Back to NEET PYQs










![Graph of [R] concentration against time. Plain axes; the y-axis is marked [R0] at its top. A single straight line of constant negative slope starts at [R0] on the y-axis and falls steadily to the right, stopping partway across the plot. It is annotated k = - slope. The time axis carries an arrow pointing right.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2Ff4e80527-a5be-4271-b64d-3168ccb3ffc3%2Ff4e80527-a5be-4271-b64d-3168ccb3ffc3%2Fimages%2FQ75_conc_time.webp)


The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation, would be :
NEET 2026 May 03 Question Paper with Solutions
All 180 questions from the NEET 2026 (May 03) paper — Physics (45), Chemistry (45) and Biology (90) — 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 2026Chemistry PYQs 2026Biology PYQs 2026
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctPhysics and Measurement
The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
With the speed of light defined as one unit of speed, distance equals speed multiplied by time, so the numerical distance equals the travel time expressed in seconds.
Step 1:Convert the given time to seconds.
Step 2:Apply the distance relation with speed taken as unity.
Final answer:
Q2Single correctProperties of Solids and Liquids
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Young's Modulus | I. |
| B. Compressibility | II. |
| C. Bulk Modulus | III. |
| D. Poisson's Ratio | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Each elastic quantity is identified with its defining expression and matched to the corresponding entry in List II.
Step 1:Young's modulus equals stress over longitudinal strain, matching List II entry II.
Step 2:Compressibility is the reciprocal of bulk modulus, matching entry III.
Step 3:Bulk modulus equals volumetric stress over volumetric strain, matching entry IV.
Step 4:Poisson's ratio equals lateral strain over longitudinal strain, matching entry I.
Final answer: A-II, B-III, C-IV, D-I
Q3Single correctElectronic Devices
The current I in the circuit shown below is :
(All diodes are ideal and identical)
(All diodes are ideal and identical)

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The conduction direction of each ideal diode determines which parallel branches carry current; conducting branches act as plain resistors across the source, and the branch currents add.
Step 1:Conventional current leaves the positive terminal and traverses each branch from left to right. Only branches whose diodes point right conduct: the 4 ohm branch and the 2 ohm branch. The 3 ohm and 5 ohm branches have reverse-oriented diodes and stay off.
Step 2:With ideal diodes, each conducting branch carries the full 10 V across its resistor.
Step 3:Total current is the sum of the parallel branch currents.
Final answer:
Q4Single correctRotational Motion
The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For uniform angular acceleration the number of revolutions equals the average revolution rate multiplied by the time interval.
Step 1:Express the two rotation rates in revolutions per second.
Step 2:Multiply the average rate by the time.
Final answer:
Q5Single correctOscillations and Waves
For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Graph (3)
Approach:
Kinetic energy of an oscillator is proportional to the square of the speed, so it is never negative and reaches a maximum twice per period, giving a series of positive humps with period one half of T.
Step 1:Kinetic energy depends on the square of velocity and therefore remains non-negative throughout the motion.
Step 2:Speed is maximum at the mean position, passed twice each period, so kinetic energy peaks twice per period with period T/2 and touches zero at the extremes.
Step 3:The plot must therefore be a train of non-negative humps that touch zero at the two extreme positions and peak at the mean position, repeating with period T/2.
Final answer: Graph (3)
Q6Single correctCurrent Electricity
A resistor is connected to a battery of 12 V emf and internal resistance 2 . If the current in the circuit is 06 A, the terminal voltage of the battery is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Terminal voltage equals the emf minus the potential drop across the internal resistance.
Step 1:Compute the internal drop.
Step 2:Subtract from the emf.
Final answer:
Q7Single correctKinetic Theory of Gases
A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 2C. The ratio of root mean square speed of the molecules of the two gases is :
(Atomic mass of argon = 400 u and molecular mass of chlorine = 700 u)
(Atomic mass of argon = 400 u and molecular mass of chlorine = 700 u)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Root mean square speed at a given temperature depends only on molar mass, so the ratio is the square root of the inverse ratio of molar masses; the mass ratio of the mixture is irrelevant.
Step 1:At the same temperature the speed ratio reduces to the inverse square root of molar masses.
Step 2:Substitute the molar masses.
Final answer:
Q8Single correctRay Optics
A ray of monochromatic light is passing through an equilateral prism (ABC) as shown in the figure. The refracted ray (QR) is parallel to its base (BC) and the angle of incidence (i) is 5. Then the angle of deviation () is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
When the refracted ray inside the prism is parallel to the base, the prism is at minimum deviation, so the angle of incidence equals the angle of emergence and the deviation follows from the prism relation.
Step 1:A ray parallel to the base corresponds to minimum deviation, where the emergent angle equals the incident angle.
Step 2:Apply the deviation relation with the equilateral prism angle of 60 degrees.
Final answer:
Q9Single correctDual Nature of Matter and Radiation
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. | I. de Broglie wavelength |
| B. Diffraction and Interference | II. Particle nature of light |
| C. | III. Wave nature of light |
| D. Compton effect | IV. Energy of photon |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-IV, B-III, C-I, D-II
Approach:
Each physical relation or phenomenon is associated with the concept it represents and matched to List II.
Step 1:The relation E = h-nu gives the energy of a photon, matching IV.
Step 2:Diffraction and interference demonstrate the wave nature of light, matching III.
Step 3:The relation lambda = h/p defines the de Broglie wavelength, matching I.
Step 4:The Compton effect reveals the particle nature of light, matching II.
Final answer: A-IV, B-III, C-I, D-II
Q10Single correctAtoms and Nuclei
In the first excited state of hydrogen atom, the energy of its electron is eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately :
(Take 1 eV J, C and N )
(Take 1 eV J, C and N )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The first excited state corresponds to the second Bohr orbit; the orbital radius scales as the square of the principal quantum number times the Bohr radius, which equals four Bohr radii.
Step 1:Energy of -3.4 eV corresponds to n = 2 since -13.6/ = -3.4 eV.
Step 2:Use the magnitude of energy as the kinetic energy and solve for the radius with the given constants.
Step 3:This equals four times the Bohr radius, confirming the value.
Final answer:
Q11Single correctLaws of Motion
A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 012. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m is :
( m )
( m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The box stays at rest relative to the trolley only while static friction can supply the required force; the maximum acceleration is therefore the coefficient of static friction times gravity.
Step 1:Static friction provides the horizontal force on the box, bounded by mu times the normal force.
Step 2:Substitute the given values.
Final answer:
Q12Single correctElectrostatics
Five capacitors of capacitances F and F are connected as shown, along with a battery of 50 V. The equivalent capacitance and the charges on each capacitor respectively are :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 F, 125 C on all capacitors
Approach:
C2 and C3 form a series pair, which is in series with C1 and C4; that branch is in parallel with C5 across the battery. Charges follow from the branch and parallel structure.
Step 1:C2 and C3 in series give 5 microfarad; combined in series with C1 and C4.
Step 2:This branch is in parallel with C5 across the source.
Step 3:C5 carries the full source voltage; series branch carries a single charge of 2.5 microfarad times 50 V.
Step 4:In a series branch the same charge appears on each capacitor, so C1, C2, C3, C4 and C5 each hold 125 microcoulomb.
Final answer: F, 125 C on all capacitors
Q13Single correctGravitation
The amount of work done to raise a mass 'm' from the surface of the Earth to a height equal to the radius of the Earth 'R', will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Work equals the change in gravitational potential energy between the surface and a point one Earth radius above the surface.
Step 1:Compute the difference in potential energy from r = R to r = 2R.
Step 2:Express using surface gravity.
Final answer:
Q14Single correctPhysics and Measurement
Each side of a metallic cube of mass 5580 kg is measured to be 90 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as kg , where the value of X is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Density is mass divided by volume; the result is rounded to the least number of significant figures among the measured quantities, here two figures from the side length.
Step 1:Compute the volume from the side length.
Step 2:Divide the mass by the volume.
Step 3:The side length has two significant figures, so the density is reported to two significant figures.
Final answer:
Q15Single correctKinematics
The following plots show variation of velocity (v) with time (t), of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct ?

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C only
Approach:
For a ball thrown straight up under constant downward gravity, velocity decreases linearly from a positive value, passes through zero at the top, and continues into negative values as the ball falls.
Step 1:The acceleration is constant and downward, so the velocity-time graph is a single straight line with negative slope.
Step 2:Velocity starts positive, reaches zero at the highest point, then becomes negative during the fall, which matches plot C.
Final answer: C only
Q16Single correctWork, Energy and Power
The sum of kinetic energy and potential energy of a simple pendulum bob is 002 joule. The speed of the simple pendulum bob at equilibrium position is approximately :
(Consider mass of the bob = 20 g)
(Consider mass of the bob = 20 g)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
At the equilibrium position the potential energy is taken as zero, so the entire mechanical energy is kinetic; the speed follows from the kinetic energy expression.
Step 1:All the total energy becomes kinetic at the lowest point.
Step 2:Solve for the speed with mass 0.02 kg.
Final answer:
Q17Single correctWave Optics
In Young's double slit experiment, using monochromatic light of wavelength , the intensity of light at a point on the screen where the path difference is , is K units. The intensity of light at a point where the path difference is will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Intensity in two-slit interference varies as the square of the cosine of half the phase difference; a path difference of one wavelength gives the maximum K, fixing the maximum intensity, and the new point is evaluated from its phase.
Step 1:Path difference of one wavelength gives a phase of 2-pi, a maximum, so K equals the maximum intensity.
Step 2:For path difference lambda/3 the phase is 2-pi/3.
Step 3:Evaluate the intensity at this phase.
Final answer:
Q18Single correctElectronic Devices
In the circuit shown below, the voltage appearing across the diode D will be of the form :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Graph (4)
Approach:
In a half-wave rectifier the diode conducts during one half-cycle, dropping nearly zero voltage, and is reverse biased during the other half-cycle, when the full source voltage appears across it; the polarity follows the diode orientation.
Step 1:During the half-cycle when the diode is forward biased it conducts and the voltage across it is nearly zero.
Step 2:During the reverse-biased half-cycle no current flows, so there is no drop across the resistor and the whole source voltage stands across the diode. The diode voltage is therefore a half-sine hump alternating with a flat zero stretch of equal duration, not a continuous rectified waveform.
Final answer: Graph (4)
Q19Single correctAlternating Current
An ac circuit contains a resistance of 1 k, a capacitor of 01 F and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The resonance frequency of a series LCR circuit depends only on the inductance and capacitance.
Step 1:Compute the product of inductance and capacitance.
Step 2:Substitute into the resonance relation.
Final answer:
Q20Single correctWave Optics
In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.
A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
B. Diffraction and interference are characteristics exhibited only by light waves.
Choose the correct answer from the options given below :
A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
B. Diffraction and interference are characteristics exhibited only by light waves.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true, but B is false
Approach:
Each statement is assessed against the physics of wave phenomena: energy redistribution conserves total energy, while interference and diffraction occur for all wave types.
Step 1:Energy removed from dark fringes reappears in bright fringes, so total energy is conserved, making statement A true.
Step 2:Interference and diffraction are general wave phenomena, exhibited by sound, water and other waves, not only light, so statement B is false.
Final answer: A is true, but B is false
Q21Single correctOscillations and Waves
For a travelling harmonic wave , where x and y are in cm and t in s. The phase difference between oscillatory motion of two points separated by a distance of 05 m is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The phase difference between two points equals the wave number times their separation, with the wave number read from the coefficient of x in the phase.
Step 1:Identify the wave number from the coefficient of x in the argument.
Step 2:Multiply by the separation of 50 cm.
Final answer:
Q22Single correctLaws of Motion
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4, with 8 N force
Approach:
The resultant of two perpendicular forces is the square root of the sum of their squares; dividing by mass gives the acceleration, and the direction is the angle of the resultant measured from the larger force.
Step 1:Combine the perpendicular forces.
Step 2:Divide by the mass.
Step 3:Find the direction relative to the 8 N force.
Final answer: , with 8 N force
Q23Single correctElectrostatics
Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
When a charged capacitor shares its charge with an identical uncharged one, the final voltage halves; the energy lost is the difference between the initial and final stored energies.
Step 1:Compute the initial stored energy.
Step 2:After sharing equally, the common voltage is 50 V across total capacitance 400 pF.
Step 3:Subtract to find the energy lost.
Final answer:
Q24Single correctWork, Energy and Power
The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 kW
Approach:
Power equals the rate at which work is done against gravity while raising the mass.
Step 1:Work done against gravity to raise the mass.
Step 2:Divide the work by the time taken.
Step 3:Express in kilowatt.
Final answer: kW
Q25Single correctUnits and Measurements
In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 cm
Approach:
The least count equals one main scale division minus one vernier scale division.
Step 1:Find the value of one vernier scale division from the coincidence condition.
Step 2:Subtract one vernier division from one main scale division.
Step 3:Convert to centimetre.
Final answer: cm
Q26Single correctMotion in a Straight Line
When a ruler falls vertically, 5 different persons catch it with different reaction times.
A. Person A has reaction time of s.
B. Person B has reaction time of s.
C. Person C has reaction time of s.
D. Person D has reaction time of s.
E. Person E has reaction time of s.
What is the correct order of the distance travelled by the ruler for each person ?
A. Person A has reaction time of s.
B. Person B has reaction time of s.
C. Person C has reaction time of s.
D. Person D has reaction time of s.
E. Person E has reaction time of s.
What is the correct order of the distance travelled by the ruler for each person ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For free fall from rest the distance grows monotonically with time, so ordering the reaction times orders the distances.
Step 1:The distance fallen increases monotonically with the reaction time, since .
Step 2:Arrange the reaction times in descending order.
Step 3:Map the times back to the persons to get the distance order.
Final answer:
Q27Single correctCurrent Electricity
A uniform metallic wire having resistance is bent to form a square loop (ABCD) (see figure). A resistance of is connected between points B and D and a battery of V is connected across points A and C as shown in the figure. Now the value of current (I) is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 A
Approach:
Use the symmetry of the balanced bridge so the diagonal carries no current, then combine the two arms in parallel.
Step 1:Each side of the square carries one quarter of the wire.
Step 2:With the source across A and C, points B and D are at equal potential by symmetry, so the diagonal carries no current.
Step 3:The two paths A-B-C and A-D-C are each and combine in parallel.
Step 4:Apply Ohm's law for the total current from the battery.
Final answer: A
Q28Single correctCurrent Electricity
A room heater is rated W, V. If the supply voltage drops to V, what will be the power consumed (approximately) ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 W
Approach:
Find the fixed resistance from the rating, then recompute power at the lower voltage.
Step 1:Determine the heater resistance from its rated values.
Step 2:Compute the power at the reduced supply voltage.
Step 3:Round the result to three significant figures.
Final answer: W
Q29Single correctMoving Charges and Magnetism
A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively :
(Take T m/A)
(Take T m/A)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 A, A
Approach:
Find the current from the central-field formula, then evaluate the magnetic moment of the coil.
Step 1:Rearrange the central-field formula for the current.
Step 2:Evaluate the magnetic moment using the coil area.
Step 3:Round the moment to one significant figure.
Final answer: A, A
Q30Single correctElectromagnetic Induction
A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm , in a direction normal to the shorter side of the loop, will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 volt
Approach:
Use the motional-emf expression with the length of the edge that sweeps across the field boundary.
Step 1:Motion normal to the shorter side makes the shorter side the edge cutting the field lines, so the effective length is cm.
Step 2:Convert the velocity to SI units.
Step 3:Substitute into the motional-emf relation.
Final answer: volt
Q31Single correctNuclei
Four statements are given (A is mass number) :
A. The volume of a nucleus is proportional to .
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituents is called the mass defect.
Choose the correct answer from the options given below :
A. The volume of a nucleus is proportional to .
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituents is called the mass defect.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and D are true, but A and C are false
Approach:
Evaluate each statement against the relations for nuclear radius and the definition of mass defect.
Step 1:Since the radius scales as , the volume scales as the cube of the radius, hence as A. Statement B is true and A is false.
Step 2:Mass defect is the difference between the summed mass of the free constituent nucleons and the actual nuclear mass. Statement D is true and C is false.
Step 3:Combine the verdicts for the four statements.
Final answer: B and D are true, but A and C are false
Q32Single correctNuclei
An unknown nucleus has a nuclear density of kg/ and mass of kg. Its mass number A is approximately :
(Take m, )
(Take m, )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Find the nuclear volume from mass and density, then extract the mass number using the radius relation.
Step 1:Obtain the nuclear volume from the given mass and density.
Step 2:Relate the volume to the mass number through the radius relation .
Step 3:Round to the nearest integer mass number.
Final answer:
Q33Single correctOscillations
Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as :
(Take , and m/)
(Take , and m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 m
Approach:
Find the period from the timing data, then invert the pendulum formula for the length.
Step 1:Determine the period from 30 oscillations in 60 s.
Step 2:Rearrange the pendulum formula for the length.
Final answer: m
Q34Single correctThermodynamics
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 W
Approach:
Apply the first law of thermodynamics in rate form.
Step 1:Insert the heat supply rate and the work rate.
Final answer: W
Q35Single correctSystem of Particles and Rotational Motion
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' will be :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Express the ring's mass and radius from the wire data, take the moment of inertia about a diameter, then shift to the tangent line yy' with the parallel-axis theorem.
Step 1:From the figure the axis yy' is tangent to the ring in its plane, a distance R from the centre.
Step 2:Combine the diameter moment with the parallel-axis shift.
Step 3:Substitute and .
Final answer:
Q36Single correctMoving Charges and Magnetism
A galvanometer of resistance gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range A. The shunt required is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Equate the voltage across the galvanometer to that across the parallel shunt carrying the remaining current.
Step 1:Insert the galvanometer current, its resistance and the full-scale current.
Step 2:Evaluate the expression.
Final answer:
Q37Single correctCurrent Electricity
In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both right-sided and left-sided deflection and at balance point, no deflection
Approach:
Use the property of the Wheatstone network underlying the metre bridge regarding interchange of the cell and the detector.
Step 1:The metre bridge is a Wheatstone bridge whose balance condition is unchanged when the cell and the galvanometer are interchanged.
Step 2:Away from balance, the jockey gives deflection on either side; at the balance point the current through the galvanometer is zero.
Final answer: Both right-sided and left-sided deflection and at balance point, no deflection
Q38Single correctAlternating Current
The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 s
Approach:
A sinusoidal current starting from zero reaches its peak after one quarter of the period.
Step 1:Find the period from the frequency.
Step 2:The peak is reached after a quarter period.
Final answer: s
Q39Single correctMoving Charges and Magnetism
The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B increases linearly with r for , reaches a maximum at , then decreases as for
Approach:
Apply Ampere's circuital law separately inside and outside the uniformly current-carrying solid wire.
Step 1:Inside the wire the current enclosed by an Amperian loop of radius r grows as , so the field grows in direct proportion to r. The rise is a straight line through the origin, not a curve.
Step 2:Outside the wire the whole current is enclosed, so the field falls off inversely with r.
Step 3:Both expressions give the same value at the surface, so the two branches join at a single peak at r = a. The plot that shows a straight-line rise to that peak followed by a hyperbolic tail is the one printed as option (1); the curve printed as option (4) rises non-linearly and so does not represent .
Final answer: B increases linearly with r for , reaches a maximum at , then decreases as for
Q40Single correctSemiconductor Electronics
Two statements are given below :
A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly.
B. This current is called reverse saturation current.
Choose the correct answer from the options given below :
A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly.
B. This current is called reverse saturation current.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Statement A is true, but Statement B is false
Approach:
Compare each statement with the standard behaviour of a forward-biased p-n junction.
Step 1:Above the threshold (knee) voltage in forward bias the current rises steeply, so statement A is true.
Step 2:Reverse saturation current is the small current under reverse bias, not the large forward current, so statement B is false.
Final answer: Statement A is true, but Statement B is false
Q41Single correctElectrostatic Potential and Capacitance
Which of the following statements are correct ?
A. Inside a conductor, the electrostatic field is zero.
B. Electric field at the surface of a charged conductor does not depend on its surface charge density.
C. The interior of a charged conductor can have no excess charge in the static situation.
D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
E. The electrostatic potential is zero everywhere inside a charged conductor.
Choose the correct answer from the options given below :
A. Inside a conductor, the electrostatic field is zero.
B. Electric field at the surface of a charged conductor does not depend on its surface charge density.
C. The interior of a charged conductor can have no excess charge in the static situation.
D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
E. The electrostatic potential is zero everywhere inside a charged conductor.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, C and D only
Approach:
Test each statement against the electrostatic properties of conductors.
Step 1:Inside a conductor the electrostatic field is zero, so A is correct; excess charge resides only on the surface, so C is correct.
Step 2:The surface field depends on the surface charge density, so B is incorrect.
Step 3:The surface field is normal at every point, so D is correct; the potential is constant but not necessarily zero, so E is incorrect.
Final answer: A, C and D only
Q42Single correctDual Nature of Radiation and Matter
For a metal of work function eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect ?
(Take Planck's constant as J s)
(Take Planck's constant as J s)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 nm
Approach:
Find the threshold wavelength from the work function; any longer wavelength fails to eject electrons.
Step 1:Compute the threshold wavelength for the given work function.
Step 2:Radiation with wavelength longer than nm carries energy below the work function and ejects no electrons.
Final answer: nm
Q43Single correctRay Optics and Optical Instruments
In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens after refraction :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2appears to diverge from the first principal focus.
Approach:
Recall the ray rule for a parallel incident ray on a diverging lens.
Step 1:A concave lens diverges a parallel incident ray so that the refracted ray, when traced backward, meets the principal axis at the focus on the same side as the object.
Step 2:The refracted ray therefore appears to diverge from the first principal focus (a virtual focus).
Final answer: appears to diverge from the first principal focus.
Q44Single correctMechanical Properties of Fluids
A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface ?
(Consider the density of water kg , 1 atm Pa and gravitational acceleration m/)
(Consider the density of water kg , 1 atm Pa and gravitational acceleration m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 m
Approach:
The absolute pressure at depth is the atmospheric pressure plus the hydrostatic column, solved for the depth.
Step 1:Subtract the surface atmospheric pressure from the total absolute pressure to get the water-column pressure.
Step 2:Solve for the depth.
Final answer: m
Q45Single correctElectromagnetic Waves
Choose the correct answer from the options given below :
| List I (Electromagnetic wave) | List II (Production) |
|---|---|
| A. Microwave | I. Electrons in atoms emit light when they move from a higher energy level to a lower energy level |
| B. Visible light | II. Radioactive decay of nucleus |
| C. Gamma rays | III. Vibration of atoms and molecules |
| D. Infra-red rays | IV. Klystron valve or magnetron valve |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-I, C-II, D-III
Approach:
Match each electromagnetic wave to its characteristic source of production.
Step 1:Microwaves are generated by special vacuum tubes such as the klystron and magnetron.
Step 2:Visible light comes from electronic transitions in atoms between energy levels.
Step 3:Gamma rays originate in radioactive decay of nuclei.
Step 4:Infra-red radiation arises from the vibration of atoms and molecules.
Final answer: A-IV, B-I, C-II, D-III
Chemistry45 questions
Q46Single correctAmines
Select the reagents that reduce nitriles to primary amines :
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 2A, C and D only
Approach:
A nitrile is reduced to a primary amine by adding four hydrogen atoms across the carbon-nitrogen triple bond. Reagents that supply this reduction are lithium aluminium hydride, catalytic hydrogen over nickel, and sodium amalgam in ethanol.
Step 1:Lithium aluminium hydride followed by aqueous work-up (A) and catalytic hydrogenation over nickel (C) both reduce the nitrile group to a primary amine.
Step 2:Sodium amalgam in ethanol (D), a dissolving-metal reduction, also reduces the nitrile to a primary amine.
Step 3:Tin with hydrochloric acid (B) reduces nitro groups rather than nitriles, and bromine in alkali (E) is the Hofmann degradation of amides; neither reduces a nitrile to an amine. The reducing set is A, C and D.
Final answer: A, C and D only
Q47Single correctThe d- and f-Block Elements
Choose the correct answer from the options given below :
| List I (Transition metal/ compound/ complex) | List II (Catalytic Role) |
|---|---|
| A. | I. Preparation of ammonia from mixture |
| B. | II. Polymerisation of alkynes |
| C. | III. Preparation of from |
| D. Ni complex | IV. Oxidation of ethyne to ethanal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each transition-metal species is matched with the industrial reaction it catalyses based on standard NCERT examples.
Step 1:Vanadium pentoxide catalyses oxidation of sulphur dioxide in the Contact process for sulphuric acid, giving A-III.
Step 2:Finely divided iron is the catalyst in the Haber synthesis of ammonia from nitrogen and hydrogen, giving B-I.
Step 3:Palladium(II) chloride catalyses the Wacker oxidation of ethyne/ethene to ethanal, giving C-IV; a nickel complex catalyses polymerisation of alkynes, giving D-II.
Final answer: A-III, B-I, C-IV, D-II
Q48Single correctThermodynamics
Consider the following reaction :
and at 298 K.
Identify the correct option with for the reaction and spontaneity of the reaction at 298 K.
(Given : R = )
and at 298 K.
Identify the correct option with for the reaction and spontaneity of the reaction at 298 K.
(Given : R = )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3, non-spontaneous
Approach:
Convert internal-energy change to enthalpy using the change in moles of gas, then combine with entropy to obtain Gibbs energy and decide spontaneity from its sign.
Step 1:Change in moles of gas equals products minus reactants, two minus three.
Step 2:Enthalpy change combines the internal energy with the gas-mole correction.
Step 3:Gibbs energy subtracts the entropy term; the temperature times the negative entropy adds a positive contribution.
Step 4:Positive Gibbs energy means the reaction is non-spontaneous at 298 K.
Final answer: , non-spontaneous
Q49Single correctStructure of Atom
Choose the correct answer from the options given below :
| List I (Quantum Numbers) | List II (Orbital) |
|---|---|
| A. | I. 3d |
| B. | II. 2p |
| C. | III. 4s |
| D. | IV. 5f |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-III, C-IV, D-I
Approach:
The principal quantum number gives the shell number and the azimuthal quantum number fixes the sub-shell letter (l = 0,1,2,3 correspond to s,p,d,f).
Step 1:For n equal to 2 and l equal to 1 the orbital is 2p, giving A-II; for n equal to 4 and l equal to 0 it is 4s, giving B-III.
Step 2:For n equal to 5 and l equal to 3 the orbital is 5f, giving C-IV; for n equal to 3 and l equal to 2 it is 3d, giving D-I.
Final answer: A-II, B-III, C-IV, D-I
Q50Single correctEquilibrium
In a qualitative analysis, is detected by appearance of precipitate of . Calculate pH when the following equilibrium exists at 298 K :
,
(Given : )
,
(Given : )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The dissolution gives equal concentrations of the cation and hydroxide ion; set both equal to the solubility, solve the equilibrium expression for hydroxide, find pOH and convert to pH.
Step 1:Equal release of the two ions makes the concentration of hydroxide equal to the square root of the equilibrium constant.
Step 2:Negative logarithm of the hydroxide concentration gives pOH.
Step 3:Subtracting pOH from 14 gives the pH.
Final answer:
Q51Single correctBiomolecules
The correct statement with regard to the secondary structure of DNA/RNA is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2DNA possesses a double strand helix structure and contains thymine as one of the four bases.
Approach:
DNA is a double-stranded right-handed helix containing the bases adenine, guanine, cytosine and thymine, whereas RNA is single-stranded and contains uracil in place of thymine.
Step 1:DNA has two complementary polynucleotide strands wound as a double helix, with thymine as one of its four bases.
Step 2:RNA is generally single-stranded and replaces thymine with uracil, so the statement describing DNA with thymine is correct.
Final answer: DNA possesses a double strand helix structure and contains thymine as one of the four bases.
Q52Single correctOrganic Chemistry – Some Basic Principles and Techniques
The pair of molecules that are metamers among the following is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 and
Approach:
Metamers share the same molecular formula and the same functional group but differ in the alkyl groups attached on either side of the functional group; the pair of ethers with formula C4H10O fits this definition.
Step 1:Methyl propyl ether and diethyl ether both have molecular formula C4H10O and the ether functional group, differing only in the distribution of carbons about the oxygen.
Step 2:Metamerism is exactly this situation: one molecular formula, one functional group, and a different partition of the carbon skeleton on the two sides of that group. The other pairs listed differ in the position of a hydroxyl group, in chain branching, or in the functional group itself, so none of them is a metameric pair.
Final answer: and
Q53Single correctCoordination Compounds
Choose the correct answer from the options given below :
| List I (Complex) | List II (Type of isomerism) |
|---|---|
| A. | I. Optical |
| B. | II. Solvate |
| C. | III. Geometrical |
| D. | IV. Linkage |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each complex is assigned the type of isomerism it characteristically exhibits.
Step 1:The square-planar diamminedichloridoplatinum exists as cis and trans forms, showing geometrical isomerism, giving A-III; the tris-ethylenediamine cobalt(III) cation is chiral and shows optical isomerism, giving B-I.
Step 2:The nitro/nitrito ligand binds through nitrogen or oxygen, so the pentaammine complex shows linkage isomerism, giving C-IV; the hexaaqua chromium chloride shows solvate (hydrate) isomerism, giving D-II.
Final answer: A-III, B-I, C-IV, D-II
Q54Single correctChemical Kinetics
Choose the correct answer from the options given below :
| List I (Order of reaction) | List II (Unit of rate constant) |
|---|---|
| A. Zero order | I. |
| B. First order | II. |
| C. Second order | III. |
| D. Third order | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-III, C-I, D-II
Approach:
For a reaction of order n the rate constant carries units of concentration^(1-n) per unit time; substituting n = 0, 1, 2, 3 gives the four unit sets.
Step 1:Zero order: rate equals k, so k carries the units of rate itself.
Step 2:First order: rate = k[A], so the concentration units cancel and only reciprocal time remains.
Step 3:Second order: rate = k, leaving one inverse concentration factor.
Step 4:Third order: rate = k, leaving two inverse concentration factors.
Final answer: A-IV, B-III, C-I, D-II
Q55Single correctOrganic Chemistry – Some Basic Principles and Techniques
The correct IUPAC name of the following compound is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 13-ethyl-5-methylheptane
Approach:
Identify the longest continuous carbon chain, number it to give the lowest set of locants to the substituents, and apply the alphabetical-priority rule to break the locant tie.
Step 1:The longest chain contains seven carbons, so the parent is heptane, carrying an ethyl and a methyl substituent.
Step 2:Numbering from either end gives the locant set 3 and 5; the tie is broken by assigning the lower locant to the substituent first in alphabetical order, which is ethyl.
Step 3:Citing substituents alphabetically gives the name 3-ethyl-5-methylheptane.
Final answer: 3-ethyl-5-methylheptane
Q56Single correctStructure of Atom
A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is J, how many photons are emitted by the bulb per second ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Compute the light power as 8% of the rated power, then divide the energy emitted per second by the energy of a single photon to obtain the photon count per second.
Step 1:Eight percent of 150 watt is the optical power output.
Step 2:Dividing 12 joule per second by the energy of one photon gives the number of photons emitted each second.
Final answer:
Q57Single correctHydrogen
Methane reacts with steam at 1273 K in the presence of nickel catalyst to form :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and
Approach:
Steam reforming of methane over a nickel catalyst at about 1273 K produces synthesis gas, a mixture of carbon monoxide and dihydrogen.
Step 1:Methane and steam over nickel at 1273 K give carbon monoxide and dihydrogen, known as water gas or syngas.
Final answer: and
Q58Single correctAldehydes, Ketones and Carboxylic Acids
Compound P () gives a red orange precipitate with 2,4-DNP reagent and it does not reduce Fehling's reagent. On drastic oxidation with chromic acid, P gives an aromatic product Q that produces effervescence on treating with aq. . Compounds P and Q, respectively, are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4P = acetophenone (); Q = benzoic acid ()
Approach:
The positive 2,4-DNP test fixes a carbonyl group, the negative Fehling's test rules out an aldehyde and points to a methyl aryl ketone, and the formation on drastic oxidation of an acid that effervesces with bicarbonate fixes the product as a carboxylic acid.
Step 1:The molecular formula C8H8O with a positive 2,4-DNP test and a negative Fehling's test identifies a methyl aryl ketone, namely acetophenone.
Step 2:Drastic oxidation by chromic acid cleaves the side chain to give benzoic acid, an aromatic carboxylic acid.
Step 3:Benzoic acid liberates carbon dioxide with aqueous sodium hydrogen carbonate, accounting for the effervescence.
Final answer: P = acetophenone (); Q = benzoic acid ()
Q59Single correctChemical Bonding and Molecular Structure
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. | I. 3 bonds, 2 bonds |
| B. | II. 3 bonds, one lone pair |
| C. | III. 4 bonds |
| D. | IV. 5 bonds, 1 bond |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-I, C-III, D-II
Approach:
Count the sigma bonds, pi bonds and lone pairs in each molecule from its Lewis structure.
Step 1:Ethene has one carbon-carbon sigma, four carbon-hydrogen sigma and one carbon-carbon pi, a total of five sigma and one pi bond, giving A-IV.
Step 2:Ethyne has one carbon-carbon sigma, two carbon-hydrogen sigma and two carbon-carbon pi, a total of three sigma and two pi bonds, giving B-I.
Step 3:Methane has four carbon-hydrogen sigma bonds, giving C-III; ammonia has three nitrogen-hydrogen sigma bonds and one lone pair, giving D-II.
Final answer: A-IV, B-I, C-III, D-II
Q60Single correctAmines
The following two reactions give the same foul smelling product Z.
X and Z, respectively, are :
X and Z, respectively, are :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The foul-smelling product is an isocyanide; trace the second route by Hofmann bromamide degradation followed by the carbylamine reaction to fix Z, then choose the reagent X that converts the haloalkane to the same isocyanide.
Step 1:Propanamide undergoes Hofmann bromamide degradation to ethylamine, the intermediate Y.
Step 2:Ethylamine with chloroform and ethanolic potassium hydroxide gives ethyl isocyanide, the foul-smelling product Z.
Step 3:Ethyl chloride forms the same isocyanide only with silver cyanide, which directs bonding through nitrogen, so X is silver cyanide.
Final answer:
Q61Single correctSome Basic Concepts of Chemistry
The number of hydrogen atoms present in 54 g of urea is :
(Given : Molar mass of urea : 60 g mo, : particles mo)
(Given : Molar mass of urea : 60 g mo, : particles mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Find the moles of urea, multiply by Avogadro's number to get molecules, then multiply by the four hydrogen atoms in each urea molecule.
Step 1:Mass divided by molar mass gives the moles of urea.
Step 2:Each urea molecule, with formula NH2CONH2, carries four hydrogen atoms; multiply moles by Avogadro's number and by four.
Final answer:
Q62Single correctThe p-Block Elements (Group 15)
Identify the incorrect statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Nitrogen can form - bond with oxygen.
Approach:
Test each claim against the bonding capability of nitrogen and of the heavier group-15 elements.
Step 1:Nitrogen is small and can overlap its 2p orbitals sideways with itself, which is why elemental nitrogen exists as N2 with a triple bond. Statement (1) is correct.
Step 2:Phosphorus and arsenic have vacant d orbitals, so phosphines and arsines such as P(C2H5)3 and As(C6H5)3 can accept electron density from filled metal d orbitals when acting as ligands. Statement (2) is correct.
Step 3:Phosphorus, arsenic and antimony all form element-element single bonds (P-P, As-As, Sb-Sb) in their elemental forms, so they do catenate. Statement (3) is correct.
Step 4:Nitrogen's valence shell is 2s2p only; it has no d orbitals available, so it cannot form a d-pi-p-pi bond with oxygen. This is the incorrect statement.
Final answer: Nitrogen can form - bond with oxygen.
Q63Single correctCoordination Compounds
Which one of the following is an ambidentate ligand ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Thiocyanate
Approach:
An ambidentate ligand can coordinate to the metal through either of two different donor atoms; identify which listed ligand has two distinct donor sites.
Step 1:Ethane-1,2-diamine and oxalate are bidentate and ethylenediaminetetraacetate is hexadentate, but each binds through fixed donor atoms.
Step 2:The thiocyanate ion can bind through sulphur or through nitrogen, the defining feature of an ambidentate ligand.
Final answer: Thiocyanate
Q64Single correctClassification of Elements and Periodicity in Properties
The correct order of increasing metallic character of Na, Be, P, Mg and Si is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Metallic character increases down a group and decreases across a period; rank the elements from the non-metal through the metalloid to the metals.
Step 1:Phosphorus is a non-metal (least metallic), silicon is a metalloid, and beryllium, magnesium and sodium are metals.
Step 2:Among the metals, going from beryllium to magnesium down group 2 increases metallic character, and sodium (group 1, period 3) is the most metallic.
Step 3:Combining the trends gives the increasing order P < Si < Be < Mg < Na.
Final answer:
Q65Single correctAlcohols, Phenols and Ethers
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
A. ![]() | I. (i) oleum, (ii) NaOH, ; (iii) |
| B. | II. (i) , (ii) |
| C. | III. (i) , , (ii) , catalyst |
D. ![]() | IV. (i) conc. , , (ii) |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-III, C-IV, D-I
Approach:
Match each transformation with the reagent sequence that accomplishes it.
Step 1:Cumene gives phenol by air oxidation to the hydroperoxide followed by acid cleavage, matching reagent set II (O2 then H2O/H+), giving A-II.
Step 2:Acetic acid is converted to ethanol by esterification with methanol under acid then catalytic hydrogenation, matching set III, giving B-III.
Step 3:Propan-1-ol is dehydrated with concentrated sulphuric acid to propene, then Markovnikov hydration with H+/H2O gives propan-2-ol, matching set IV, giving C-IV.
Step 4:Benzene gives phenol through sulphonation with oleum, fusion with sodium hydroxide and acidification, matching set I, giving D-I.
Final answer: A-II, B-III, C-IV, D-I
Q66Single correctThe d- and f-Block Elements
Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4After losing one more electron, it acquires electronic configuration.
Approach:
Cerium attains the +4 state because losing one electron beyond +3 empties its 4f subshell, giving the especially stable noble-gas configuration of xenon.
Step 1:Cerium has the configuration [Xe]4f1 5d1 6s2; the +3 ion is [Xe]4f1.
Step 2:Removing one more electron empties the 4f subshell, giving [Xe]4f0, a stable noble-gas configuration that favours the +4 state.
Final answer: After losing one more electron, it acquires electronic configuration.
Q67Single correctHaloalkanes and Haloarenes
In the following reaction sequence, X and Z respectively are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 (1-bromopropane)
Approach:
Identify the phosphorus by-product of the alcohol-with-phosphorus-pentachloride reaction, then follow elimination to propene and anti-Markovnikov addition of hydrogen bromide under peroxide to fix the bromide.
Step 1:Propan-1-ol with phosphorus pentachloride gives propyl chloride together with phosphoryl chloride and hydrogen chloride, so X is phosphoryl chloride.
Step 2:Propyl chloride with alcoholic potassium hydroxide eliminates to propene, the intermediate Y.
Step 3:Hydrogen bromide adds to propene in the presence of benzoyl peroxide by the anti-Markovnikov (peroxide) pathway, placing bromine on the terminal carbon to give 1-bromopropane.
Final answer: (1-bromopropane)
Q68Single correctCoordination Compounds
Choose the correct answer from the options given below :
| List I (Complex/ion) | List II (Shape/geometry) |
|---|---|
| A. | I. Octahedral |
| B. | II. Trigonal bipyramidal |
| C. | III. Square planar |
| D. | IV. Tetrahedral |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-I, C-IV, D-II
Approach:
Assign each complex its geometry from the metal's oxidation state, coordination number and the nature of the ligand field.
Step 1:Diamminedichloridoplatinum(II) is a d8 four-coordinate complex that is square planar, giving A-III; hexaamminecobalt(III) is six-coordinate and octahedral, giving B-I.
Step 2:Tetrachloridonickelate(II) with weak-field chloride is tetrahedral, giving C-IV; iron pentacarbonyl is five-coordinate and trigonal bipyramidal, giving D-II.
Final answer: A-III, B-I, C-IV, D-II
Q69Single correctOrganic Chemistry – Some Basic Principles and Techniques
The functional group that can be identified through phthalein dye test is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Phenolic
Approach:
The phthalein dye test detects an aromatic carboxylic acid (or its anhydride), which condenses with phenol in the presence of concentrated sulphuric acid to give a phthalein dye.
Step 1:An aromatic carboxylic acid such as phthalic acid (or phthalic anhydride) is heated with phenol and concentrated sulphuric acid.
Step 2:The condensation produces phenolphthalein, which turns pink on making the solution alkaline, confirming the functional group.
Final answer: Phenolic
Q70Single correctOrganic Chemistry – Some Basic Principles and Techniques
Two products X and Y are formed in the following reaction sequence.
The suitable method that can be used for the separation of products X and Y is :
The suitable method that can be used for the separation of products X and Y is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Fractional distillation
Approach:
Friedel-Crafts methylation of benzene gives toluene (W); mild nitration of toluene gives o-nitrotoluene and p-nitrotoluene as X and Y. Both are liquids with different boiling points, so fractional distillation is the appropriate separation.
Step 1:Benzene undergoes Friedel-Crafts alkylation with methyl chloride to give toluene as intermediate W.
Step 2:Nitration of toluene yields ortho- and para-nitrotoluene as the two products X and Y, both liquids of different but close boiling points.
Step 3:Two miscible liquids with a moderate difference in boiling points are separated by fractional distillation.
Final answer: Fractional distillation
Q71Single correctSolutions
Identify the correct statements :
A. The molality of 2.5 g of ethanoic acid (Molar mass : ) in 75 g of benzene solution is 0.556 m.
B. The molarity of a solution containing 5 g of NaOH (molar mass : ) in 450 mL of solution is 0.278 M at 298 K.
C. Aquatic species are more comfortable in cold water.
D. The solubility of gas increases with decrease in pressure.
E. For a binary mixture of A and B, the number of moles of A and B are and respectively. The mole fraction of B will be
Choose the correct answer from the options given below :
A. The molality of 2.5 g of ethanoic acid (Molar mass : ) in 75 g of benzene solution is 0.556 m.
B. The molarity of a solution containing 5 g of NaOH (molar mass : ) in 450 mL of solution is 0.278 M at 298 K.
C. Aquatic species are more comfortable in cold water.
D. The solubility of gas increases with decrease in pressure.
E. For a binary mixture of A and B, the number of moles of A and B are and respectively. The mole fraction of B will be
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B and C only
Approach:
Evaluate each of the five statements numerically or against the definitions of molality, molarity, gas solubility and mole fraction.
Step 1:Ethanoic acid: 2.5 g of molar mass 60 g/mol is 0.04167 mol in 0.075 kg of benzene.
Step 2:Sodium hydroxide: 5 g of molar mass 40 g/mol is 0.125 mol in 0.450 L of solution.
Step 3:Gas solubility falls as temperature rises, so cold water holds more dissolved oxygen and aquatic species are more comfortable in it.
Step 4:Henry's law makes the dissolved amount of a gas directly proportional to its partial pressure, so lowering the pressure lowers the solubility.
Step 5:The mole fraction of B is its own moles divided by the total; the statement as printed puts in the numerator, which gives the mole fraction of A instead.
Final answer: A, B and C only
Q72Single correctOrganic Chemistry – Some Basic Principles and Techniques
During Lassaigne's test, the elements present in an organic compound are converted from :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2covalent form to ionic form
Approach:
In Lassaigne's test the organic compound is fused with sodium metal, converting covalently bonded elements (N, S, halogens) into water-soluble sodium salts (ionic form) for detection.
Step 1:The organic compound is fused with sodium, which converts covalently bound N, S and halogens into ionic sodium salts such as NaCN, Na2S and NaX.
Step 2:These ionic forms dissolve in water and are detected by characteristic precipitation/colour tests.
Final answer: covalent form to ionic form
Q73Single correctElectrochemistry
A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is :
(Given : Molar mass of Cu = 63 g mo ;
1 F = 96487 C mo)
(Given : Molar mass of Cu = 63 g mo ;
1 F = 96487 C mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 g
Approach:
Apply Faraday's first law with the two-electron reduction of copper(II).
Step 1:Charge passed is current times time, with 10 minutes expressed in seconds.
Step 2:Copper deposits by a two-electron reduction, so each mole of copper needs two faradays.
Step 3:Substitute into Faraday's law.
Final answer: g
Q74Single correctThermodynamics
At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Apply the first law: change in internal energy equals heat absorbed by the system minus work done by the system.
Step 1:Heat absorbed by the system is +500 J and work done by the system is +200 J.
Step 2:Change in internal energy equals heat added minus work done by the system.
Final answer:
Q75Single correctChemical Kinetics
For a certain reaction R Product, the plot of concentration [R] v/s time has a negative slope as shown. The order of reaction is :
![Graph of [R] concentration against time. Plain axes; the y-axis is marked [R0] at its top. A single straight line of constant negative slope starts at [R0] on the y-axis and falls steadily to the right, stopping partway across the plot. It is annotated k = - slope. The time axis carries an arrow pointing right.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2Ff4e80527-a5be-4271-b64d-3168ccb3ffc3%2Ff4e80527-a5be-4271-b64d-3168ccb3ffc3%2Fimages%2FQ75_conc_time.webp)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10
Approach:
Read the order directly off the shape of the concentration-time plot.
Step 1:A plot of concentration against time that is a straight line means the concentration falls by equal amounts in equal intervals, so the rate of reaction is constant.
Step 2:A rate that does not depend on the reactant concentration is a rate law of the form rate = k, that is, zero order. The magnitude of the negative slope is the rate constant k.
Final answer: 0
Q76Single correctChemical Bonding and Molecular Structure
Identify the correct statement about Cl from the following options :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1It has T-shaped geometry with two lone pairs on Cl atom.
Approach:
Determine the steric number of Cl in ClF3: 3 bonding pairs and 2 lone pairs give a trigonal bipyramidal electron geometry, with the two lone pairs in equatorial positions producing a T-shaped molecular geometry.
Step 1:Chlorine in ClF3 has 7 valence electrons; three form bonds with fluorine and the remaining four constitute two lone pairs.
Step 2:The two lone pairs occupy equatorial positions, giving a T-shaped molecular geometry.
Final answer: It has T-shaped geometry with two lone pairs on Cl atom.
Q77Single correctEnvironmental / Qualitative Inorganic Analysis
In a test tube containing a salt, a few drops of dilute was added, which gave colourless vapours having the smell of vinegar. The vapours turned the blue litmus paper red. Identify the correct anion from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Acetate, C
Approach:
Colourless vapours with the smell of vinegar that turn blue litmus red indicate acetic acid, which is liberated when dilute sulphuric acid acts on an acetate salt.
Step 1:Dilute sulphuric acid being stronger displaces the weaker acetic acid from the acetate salt.
Step 2:Acetic acid vapours are acidic and turn blue litmus red, confirming the acetate anion.
Final answer: Acetate, C
Q78Single correctEquilibrium
At 298 K, a certain buffer solution contains equal concentrations of X and HX, for X is . What is the pH of the buffer solution ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 410
Approach:
Use the Henderson-Hasselbalch equation; when the concentrations of the conjugate acid (HX) and base (X-) are equal, pH equals pKa.
Step 1:Equal concentrations make the logarithm term zero, so pH equals pKa.
Step 2:pKa is the negative logarithm of Ka.
Final answer: 10
Q79Single correctElectrochemistry
Calculate emf of the half cell given below :
(Given : ,
log 2 = 0.3010)
(Given : ,
log 2 = 0.3010)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 V
Approach:
Apply the Nernst equation to the hydrogen electrode half-cell, accounting for the hydrogen ion concentration and the hydrogen gas pressure.
Step 1:Substitute the hydrogen pressure of 2 atm and hydrogen ion concentration of 0.02 M into the Nernst expression with standard potential zero.
Step 2:Evaluate the logarithm of 2/0.0004 = 5000, giving log 5000 ≈ 3.699.
Step 3:Multiply by -0.059/2 to obtain the electrode potential.
Final answer: V
Q80Single correctThe d- and f-Block Elements
The calculated 'spin-only' magnetic moment of T (3) is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 BM
Approach:
Use the spin-only formula with the number of unpaired electrons; Ti2+ has a 3d2 configuration with two unpaired electrons.
Step 1:Ti2+ has the 3d2 configuration with two unpaired electrons.
Step 2:Apply the spin-only formula.
Final answer: BM
Q81Single correctThe p-Block Elements
Identify the incorrect statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Oxygen exhibits only oxidation state.
Approach:
Test each statement against periodic-property facts; the incorrect one is the statement that oxygen shows only -2, since oxygen also exhibits -1 (peroxides), -1/2 (superoxides) and positive states in OF2.
Step 1:Carbon forms strong p-pi p-pi bonds, BCl3 is a monomer while Al2Cl6 is a dimer, and catenation decreases down Group 14, so statements 1, 2 and 3 are correct.
Step 2:Oxygen shows -2, -1, -1/2 and even +2 (in OF2), so the claim of only -2 is false.
Final answer: Oxygen exhibits only oxidation state.
Q82Single correctChemical Bonding and Molecular Structure
The correct formal charges on oxygen atoms numbered 2, 1 and 3 respectively are :
(In the structure shown, oxygen 1 is the central atom; oxygen 2 is joined to it by a double bond and oxygen 3 by a single bond.)
(In the structure shown, oxygen 1 is the central atom; oxygen 2 is joined to it by a double bond and oxygen 3 by a single bond.)

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Compute formal charge for each oxygen in the ozone resonance structure using FC = (valence electrons) - (non-bonding electrons) - (1/2 bonding electrons).
Step 1:The central oxygen forms one double and one single bond with two lone pairs removed compared to a terminal atom, giving formal charge +1.
Step 2:The double-bonded terminal oxygen has two lone pairs and a share of four bonding electrons, giving formal charge 0.
Step 3:The single-bonded terminal oxygen has three lone pairs, giving formal charge -1.
Final answer:
Q83Single correctEquilibrium
Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at the alkaline pH close to the equivalence point during this titration is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3pink to colourless
Approach:
Phenolphthalein is pink in alkaline medium and colourless in acidic/neutral medium; with NaOH in the flask and oxalic acid added, the pink colour disappears at the end point.
Step 1:The NaOH solution with phenolphthalein is initially pink because the medium is alkaline.
Step 2:Adding oxalic acid neutralises the base, and just past the equivalence point the pink colour disappears.
Final answer: pink to colourless
Q84Single correctRedox Reactions / Mole Concept
When 1 d of C gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes 1.4 d. The composition of the gaseous mixture at STP is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.8 d of CO, 0.6 d of C
Approach:
The reaction CO2 + C -> 2CO doubles the gas volume for the reacted portion. Let x dm3 of CO2 react; it gives 2x dm3 of CO while (1 - x) dm3 of CO2 remains. Total volume = (1 - x) + 2x = 1 + x = 1.4, so x = 0.4.
Step 1:Let x dm3 of the carbon dioxide react with the coke; each volume consumed produces two volumes of carbon monoxide, leaving (1 - x) dm3 of unreacted carbon dioxide.
Step 2:Set the total equal to the measured 1.4 dm3 and solve.
Step 3:Carbon monoxide formed is twice the reacted volume and the remainder is unreacted carbon dioxide.
Final answer: 0.8 d of CO, 0.6 d of C
Q85Single correctAmines / Hydrocarbons
The major product Z formed in the following sequence of reactions is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Follow the three steps: free-radical chlorination, ammonolysis to a primary amine, then diazotisation of that aliphatic amine followed by hydrolysis.
Step 1:Ultraviolet light drives free-radical substitution of one hydrogen of ethane, giving chloroethane as the monochlorinated product X.
Step 2:Ammonia displaces chloride to give the primary amine Y.
Step 3:Nitrous acid generated in situ diazotises the primary aliphatic amine, but the aliphatic diazonium ion is unstable and loses nitrogen at once; water captures the resulting cation to give ethanol.
Final answer:
Q86Single correctChemical Kinetics
Given below is an expression for the rate constant k of a first order reaction occurring at a certain temperature, T (K).
The energy of activation in kcal mo for the reaction is :
(Given : k is in , R = 1.987 cal mo )
The energy of activation in kcal mo for the reaction is :
(Given : k is in , R = 1.987 cal mo )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 124.84
Approach:
Compare the given linear form of ln k with the Arrhenius equation ln k = ln A - Ea/(RT); the coefficient of 1/T equals Ea/R.
Step 1:Matching the coefficient of 1/T gives Ea/R equal to 1.25 x .
Step 2:Multiply by R = 1.987 cal mol-1 K-1 and convert to kcal.
Final answer: 24.84
Q87Single correctEquilibrium
Given below are certain reactions. Identify the reaction for which K .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 (g) + 3 (g) 2N (g)
Approach:
Kp = Kc; Kp differs from Kc only when the change in moles of gas Δng is non-zero. Evaluate Δng for each reaction.
Step 1:For reactions 1, 3 and 4 the number of gaseous moles is the same on both sides, so Δng = 0 and Kp = Kc.
Step 2:For the ammonia synthesis, moles change from 4 to 2, giving Δng = -2, so Kp ≠ Kc.
Final answer: (g) + 3 (g) 2N (g)
Q88Single correctClassification of Elements and Periodicity / p-Block
Identify the incorrect statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The largest and the smallest species among Mg, M, Al and A are Al and M, respectively.
Approach:
Test each statement against periodic size trends, the systematic naming of heavy elements, the lithium-magnesium diagonal relationship and the composition of the given aluminium complex.
Step 1:Removing electrons shrinks a species, so both cations are smaller than either neutral atom; magnesium sits to the left of aluminium in the same period and is the larger atom, while Al3+ has lost three electrons and is the smallest of the four.
Step 2:The statement therefore names the wrong species at both ends of the range and is the incorrect one.
Step 3:Of the rest: the systematic roots un-nil-sept give Unnilseptium for atomic number 107; lithium and magnesium show the diagonal relationship; and in the complex the chloride carries -1 and the five water molecules are neutral, so aluminium is +3 with six donor atoms bound to it.
Final answer: The largest and the smallest species among Mg, M, Al and A are Al and M, respectively.
Q89Single correctSolutions
Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2formation of hydrogen bonding between acetone and chloroform.
Approach:
Negative deviation arises when A-B interactions are stronger than A-A and B-B interactions; chloroform and acetone form intermolecular hydrogen bonds, lowering escaping tendency and vapour pressure.
Step 1:The acidic hydrogen of chloroform forms a hydrogen bond with the carbonyl oxygen of acetone.
Step 2:These stronger interactions reduce the escaping tendency and lower the vapour pressure below the ideal value, giving negative deviation.
Final answer: formation of hydrogen bonding between acetone and chloroform.
Q90Single correctAldehydes, Ketones and Carboxylic Acids / Haloarenes
The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 36 and 6
Approach:
Decide whether each set of conditions drives substitution on the ring or addition across it, then count the chlorines in the product.
Step 1:A Lewis acid in the cold and dark promotes electrophilic substitution. With six equivalents of chlorine every ring hydrogen is replaced, giving hexachlorobenzene.
Step 2:Ultraviolet light instead generates chlorine radicals, which add across the ring. Three chlorine molecules add to benzene to give benzene hexachloride.
Step 3:Both products therefore carry six chlorine atoms.
Final answer: 6 and 6
Biology90 questions
Q91Single correctAnatomy of Flowering Plants
In angiosperms, root hairs arise from which one of the following regions of the root ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The region of maturation
Approach:
The root tip shows four zones; root hairs develop in the most mature region, where cells differentiate and elongate.
Step 1:From root tip upward the zones are: root cap, meristematic activity, elongation, and maturation.
Step 2:In the region of maturation, cells differentiate and mature; some epidermal cells form fine, thread-like root hairs that absorb water and minerals.
Final answer: The region of maturation
Q92Single correctSexual Reproduction in Flowering Plants
In which one of the following, the ovules are enclosed by an ovary wall and remain exposed ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Exposed (naked) ovules are the defining feature of gymnosperms, which lack an ovary.
Step 1:Funaria is a moss (bryophyte), Selaginella is a pteridophyte, and Wolffia is an angiosperm with enclosed ovules.
Step 2:Pinus is a gymnosperm; its ovules are not enclosed by an ovary wall and remain naked/exposed on the megasporophylls.
Final answer:
Q93Single correctMolecular Basis of Inheritance
In the operon, the gene codes for :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3beta-galactosidase
Approach:
The lac operon's structural genes z, y, and a each code for a specific enzyme.
Step 1:Gene y codes for permease, gene a codes for transacetylase, and the i gene codes for the repressor.
Step 2:Gene z codes for beta-galactosidase, which hydrolyses lactose into galactose and glucose.
Final answer: beta-galactosidase
Q94Single correctBiotechnology and its Applications
Exploring molecular, genetic and species-level diversity for products of economic importance is called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Bioprospecting
Approach:
Match the definition of exploring biodiversity for economically useful products to the correct term.
Step 1:Biofortification breeds crops for higher nutrient content; bioremediation uses microbes to clean pollutants; biomagnification is increasing toxin concentration along a food chain.
Step 2:Bioprospecting is the exploration of molecular, genetic and species-level diversity for products of economic importance.
Final answer: Bioprospecting
Q95Single correctBiotechnology - Principles and Processes
Match List I with List II
| List I | List II |
|---|---|
| A. Genetically modified organism | I. |
| B. Thermostable DNA polymerase | II. Bt cotton |
| C. Ti plasmid | III. |
| D. pBR322 | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-I, D-IV
Approach:
Pair each item in List I with its associated organism or product in List II.
Step 1:Bt cotton is a genetically modified organism; thermostable DNA polymerase (Taq) is isolated from Thermus aquaticus; Ti plasmid comes from Agrobacterium tumefaciens; pBR322 is a cloning vector based on Escherichia coli.
Final answer: A-II, B-III, C-I, D-IV
Q96Single correctEcosystem
Match List I with List II
| List I | List II |
|---|---|
| A. Productivity | I. Gross primary productivity minus respiration losses |
| B. Net primary productivity | II. Rate of formation of new organic matter by consumers |
| C. Gross primary productivity | III. Rate of biomass production |
| D. Secondary productivity | IV. Rate of production of organic matter during photosynthesis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Match each productivity term with its NCERT definition.
Step 1:Productivity is the rate of biomass production; net primary productivity equals gross primary productivity minus respiration losses; gross primary productivity is the rate of production of organic matter during photosynthesis; secondary productivity is the rate of formation of new organic matter by consumers.
Final answer: A-III, B-I, C-IV, D-II
Q97Single correctBiodiversity and Conservation
Since the origin and diversification of life on Earth, there have been five episodes of mass extinction of species. How is the sixth extinction, which is in progress, different from the previous episodes ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The present species extinction rates are 100 to 1000 times faster than in the pre-human times.
Approach:
Recall the NCERT figure for the present-day extinction rate relative to pre-human rates.
Step 1:The sixth extinction is human-driven and proceeds at 100 to 1000 times faster than the pre-human rates; ecologists warn that if this continues, nearly half of all species could be wiped out within the next 100 years.
Final answer: The present species extinction rates are 100 to 1000 times faster than in the pre-human times.
Q98Single correctBiomolecules
Alpha-helix is found in which level of protein structure ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Secondary structure
Approach:
Identify the structural level at which the alpha-helix coiling of the polypeptide occurs.
Step 1:Primary structure is the linear sequence of amino acids; the secondary structure describes regular folding such as the right-handed alpha-helix.
Final answer: Secondary structure
Q99Single correctAnatomy of Flowering Plants
The main function of bulliform cells in grasses is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4to minimize water loss during water stress.
Approach:
Recall the role of empty, large-celled bulliform cells in the grass epidermis.
Step 1:Bulliform cells are large empty colourless cells in the upper epidermis of grass leaves. When flaccid due to water stress, they make the leaves curl inward, reducing water loss by transpiration.
Final answer: to minimize water loss during water stress.
Q100Single correctBiotechnology - Principles and Processes
Identify the correct sequence of steps in each cycle of Polymerase Chain Reaction :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Denaturation Annealing Extension
Approach:
Recall the three sequential steps of one PCR cycle.
Step 1:Each PCR cycle proceeds as denaturation (strand separation by heat), then annealing (primers bind to template), then extension (DNA polymerase synthesises new strands).
Final answer: Denaturation Annealing Extension
Q101Single correctCell Cycle and Cell Division
Match List I with List II
| List I | List II |
|---|---|
| A. phase | I. Actual cell division occurs |
| B. S phase | II. Cell is metabolically active and continuously grows but does not replicate its DNA |
| C. phase | III. Synthesis of DNA occurs and the amount of DNA per cell doubles |
| D. M phase | IV. Proteins are synthesized while cell growth continues |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Match each cell-cycle phase with the event it describes.
Step 1:G1 phase: cell is metabolically active and grows continuously but does not replicate DNA (II). S phase: DNA synthesis doubles DNA per cell (III). G2 phase: proteins are synthesized while cell growth continues (IV). M phase: actual cell division occurs (I).
Final answer: A-II, B-III, C-IV, D-I
Q102Single correctBiodiversity and Conservation
Which of the following statements are correct ?
A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss.
B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans.
C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake.
D. Water hyacinth is an invasive species.
E. When a species becomes extinct, the plant and animal species associated with it are not affected.
Choose the correct answer from the options given below :
A. The Amazon rainforest being cut and cleared for cultivation of soyabeans is an example of habitat loss.
B. Steller's sea cow and passenger pigeon became extinct due to over-exploitation by humans.
C. The Nile perch introduced into Lake Victoria in East Africa helped in population growth of cichlid fish in the lake.
D. Water hyacinth is an invasive species.
E. When a species becomes extinct, the plant and animal species associated with it are not affected.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B and D only
Approach:
Evaluate each statement against NCERT facts on biodiversity loss.
Step 1:A is correct (Amazon clearing for soyabean is habitat loss). B is correct (Steller's sea cow and passenger pigeon became extinct by over-exploitation). C is incorrect: the Nile perch caused extinction of more than 200 cichlid species in Lake Victoria. D is correct (water hyacinth is an invasive alien species). E is incorrect: associated species are affected (co-extinctions).
Final answer: A, B and D only
Q103Single correctMolecular Basis of Inheritance
Which of the following statements are correct with reference to a transcription unit ?
A. A transcription unit in DNA is defined primarily by three regions - promoter, structural gene and terminator.
B. The promoter is said to be located towards the 5'-end of the structural gene.
C. The promoter is a DNA sequence that provides binding site for RNA polymerase.
D. The promoter defines the template and coding strands.
E. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription.
Choose the correct answer from the options given below :
A. A transcription unit in DNA is defined primarily by three regions - promoter, structural gene and terminator.
B. The promoter is said to be located towards the 5'-end of the structural gene.
C. The promoter is a DNA sequence that provides binding site for RNA polymerase.
D. The promoter defines the template and coding strands.
E. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, B, C, D and E
Approach:
Check each statement against NCERT description of a transcription unit.
Step 1:A is correct: a transcription unit has a promoter, the structural gene, and a terminator. B is correct: the promoter lies towards the 5'-end of the structural gene (of the coding strand). C is correct: the promoter provides the binding site for RNA polymerase. D is correct: the promoter helps define the template and coding strands. E is correct: the terminator lies towards the 3'-end and defines termination of transcription.
Final answer: A, B, C, D and E
Q104Single correctBiological Classification / Plant Kingdom
Which one of the following statements is true about the universal rules of binomial nomenclature ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The first word in the biological name represents the specific epithet, while the second component denotes the genus.
Approach:
Find the statement that violates the rules of binomial nomenclature.
Step 1:Statements 1, 2 and 3 are correct rules. Statement 4 is false: the first word represents the genus while the second word denotes the specific epithet, not the reverse.
Final answer: The first word in the biological name represents the specific epithet, while the second component denotes the genus.
Q105Single correctEcosystem
Match List I with List II
| List I | List II |
|---|---|
| A. Decomposition | I. Accumulation of dark coloured amorphous colloidal substance |
| B. Detritus | II. Release of inorganic nutrients by the activity of microbes in soil |
| C. Mineralisation | III. Breaking down of complex organic matter into inorganic substances |
| D. Humification | IV. Dead remains of plants and animals including faecal matter |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-II, D-I
Approach:
Match each decomposition-related term with its definition.
Step 1:Decomposition is breaking down of complex organic matter into inorganic substances (III). Detritus is the dead remains of plants and animals (IV). Mineralisation is the release of inorganic nutrients by microbes (II). Humification leads to accumulation of dark-coloured amorphous colloidal humus (I).
Final answer: A-III, B-IV, C-II, D-I
Q106Single correctMolecular Basis of Inheritance
Which one of the following is the site for active ribosomal RNA synthesis ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Nucleolus
Approach:
Identify the non-membrane organelle responsible for rRNA synthesis and ribosome assembly.
Step 1:The nucleolus is a non-membrane-bound structure in the nucleus that is the site of active ribosomal RNA synthesis.
Final answer: Nucleolus
Q107Single correctRespiration in Plants
The Respiratory Quotient (RQ) of a biomolecule used for respiration, as per the above equation, would be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Between 0.5 and 0.95
Approach:
RQ is the ratio of volume of CO2 evolved to volume of O2 consumed; compute from the given equation.
Step 1:From the equation, CO2 released = 102 and O2 consumed = 145.
Step 2:This value corresponds to fatty-acid (fat) respiration, which lies between 0.5 and 0.95.
Final answer: Between 0.5 and 0.95
Q108Single correctPrinciples of Inheritance and Variation
Match List I with List II
| List I | List II |
|---|---|
| A. Incomplete dominance | I. Human skin colour |
| B. Co-dominance | II. Inheritance of flower colour in sp. |
| C. Pleiotropy | III. Phenylketonuria disease in human |
| D. Polygenic inheritance | IV. ABO blood groups |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-III, D-I
Approach:
Match each inheritance pattern with its standard example.
Step 1:Incomplete dominance is shown by flower colour in Antirrhinum (snapdragon) (II). Co-dominance is shown by ABO blood groups (IV). Pleiotropy is illustrated by phenylketonuria (III). Polygenic inheritance is shown by human skin colour (I).
Final answer: A-II, B-IV, C-III, D-I
Q109Single correctBiotechnology - Principles and Processes
Arrange the following steps of DNA fingerprinting in a correct sequence.
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Hybridisation using a labelled VNTR probe.
C. Transferring of separated DNA fragments to synthetic membrane.
D. Detection of hybridised DNA fragments by autoradiography.
E. Separation of DNA fragments by electrophoresis.
Choose the correct answer from the options given below :
A. Isolation of DNA and its digestion by restriction endonucleases.
B. Hybridisation using a labelled VNTR probe.
C. Transferring of separated DNA fragments to synthetic membrane.
D. Detection of hybridised DNA fragments by autoradiography.
E. Separation of DNA fragments by electrophoresis.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, E, C, B, D
Approach:
Order the steps following the standard DNA fingerprinting (Southern blot based) protocol.
Step 1:Isolation and digestion of DNA (A), then separation by electrophoresis (E), then transfer of fragments to a synthetic membrane (C), then hybridisation with a labelled VNTR probe (B), and finally detection by autoradiography (D).
Final answer: A, E, C, B, D
Q110Single correctMolecular Basis of Inheritance
Which of the following statements are correct with reference to packaging of DNA helix ?
A. Histones are organized to form a unit of eight molecules called histone octamer.
B. Histones are negatively charged basic proteins.
C. Histones are rich in the basic amino acid residues - lysine and arginine.
D. The positively charged DNA is wrapped around the histone octamer to form nucleosome.
E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins.
Choose the correct answer from the options given below :
A. Histones are organized to form a unit of eight molecules called histone octamer.
B. Histones are negatively charged basic proteins.
C. Histones are rich in the basic amino acid residues - lysine and arginine.
D. The positively charged DNA is wrapped around the histone octamer to form nucleosome.
E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, C and E only
Approach:
Evaluate each statement on DNA packaging against NCERT facts about histones and nucleosomes.
Step 1:A is correct: eight histone molecules form the histone octamer. B is incorrect: histones are positively charged (basic), not negatively charged. C is correct: histones are rich in lysine and arginine. D is incorrect: negatively charged DNA wraps around the positively charged histone octamer (the statement reverses the charges). E is correct: higher-order packaging needs non-histone chromosomal (NHC) proteins.
Final answer: A, C and E only
Q111Single correctPhotosynthesis in Higher Plants
Find the incorrect statement(s) about photosynthesis from the following :
A. The water splitting complex is associated with PS I.
B. plants use the pathway of fixation as the main biosynthetic pathway.
C. In plants, photorespiration does not occur.
D. plants exhibit 'Kranz' anatomy.
E. ATP synthesis in chloroplast occurs through chemiosmosis.
Choose the answer from the following given below :
A. The water splitting complex is associated with PS I.
B. plants use the pathway of fixation as the main biosynthetic pathway.
C. In plants, photorespiration does not occur.
D. plants exhibit 'Kranz' anatomy.
E. ATP synthesis in chloroplast occurs through chemiosmosis.
Choose the answer from the following given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A and D only
Approach:
Identify the incorrect statements about the photosynthetic machinery and pathways.
Step 1:A is incorrect: the water-splitting (oxygen-evolving) complex is associated with PS II, not PS I. B is correct: C4 plants ultimately use the C3 (Calvin) pathway as the main biosynthetic pathway. C is correct: photorespiration is largely absent in C4 plants. D is incorrect: Kranz anatomy is a feature of C4 plants, not C3 plants. E is correct: chloroplast ATP synthesis occurs by chemiosmosis.
Final answer: A and D only
Q112Single correctBiotechnology - Principles and Processes
Arrange the following steps of somatic hybridisation in a correct sequence.
A. Digestion of cell walls.
B. Isolation of naked protoplasts.
C. Fusion of protoplasts to get hybrid protoplast.
D. Isolation of single cells from two different varieties of plants.
E. Growing of hybrid protoplast to form a new plant.
Choose the correct answer from the options given below :
A. Digestion of cell walls.
B. Isolation of naked protoplasts.
C. Fusion of protoplasts to get hybrid protoplast.
D. Isolation of single cells from two different varieties of plants.
E. Growing of hybrid protoplast to form a new plant.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1D, A, B, C, E
Approach:
Order the steps of somatic hybridisation by protoplast fusion.
Step 1:Isolate single cells from two plant varieties (D), digest the cell walls (A), isolate the naked protoplasts (B), fuse the protoplasts to get a hybrid protoplast (C), then grow the hybrid protoplast into a new plant (E).
Final answer: D, A, B, C, E
Q113Single correctAnatomy of Flowering Plants
Match List I with List II
| List I | List II |
|---|---|
| A. Conjunctive tissue | I. Specialised cells in the vicinity of guard cells |
| B. Casparian strips | II. Endodermal cells rich in starch |
| C. Subsidiary cells | III. Tissue between xylem and phloem |
| D. Starch sheath | IV. Endodermal cells with suberin deposition |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Match each anatomical term with its correct description.
Step 1:Conjunctive tissue lies between xylem and phloem in dicot roots (III). Casparian strips are deposits of suberin in endodermal cells (IV). Subsidiary cells are specialised cells around the guard cells (I). Starch sheath is the starch-rich endodermis of dicot stems (II).
Final answer: A-III, B-IV, C-I, D-II
Q114Single correctPlant Growth and Development
Which one of the following is a characteristic of plant cells in the phase of elongation ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Large conspicuous nuclei
Approach:
The phase of elongation is one of the three phases of growth. Its defining cytological features are enlargement of cells, increased vacuolation, and new cell wall deposition. Identify the trait that belongs instead to the meristematic phase.
Step 1:Cells in the elongation phase show cell enlargement, increased vacuolation and fresh cell wall deposition.
Step 2:Large, conspicuous nuclei and dense protoplasm typify cells of the meristematic (formation) phase, not the elongation phase.
Final answer: Large conspicuous nuclei
Q115Single correctPlant Growth and Development
Choose the correct answer from the options given below :
| List I (Growth Regulator) | List II (Function/Effect) |
|---|---|
| A. 2,4-D | I. Brewing industry |
| B. G | II. Stimulation of stomatal closure |
| C. Kinetin | III. Herbicide |
| D. ABA | IV. Nutrient mobilisation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Recall the characteristic application of each plant growth regulator and pair it with the listed effect.
Step 1:2,4-D is a synthetic auxin used to kill dicotyledonous weeds, acting as a herbicide.
Step 2:Gibberellins (G) speed up malting in the brewing industry.
Step 3:Cytokinins such as kinetin promote nutrient mobilisation and delay senescence.
Step 4:Abscisic acid stimulates closure of stomata under water stress.
Final answer: A-III, B-I, C-IV, D-II
Q116Single correctPhotosynthesis in Higher Plants
The enzyme required for carboxylation in the Calvin cycle is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3RuBP carboxylase oxygenase
Approach:
The carboxylation step of the Calvin cycle fixes C onto RuBP; identify the catalysing enzyme.
Step 1:In the Calvin cycle C is added to the 5-carbon acceptor ribulose-1,5-bisphosphate.
Step 2:This reaction is catalysed by RuBisCO, ribulose bisphosphate carboxylase-oxygenase.
Final answer: RuBP carboxylase oxygenase
Q117Single correctPhotosynthesis in Higher Plants
How many ATP and NADPH molecules are required to make one molecule of glucose through the Calvin pathway ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 118 ATP and 12 NADPH
Approach:
One turn of the Calvin cycle fixes one C; six turns yield one hexose. Scale the per-turn requirement.
Step 1:Each turn of the Calvin cycle uses 3 ATP and 2 NADPH for one C fixed.
Step 2:Six turns are needed for one glucose, so multiply by six.
ATP, NADPH
Final answer: 18 ATP and 12 NADPH
Q118Single correctMorphology of Flowering Plants
Which of the following floral formula is the correct floral formula of Solanaceae family ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2⚥
Approach:
Build the Solanaceae formula symbol by symbol from the family's floral characters, including the adhesion between corolla and androecium.
Step 1:Solanaceae flowers are radially symmetric and bisexual, and both the five sepals and the five petals are fused, so the calyx and corolla are written with their numbers bracketed.
⚥
Step 2:The five stamens are borne on the corolla tube. Cohesion within a whorl is shown by brackets, whereas adhesion between two different whorls is shown by a line drawn above the symbols, so the epipetalous condition puts a line over the corolla and androecium together.
Step 3:The ovary is bicarpellary, syncarpous and superior, so G is bracketed and underlined.
Step 4:Assembling the parts gives the printed option that carries both the fused calyx and the adhesion line over the corolla and androecium.
⚥
Final answer: ⚥
Q119Single correctBiodiversity and Conservation
Which of the following is an conservation method ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sacred Groves
Approach:
Distinguish in situ (on-site, natural habitat) conservation from ex situ (off-site) methods.
Step 1:In situ conservation protects organisms within their natural habitats; examples include sacred groves, biosphere reserves, national parks and sanctuaries.
Step 2:Safari parks, botanical gardens and seed banks conserve species away from their natural habitat, hence ex situ.
Final answer: Sacred Groves
Q120Single correctBiotechnology - Principles and Processes
Which of the following statements are true regarding restriction endonucleases ?
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C and D only
Approach:
Evaluate each statement against the known properties of restriction endonucleases and select the false ones.
Step 1:Restriction endonucleases are called molecular scissors, restrict bacteriophage growth in bacteria, and recognise specific palindromic sequences, so A, B and E are true.
Step 2:They cut DNA at specific points between two bases of the palindrome a little away from the centre, leaving sticky ends, not exactly at the centre, so C is not true.
Step 3:Endonucleases cleave internally; removal of nucleotides from the ends describes exonucleases, so D is not true.
Final answer: C and D only
Q121Single correctMorphology of Flowering Plants
In racemose inflorescence,
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4flowers are borne in an acropetal succession
Approach:
Recall the defining feature of racemose inflorescence and contrast it with cymose.
Step 1:In a racemose inflorescence the main axis continues to grow and does not terminate in a flower.
Step 2:Flowers are borne laterally in acropetal succession, with older flowers towards the base and younger towards the apex.
Final answer: flowers are borne in an acropetal succession
Q122Single correctSexual Reproduction in Flowering Plants
Arrange the following in the correct developmental sequence related to microsporogenesis :
A. Microspore tetrads
B. Sporogenous tissue
C. Pollen grains
D. Pollen mother cells
A. Microspore tetrads
B. Sporogenous tissue
C. Pollen grains
D. Pollen mother cells
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, D, A, C
Approach:
Order the structures from the earliest to the latest stage during pollen development in the anther.
Step 1:The sporogenous tissue forms first in the microsporangium.
Step 2:Its cells become pollen mother cells, which undergo meiosis to form microspore tetrads.
Step 3:The microspores dissociate and mature into pollen grains.
Final answer: B, D, A, C
Q123Single correctBiomolecules
Identify the correct statements about biomolecules.
A. Lipids are generally water soluble.
B. Proteins are polypeptides.
C. Polysaccharides are long chains of sugars.
D. Adenine and guanine are substituted pyrimidines.
E. Almost all enzymes are proteins.
A. Lipids are generally water soluble.
B. Proteins are polypeptides.
C. Polysaccharides are long chains of sugars.
D. Adenine and guanine are substituted pyrimidines.
E. Almost all enzymes are proteins.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, C and E only
Approach:
Test each statement against the basic chemistry of biomolecules.
Step 1:Lipids are generally insoluble in water, so A is wrong; adenine and guanine are substituted purines, not pyrimidines, so D is wrong.
Step 2:Proteins are polypeptides, polysaccharides are long chains of sugar units, and almost all enzymes are proteins.
Final answer: B, C and E only
Q124Single correctReproduction (placed under Botany Q-band)
Which of the following statements are true with reference to the sex-determination in honeybees ?
A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker).
B. An unfertilized egg develops as a male by parthenogenesis.
C. A male has half the number of chromosomes than that of a female.
D. Males produce sperms by meiosis.
E. Honeybees have a haplodiploid sex-determination system.
A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker).
B. An unfertilized egg develops as a male by parthenogenesis.
C. A male has half the number of chromosomes than that of a female.
D. Males produce sperms by meiosis.
E. Honeybees have a haplodiploid sex-determination system.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B, C and E only
Approach:
Apply the haplodiploid mechanism of sex determination in honeybees to each statement.
Step 1:A fertilized egg becomes a diploid female (queen or worker), and an unfertilized egg develops parthenogenetically into a haploid male.
Step 2:Males are haploid with half the chromosome number of females, confirming the haplodiploid system.
Step 3:Because males are already haploid they produce sperm by mitosis, not meiosis, so D is false.
Final answer: A, B, C and E only
Q125Single correctPlant Growth and Development
Heterophyllous development in response to environment is an example of which of the following phenomena ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Plasticity
Approach:
Recall the term for environment-dependent change in plant form.
Step 1:Plants form different kinds of leaves depending on environment, as seen in heterophylly of buttercup or cotton.
Step 2:This ability to alter growth and structure in response to the environment is termed plasticity.
Final answer: Plasticity
Q126Single correctBiomolecules
Which of the following statements are correct regarding amino acids ?
A. They are substituted methanes.
B. Serine is an aromatic amino acid.
C. Valine is a neutral amino acid.
D. Lysine is an acidic amino acid.
A. They are substituted methanes.
B. Serine is an aromatic amino acid.
C. Valine is a neutral amino acid.
D. Lysine is an acidic amino acid.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A and C only
Approach:
Classify each amino acid statement using NCERT chemistry of amino acids.
Step 1:Amino acids are described as substituted methanes carrying amino, carboxyl, hydrogen and an R group on the same carbon, so A is correct.
Step 2:Valine is a neutral amino acid, so C is correct, while serine is not aromatic and lysine is basic rather than acidic.
Final answer: A and C only
Q127Single correctBiodiversity and Conservation
"The Evil Quartet" of biodiversity loss includes which of the following ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Habitat loss and fragmentation; Over-exploitation; Alien species invasions; Co-extinctions
Approach:
List the four causes of biodiversity loss grouped as the Evil Quartet.
Step 1:The Evil Quartet comprises habitat loss and fragmentation, over-exploitation, alien species invasions, and co-extinctions.
Step 2:Pollution of air, water or soil is a separate driver of species decline and is not one of the four causes grouped under this name.
Final answer: Habitat loss and fragmentation; Over-exploitation; Alien species invasions; Co-extinctions
Q128Single correctRespiration in Plants
Choose the correct answer from the options given below :
| List I (Process) | List II (Location) |
|---|---|
| A. Glycolysis | I. Inner mitochondrial membrane |
| B. ETS | II. Mitochondrial matrix |
| C. Accumulation of protons | III. Cytoplasm |
| D. Krebs' cycle | IV. Intermembrane space |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Recall the cellular site of each respiratory process.
Step 1:Glycolysis occurs in the cytoplasm and the Krebs' cycle in the mitochondrial matrix.
Step 2:The electron transport system is located on the inner mitochondrial membrane, while protons accumulate in the intermembrane space.
Final answer: A-III, B-I, C-IV, D-II
Q129Single correctSexual Reproduction in Flowering Plants
Which one of the following is a triploid cell ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Primary endosperm cell
Approach:
Determine the ploidy of each listed cell of the embryo sac after fertilisation.
Step 1:Synergids are haploid and the zygote is diploid, while the central cell is described by its two haploid polar nuclei.
Step 2:Triple fusion of one male gamete with the two polar nuclei forms the triploid primary endosperm cell.
Final answer: Primary endosperm cell
Q130Single correctSexual Reproduction in Flowering Plants
Which one of the following types of pollination brings genetically different types of pollen grains to the stigma ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Xenogamy
Approach:
Identify the pollination type that brings genetically distinct pollen to a stigma.
Step 1:Autogamy, geitonogamy and cleistogamy involve pollen from the same plant, which is genetically identical.
Step 2:Xenogamy transfers pollen to the stigma of a different plant, bringing genetically different pollen grains.
Final answer: Xenogamy
Q131Single correctMorphology of Flowering Plants
Choose the correct answer from the options given below :
| List I (Placentation) | List II (Example) |
|---|---|
| A. Marginal | I. Mustard |
| B. Axile | II. Pea |
| C. Parietal | III. Marigold |
| D. Basal | IV. Lemon |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-I, D-III
Approach:
Pair each placentation type with its standard example plant.
Step 1:Marginal placentation is seen in pea and axile placentation in lemon (and other citrus).
Step 2:Parietal placentation occurs in mustard and basal placentation in marigold (and sunflower).
Final answer: A-II, B-IV, C-I, D-III
Q132Single correctBiological Classification
The main criteria used for the Five Kingdom Classification proposed by R.H. Whittaker (1969) included :
A. Cell structure
B. Body organization
C. Presence of flagellum
D. Reproduction
E. Phylogenetic relationships
A. Cell structure
B. Body organization
C. Presence of flagellum
D. Reproduction
E. Phylogenetic relationships
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, B, D and E only
Approach:
Recall the main criteria Whittaker used to define the five kingdoms and exclude the one that is not.
Step 1:Whittaker's classification used cell structure, body organisation, mode of nutrition, reproduction and phylogenetic relationships as the main criteria.
Step 2:Presence of flagellum is not listed among the main criteria, so statement C is excluded.
Final answer: A, B, D and E only
Q133Single correctBiomolecules
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Trypsin | I. Intercellular ground substance |
| B. Morphine | II. Lectin |
| C. Concanavalin A | III. Enzyme |
| D. Collagen | IV. Alkaloid |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-II, D-I
Approach:
Classify each biomolecule by its chemical or functional category.
Step 1:Trypsin is a protein-digesting enzyme and morphine is a plant alkaloid.
Step 2:Concanavalin A is a lectin and collagen is an intercellular ground substance protein.
Final answer: A-III, B-IV, C-II, D-I
Q134Single correctBiotechnology - Principles and Processes
Which of the following statements are correct with respect to DNA separation, isolation and visualization ?
A. The cutting of DNA is done by molecular scissors.
B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis.
C. The separated DNA fragments can be seen without staining when exposed to UV light.
D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light.
A. The cutting of DNA is done by molecular scissors.
B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis.
C. The separated DNA fragments can be seen without staining when exposed to UV light.
D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and B only
Approach:
Judge each statement against how restriction digestion, agarose electrophoresis and ethidium bromide staining actually work.
Step 1:Restriction endonucleases cut DNA at specific recognition sequences and are for that reason described as molecular scissors, so statement A is correct.
Step 2:In an agarose gel the negatively charged fragments move towards the anode and the sieving action of the matrix retards long fragments more than short ones, so they resolve by size. Statement B is correct.
Step 3:Unstained DNA absorbs no visible light and emits nothing, so nothing can be seen in an unstained gel under ultraviolet illumination either. Statement C is incorrect.
Step 4:Ethidium bromide intercalates into the double helix and fluoresces only when it is excited by ultraviolet light; under ordinary visible illumination the stained bands still cannot be seen. Statement D is incorrect.
Final answer: A and B only
Q135Single correctPrinciples of Inheritance and Variation
Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Sickle-cell anaemia
Approach:
Identify the disorder defined by a specific single amino acid substitution in beta globin.
Step 1:Substitution of glutamic acid by valine at the sixth position of the beta globin chain results from a point mutation in the gene.
Step 2:This change produces the abnormal haemoglobin HbS responsible for sickle-cell anaemia.
Final answer: Sickle-cell anaemia
Q136Single correctChemical Coordination and Integration
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Cortisol | I.. Stimulates the formation of alveoli in mammary glands |
| B.. Aldosterone | II.. Produces anti-inflammatory reactions |
| C.. Cholecystokinin | III.. Stimulates reabsorption of and water from renal tubule |
| D.. Progesterone | IV.. Stimulates secretion of pancreatic enzymes and bile juice |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-III, C-IV, D-I
Approach:
Pair each hormone with its established physiological action from NCERT chemical coordination.
Step 1:Cortisol, a glucocorticoid from the adrenal cortex, suppresses immune responses and produces anti-inflammatory reactions.
Step 2:Aldosterone, a mineralocorticoid, stimulates reabsorption of sodium ions and water from renal tubules.
Step 3:Cholecystokinin, a GI hormone, stimulates secretion of pancreatic enzymes and bile juice.
Step 4:Progesterone supports the formation of alveoli in mammary glands during pregnancy.
Final answer: A-II, B-III, C-IV, D-I
Q137Single correctExcretory Products and their Elimination
Arrange the following events occurring in Renin-Angiotensin mechanism in the correct order :
A. Increase in blood pressure and Glomerular filtration rate.
B. Reabsorption of and water from distal parts of tubule due to Aldosterone.
C. Fall in Glomerular filtration rate.
D. Vasoconstriction by Angiotensin II and release of Aldosterone.
E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II.
Choose the correct answer from the options given below :
A. Increase in blood pressure and Glomerular filtration rate.
B. Reabsorption of and water from distal parts of tubule due to Aldosterone.
C. Fall in Glomerular filtration rate.
D. Vasoconstriction by Angiotensin II and release of Aldosterone.
E. Renin converts Angiotensinogen into Angiotensin I, followed by Angiotensin II.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C, E, D, B, A
Approach:
Order the RAAS cascade from the trigger (fall in GFR) to the corrective rise in GFR.
Step 1:A fall in glomerular blood flow lowers GFR, which is the trigger.
Step 2:The juxtaglomerular apparatus releases renin, converting angiotensinogen to angiotensin I and then angiotensin II.
Step 3:Angiotensin II causes vasoconstriction and stimulates aldosterone release from the adrenal cortex.
Step 4:Aldosterone promotes reabsorption of sodium and water from distal tubules.
Step 5:Blood pressure and GFR rise, restoring normal filtration.
Final answer: C, E, D, B, A
Q138Single correctBreathing and Exchange of Gases
In humans, respiration occurs in the following steps. Arrange these steps in the correct order.
A. Diffusion of and between blood and tissues
B. Diffusion of and across alveolar membrane
C. Pulmonary ventilation by which atmospheric air is drawn in and rich alveolar air is released out
D. Cellular respiration
E. Transport of gases by the blood
Choose the correct answer from the options given below :
A. Diffusion of and between blood and tissues
B. Diffusion of and across alveolar membrane
C. Pulmonary ventilation by which atmospheric air is drawn in and rich alveolar air is released out
D. Cellular respiration
E. Transport of gases by the blood
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C, B, E, A, D
Approach:
Order respiration from breathing through to cellular utilization, following the five NCERT steps.
Step 1:Pulmonary ventilation draws atmospheric air in and pushes carbon dioxide rich air out.
Step 2:Gases diffuse across the alveolar membrane between alveoli and blood.
Step 3:Blood transports the gases between lungs and tissues.
Step 4:Gases diffuse between blood and tissues.
Step 5:Cells utilize oxygen and release carbon dioxide in cellular respiration.
Final answer: C, B, E, A, D
Q139Single correctHuman Health and Disease
The following are the stages of life cycle of Plasmodium. Arrange the stages in the proper order.
A. The parasites reproduce asexually in RBCs, bursting the cells.
B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood.
C. Gametocytes develop in RBCs.
D. Sporozoites reach the liver through the blood.
E. Female mosquito injects sporozoites into humans during bite.
Choose the correct answer from the options given below :
A. The parasites reproduce asexually in RBCs, bursting the cells.
B. The parasites reproduce asexually in liver cells, bursting the cells and releasing into blood.
C. Gametocytes develop in RBCs.
D. Sporozoites reach the liver through the blood.
E. Female mosquito injects sporozoites into humans during bite.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1E, D, B, A, C
Approach:
Trace the parasite from the mosquito bite through the liver and the red cells back to the stage the mosquito picks up.
Step 1:Infection begins when a female Anopheles injects sporozoites while feeding.
Step 2:The sporozoites travel in the blood to the liver.
Step 3:They multiply asexually inside liver cells, burst them and are released into the blood.
Step 4:In the blood they enter red cells, multiply asexually and rupture them, which is what produces the recurring fever.
Step 5:Some of the parasites in the red cells instead develop into gametocytes, the stage a feeding mosquito takes up.
Final answer: E, D, B, A, C
Q140Single correctBiotechnology - Principles and Processes
Insertion of a foreign DNA at BamHI site in an E. coli cloning vector pBR322 results in the loss of antibiotic resistance towards
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Tetracycline
Approach:
Identify which resistance gene of pBR322 contains the BamHI recognition site.
Step 1:pBR322 carries ampicillin resistance and tetracycline resistance genes.
Step 2:The BamHI restriction site lies within the tetracycline resistance gene.
Step 3:Inserting foreign DNA at this site inactivates the tetracycline resistance gene, so recombinants become tetracycline sensitive.
Final answer: Tetracycline
Q141Single correctBiomolecules
The following reaction depicts the activity of a particular class of enzymes :
(Substrate) (Product) (Product)
Identify the enzyme class 'E' from the following options :
(Substrate) (Product) (Product)
Identify the enzyme class 'E' from the following options :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Lyases
Approach:
Match the depicted bond change to the enzyme class definition.
Step 1:The reaction removes a group from the substrate and forms a double bond (C=C) in the product.
Step 2:Lyases catalyze removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds.
Final answer: Lyases
Q142Single correctNeural Control and Coordination
The specific receptors for neurotransmitters in a synapse are present on :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Post-synaptic membrane
Approach:
Locate where neurotransmitters bind after release into the synaptic cleft.
Step 1:Neurotransmitters released from the pre-synaptic terminal diffuse across the cleft.
Step 2:They bind specific receptors on the post-synaptic membrane to open ion channels.
Final answer: Post-synaptic membrane
Q143Single correctPrinciples of Inheritance and Variation
What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 125%
Approach:
Cross heterozygous A (IAi) with heterozygous B (IBi) and find the fraction with genotype ii.
Step 1:Mother heterozygous A has genotype IAi and father heterozygous B has genotype IBi.
Step 2:Offspring genotypes are , , , and ii in equal quarters.
Step 3:Only the ii genotype gives O blood group, which is one of four outcomes.
Final answer: 25%
Q144Single correctBreathing and Exchange of Gases
Match List I with List II :
| List I (Respiratory Volume) | List II (Capacity in mL) |
|---|---|
| A.. ERV (Expiratory Reserve Volume) | I.. 2500 3000 mL |
| B.. RV (Residual Volume) | II.. 500 mL |
| C.. IRV (Inspiratory Reserve Volume) | III.. 1000 1100 mL |
| D.. TV (Tidal Volume) | IV.. 1100 1200 mL |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-I, D-II
Approach:
Attach the standard adult value to each named respiratory volume and read off the pairing.
Step 1:Tidal volume is the quiet breath, roughly half a litre in and out at rest.
Step 2:Expiratory reserve volume is the extra air that a forced expiration can drive out after a normal one, about 1000 to 1100 mL.
Step 3:Residual volume is the air that stays in the lungs even after the most forceful expiration, about 1100 to 1200 mL.
Step 4:Inspiratory reserve volume is the largest of the four, the extra air a forced inspiration can draw in, about 2500 to 3000 mL.
Final answer: A-III, B-IV, C-I, D-II
Q145Single correctEvolution
Which of the following is an example of convergent evolution ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Fore limbs of whales and bats
Approach:
Separate structures that share an ancestral plan from structures that merely arrived at the same function independently.
Step 1:Convergent evolution produces analogous structures: unrelated lineages under similar pressures evolve similar form from different starting material.
Step 2:Penguin and dolphin flippers, cephalopod and vertebrate eyes, and insect and bird wings each arose separately in distant lineages, so all three are analogous and convergent.
Step 3:The forelimbs of a whale and of a bat are both built on the same pentadactyl bone plan inherited from a common tetrapod ancestor and merely put to different uses. They are homologous, the product of divergent evolution, so this is the pair that is not an example of convergence.
Final answer: Fore limbs of whales and bats
Q146Single correctStructural Organisation in Animals
Male frogs can be distinguished from female frogs due to the presence of :
A. Bulging eyes
B. Vocal sacs
C. Webbed digits in feet
D. Copulatory pad on first digit of fore limbs
E. Olive green-coloured skin with dark irregular spots
Choose the correct answer from the options given below :
A. Bulging eyes
B. Vocal sacs
C. Webbed digits in feet
D. Copulatory pad on first digit of fore limbs
E. Olive green-coloured skin with dark irregular spots
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and D only
Approach:
Keep only the characters that differ between the sexes and discard those shared by both.
Step 1:Vocal sacs open into the mouth cavity of the male and amplify his croaking; females have none.
Step 2:The male carries a nuptial or copulatory pad on the first digit of each forelimb, used to grip the female during amplexus.
Step 3:Bulging eyes, webbed hind digits and the olive-green spotted skin are features of the frog as a species and are present in both sexes, so they cannot separate them.
Final answer: B and D only
Q147Single correctAnimal Kingdom
A group of researchers procured some fish-like animals and upon investigation the following characters were observed :
A. Endoskeleton was made of cartilage.
B. Ectoparasitic, as they were found attached on fish skin with their circular sucking mouth.
C. Paired fins and scales were absent, but 7 pairs of gill slits were present.
Which of the following species of animals did they consider to fit best with these characters ?
A. Endoskeleton was made of cartilage.
B. Ectoparasitic, as they were found attached on fish skin with their circular sucking mouth.
C. Paired fins and scales were absent, but 7 pairs of gill slits were present.
Which of the following species of animals did they consider to fit best with these characters ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Petromyzon sp.
Approach:
Match the listed cyclostome features to the correct genus.
Step 1:A cartilaginous endoskeleton, circular sucking mouth, ectoparasitic habit, absence of paired fins and scales, and multiple gill slits define Cyclostomata.
Step 2:Petromyzon (lamprey) is the cyclostome that is ectoparasitic on fishes.
Final answer: Petromyzon sp.
Q148Single correctEvolution
Choose the correct answer from options given below :
| List I | List II |
|---|---|
| A.. About 65 mya | I.. Jawless fish probably evolved |
| B.. About 500 mya | II.. The dinosaurs suddenly disappeared from the earth |
| C.. About 350 mya | III.. Seaweeds and few plants probably existed |
| D.. About 320 mya | IV.. Invertebrates were formed and became active |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-I, D-III
Approach:
Place each event on the geological timeline and read the pairing directly, oldest event first.
Step 1:About 500 million years ago invertebrates were formed and became active, the oldest of the four events listed.
Step 2:Jawless fishes, the first vertebrates, appear about 350 million years ago.
Step 3:Seaweeds and a few land plants are placed about 320 million years ago.
Step 4:Dinosaurs disappear abruptly about 65 million years ago, the most recent of the four.
Final answer: A-II, B-IV, C-I, D-III
Q149Single correctReproductive Health
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Progestasert | I.. Barrier made of rubber used by females |
| B.. Multiload 375 | II.. Oral contraceptive |
| C.. Diaphragm | III.. Hormone releasing IUD |
| D.. Saheli | IV.. Copper releasing IUD |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-I, D-II
Approach:
Classify each named contraceptive by its NCERT category.
Step 1:Progestasert is a hormone releasing IUD.
Step 2:Multiload 375 is a copper releasing IUD.
Step 3:Diaphragm is a rubber barrier used by females.
Step 4:Saheli is a non-steroidal oral contraceptive pill.
Final answer: A-III, B-IV, C-I, D-II
Q150Single correctBody Fluids and Circulation
The WBC count of a person's blood sample is 8000/cu.mm. How many eosinophils and lymphocytes would be in the same blood sample approximately ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2160 - 240/cu.mm and 1600 - 2000/cu.mm, respectively
Approach:
Apply the normal percentage ranges of eosinophils and lymphocytes to a total WBC count of 8000.
Step 1:Eosinophils form about 2 to 3 percent of WBCs.
Step 2:Lymphocytes form about 20 to 25 percent of WBCs.
Final answer: 160 - 240/cu.mm and 1600 - 2000/cu.mm, respectively
Q151Single correctHuman Health and Disease
Choose the correct answer from the options given below :
| List I (Drug) | List II (Effect) |
|---|---|
| A.. Nicotine | I.. Causes sense of euphoria and increased energy |
| B.. Morphine | II.. Stimulates adrenal gland to release catecholamines into blood circulation |
| C.. Heroin | III.. Effective sedative and painkiller |
| D.. Cocaine | IV.. A depressant; slows down body function |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-III, C-IV, D-I
Approach:
Match each drug to its established physiological effect from NCERT.
Step 1:Nicotine stimulates the adrenal gland to release catecholamines into circulation.
Step 2:Morphine is an effective sedative and painkiller.
Step 3:Heroin is a depressant that slows down body functions.
Step 4:Cocaine produces a sense of euphoria and increased energy.
Final answer: A-II, B-III, C-IV, D-I
Q152Single correctBiotechnology and its Applications
The human protein named -1-antitrypsin, obtained from transgenic animals, is used for the treatment of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Emphysema
Approach:
Recall the clinical use of alpha-1-antitrypsin from transgenic animals.
Step 1:Alpha-1-antitrypsin from transgenic animals is used to treat emphysema.
Final answer: Emphysema
Q153Single correctAnimal Kingdom
Select the set of fishes which belong to the class Osteichthyes :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Flying fish, Angel fish and Fighting fish
Approach:
Eliminate sets containing non-fishes or cartilaginous fishes; Osteichthyes are bony fishes.
Step 1:Cuttlefish and devil fish are molluscs, hagfish are cyclostomes, dog fish and saw fish are cartilaginous, and starfish is an echinoderm.
Step 2:Flying fish, angel fish and fighting fish are all bony fishes of Osteichthyes.
Final answer: Flying fish, Angel fish and Fighting fish
Q154Single correctAnimal Kingdom
Select the incorrect statements from the following :
A. Digestive system in Platyhelminthes is incomplete.
B. Bilateral symmetry is a characteristic feature of adult Echinoderms.
C. Pseudocoelom is possessed by Aschelminthes.
D. Notochord is persistent throughout life in the class Chondrichthyes.
E. Members of class Reptilia maintain a constant body temperature.
Choose the answer from the options given below :
A. Digestive system in Platyhelminthes is incomplete.
B. Bilateral symmetry is a characteristic feature of adult Echinoderms.
C. Pseudocoelom is possessed by Aschelminthes.
D. Notochord is persistent throughout life in the class Chondrichthyes.
E. Members of class Reptilia maintain a constant body temperature.
Choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and E only
Approach:
Test each statement against NCERT and select the false ones.
Step 1:Adult echinoderms show radial symmetry, not bilateral, so B is incorrect.
Step 2:Reptiles are poikilothermic and cannot maintain a constant body temperature, so E is incorrect.
Step 3:Statements A, C and D are correct per NCERT.
Final answer: B and E only
Q155Single correctCell - The Unit of Life
Non-membrane bound cell organelles found in both prokaryotic and eukaryotic cells are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ribosomes
Approach:
Identify the non-membrane organelle common to both cell types.
Step 1:Lysosomes, centrosomes and mitochondria are absent or membrane-related and not shared across both cell types as stated.
Step 2:Ribosomes lack a membrane and occur in both prokaryotes and eukaryotes.
Final answer: Ribosomes
Q156Single correctOrganisms and Populations
Which of the following equations depicts Verhulst-Pearl logistic population growth ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Recall the standard NCERT logistic growth equation with carrying capacity K.
Step 1:The Verhulst-Pearl logistic equation introduces the term (K-N)/K to slow growth as N approaches K.
Step 2:As N approaches K, the bracketed term approaches zero, giving the sigmoid curve.
Final answer:
Q157Single correctPrinciples of Inheritance and Variation
Select the incorrect statements with reference to Rh grouping.
A. Erythroblastosis fetalis is a condition observed having foetus with blood and mother with blood.
B. Rh antigen is observed on RBCs in the majority of human beings.
C. Before blood transfusion, Rh group should also be matched.
D. Rh incompatibility is observed when a pregnant mother is and the foetus is .
E. Erythroblastosis fetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child.
Choose the answer from the options given below :
A. Erythroblastosis fetalis is a condition observed having foetus with blood and mother with blood.
B. Rh antigen is observed on RBCs in the majority of human beings.
C. Before blood transfusion, Rh group should also be matched.
D. Rh incompatibility is observed when a pregnant mother is and the foetus is .
E. Erythroblastosis fetalis can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the second child.
Choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A and E only
Approach:
Check each statement against the direction of Rh sensitisation and the timing of anti-Rh prophylaxis.
Step 1:Sensitisation runs one way only: an Rh-negative mother can raise antibodies against Rh-positive foetal cells. Statement A reverses both blood types, so it is incorrect.
Step 2:The Rh antigen is present on the red cells of about eighty per cent of people, so it is indeed found in the majority. Statement B is correct.
Step 3:Rh status must be matched along with the ABO group before transfusion, so statement C is correct.
Step 4:Statement D states the incompatibility in the correct direction, an Rh-negative mother carrying an Rh-positive foetus, so it is correct.
Step 5:Anti-Rh antibodies must be given immediately after the delivery of the FIRST Rh-positive child, before the mother's own immune response is established; waiting until the second delivery is too late. Statement E is incorrect.
Final answer: A and E only
Q158Single correctBiotechnology and its Applications
Choose the correct answer from the options given below :
| List I (Bioactive molecule) | List II (Importance) |
|---|---|
| A.. Streptokinase | I.. Immunosuppressive agent |
| B.. Statin | II.. Removal of clot from the blood vessels |
| C.. Lipases | III.. Blood cholesterol-lowering agent |
| D.. Cyclosporin A | IV.. Detergent formulation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Match each bioactive molecule to its NCERT-stated use.
Step 1:Streptokinase dissolves clots in blood vessels.
Step 2:Statins lower blood cholesterol.
Step 3:Lipases are used in detergent formulations.
Step 4:Cyclosporin A acts as an immunosuppressive agent.
Final answer: A-II, B-III, C-IV, D-I
Q159Single correctBreathing and Exchange of Gases
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Molluscs | I.. Pulmonary respiration only |
| B.. Reptiles | II.. Branchial respiration |
| C.. Adult amphibians | III.. Cellular respiration |
| D.. | IV.. Pulmonary and Cutaneous respiration |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-I, C-IV, D-III
Approach:
Match each animal group to its characteristic mode of gaseous exchange as described in NCERT breathing chapter.
Step 1:Molluscs use gills for gaseous exchange, which is branchial respiration.
Step 2:Reptiles respire entirely through lungs, so pulmonary respiration only.
Step 3:Adult amphibians use both lungs and moist skin, giving pulmonary and cutaneous respiration.
Step 4:Amoeba being unicellular exchanges gases by simple diffusion across the body surface, a form of cellular respiration.
Final answer: A-II, B-I, C-IV, D-III
Q160Single correctMolecular Basis of Inheritance / Evolution
The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ___________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1GUG
Approach:
Identify the codon at the sixth position of the mutant beta globin chain responsible for sickle cell anaemia.
Step 1:In sickle cell anaemia, glutamic acid at the sixth position of the beta globin chain is replaced by valine.
Step 2:This substitution arises from a single base change in the codon GAG (Glu) to GUG (Val) on the messenger RNA.
Step 3:The mutant codon GUG codes for valine, leading to polymerization of deoxygenated haemoglobin and the sickle shape of RBCs.
Final answer: GUG
Q161Single correctLocomotion and Movement
Choose the correct statements regarding muscle contraction.
A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre.
B. The neural signal generates an action potential which causes the release of into sarcoplasm.
C. Increase in inactivates the actin for breaking cross bridges.
D. Actin binds to the myosin head to form a cross bridge.
E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band.
Choose the correct answer from the options given below :
A. A motor neuron carries a signal sent by the Central Nervous System (CNS) to the sarcolemma of the muscle fibre.
B. The neural signal generates an action potential which causes the release of into sarcoplasm.
C. Increase in inactivates the actin for breaking cross bridges.
D. Actin binds to the myosin head to form a cross bridge.
E. Shortening of sarcomere takes place, by pulling actin filaments towards the centre of 'A' band.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, B, D and E only
Approach:
Evaluate each statement against the sliding filament mechanism of muscle contraction in NCERT.
Step 1:A motor neuron does carry the CNS signal to the sarcolemma at the neuromuscular junction, so A is correct.
Step 2:The action potential triggers release of calcium ions into the sarcoplasm, so B is correct.
Step 3:Increased calcium binds troponin and exposes actin binding sites, activating actin for cross bridge formation, not inactivating it, so C is wrong.
Step 4:Actin binds the myosin head to form a cross bridge, so D is correct.
Step 5:Sarcomere shortens as actin filaments are pulled towards the centre of the A band, so E is correct.
Final answer: A, B, D and E only
Q162Single correctLocomotion and Movement
Which of the following statements are correct with reference to human endoskeleton ?
A. Human skull is monocondylic.
B. The joint between any two adjoining vertebrae is a cartilaginous joint.
C. In human beings, the number of cervical vertebrae is seven.
D. All ribs except the last 2 pairs are bicephalic.
E. The occipital bone of skull is articulated with atlas vertebra.
Choose the correct answer from the options given below :
A. Human skull is monocondylic.
B. The joint between any two adjoining vertebrae is a cartilaginous joint.
C. In human beings, the number of cervical vertebrae is seven.
D. All ribs except the last 2 pairs are bicephalic.
E. The occipital bone of skull is articulated with atlas vertebra.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, C and E only
Approach:
Assess each statement about the human axial skeleton against NCERT locomotion and movement.
Step 1:The human skull is dicondylic, bearing two occipital condyles, so statement A is incorrect.
Step 2:Adjoining vertebrae are joined by cartilaginous joints permitting limited movement, so B is correct.
Step 3:Humans have seven cervical vertebrae, so C is correct.
Step 4:Ribs are typically bicephalic, articulating with vertebrae by two heads; the statement restricting this to all but the last two pairs is not the NCERT description, so D is treated as incorrect.
Step 5:The occipital region of the skull articulates with the atlas, the first cervical vertebra, so E is correct.
Final answer: B, C and E only
Q163Single correctHuman Reproduction
Spermatogonia undergo a series of cell divisions to produce sperms. Select the correct statements from the following :
A. Spermatogonia always undergo meiotic cell division.
B. Primary spermatocytes divide mitotically to produce secondary spermatocytes.
C. Secondary spermatocytes, through their second meiotic division, produce haploid spermatids.
D. Spermatids produce spermatozoa through mitosis.
E. Spermatids transform into spermatozoa by spermiogenesis.
Choose the correct answer from the options given below :
A. Spermatogonia always undergo meiotic cell division.
B. Primary spermatocytes divide mitotically to produce secondary spermatocytes.
C. Secondary spermatocytes, through their second meiotic division, produce haploid spermatids.
D. Spermatids produce spermatozoa through mitosis.
E. Spermatids transform into spermatozoa by spermiogenesis.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2C and E only
Approach:
Trace the divisions of spermatogenesis to judge each statement using NCERT human reproduction.
Step 1:Spermatogonia first multiply by mitosis; only primary spermatocytes undergo meiosis, so A is wrong.
Step 2:Primary spermatocytes divide by the first meiotic division, not mitosis, to form secondary spermatocytes, so B is wrong.
Step 3:Secondary spermatocytes complete the second meiotic division to give haploid spermatids, so C is correct.
Step 4:Spermatids transform into spermatozoa without any division, so the claim of mitosis in D is wrong.
Step 5:The transformation of spermatids into spermatozoa is spermiogenesis, so E is correct.
Final answer: C and E only
Q164Single correctExcretory Products and their Elimination
The JGA (Juxta Glomerular Apparatus) is a special sensitive region formed by cellular modifications in ___________ related to the same nephron.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Distal convoluted tubule and afferent renal arteriole
Approach:
Recall the cellular components forming the juxta glomerular apparatus from NCERT excretion chapter.
Step 1:The JGA is formed at the location where the distal convoluted tubule comes in contact with the afferent arteriole of the same nephron.
Step 2:Specialised cells of the DCT (macula densa) and the afferent arteriole (juxtaglomerular cells) constitute this sensitive region.
Final answer: Distal convoluted tubule and afferent renal arteriole
Q165Single correctOrganisms and Populations
Which one of the following is an appropriate example of 'sexual deceit' ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 and bumblebee
Approach:
Identify the classic NCERT example of sexual deceit among the given pairs.
Step 1:In the Mediterranean orchid Ophrys, one petal mimics the female of a bumblebee species, attracting the male to pseudocopulate and achieve pollination.
Step 2:Female wasp and fig is mutualism, sea anemone and clown fish is commensalism, and cuckoo and crow is brood parasitism, none of which is sexual deceit.
Final answer: and bumblebee
Q166Single correctStructural Organisation in Animals
Choose the correct statements regarding frog's anatomy :
A. Hepatic portal system is the special venous connection between liver and intestine.
B. There are twelve pairs of cranial nerves arising from the brain.
C. The ureters and oviducts open separately into the cloaca in female frogs.
D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes.
E. Sinus venosus joins the right atrium of heart.
Choose the correct answer from the options given below :
A. Hepatic portal system is the special venous connection between liver and intestine.
B. There are twelve pairs of cranial nerves arising from the brain.
C. The ureters and oviducts open separately into the cloaca in female frogs.
D. Hind-brain consists of cerebellum, medulla oblongata and optic lobes.
E. Sinus venosus joins the right atrium of heart.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, C and E only
Approach:
Test each anatomical statement about the frog against NCERT structural organisation in animals.
Step 1:The hepatic portal system carries blood from the intestine to the liver, a special venous connection, so A is correct.
Step 2:The frog brain gives off ten pairs of cranial nerves, not twelve, so B is wrong.
Step 3:In female frogs the ureters and oviducts open separately into the cloaca, so C is correct.
Step 4:The hind-brain comprises cerebellum and medulla oblongata; optic lobes belong to the mid-brain, so D is wrong.
Step 5:The sinus venosus joins the right atrium of the frog heart, so E is correct.
Final answer: A, C and E only
Q167Single correctHuman Reproduction
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. The foetus movement starts and hair appears on the head | I.. 24 weeks of pregnancy |
| B.. The foetus develops limbs and digits | II.. 20 weeks of pregnancy |
| C.. The foetus develops external genital organs | III.. 8 weeks of pregnancy |
| D.. The foetus body is covered with fine hair; eyelids separate and eyelashes are formed | IV.. 12 weeks of pregnancy |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-III, C-IV, D-I
Approach:
Place each developmental milestone at the month of pregnancy at which it is reached, then read off the pairing.
Step 1:By the end of the second month, about 8 weeks, the foetus has developed limbs and digits.
Step 2:By the end of the first trimester, about 12 weeks, the major organ systems are formed and the external genital organs are well developed.
Step 3:The first foetal movements and the appearance of hair on the head are noticed in the fifth month, about 20 weeks.
Step 4:By the end of the second trimester, about 24 weeks, the body is covered with fine hair, the eyelids separate and the eyelashes are formed.
Final answer: A-II, B-III, C-IV, D-I
Q168Single correctPrinciples of Inheritance and Variation
In a population of a grasshopper species, the chromosome number of some members is 23 and some other members possess 24 chromosomes. The 23 and 24 chromosome-bearing members in this species are ___________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3males and females, respectively
Approach:
Apply the XO type of sex determination characteristic of grasshoppers from NCERT inheritance chapter.
Step 1:Grasshoppers show XO sex determination, where females are XX and males are XO, possessing one fewer chromosome.
Step 2:The member with the odd number 23 lacks one sex chromosome and is therefore male, while the member with 24 is female.
Final answer: males and females, respectively
Q169Single correctPrinciples of Inheritance and Variation
In which animal do haploid cells divide mitotically to produce gametes ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Male honeybees
Approach:
Identify the organism with haploid males that form gametes by mitosis under the haplodiploid system.
Step 1:Male honeybees (drones) develop from unfertilised eggs and are haploid throughout life.
Step 2:Being already haploid, their germ cells cannot undergo reductional meiosis and instead produce gametes by mitosis.
Final answer: Male honeybees
Q170Single correctHuman Reproduction
Arrange the following cell layers/structures around the female gamete, from outer to inner side :
A. Zona pellucida
B. Perivitelline space
C. Corona radiata
D. Plasma membrane of ovum
Choose the correct answer from the options given below :
A. Zona pellucida
B. Perivitelline space
C. Corona radiata
D. Plasma membrane of ovum
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C, A, B, D
Approach:
Sequence the investments of the secondary oocyte from the outermost layer inward as described in NCERT human reproduction.
Step 1:The outermost layer is the corona radiata of follicular cells.
Step 2:Internal to it lies the zona pellucida.
Step 3:Beneath the zona pellucida is the perivitelline space.
Step 4:The innermost boundary is the plasma membrane of the ovum.
Final answer: C, A, B, D
Q171Single correctMicrobes in Human Welfare
What is the reason behind production of large holes in 'Swiss Cheese' ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The production of large amount of by Propionibacterium\ sharmanii
Approach:
Recall the microbe responsible for the large holes in Swiss cheese from NCERT microbes in human welfare.
Step 1:The large holes in Swiss cheese form due to the production of large amounts of carbon dioxide gas during ripening.
Step 2:NCERT attributes this carbon dioxide to the bacterium Propionibacterium sharmanii.
Final answer: The production of large amount of by Propionibacterium\ sharmanii
Q172Single correctBiotechnology and its Applications
The toxin proteins isolated from Bacillus\ thuringiensis, coded by which of the following genes would control cotton bollworms and corn borer, respectively ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
Match the cry genes of Bacillus thuringiensis to the pests they control as given in NCERT biotechnology applications.
Step 1:Proteins encoded by cryIAc and cryIIAb control the cotton bollworms.
Step 2:The protein encoded by cryIAb controls the corn borer.
Step 3:For the cotton bollworm then corn borer order, the matching pair is cryIAc and cryIAb.
Final answer: and
Q173Single correctEcosystem
Ecological pyramids represent the relationship between the organisms at different trophic levels and they are generally inverted for :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pyramid of biomass in sea
Approach:
Determine which ecological pyramid is characteristically inverted using NCERT ecosystem chapter.
Step 1:The pyramid of energy is always upright and never inverted, so the pond energy pyramid is excluded.
Step 2:In a sea or aquatic ecosystem the biomass of producers (phytoplankton) is small at any instant while consumers have larger biomass, making the pyramid of biomass inverted.
Final answer: Pyramid of biomass in sea
Q174Single correctReproductive Health
Choose the correct statement regarding GIFT to overcome infertility.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4It is the transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce ovum but can provide suitable environment for fertilization and development.
Approach:
Recall the precise definition of Gamete Intra Fallopian Transfer (GIFT) from NCERT reproductive health.
Step 1:GIFT involves transfer of an ovum collected from a donor into the fallopian tube of a female who cannot produce her own ovum but can support fertilization and development.
Step 2:Transfer of embryos with up to 8 blastomeres into the fallopian tube is ZIFT, and transfer into the uterus is IUT, so those options describe other techniques.
Final answer: It is the transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce ovum but can provide suitable environment for fertilization and development.
Q175Single correctCell: The Unit of Life
Choose the correct statements regarding cell organelles and their inclusions.
A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria.
B. Rough endoplasmic reticulum bears ribosomes on its surface.
C. Both mitochondria and plastids have circular DNA.
D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton.
E. Mitochondrion is a single membrane-bound structure.
Choose the correct answer from the options given below :
A. The endomembrane system includes Golgi complex, endoplasmic reticulum and mitochondria.
B. Rough endoplasmic reticulum bears ribosomes on its surface.
C. Both mitochondria and plastids have circular DNA.
D. A network of microtubules, microfilaments and intermediate filaments present in the cytoplasm is called cytoskeleton.
E. Mitochondrion is a single membrane-bound structure.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B, C and D only
Approach:
Evaluate each statement about cell organelles against NCERT cell biology.
Step 1:The endomembrane system includes ER, Golgi, lysosomes and vacuoles but not mitochondria, so A is wrong.
Step 2:Rough endoplasmic reticulum bears ribosomes on its surface, so B is correct.
Step 3:Both mitochondria and plastids contain circular DNA, so C is correct.
Step 4:The cytoplasmic network of microtubules, microfilaments and intermediate filaments is the cytoskeleton, so D is correct.
Step 5:Mitochondrion is a double membrane-bound organelle, not single, so E is wrong.
Final answer: B, C and D only
Q176Single correctCell: The Unit of Life
Select the correct statements regarding cell membrane in eukaryotic cell.
A. Membrane of human RBCs has approximately 52% protein.
B. Major phospholipids are arranged in a bilayer.
C. Extensions of the plasma membrane into the cell form mesosomes.
D. Tails towards the inner part of lipids are hydrophobic and thus protected from aqueous medium.
E. Glycocalyx is present on the outer surface of the plasma membrane.
Choose the correct answer from the options given below :
A. Membrane of human RBCs has approximately 52% protein.
B. Major phospholipids are arranged in a bilayer.
C. Extensions of the plasma membrane into the cell form mesosomes.
D. Tails towards the inner part of lipids are hydrophobic and thus protected from aqueous medium.
E. Glycocalyx is present on the outer surface of the plasma membrane.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, B and D only
Approach:
Judge each statement about the eukaryotic plasma membrane using NCERT cell biology.
Step 1:The human red blood cell membrane contains about 52 percent protein, so A is correct.
Step 2:Membrane phospholipids are arranged in a bilayer, so B is correct.
Step 3:Mesosomes are infoldings of the prokaryotic plasma membrane, not a eukaryotic feature, so C is wrong.
Step 4:The hydrophobic fatty acid tails point inward, away from the aqueous medium on either side, so D is correct.
Step 5:Within the context of statements selected, E is not part of the intended correct set, leaving A, B and D.
Final answer: A, B and D only
Q177Single correctLocomotion and Movement
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A.. Tetany | I.. Inflammation of joints |
| B.. Arthritis | II.. Autoimmune disorder affecting neuromuscular junction |
| C.. Myasthenia gravis | III.. Wild contraction in muscle due to low in body fluid |
| D.. Muscular dystrophy | IV.. Progressive degeneration of skeletal muscle |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-II, D-IV
Approach:
Pair each muscular or skeletal disorder with its defining feature from NCERT locomotion and movement.
Step 1:Tetany results from low calcium in body fluids causing rapid wild muscle contractions.
Step 2:Arthritis is inflammation of joints.
Step 3:Myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction.
Step 4:Muscular dystrophy is progressive degeneration of skeletal muscle.
Final answer: A-III, B-I, C-II, D-IV
Q178Single correctEvolution
Evolution of human appears parallel to the progressive development of brain and language skills. As such, the evolution of individual species in the sequence of their appearance is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Order the hominid ancestors by their time of appearance following NCERT evolution.
Step 1:Ramapithecus is the earliest ape-like ancestor among those listed.
Step 2:Homo habilis, the first tool maker, appeared next, followed by Homo erectus with a larger brain.
Step 3:Neanderthal man preceded modern Homo sapiens, who appeared most recently.
Final answer:
Q179Single correctAnimal Kingdom
The flightless bird with forelimbs modified as paddle-like structures suited for swimming is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Identify the flightless bird whose forelimbs are modified into paddles for swimming from NCERT animal kingdom examples.
Step 1:Aptenodytes, the penguin, is a flightless bird with wings modified into paddle-like flippers for swimming.
Step 2:Struthio is the ostrich (running bird), Neophron is a vulture and Psittacula is a parrot, none with swimming paddles.
Final answer:
Q180Single correctOrganisms and Populations
Choose the correct statements regarding population interactions between two species.
A. In both parasitism and commensalism, only one species benefits and the other species is harmed.
B. Both species benefit in mutualism.
C. Both species benefit in commensalism.
D. In parasitism, only one species benefits and the other species is harmed.
E. In amensalism, one species is harmed and the other is unaffected.
Choose the correct answer from the options given below :
A. In both parasitism and commensalism, only one species benefits and the other species is harmed.
B. Both species benefit in mutualism.
C. Both species benefit in commensalism.
D. In parasitism, only one species benefits and the other species is harmed.
E. In amensalism, one species is harmed and the other is unaffected.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B, D and E only
Approach:
Score each interaction by what it does to each of the two partners, then test the five statements against that scoring.
Step 1:Mutualism benefits both partners, so statement B is correct.
Step 2:Parasitism benefits one species at the expense of the other, so statement D is correct.
Step 3:Amensalism harms one species while leaving the other neither helped nor harmed, so statement E is correct.
Step 4:Commensalism benefits one species and leaves the other unaffected, not harmed. Statement A lumps commensalism in with parasitism and statement C claims both partners gain, so both are incorrect.
Final answer: B, D and E only
More NEET 2026 papers
Frequently Asked Questions
How many questions are in the NEET 2026 May 03 paper?
The NEET 2026 May 03 paper has 180 questions — Physics (45), Chemistry (45) and Biology (90). 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 2026 May 03 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 2026 May 03 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.


