approach:
Each stage of cellular respiration is matched with its site within the cell.
Step 1:The citric acid cycle occurs in the mitochondrial matrix.
Step 2:Glycolysis occurs in the cytosol of most cells.
Step 3:The electron transport system is present in the inner mitochondrial membrane.
Step 4:The proton gradient is formed across the inner membrane into the intermembrane space of mitochondria.
Final answer: A-II, B-I, C-IV, D-III
Q150Single correctMolecular Basis of Inheritance
Which of the following statement is correct regarding the process of replication in ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The DNA dependent DNA polymerase catalyses polymerization in direction
Approach:
DNA polymerases add nucleotides in a fixed chemical direction during replication.
Step 1:In prokaryotes such as E. coli the DNA dependent DNA polymerase catalyses polymerization only in the direction from the five prime end to the three prime end.
Final answer: The DNA dependent DNA polymerase catalyses polymerization in direction
Q151Single correctChemical Coordination and Integration
Which of the following is not a steroid hormone?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Glucagon
Approach:
Each option is classified by its chemical nature to identify the one that is not a steroid.
Step 1:Cortisol, testosterone and progesterone are steroid hormones derived from cholesterol.
Step 2:Glucagon is a polypeptide (proteinaceous) hormone secreted by the alpha cells of the pancreatic islets and is not a steroid.
Final answer: Glucagon
Q152Single correctHuman Reproduction
Given below are two statements:
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
Each statement about the hymen is assessed for biological accuracy.
Step 1:The presence or absence of the hymen can also be broken by a sudden jolt, insertion of a vaginal tampon or active participation in some sports, so it is not a reliable indicator of virginity.
Step 2:The hymen can be torn by causes other than first coitus, and in some women it persists even after coitus, so it is not torn during the first coitus only.
Final answer: Statement I is true but Statement II is false
Q153Single correctBiotechnology and its Applications
Which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Bio-reactors are used to produce small scale bacterial cultures
Approach:
Each statement about bioreactors is checked against their function in large-scale production.
Step 1:A bioreactor is the vessel used to turn raw material into a product on an industrial scale, routinely processing cultures of 100 to 1000 litres.
Step 2:Small-scale bacterial cultures are grown in shaking flasks; describing a bioreactor as a device for small-scale cultures contradicts its purpose, so that statement is the incorrect one.
Final answer: Bio-reactors are used to produce small scale bacterial cultures
Q154Single correctMolecular Basis of Inheritance
Which one is the correct product of DNA dependent RNA polymerase to the given template?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The RNA transcript is built complementary and antiparallel to the template strand, replacing thymine with uracil.
Step 1:The template strand reads 3' TACATGGCAAATATCCATTCA 5', so transcription proceeds from its 3' end to give a 5' to 3' RNA product.
Step 2:Pairing each template base with its RNA complement (A with U, T with A, G with C, C with G) yields the messenger RNA.
Final answer:
Q155Single correctHuman Reproduction
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male.
Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male.
Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A is false but R is true
Approach:
The Assertion and Reason are evaluated separately against the hormonal control of gonads.
Step 1:FSH affects ovarian follicles in females but in males LH, not FSH, acts on the Leydig cells to stimulate androgen secretion, so the Assertion is false.
Step 2:Growing ovarian follicles secrete estrogen in females while interstitial cells secrete androgen in males, so the Reason is true.
Final answer: A is false but R is true
Q156Single correctHuman Health and Disease
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Typhoid | I. Fungus |
| B. Leishmaniasis | II. Nematode |
| C. Ringworm | III. Protozoa |
| D. Filariasis | IV. Bacteria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-III, C-I, D-II
Approach:
Each disease is paired with the taxonomic group of its causative agent.
Step 1:Typhoid is caused by Salmonella typhimurium, a bacterium, so A pairs with IV.
Step 2:Leishmaniasis is caused by the protozoan Leishmania donovani, so B pairs with III.
Step 3:Ringworm is caused by fungi of the genera Microsporum, Trichophyton and Epidermophyton, so C pairs with I.
Step 4:Filariasis is caused by the nematodes Wuchereria bancrofti and Wuchereria malayi, so D pairs with II.
Final answer: A-IV, B-III, C-I, D-II
Q157Single correctBody Fluids and Circulation
Following are the stages of pathway for conduction of an action potential through the heart
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1E-C-A-D-B
Approach:
The cardiac impulse is traced through its conduction structures in order.
Step 1:The impulse originates at the SA node, passes to the AV node, then to the AV bundle, on to the bundle branches and finally to the Purkinje fibres.
Step 2:Substituting the letters gives E to C to A to D to B.
Final answer: E-C-A-D-B
Q158Single correctCell Cycle and Cell Division
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Diakinesis | I. Synaptonemal complex formation |
| B. Pachytene | II. Completion of terminalisation of chiasmata |
| C. Zygotene | III. Chromosomes look like thin threads |
| D. Leptotene | IV. Appearance of recombination nodules |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-IV, C-I, D-III
Approach:
Each sub-stage of prophase I is paired with its characteristic event.
Step 1:Diakinesis marks the completion of terminalisation of chiasmata, so A pairs with II.
Step 2:Pachytene shows the appearance of recombination nodules, so B pairs with IV.
Step 3:Zygotene features synaptonemal complex formation, so C pairs with I.
Step 4:Leptotene shows chromosomes looking like thin threads, so D pairs with III.
Final answer: A-II, B-IV, C-I, D-III
Q159Single correctStructural Organisation in Animals
Three types of muscles are given as a, b and c. Identify the correct matching pair along with their location in human body:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) Skeletal - Triceps
(b) Smooth - Stomach
(c) Cardiac - Heart
(b) Smooth - Stomach
(c) Cardiac - Heart
Approach:
Each labelled muscle image is identified by its structure and matched to a correct body location.
Step 1:Figure (a) represents skeletal muscle fibres closely attached to skeletal bones, such as the triceps and biceps, with striated fibres bundled together in parallel fashion.
Step 2:Figure (b) represents smooth muscle fibres present in the wall of internal organs such as blood vessels, stomach and intestine.
Step 3:Figure (c) represents cardiac muscle fibres exclusively present in the heart.
Final answer: (a) Skeletal - Triceps
(b) Smooth - Stomach
(c) Cardiac - Heart
(b) Smooth - Stomach
(c) Cardiac - Heart
Q160Single correctReproductive Health
Choose the correct answer from the option given below:
| List I | List II |
|---|---|
| A. Non-medicated IUD | I. Multiload 375 |
| B. Copper releasing IUD | II. Progestogens |
| C. Hormone releasing IUD | III. Lippes loop |
| D. Implants | IV. LNG-20 |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each contraceptive category is paired with its representative example.
Step 1:Lippes loop is a non-medicated IUD, so A pairs with III.
Step 2:Multiload 375 is a copper releasing IUD, so B pairs with I.
Step 3:LNG-20 is a hormone releasing IUD, so C pairs with IV.
Step 4:Progestogens are used as implants, so D pairs with II.
Final answer: A-III, B-I, C-IV, D-II
Q161Single correctBiomolecules
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A. Lipase | I. Peptide bond |
| B. Nuclease | II. Ester bond |
| C. Protease | III. Glycosidic bond |
| D. Amylase | IV. Phosphodiester bond |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-IV, C-I, D-III
Approach:
Each enzyme is paired with the chemical bond it acts upon.
Step 1:Lipase digests ester bonds found in lipids, so A pairs with II.
Step 2:Nuclease helps in digestion of phosphodiester bonds found in nucleic acids, so B pairs with IV.
Step 3:Protease helps in digestion of peptide bonds found in proteins, so C pairs with I.
Step 4:Amylase digests or breaks the glycosidic bonds found in carbohydrates, releasing smaller molecules that maltase ultimately yields as glucose, so D pairs with III.
Final answer: A-II, B-IV, C-I, D-III
Q162Single correctHuman Reproduction
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are correct and R is the correct explanation of A
Approach:
Both the Assertion and the Reason are evaluated, and the causal link between them is examined.
Step 1:Breast-feeding during the initial period of infant growth is recommended by doctors for bringing a healthy baby, so the Assertion is correct.
Step 2:Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby, which explains why breast-feeding is recommended, 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
Q163Single correctEvolution
Given below are some stages of human evolution.
Arrange them in correct sequence. (Past to Recent)
A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus
Choose the correct sequence of human evolution from the options given below:
Arrange them in correct sequence. (Past to Recent)
A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus
Choose the correct sequence of human evolution from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-D-C-B
Approach:
The listed hominins are ordered by their appearance in time from past to recent.
Step 1:The correct sequence of stages of human evolution from past to recent runs Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens.
Step 2:Substituting the letters gives A to D to C to B.
Final answer: A-D-C-B
Q164Single correctCell: The Unit of Life
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Axoneme | I. Centriole |
| B. Cartwheel pattern | II. Cilia and flagella |
| C. Crista | III. Chromosome |
| D. Satellite | IV. Mitochondria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-I, C-IV, D-III
Approach:
Each structural feature is paired with the organelle it characterises.
Step 1:Axoneme is seen in cilia and flagella, so A pairs with II.
Step 2:The centriole shows the cartwheel appearance, so B pairs with I.
Step 3:Crista is found in mitochondria, so C pairs with IV.
Step 4:Satellite is present in chromosomes, so D pairs with III.
Final answer: A-II, B-I, C-IV, D-III
Q165Single correctAnimal Kingdom
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. | I. Hag fish |
| B. | II. Saw fish |
| C. | III. Angel fish |
| D. | IV. Flying fish |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-I, C-II, D-IV
Approach:
Each scientific name is paired with its common fish name.
Step 1:Pterophyllum is the scientific name for Angel fish, so A pairs with III.
Step 2:Myxine is the scientific name for Hag fish, so B pairs with I.
Step 3:Pristis is the scientific name for Saw fish, so C pairs with II.
Step 4:Exocoetus is the scientific name for Flying fish, so D pairs with IV.
Final answer: A-III, B-I, C-II, D-IV
Q166Single correctBiotechnology and its Applications
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. antitrypsin | I. Cotton bollworm |
| B. Cry IAb | II. ADA deficiency |
| C. Cry IAc | III. Emphysema |
| D. Enzyme replacement therapy | IV. Corn borer |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Each biotechnological agent is paired with its associated condition or pest.
Step 1:Alpha-1 antitrypsin is used for treatment of emphysema, so A pairs with III.
Step 2:Cry IAb controls corn borer, so B pairs with IV.
Step 3:Cry IAc controls cotton bollworm, so C pairs with I.
Step 4:Enzyme replacement therapy can be used as a treatment option in ADA deficiency, so D pairs with II.
Final answer: A-III, B-IV, C-I, D-II
Q167Single correctBiotechnology: Principles and Processes
The 'Ti plasmid' of Agrobacterium tumefaciens stands for
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Tumor inducing plasmid
Approach:
The abbreviation Ti is expanded based on the biology of Agrobacterium tumefaciens.
Step 1:The Ti plasmid of Agrobacterium tumefaciens is the tumor inducing plasmid.
Step 2:Its T-DNA causes tumours in several dicot plants when transferred into the host genome.
Final answer: Tumor inducing plasmid
Q168Single correctAnimal Kingdom
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Pleurobrachia | I. Mollusca |
| B. Radula | II. Ctenophora |
| C. Stomochord | III. Osteichthyes |
| D. Air bladder | IV. Hemichordata |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-I, C-IV, D-III
Approach:
Each animal or structure is paired with the phylum or class to which it belongs.
Step 1:Pleurobrachia is a member of phylum Ctenophora, so A pairs with II.
Step 2:Radula is a rasping feeding organ present in phylum Mollusca, so B pairs with I.
Step 3:Stomochord is a rudimentary structure similar to notochord found in the collar region of members of phylum Hemichordata, so C pairs with IV.
Step 4:Air bladder is found in Osteichthyes which provides them buoyancy, so D pairs with III.
Final answer: A-II, B-I, C-IV, D-III
Q169Single correctReproductive Health
Which of the following is not a natural/traditional contraceptive method?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Vaults
Approach:
Each method is classified as natural or as a barrier or other device to find the exception.
Step 1:Coitus interruptus, periodic abstinence and lactational amenorrhea are natural methods of contraception that avoid chance of meeting of sperm and ovum.
Step 2:A vault is a barrier method made of rubber that is inserted into the female reproductive tract to cover the cervix during coitus, so it is not a natural method.
Final answer: Vaults
Q170Single correctCell Cycle and Cell Division
Following are the stages of cell division :
A. Gap 2 phase
B. Cytokinesis
C. Synthesis phase
D. Karyokinesis
E. Gap 1 phase
Choose the correct sequence of stages from the options given below :
A. Gap 2 phase
B. Cytokinesis
C. Synthesis phase
D. Karyokinesis
E. Gap 1 phase
Choose the correct sequence of stages from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4E-C-A-D-B
Approach:
The phases of the cell cycle are placed in their natural order of occurrence.
Step 1:The cell cycle proceeds as Gap 1 phase, then Synthesis phase, then Gap 2 phase, then Karyokinesis, then Cytokinesis.
Step 2:Substituting the letters gives E to C to A to D to B.
Final answer: E-C-A-D-B
Q171Single correctEvolution
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Constant gene pool
Approach:
The Hardy-Weinberg principle holds that allele frequencies stay constant across generations when no disturbing force acts, so the factor that maintains a constant gene pool leaves the equilibrium undisturbed.
Step 1:Genetic equilibrium is maintained when the gene pool remains constant, which is the defining condition of the Hardy-Weinberg principle.
Step 2:Gene migration, genetic drift and genetic recombination each alter allele frequencies and so disturb the equilibrium, driving evolutionary change.
Step 3:A constant gene pool introduces no change in allele frequencies, so it is the only listed factor that does not affect the Hardy-Weinberg equilibrium.
Final answer: Constant gene pool
Q172Single correctNeural Control and Coordination
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Pons | I. Provides additional space for Neurons, regulates posture and balance. |
| B. Hypothalamus | II. Controls respiration and gastric secretions. |
| C. Medulla | III. Connects different regions of the brain. |
| D. Cerebellum | IV. Neuro secretory cells |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-II, D-I
Approach:
Each brain region is paired with its principal function.
Step 1:Pons forms part of the hindbrain and connects different regions of the brain, so A pairs with III.
Step 2:Hypothalamus contains neuro secretory cells which secrete hormones, so B pairs with IV.
Step 3:Medulla oblongata is part of the hindbrain which controls respiration and gastric secretions, so C pairs with II.
Step 4:Cerebellum has a convoluted surface that provides additional space for neurons and also regulates posture and balance, so D pairs with I.
Final answer: A-III, B-IV, C-II, D-I
Q173Single correctBreathing and Exchange of Gases
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Expiratory capacity | I. Expiratory reserve volume + Tidal volume + Inspiratory reserve volume |
| B. Functional residual capacity | II. Tidal volume + Expiratory reserve volume |
| C. Vital capacity | III. Tidal volume + Inspiratory reserve volume |
| D. Inspiratory capacity | IV. Expiratory reserve volume + Residual volume |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-I, D-III
Approach:
Each respiratory capacity is expressed as the sum of its component lung volumes.
Step 1:Expiratory capacity equals tidal volume plus expiratory reserve volume, so A pairs with II.
Step 2:Functional residual capacity equals expiratory reserve volume plus residual volume, so B pairs with IV.
Step 3:Vital capacity equals expiratory reserve volume plus tidal volume plus inspiratory reserve volume, so C pairs with I.
Step 4:Inspiratory capacity equals tidal volume plus inspiratory reserve volume, so D pairs with III.
Final answer: A-II, B-IV, C-I, D-III
Q174Single correctHuman Health and Disease
Which of the following are Autoimmune disorders?
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B & E only
Approach:
Each disorder is classified by its underlying cause to identify the autoimmune ones.
Step 1:Myasthenia gravis, rheumatoid arthritis and systemic lupus erythematosus are autoimmune disorders in which the body's immune system attacks self tissues.
Step 2:Muscular dystrophy is a genetic disorder that progressively affects the skeletal muscles, and gout is the inflammation of joints due to deposition of uric acid crystals, so neither is autoimmune.
Final answer: A, B & E only
Q175Single correctHuman Reproduction
Which of the following is not a component of Fallopian tube?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Uterine fundus
Approach:
The parts of the Fallopian tube are recalled to identify the structure that does not belong to it.
Step 1:The Fallopian tube consists of the infundibulum bearing fimbriae, the ampulla and the isthmus.
Step 2:The uterine fundus is the upper rounded portion of the uterus and is not a part of the Fallopian tube.
Final answer: Uterine fundus
Q176Single correctLocomotion and Movement
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Fibrous joints | I. Adjacent vertebrae, limited movement |
| B. Cartilaginous joints | II. Humerus and Pectoral girdle, rotational movement |
| C. Hinge joints | III. Skull, don't allow any movement |
| D. Ball and socket joints | IV. Knee, help in locomotion |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each joint type is matched to its anatomical example based on the structure connecting the bones and the degree of movement permitted.
Step 1:Fibrous joints are present between the flat skull bones, which fuse end-to-end with the help of dense fibrous connective tissues and allow no movement.
Step 2:Cartilaginous joints are present between the adjacent vertebrae of the vertebral column and permit limited movement.
Step 3:Hinge joints, a type of synovial joint, are present in the knee and help in locomotion.
Step 4:Ball and socket joints, also synovial, occur between the humerus and pectoral girdle and allow rotational movement.
Final answer: A-III, B-I, C-IV, D-II
Q177Single correctEvolution
The flippers of the Penguins and Dolphins are the example of the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Convergent evolution
Approach:
The flippers of Penguins and Dolphins perform a similar function while the animals are not anatomically similar, which identifies the type of evolution involved.
Step 1:When two unrelated species develop similar structures to perform the same function, the structures are analogous and the process is termed convergent evolution.
Step 2:Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas.
Step 3:Natural selection is a key mechanism of evolution, and divergent evolution results in the formation of homologous structures, so neither applies here.
Final answer: Convergent evolution
Q178Single correctExcretory Products and their Elimination
Given below are two statements :
Statement I : In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II : The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements, choose the correct answer from the option given below :
Statement I : In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II : The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements, choose the correct answer from the option given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are false
Approach:
Each statement is checked against the structure and permeability properties of the nephron, then the truth combination is matched to the options.
Step 1:The descending limb of the loop of Henle is permeable to water and almost impermeable to electrolytes, so water leaves and the filtrate is concentrated; Statement I reverses these permeabilities and is therefore false.
Step 2:The proximal convoluted tubule is lined by simple cuboidal brush border epithelium that increases the surface area for reabsorption; Statement II describes the lining as columnar and is therefore false.
Step 3:With both statements false, the matching choice is option 2.
Final answer: Both Statement I and Statement II are false
Q179Single correctStructural Organisation in Animals
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 segment
Approach:
The abdominal segment bearing the anal cerci in both sexes of cockroach is identified.
Step 1:In both sexes of cockroach, the tenth segment bears a pair of jointed filamentous structures called anal cerci.
Step 2:Adult cockroaches have ten abdominal segments; the 11th segment is absent, so the other options do not bear the anal cerci.
Final answer: segment
Q180Single correctBreathing and Exchange of Gases
Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2High and Lesser concentration
Approach:
The conditions in the alveoli that promote binding of oxygen to haemoglobin are identified from the oxygen dissociation behaviour.
Step 1:Conditions favourable for formation of oxyhaemoglobin in alveoli are high pO2, less H+ concentration, low pCO2 and low temperature.
Step 2:Options 1, 3 and 4 contain factors such as high pCO2, high H+ concentration or high temperature that do not favour formation of oxyhaemoglobin.
Final answer: High and Lesser concentration
Q181Single correctBiotechnology - Principles and Processes
The following diagram showing restriction sites in coli cloning vector pBR322. Find the role of '' and '' genes :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The gene '' is responsible for controlling the copy number of the linked DNA and '' for protein involved in the replication of Plasmid.
Approach:
The functions of the genes labelled X and Y in the pBR322 cloning vector are identified from their standard roles.
Step 1:In the diagram, 'X' is rop while 'Y' is ori.
Step 2:The gene 'X', rop, is responsible for controlling the copy number of the linked DNA, and 'Y', ori, codes for protein involved in the replication of the plasmid.
Step 3:Options 1, 3 and 4 are incorrect because they assign X and Y to functions unrelated to their actual roles.
Final answer: The gene '' is responsible for controlling the copy number of the linked DNA and '' for protein involved in the replication of Plasmid.
Q182Single correctPrinciples of Inheritance and Variation
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Down's syndrome | I. chromosome |
| B. -Thalassemia | II. 'X' chromosome |
| C. -Thalassemia | III. chromosome |
| D. Klinefelter's syndrome | IV. chromosome |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Each disorder is matched to the chromosome on which its causative gene or extra copy is located.
Step 1:Down's syndrome is due to the presence of an additional copy of chromosome number 21.
Step 2:alpha-Thalassemia is caused due to presence of an additional copy of X-chromosome, while it is controlled by two closely linked genes on chromosome 16 of each parent.
Step 3:beta-Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent.
Step 4:Klinefelter's syndrome is associated with an additional X chromosome.
Final answer: A-III, B-IV, C-I, D-II
Q183Single correctAnimal Kingdom
Consider the following statements :
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below :
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A only
Approach:
Each statement is checked against the coelomic classification of the listed phyla.
Step 1:Annelids are true coelomate animals, so statement A is correct.
Step 2:Poriferans are acoelomate, Aschelminthes are pseudocoelomate, and Platyhelminthes are acoelomate, so statements B, C and D are incorrect.
Final answer: A only
Q184Single correctHuman Health and Disease
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Common cold | I. |
| B. Haemozoin | II. Typhoid |
| C. Widal test | III. Rhinoviruses |
| D. Allergy | IV. Dust mites |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-II, D-IV
Approach:
Each item is matched to its causative agent, associated organism or diagnostic relevance.
Step 1:Common cold is caused by Rhinoviruses.
Step 2:Haemozoin is released in blood due to ruptured RBCs after Plasmodium infection.
Step 3:Widal test is used to confirm the typhoid fever.
Step 4:Allergy is caused due to dust mites.
Final answer: A-III, B-I, C-II, D-IV
Q185Single correctHuman Health and Disease
Match List I with List II : Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Cocaine | I. Effective sedative in surgery |
| B. Heroin | II. |
| C. Morphine | III. |
| D. Marijuana | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-I, D-II
Approach:
Each drug is matched to its plant source or pharmacological property.
Step 1:Cocaine is obtained from plant Erythroxylum coca, stimulating action on CNS.
Step 2:Heroin is formed by the acetylation of morphine, which is obtained from plant Papaver somniferum.
Step 3:Morphine is obtained from Papaver somniferum and is an effective sedative used in surgery.
Step 4:Marijuana is obtained from Cannabis sativa, produces hallucinogenic effect and affects the cardiovascular system of the body.
Final answer: A-III, B-IV, C-I, D-II
Q186Single correctOrganisms and Populations
Given below are two statements:
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is false but Statement II is true.
Approach:
Each statement of Gause's competitive exclusion principle is evaluated against its correct formulation.
Step 1:Gause's competitive exclusion principle states that two closely related species competing for the same resources cannot exist indefinitely, so Statement I is false because it says different resources.
Step 2:The inferior competitor will be eliminated eventually, which holds true if resources are limiting, so Statement II is true.
Final answer: Statement I is false but Statement II is true.
Q187Single correctAnimal Kingdom
The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post anal tail is absent.
Choose the most appropriate answer from the options given below:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post anal tail is absent.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B, D & E only
Approach:
Each statement is checked to identify which features belong to non-chordates rather than chordates.
Step 1:In non-chordates the notochord is absent, the heart is dorsal if present, and the post anal tail is absent, so statements B, D and E are features of non-chordates.
Step 2:A perforated pharynx with gill slits and a dorsal central nervous system are features of chordates, so statements A and C are incorrect.
Final answer: B, D & E only
Q188Single correctNeural Control and Coordination
Given below are two statements:
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
Each statement about the connection of cerebral hemispheres and the composition of the brain stem is verified.
Step 1:A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres, and these hemispheres are connected by a tract of nerve fibres called corpus callosum, so Statement I is correct.
Step 2:The three major regions making up the brain stem are mid brain, pons and medulla oblongata; the cerebrum is a part of forebrain and does not form the brain stem, so Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect.
Q189Single correctStructural Organisation in Animals
Match List I with List II related to digestive system of cockroach. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. The structures used for storing of food | I. Gizzard |
| B. Ring of 6-8 blind tubules at junction of foregut and midgut. | II. Gastric Caeca |
| C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut. | III. Malpighian tubules |
| D. The structures used for grinding the food. | IV. Crop |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-II, C-III, D-I
Approach:
Each described structure of the cockroach alimentary canal is matched to its name.
Step 1:The crop is the structure used for storing of food.
Step 2:A ring of 6-8 blind tubules at the junction of foregut and midgut forms the gastric caeca, which secrete digestive juices.
Step 3:A ring of 100-150 yellow coloured thin filaments at the junction of midgut and hindgut forms the Malpighian tubules, which help in the elimination of nitrogenous wastes.
Step 4:The gizzard is used for grinding the food particles.
Final answer: A-IV, B-II, C-III, D-I
Q190Single correctChemical Coordination and Integration
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Exophthalmic goiter | I. Excess secretion of cortisol, moon face & hyperglycemia. |
| B. Acromegaly | II. Hypo-secretion of thyroid hormone and stunted growth. |
| C. Cushing's syndrome | III. Hyper secretion of thyroid hormone & protruding eye balls. |
| D. Cretinism | IV. Excessive secretion of growth hormone. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-I, D-II
Approach:
Each endocrine disorder is matched to its hormonal cause and clinical features.
Step 1:Exophthalmic goiter results from hyper secretion of thyroid hormone and is characterized by protruding eye balls.
Step 2:Acromegaly results from excessive secretion of growth hormone.
Step 3:Cushing's syndrome results from excess secretion of cortisol with moon face and hyperglycaemia.
Step 4:Cretinism results from hypo-secretion of thyroid hormone with stunted growth.
Final answer: A-III, B-IV, C-I, D-II
Q191Single correctExcretory Products and their Elimination
Choose the correct statement given below regarding juxta medullary nephron.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Loop of Henle of juxta medullary nephron runs deep into medulla.
Approach:
Each statement about juxta medullary nephrons is checked against their location and structure.
Step 1:The length of loop of Henle of juxta medullary nephron is longer than the length of loop of Henle of cortical nephron and runs deep into medulla, so statement 3 is correct.
Step 2:Juxta medullary nephrons are located in the columns of Bertini in statement 1, which is incorrect.
Step 3:The renal corpuscle of juxta medullary nephron lies in the inner cortical region, not the renal medulla, so statement 2 is incorrect.
Step 4:Juxta medullary nephrons are lesser in number than cortical nephrons, so statement 4 is incorrect.
Final answer: Loop of Henle of juxta medullary nephron runs deep into medulla.
Q192Single correctBiomolecules
Regarding catalytic cycle of an enzyme action, select the correct sequential steps :
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below :
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1E, A, D, C, B
Approach:
The catalytic cycle of enzyme action is arranged in its correct sequence of events.
Step 1:First, the substrate binds to the active site of the enzyme, fitting into the active site.
Step 2:The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate, forming the enzyme-substrate complex.
Step 3:The active site of the enzyme, now in close proximity of the substrate, breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
Step 4:The enzyme releases the products of the reaction, and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.
Final answer: E, A, D, C, B
Q193Single correctStructural Organisation in Animals
Match List I with List II : Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Unicellular glandular epithelium | I. Salivary glands |
| B. Compound epithelium | II. Pancreas |
| C. Multicellular glandular epithelium | III. Goblet cells of alimentary canal |
| D. Endocrine glandular epithelium | IV. Moist surface of buccal cavity |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Each type of glandular or compound epithelium is matched to its representative example.
Step 1:Unicellular glandular epithelium is represented by the goblet cells of the alimentary canal.
Step 2:Compound epithelium lines the moist surface of the buccal cavity.
Step 3:Multicellular glandular epithelium forms the salivary glands.
Step 4:Endocrine glandular epithelium is represented by the pancreas.
Final answer: A-III, B-IV, C-I, D-II
Q194Single correctEvolution
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Mesozoic Era | I. Lower invertebrates |
| B. Proterozoic Era | II. Fish & Amphibia |
| C. Cenozoic Era | III. Birds & Reptiles |
| D. Paleozoic Era | IV. Mammals |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each geological era is matched to the dominant life forms that appeared during it.
Step 1:The Mesozoic Era is associated with birds and reptiles.
Step 2:The Proterozoic Era is associated with lower invertebrates.
Step 3:The Cenozoic Era is associated with mammals.
Step 4:The Paleozoic Era is associated with fish and amphibia.
Final answer: A-III, B-I, C-IV, D-II
Q195Single correctCell - The Unit of Life
Given below are two statements:
Statement I: Mitochondria and chloroplasts both double membranes bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: Mitochondria and chloroplasts both double membranes bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast.
In the light of the above statements, choose the 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 about the membranes of mitochondria and chloroplasts is verified.
Step 1:Both mitochondria and chloroplasts are double membrane bound cell organelles, so Statement I is correct.
Step 2:Transport of ions occurs across the inner membrane of mitochondria, while the inner membrane of chloroplast is impermeable to ions and metabolites, so the inner membrane of mitochondria is relatively more permeable than that of chloroplast, making Statement II incorrect.
Final answer: Statement I is correct but Statement II is incorrect.
Q196Single correctBody Fluids and Circulation
Match List I with List II : Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. P wave | I. Heart muscles are electrically silent. |
| B. QRS complex | II. Depolarisation of ventricles. |
| C. T wave | III. Depolarisation of atria. |
| D. T-P gap | IV. Repolarisation of ventricles. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-II, C-IV, D-I
Approach:
Each component of the electrocardiogram is matched to the cardiac electrical event it represents.
Step 1:The P wave represents the depolarisation of the atria.
Step 2:The QRS complex represents the depolarisation of the ventricles.
Step 3:The T wave represents the repolarisation of the ventricles.
Step 4:During the T-P gap the heart muscles are electrically silent.
Final answer: A-III, B-II, C-IV, D-I
Q197Single correctHuman Health and Disease
Given below are two statements :
Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes.
In the light of above statements, choose the most appropriate answer from the options given below :
Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes.
In the light of above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct.
Approach:
Each statement about the role of bone marrow and thymus as lymphoid organs is evaluated.
Step 1:Bone marrow is the primary lymphoid organ in which all blood cells, lymphocytes included, are produced, so Statement I is correct.
Step 2:Bone marrow and thymus are both primary lymphoid organs: immature lymphocytes arise in the bone marrow and T-lymphocytes then migrate to the thymus, so both organs supply the micro-environments in which T-lymphocytes develop and mature. Statement II is correct as well.
Final answer: Both Statement I and Statement II are correct.
Q198Single correctHuman Reproduction
Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
The flow chart begins with GnRH branching into LH and (A). LH leads to (B), which leads to Androgens, which lead to Formation of spermatids. (A) leads to (C), which leads to Factors, which lead to (D).
The flow chart begins with GnRH branching into LH and (A). LH leads to (B), which leads to Androgens, which lead to Formation of spermatids. (A) leads to (C), which leads to Factors, which lead to (D).
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1FSH, Leydig cells, Sertoli cells, spermiogenesis.
Approach:
The flow chart of hormonal control of spermatogenesis is completed by identifying each labelled component.
Step 1:GnRH from the hypothalamus drives the anterior pituitary to release the two gonadotropins, LH and FSH. LH is already named on the chart, so the unlabelled branch (A) is FSH.
Step 2:LH acts on the Leydig cells lying in the interstitial spaces of the seminiferous tubules, and those cells secrete the androgens that drive formation of spermatids.
Step 3:FSH acts on the Sertoli cells inside the seminiferous tubules, which secrete the factors needed for the final stage of sperm production.
Step 4:That final stage is spermiogenesis, the transformation of spermatids into spermatozoa.
Final answer: FSH, Leydig cells, Sertoli cells, spermiogenesis.
Q199Single correctPrinciples of Inheritance and Variation
As per ABO blood grouping system, the blood group of father is B, mother is A and child is O. Their respective genotype can be
(A)
(B)
(C)
(D)
(E)
Choose the most appropriate answer from the options given below :
(A)
(B)
(C)
(D)
(E)
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A only
Approach:
The genotypes of the parents and child are deduced from their blood groups in the ABO multiple allele system.
Step 1:A father of blood group B carries at least one allele, so his genotype is or i.
Step 2:A mother of blood group A carries at least one allele, so her genotype is or i.
Step 3:Blood group O has no functional allele, so the child must be ii and must have received one i allele from each parent. Each parent is therefore heterozygous.
Step 4:The only listed row with a heterozygous B father, a heterozygous A mother and an ii child is row A.
Final answer: A only
Q200Single correctMolecular Basis of Inheritance
Match List I with List II: Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. RNA polymerase III | I. snRNPs |
| B. Termination of transcription | II. Promotor |
| C. Splicing of Exons | III. Rho factor |
| D. TATA box | IV. SnRNAs, tRNA |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-IV, B-III, C-I, D-II
Approach:
Each transcription-related term is matched to its associated molecule or element.
Step 1:In eukaryotes, RNA polymerase III codes for snRNAs, tRNA and 5s rRNA.
Step 2:Rho factor is responsible for the termination of transcription.
Step 3:Splicing of exons is performed by snRNPs.
Step 4:The TATA box is present in the promoter region of transcription unit.
Final answer: A-IV, B-III, C-I, D-II
More NEET 2024 papers
Frequently Asked Questions
How many questions are in the NEET 2024 May 05 paper?
The NEET 2024 May 05 paper has 200 questions — Physics (50), Chemistry (50) and Biology (100). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for NEET?
NEET awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Biology across 180 questions for 720 marks.
Are the answer key and step-by-step solutions provided for the 2024 May 05 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
Can I take the NEET 2024 May 05 paper as a timed mock test?
Yes. With a free NEETnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.
Solved this paper? Calculate your NEET score · most important chapters · formula sheets · all free tools.





