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NEET 2026 Jun 21 Question Paper with Solutions
All 180 questions from the NEET 2026 (Jun 21) 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 correctLaws of Motion
A particle of mass M moves along a horizontal x axis from to . The coefficient of kinetic friction varies as a function of x as , where are constants of appropriate dimensions, so that . The total work done by the frictional force during the motion is , where g is the acceleration due to gravity. The value of n is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The kinetic friction coefficient decreases linearly with position, so the frictional force is integrated over the displacement to obtain the total work done.
Step 1:The condition at the end point fixes the constant.
Step 2:The magnitude of work done by friction is the integral of the friction force over the path.
Step 3:Evaluating the integral and substituting gives the magnitude.
Step 4:Comparing with identifies the coefficient.
Final answer:
Q2Single correctKinetic Theory of Gases
The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are and , respectively, the correct option is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The dependence of mean free path on molecular diameter and number density is used; with the given ratio of mean free paths and the diameter relation, the ratio of number densities is obtained.
Step 1:The mean free path of gas is half that of gas , and the diameter of molecules of gas is twice that of gas .
Step 2:The mean free path for each gas is written in terms of its diameter and number density.
Step 3:Forming the ratio of the mean free paths gives a relation between the diameters and number densities.
Step 4:Substituting the diameter relation gives the ratio of number densities.
Final answer:
Q3Single correctElectromagnetic Induction
Two identical inductors are connected in two different configurations P and Q, where a time varying current I(t) is flowing, as shown in the figure. The induced emf between points a and b for configuration P is and that for configuration Q is . The ratio is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The equivalent inductance of each configuration between the terminals is determined from how the two identical coils are connected; the induced emf for each is the equivalent inductance times the rate of change of current, and the ratio of the two emfs is taken.
Step 1:In configuration the arrangement of the two identical coils gives an equivalent inductance between terminals and , so the induced emf has magnitude times the rate of change of current.
Step 2:In configuration the two coils combine in parallel, giving an equivalent inductance of half the single value, so the induced emf has magnitude half that of configuration .
Step 3:Taking the ratio of the two induced emfs eliminates the common rate of change of current.
Final answer:
Q4Single correctWaves
For sound waves, if the number of nodes for the harmonic of an open-ended pipe is n and that for the harmonic of the same pipe with one of its ends closed is m, the ratio is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The number of nodes for a given harmonic equals the harmonic number for an open pipe, while a closed pipe supports only odd harmonics; counting nodes in each case yields the ratio.
Step 1:For an open organ pipe the number of nodes equals the harmonic number, so the harmonic has five nodes.
Step 2:A closed organ pipe supports only odd harmonics, numbered as the first, third, fifth, seventh and ninth, with the number of nodes increasing by one for each successive odd harmonic.
Step 3:The ratio of the node counts is taken.
Final answer:
Q5Single correctElectromagnetic Induction
Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and is the permeability of free space, the inductance of the solenoid is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The self inductance of a long solenoid is derived from the flux linkage per unit current, with the field expressed through the turns per unit length and the cross-sectional area written in terms of the radius.
Step 1:The flux linkage is the product of the number of turns and the flux through one turn.
Step 2:The total number of turns is the turns per unit length times the length, and the field is expressed through the turns per unit length.
Step 3:Substituting these expressions and the cross-sectional area gives the inductance.
Final answer:
Q6Single correctMotion in a Straight Line
Consider a particle moving along a straight line, whose position as a function of time is given by , where , and . The average speed of the particle, in from to s is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The velocity is obtained by differentiating the position; the instant of velocity reversal is located and the total distance is computed as the sum of the magnitudes of the displacements in each interval, then divided by the elapsed time.
Step 1:Differentiating the position gives the velocity as a function of time.
Step 2:The velocity vanishes at the turning point.
Step 3:The distances in the two intervals are found from the area under the velocity-time graph and added.
Step 4:The average speed is the total distance divided by the total time.
Final answer:
Q7Single correctNuclei
Consider the following nuclear reaction: Take masses of , and as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: 1 u = 931.5 MeV ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The Q value equals the mass defect between reactant and products converted to energy through the given mass-energy equivalence.
Step 1:The mass defect is the reactant mass minus the total product mass.
Step 2:The mass defect is converted to energy using the given equivalence.
Step 3:Expressing the result in kiloelectronvolts gives the final value.
Final answer:
Q8Single correctDual Nature of Radiation and Matter
A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time t, then variation of the photocurrent I and magnitude of the stopping potential with time can be best represented by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(graph)
Approach:
The photocurrent is proportional to the number of incident photons per second and thus to the source power, while the stopping potential depends only on photon energy, which is unchanged.
Step 1:Decreasing power at constant photon energy means the number of photons per second decreases linearly with time, so the photocurrent decreases linearly.
Step 2:The energy of each photon is constant because the frequency is unchanged, so the stopping potential remains constant in time.
Step 3:The photocurrent therefore falls linearly to zero as the power decays, while the stopping potential holds a fixed value throughout.
Final answer: The photocurrent decreases linearly with time while the magnitude of the stopping potential stays constant.
Q9Single correctElectrostatic Potential and Capacitance
Three identical capacitors, P, Q and S, each of the capacitance C, are connected to a battery of voltage V, as shown in the figure. If the energy stored in the capacitor P and total energy stored in the system are and , respectively, then the ratio is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
From the circuit the voltage across each capacitor is determined, the energy stored in capacitor and the total energy are computed, and their ratio is taken.
Step 1:The three capacitors are identical, each of capacitance , and from the network the voltage across is half the battery voltage.
Step 2:The energy in capacitor is computed for a potential difference of half the battery voltage.
Step 3:The total energy is the sum of the energies in the series branch and the parallel capacitor.
Step 4:The required ratio is obtained by dividing the two energies.
Final answer:
Q10Single correctMechanical Properties of Fluids
Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is . The area of cross-section at P and Q are and , respectively. The rate of flow of water through the pipe, in , is: [Take density of water = ]

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The continuity equation relates the speeds at the two cross-sections, Bernoulli's equation links the pressure difference to the speeds, and the resulting speed gives the volume flow rate.
Step 1:Continuity gives the speed at the narrower section in terms of the wider one.
Step 2:Bernoulli's equation relates the pressure difference to the difference in the squares of the speeds.
Step 3:Solving for the speed at gives a value of about a tenth of a metre per second, and the speed at is twice this.
Step 4:The volume flow rate is the product of area and speed at the wider section, converted to cubic centimetres per second.
Final answer:
Q11Single correctMoving Charges and Magnetism
A current flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the centre O of the loop is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The current divides between the two arcs in inverse proportion to their resistances; the magnetic fields produced by the two arcs at the centre oppose each other and partially cancel, leaving a net field.
Step 1:The resistance of arc is half that of arc , so the currents divide accordingly.
Step 2:The total current is the sum of the two branch currents, fixing each branch current.
Step 3:The net field is the difference of the two arc fields, each carrying its branch current over its subtended angle, giving a residual field.
Final answer:
Q12Single correctMechanical Properties of Fluids
In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity (v) with the ratio of density of spherical ball () to density of the liquid (), is best represented by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(graph)
Approach:
The terminal velocity is expressed in terms of the density ratio; rewriting it shows a straight-line dependence on the ratio with a negative intercept, identifying the correct graph.
Step 1:The terminal velocity is rewritten by factoring the liquid density to express it through the density ratio.
Step 2:This has the form of a straight line in the density ratio with a negative intercept.
Step 3:Plotted against the density ratio, the terminal velocity is therefore a straight line whose intercept on the velocity axis is negative and which crosses zero when the ratio equals one.
Final answer: A straight line in with a negative intercept, crossing zero at .
Q13Single correctGravitation
Two planets and with equal mass have radii and , respectively, where . The escape speeds of and are and , respectively. Then is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The escape speed depends inversely on the square root of the radius for equal masses; the ratio of escape speeds follows from the ratio of radii.
Step 1:For equal masses the escape speed varies inversely with the square root of the radius.
Step 2:The ratio of escape speeds equals the square root of the inverse ratio of radii.
Final answer:
Q14Single correctGravitation
In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius is proportional to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Kepler's third law states that the square of the orbital period is proportional to the cube of the orbital radius, from which the proportionality of period to radius follows.
Step 1:For planetary motion the square of the time period is proportional to the cube of the orbital radius.
Step 2:Taking the square root gives the dependence of the period on radius.
Final answer:
Q15Single correctMoving Charges and Magnetism
Two infinitely long parallel conducting wires A and B carry currents I and , respectively, in the same direction. The wire A has uniform mass per unit length and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does not rise from the floor is: [g is the acceleration due to gravity and is the permeability of free space.]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The magnetic force per unit length between the parallel wires is set equal to the weight per unit length of the lower wire to find the maximum separation at which the wire just does not rise.
Step 1:The attractive force per unit length on wire due to wire is computed with currents and .
Step 2:For the wire not to rise, the upward magnetic force per unit length must not exceed the weight per unit length.
Step 3:The minimum height corresponds to equality, which on rearrangement gives the height.
Final answer:
Q16Single correctSemiconductor Electronics
An ideal Zener diode with breakdown voltage of V is reverse biased with a negative input voltage V. The magnitude of voltage difference between point B and A is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 V
Approach:
The Zener diode clamps the voltage across itself at the breakdown value, so the remaining input voltage appears across the series resistor; the magnitude of the potential difference between the two points is the difference of these.
Step 1:The magnitude of the input voltage exceeds the breakdown voltage, so the diode is in breakdown and holds three volts across itself.
Step 2:The Zener clamps its own voltage at the breakdown value.
Step 3:The full input magnitude is five volts, so the voltage across the resistor between the two points is the difference.
Final answer: V
Q17Single correctThermodynamics
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas () decreases from 60 K to 50 K. The work done by the gas in the process is: [Take the universal gas constant as ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 J
Approach:
The work done in an adiabatic process for one mole is expressed through the gas constant, the temperature change and the ratio of specific heats, then evaluated numerically.
Step 1:The temperature change is the final minus the initial temperature.
Step 2:Substituting the values into the adiabatic work expression with one mole and the given ratio of specific heats.
Step 3:Evaluating the expression gives the work done by the gas.
Final answer: J
Q18Single correctDual Nature of Radiation and Matter
A ray of light with wavelength is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are , and , respectively and the magnitude of stopping potentials of these cells are , and , respectively. The relation between and threshold wavelengths are , and . The correct option is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Emission occurs only when the incident wavelength is shorter than the threshold wavelength; the stopping potential increases as the threshold wavelength increases, which orders the stopping potentials for the cells.
Step 1:For emission the energy of the incident photon must exceed the work function.
Step 2:For cell 1 the threshold wavelength is shorter than the incident wavelength, so the photon energy is below the work function and no electron is ejected, giving zero stopping potential.
Step 3:For cells 2 and 3 the threshold wavelengths exceed the incident wavelength, with cell 3 having a much larger threshold and hence a smaller work function, so its stopping potential is greater.
Final answer:
Q19Single correctDual Nature of Radiation and Matter
A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron to that of photon , is: [Take speed of light = , charge of electron = C and mass of electron = kg]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The electron momentum follows from its kinetic energy and mass, the photon momentum from its energy and the speed of light; their ratio is then evaluated numerically.
Step 1:The two energies are equal, each twenty electronvolts.
Step 2:The ratio of momenta is the electron momentum divided by the photon momentum, with the photon momentum written as its energy over the speed of light.
Step 3:Substituting the numerical values for mass, energy and the speed of light gives the ratio.
Step 4:Evaluating the expression gives the required ratio.
Final answer:
Q20Single correctExperimental Physics
Which of the following measurements require 'index correction'?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Measurement of focal length of lenses using optical bench
Approach:
Index correction is an adjustment required when the observed positions differ from the actual positions of the optical components in optical bench experiments.
Step 1:Index correction is an adjustment applied in experiments related to the optical bench, where the index marks do not coincide with the actual positions of the lens or object.
Step 2:The measurement of focal length of lenses on an optical bench thus requires index correction.
Final answer: Measurement of focal length of lenses using optical bench
Q21Single correctElectrostatic Potential and Capacitance
A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density , as shown in the figure. The initial and final positions of the charge are marked by A and B at distance and respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is . The value of n is: [ is the permittivity of vacuum]

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The work done equals the unit charge times the potential difference between the two external points of the uniformly charged sphere, with the total charge expressed through the charge density and volume.
Step 1:The work done equals the unit charge times the difference of potentials at the final and initial external points.
Step 2:The total charge is the product of the charge density and the volume of the sphere.
Step 3:Simplifying the expression with the unit charge gives the magnitude of the work done.
Step 4:Comparing with the given form identifies the value of the constant.
Final answer:
Q22Single correctRay Optics and Optical Instruments
Consider three media P, Q and R with refractive indices 1, 1.25, and 1.5 respectively. The medium Q having a thickness of 5 cm is placed between extended media P and R as shown in the figure. An object O is placed at the centre of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is . For similar observation from medium R, the apparent depth is . The value of , in cm, is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The apparent depth seen from each side is the real depth scaled by the ratio of the refractive index of the viewing medium to that of the medium containing the object; the difference of the two apparent depths is taken.
Step 1:The object lies five centimetres from the relevant surface, and viewing from medium scales the real depth by the ratio of the object-medium index to the viewing-medium index.
Step 2:Viewing from medium scales the real depth by the corresponding index ratio.
Step 3:The magnitude of the difference of the two apparent depths is computed.
Final answer:
Q23Single correctSystems of Particles and Rotational Motion
A frictionless circular wire of unit radius is fixed on the horizontal plane. Two-point particles of unit mass start moving simultaneously from point with identical uniform angular speeds in opposite directions, and meet again at point . During this time, which of the following figures schematically represent the magnitude of the total linear momentum P of the system, as a function of ?

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(graph)
Approach:
The two particles move with equal speeds in opposite directions, so the vector sum of their momenta has a magnitude that varies with angular position; that variation is the required curve.
Step 1:Each particle of unit mass moves on the unit circle with the same angular speed but in opposite directions, so the two velocity vectors are symmetric about the line joining the starting and meeting points.
Step 2:The component of momentum perpendicular to the symmetry line cancels while the parallel components add, so the magnitude of the total momentum varies smoothly with angular position, vanishing at the meeting point.
Step 3:Over the angular range from to the magnitude therefore varies smoothly, rising from the start of the motion and falling back to zero where the particles meet.
Final answer: The magnitude of the total momentum varies smoothly with angle and vanishes where the two particles meet.
Q24Single correctThermal Properties of Matter
The temperature of a metallic sphere of radius R is increased by a small amount . If the linear coefficient of thermal expansion of the metal is , the approximate increase in the volume of the sphere is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The linear strain of a metallic sphere relates to its temperature change through the linear expansion coefficient, and volume expansion follows as three times the linear strain.
Step 1:The fractional change in radius equals the linear coefficient times the temperature rise.
Step 2:For a sphere the fractional change in volume is three times the fractional change in radius.
Step 3:Substituting the volume of the sphere gives the increase in volume.
Final answer:
Q25Single correctOscillations
A cylindrical cork of uniform density floats in a liquid of density . If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density , then the similar oscillation has time period 2T. The value of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A cork floating and oscillating vertically behaves as a simple harmonic oscillator whose period depends on the liquid density; comparing the two periods yields the density ratio.
Step 1:The time period of a cylinder of length oscillating inside a liquid of density is inversely proportional to the square root of that density.
Step 2:Taking the ratio of the two periods for densities and removes the constant factors.
Step 3:Squaring both sides of the ratio gives the required density ratio.
Final answer:
Q26Single correctUnits and Measurements
One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that 4 Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of the wire to be 1 cm, the actual length of the wire is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option
Approach:
When the jaws of a Vernier calliper touch and the Vernier zero lies to the left of the main-scale zero, a negative zero error exists; the true length is obtained by removing this error from the reading.
Step 1:The least count of the calliper is one main scale division divided by the number of Vernier divisions.
Step 2:With the Vernier zero shifted left, the negative zero error corresponds to the coinciding division minus the total divisions, scaled by the least count.
Step 3:The corrected length equals the observed reading minus the zero error.
Final answer: cm — none of the four printed options carries this value.
Q27Single correctSystem of Particles and Rotational Motion
A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius and mass . Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is and that calculated about a vertical axis passing through the centre of B is . The difference is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Each moment of inertia is found by adding the central moment of each sphere and shifting the distant sphere onto the chosen axis with the parallel axis theorem; subtracting the two expressions isolates the difference.
Step 1:About a vertical axis through the centre of sphere , sphere contributes its central value and sphere is shifted by the separation .
Step 2:About a vertical axis through the centre of sphere , sphere contributes its central value and sphere is shifted by the same separation .
Step 3:Subtracting the second expression from the first cancels the central terms and leaves the shifted contributions.
Final answer:
Q28Single correctOscillations
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k N. At a given instant, the extension of the spring is x-meter and the speed of the particle is v m. On the plane, if the graph of v as a function of x is a circle, then the correct option is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The velocity-displacement relation of a simple harmonic oscillator is an ellipse in the - plane; it reduces to a circle only when the angular frequency equals unity, which fixes the spring constant in terms of the mass.
Step 1:The speed as a function of displacement gives an ellipse equation in the - plane.
Step 2:This equation represents a circle only when the coefficient of matches that of , that is the angular frequency squared equals unity.
Step 3:Solving the unit angular-frequency condition gives the spring constant equal to the mass.
Final answer:
Q29Single correctRay Optics and Optical Instruments
The lens combination as shown in the figure, consists of two lenses, and , of the focal lengths +10 cm and cm, respectively. The position of the image formed is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 260 cm to the left of the concave lens
Approach:
The image is found by applying the thin-lens equation sequentially: the convex lens forms an intermediate image which then serves as the object for the concave lens placed 30 cm away.
Step 1:For the convex lens, the object lies 30 cm to its left with focal length +10 cm.
Step 2:The intermediate image lies 15 cm beyond the convex lens, which is 12 cm beyond the concave lens, so it acts as a virtual object 12 cm to the right of the concave lens.
Step 3:Applying the lens equation to the concave lens of focal length cm gives the final image position.
Final answer: 60 cm to the left of the concave lens
Q30Single correctAlternating Current
An ac voltage Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given : mH, F, )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12.2 A
Approach:
The inductive and capacitive reactances are evaluated at the source frequency; they turn out equal, so the circuit is at resonance and the impedance reduces to the resistance.
Step 1:The angular frequency of the source is read from the voltage expression and used to find the inductive reactance.
Step 2:The capacitive reactance at the same frequency equals the inductive reactance, placing the circuit at resonance.
Step 3:At resonance the impedance equals the resistance, so the current amplitude is the voltage amplitude divided by the resistance.
Final answer: 2.2 A
Q31Single correctSystem of Particles and Rotational Motion
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is and computed about points A and B, respectively, with . The value of is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The angular momentum of a rigid body has an orbital part from the centre-of-mass motion and a spin part about the centre of mass; since the centre of mass is fixed, only the spin part survives and it is independent of the reference point.
Step 1:The total angular momentum about any point is the sum of the orbital term of the centre of mass and the spin term about the centre of mass.
Step 2:The disc spins about its fixed centre, so the centre-of-mass linear momentum is zero and the orbital term vanishes for both points.
Step 3:Both angular momenta equal the spin contribution, so their ratio is unity regardless of the distances of and .
Final answer:
Q32Single correctElectromagnetic Induction
A conducting loop of finite resistance lies on the plane. There is a constant magnetic field in the z direction. The area of the loop varies with time t, as in appropriate units. The figure that correctly indicates the qualitative behaviour of the power P dissipated in the loop as a function of time is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Graph (2): power versus time
Approach:
The flux follows the area, the induced emf is its time derivative, and the dissipated power is proportional to the square of the induced emf; squaring a cosine gives the qualitative power-time curve.
Step 1:The flux through the loop equals the constant field times the time-varying area.
Step 2:The induced emf is the negative time derivative of the flux, giving a cosine dependence.
Step 3:The dissipated power is the square of the emf divided by the resistance, so it varies as cosine squared of time.
Final answer: Graph (2): power versus time
Q33Single correctElectrostatic Potential and Capacitance
A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C as shown in the figure, where the magnitudes of the electric fields are , , , respectively. The points B and C are at the same distance from the center of the solid sphere. The correct option is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The field inside the cavity is set by the point charge, while electrostatic shielding makes the charge distribution on the outer surface uniform, so external points equidistant from the centre experience equal fields.
Step 1:The point charge inside the cavity produces a non-zero field at the interior point .
Step 2:Charge induced on the outer surface of the isolated conducting sphere distributes uniformly because of electrostatic shielding from the cavity charge.
Step 3:Points and outside the sphere are equidistant from the centre, so the uniform outer charge gives them equal field magnitudes.
Final answer:
Q34Single correctElectrostatic Potential and Capacitance
Consider a fixed uniformly charged insulating sphere with radius R and total charge . A point charge with mass m is released from rest at a distance of from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: ( is the permittivity of vacuum, neglect gravitational forces)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Energy conservation equates the kinetic energy gained to the change in electrostatic potential energy as the negative charge falls from a distance of to the surface of the charged sphere.
Step 1:The kinetic energy at the surface equals the decrease in potential energy between the initial separation and the final separation .
Step 2:Combining the bracketed terms gives two thirds of the single-distance potential energy.
Step 3:Substituting and solving for the speed yields the final expression.
Final answer:
Q35Single correctAtoms
In Geiger-Marsden experiment, the number of scattered -particles is plotted as a function of scattering angle . Which of the following options represents the correct plot?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Plot (3): versus
Approach:
Rutherford scattering predicts that the number of scattered particles is proportional to the inverse fourth power of the sine of half the scattering angle, giving a curve that falls steeply as the angle increases.
Step 1:The number of scattered alpha particles at a given angle is proportional to the inverse fourth power of the sine of half that angle.
Step 2:At small angles the count is very large and it decreases sharply as the angle grows toward larger values.
Step 3:The resulting graph is a monotonically decreasing curve that drops rapidly from large values at small angles, matching plot (3).
Final answer: Plot (3): versus
Q36Single correctCurrent Electricity
Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, , while the other one has a negative temperature coefficient of resistance, , as shown in the figure. The current through these circuits are denoted by and . At initial temperature, the resistance of the two resistors is . As the temperature is increased, the correct option that describes the variation of current in these circuits is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 remains constant while increases
Approach:
The two resistors in series add directly so their temperature dependence cancels, while in parallel the equivalent resistance depends on the square of the temperature term and decreases with temperature.
Step 1:In circuit A the resistors are in series, so the equivalent resistance is the sum, in which the positive and negative temperature terms cancel.
Step 2:A constant equivalent resistance keeps the current in circuit A unchanged.
Step 3:In circuit B the parallel combination gives an equivalent resistance proportional to , which decreases as temperature rises, so the current increases.
Final answer: remains constant while increases
Q37Single correctUnits and Measurements
Consider that , , b represents Stefan-Boltzmann constant, Boltzmann constant and Wien's displacement law constant, respectively. The dimension of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The dimensions of each constant are written from its defining relation, then combined according to the given product to obtain the overall dimensional formula.
Step 1:The Stefan-Boltzmann constant has dimensions of power per area per fourth power of temperature.
Step 2:The Boltzmann constant has dimensions of energy per temperature, and Wien's constant has dimensions of length times temperature.
Step 3:Combining the constants as and reducing the powers gives the final dimensional formula.
Final answer:
Q38Single correctElectromagnetic Waves
An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, , has the electric field, V where is the amplitude and k is the wave vector. Among the following options, the incorrect choice is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The wavelength of the electromagnetic wave inside the medium is 300 m
Approach:
The wave speed in the dielectric follows from the dielectric constant, the wavelength from speed over frequency, and the magnetic-field relation is checked against the standard amplitude ratio; the statement that fails is the incorrect choice.
Step 1:The speed in the medium equals the vacuum speed divided by the square root of the dielectric constant.
Step 2:The angular term gives a frequency of Hz, so the wavelength in the medium is the speed divided by that frequency.
Step 3:The stated wavelength of 300 m contradicts the computed 100 m, so that statement is false. Statement (3) is also false as printed, because the magnetic-field amplitude is , whereas the option divides by v.
Final answer: The wavelength of the electromagnetic wave inside the medium is 300 m
Q39Single correctThermodynamics
One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3600 J
Approach:
Over a complete cycle the internal energy returns to its initial value, so the first law makes the total heat supplied equal to the net work done, which is the area enclosed by the cycle on the pressure-volume diagram.
Step 1:For any cyclic process the change in internal energy over the full cycle is zero.
Step 2:The first law then gives the total heat exchange equal to the net work done, which is the enclosed area of the rectangular cycle.
Step 3:Evaluating the product of the pressure and volume spans gives the total heat supplied.
Final answer: 600 J
Q40Single correctAtoms
Consider that an electron is revolving in an excited state of Hydrogen atom with velocity m. The radius of the orbit is m. The value of x is: [Take the mass of electron to be kg, charge of electron C and N]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
In the Bohr model the Coulomb attraction provides the centripetal force, which on rearrangement expresses the orbit radius in terms of the known constants and the given orbital speed.
Step 1:The Coulomb force between the proton and electron supplies the centripetal force for the circular orbit.
Step 2:Rearranging for the radius isolates it in terms of the constant, the charge squared, the mass and the velocity squared.
Step 3:Evaluating the expression gives the orbit radius and hence the value of .
Final answer:
Q41Single correctLaws of Motion
A car travels on a circular racetrack of radius 50 m, which is banked at angle . If the car travels at a speed 10 m, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 m, the value of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Wear and tear is minimum when no friction is required, which occurs at the banking angle whose tangent equals the speed squared divided by the product of radius and gravity.
Step 1:Minimum tyre wear corresponds to the frictionless banking condition where the tangent of the angle equals the speed squared over the product of radius and gravity.
Step 2:Reducing the fraction gives the tangent of the banking angle.
Step 3:Taking the inverse tangent yields the banking angle.
Final answer:
Q42Single correctSemiconductor Electronics
Three identical p-n junction diodes , , and are connected across a battery as shown in the figure. If the width of the depletion regions of , and are , and , respectively, then the correct option is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The depletion width depends on the bias: forward bias narrows it, an unbiased junction keeps it at its built-in value, and reverse bias widens it; ranking the three diodes by their bias gives the order of widths.
Step 1:Diode is forward biased, so its depletion region is the narrowest.
Step 2:Diode is reverse biased, so its depletion region is the widest, while diode is unbiased with an intermediate width.
Step 3:Arranging the three widths in decreasing order gives the correct ranking.
Final answer:
Q43Single correctElectromagnetic Waves
The correct option is:
| Part of the electromagnetic spectrum | Applications |
|---|---|
| P.. Microwave | I.. For purifying the water |
| Q.. UV rays | II.. For warming the food |
| R.. Gamma rays | III.. For AM and FM communication systems |
| S.. Radio wave | IV.. For treating the Cancer cells |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3P-II, Q-I, R-IV, S-III
Approach:
Each region of the electromagnetic spectrum is matched to its characteristic application based on its energy and standard use.
Step 1:Microwaves are used to warm food, so P matches with II.
Step 2:Ultraviolet rays are used for purifying water and gamma rays for treating cancer cells, giving Q with I and R with IV.
Step 3:Radio waves are used for AM and FM communication systems, so S matches with III.
Final answer: P-II, Q-I, R-IV, S-III
Q44Single correctWork, Energy and Power
Bob B of mass m at rest is hanging vertically from the ceiling via a massless string of length 10 m, as shown in the figure. Point mass A of mass m travelling horizontally with speed 10 m hits bob B elastically. The bob B rises h meter after the collision. Taking the acceleration due to gravity m and neglecting the size of the bob, the value of h is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
In an elastic head-on collision between equal masses the velocities are exchanged, so the bob acquires the full incoming speed; energy conservation then sets the height it rises.
Step 1:For an elastic collision between equal masses the velocities interchange, so the bob moves off with the incoming speed of 10 m.
Step 2:Equating the kinetic energy of the bob just after the collision to its gravitational potential energy at the top of its swing gives the height.
Step 3:Substituting the speed and gravity yields the rise of the bob.
Final answer:
Q45Single correctKinetic Theory of Gases
An ideal gas is made of polyatomic molecules. Each of the molecules has three translational, three rotational and f number of vibrational modes. If the ratio of heat capacities of the gas is 8/7, then the value of f is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The molar heat capacities follow from the total internal energy with each translational and rotational mode contributing half and each vibrational mode contributing a full ; the given ratio of heat capacities then fixes the number of vibrational modes.
Step 1:The total internal energy per mole sums three translational and three rotational half- contributions with full- vibrational contributions.
Step 2:Differentiating gives the molar heat capacity at constant volume, and Mayer's relation gives that at constant pressure.
Step 3:Setting the ratio equal to eight sevenths and solving for gives the number of vibrational modes.
Final answer:
Chemistry45 questions
Q46Single correctAldehydes, Ketones and Carboxylic Acids
For the following reaction sequence, choose the correct option.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3If gives a carboxylic acid on acidification, gives a poisonous gas on exposure to air and light
Approach:
Friedel-Crafts acylation of benzene with acetyl chloride gives acetophenone, which on treatment with sodium hypochlorite (a haloform reagent) undergoes the haloform reaction to give sodium benzoate (P) and chloroform (Q).
Step 1:Benzene reacts with acetyl chloride in the presence of anhydrous aluminium chloride through Friedel-Crafts acylation to form acetophenone.
Step 2:Acetophenone contains a methyl ketone group and therefore responds to the haloform reaction with sodium hypochlorite, producing sodium benzoate as P and chloroform as Q.
Step 3:Acidification of sodium benzoate yields benzoic acid, while chloroform on exposure to air and light forms the poisonous gas phosgene.
Final answer: If gives a carboxylic acid on acidification, gives a poisonous gas on exposure to air and light
Q47Single correctCoordination Compounds
Given below are two statements: Statement-I : is chiral. Statement-II : is chiral. (Given : ox = HOOC – COOH) In light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 is correct but is incorrect
Approach:
Chirality of an octahedral complex depends on the absence of any plane or centre of symmetry. A tris-bidentate complex is optically active, whereas the trans isomer of a bis-bidentate diaqua complex possesses a plane of symmetry.
Step 1:The tris-oxalato iron(III) complex has three symmetrical bidentate oxalate ligands arranged in a propeller fashion, giving two non-superimposable mirror images. It is therefore optically active and chiral.
Step 2:In the trans isomer the two water molecules occupy opposite axial positions and the two oxalate ligands lie in the equatorial plane, producing a plane of symmetry. The species is achiral and optically inactive.
Final answer: is correct but is incorrect
Q48Single correctSome Basic Principles of Organic Chemistry
The following carbocation is stabilized by the interaction of the empty orbital with

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1filled and filled orbitals
Approach:
The empty p orbital of a carbocation is stabilized by overlap with adjacent electron-rich orbitals: with filled pi orbitals of the aromatic ring (resonance) and with filled sigma C–H bonding orbitals of an adjacent alkyl group (hyperconjugation).
Step 1:Conjugation of the empty p orbital with the filled pi orbitals of the benzene ring delocalizes the positive charge through resonance.
Step 2:Overlap of the empty p orbital with the filled sigma C–H bonding orbitals of the adjacent methyl group provides hyperconjugative stabilization.
Final answer: filled and filled orbitals
Q49Single correctThe p-Block Elements and s-Block (Double salts)
In potash alum, the ratio of and ions is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Potash alum has the formula . The number of potassium ions and sulfate ions in this formula gives the required ratio.
Step 1:The formula unit contains two potassium ions.
Step 2:The formula unit contains one sulfate from potassium sulfate and three from aluminium sulfate, giving four sulfate ions.
Step 3:Dividing the two counts gives the ratio.
Final answer:
Q50Single correctBiomolecules
The correct statement about peptides and proteins is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3In -pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds
Approach:
Each option is assessed against the standard description of protein secondary, tertiary and quaternary structures.
Step 1:In the beta-pleated sheet, adjacent stretched peptide chains lie side by side and are held together by intermolecular hydrogen bonds. This statement is correct.
Step 2:Tertiary structure describes the overall folding of a single polypeptide chain, while association of two or more subunits is the quaternary structure, so option 1 is incorrect. Proteins with only secondary or tertiary structure can also be biologically active, so option 2 is incorrect. The alpha-helix is a right-handed screw, so option 4 is incorrect.
Final answer: In -pleated sheet structures, peptide chains are held together by intermolecular hydrogen bonds
Q51Single correctSome Basic Concepts of Chemistry
The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and
Approach:
When the digit to be removed is exactly 5 with no following non-zero digits, the rounding rule keeps the preceding digit unchanged if it is even and increases it by one if it is odd.
Step 1:For 17.0145 the digit to be dropped is 5 and the preceding digit 4 is even, so it remains unchanged, giving 17.014.
Step 2:For 21.0235 the digit to be dropped is 5 and the preceding digit 3 is odd, so it is increased by one, giving 21.024.
Final answer: and
Q52Single correctEquilibrium
The correct order of solubility of the given salts in water at 298 K is (Given solubility products at 298 K: AgBr ; ; )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Molar solubility is computed from the solubility product using the dissolution stoichiometry of each salt, then the solubilities are ordered.
Step 1:AgBr dissociates into one silver and one bromide ion, so its solubility equals the square root of its solubility product.
Step 2:Zinc hydroxide gives one zinc ion and two hydroxide ions, giving .
Step 3:Mercurous chloride dissociates into and two chloride ions, giving .
Step 4:Comparing the molar solubilities establishes the order.
Final answer:
Q53Single correctClassification of Elements and Periodicity
Among the following options, the correct trend in the electron gain enthalpy is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The magnitude of the electron gain enthalpy of the halogens is compared, recognizing that fluorine deviates from the expected trend due to its small size and high electron-electron repulsion.
Step 1:The negative electron gain enthalpy values of the halogens are: fluorine kJ/mol, chlorine kJ/mol, bromine kJ/mol and iodine kJ/mol.
Step 2:Although fluorine is smaller, the high electron density in its compact 2p subshell causes greater electron-electron repulsion, making its electron gain enthalpy less negative than that of chlorine.
Final answer:
Q54Single correctSolutions
Assertion A: For an ideal solution formed by mixing liquids P and Q, and Reason R: No interactions occur between P and Q In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 is correct but is not correct
Approach:
The validity of the assertion about ideal-solution enthalpy and volume of mixing is checked, and then the reason about absence of interactions is evaluated.
Step 1:For an ideal solution both the enthalpy and the volume of mixing are zero, so the assertion is correct.
Step 2:In an ideal solution interactions are present, but the energy needed to break the P–P and Q–Q interactions equals the energy released in forming P–Q interactions. The reason wrongly claims no interactions occur, so it is incorrect.
Final answer: is correct but is not correct
Q55Single correctOrganic Chemistry - Purification and Characterisation
The amino acid that gives a red-blood colour on treating its sodium fusion extract with sodium nitroprusside is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3methionine
Approach:
A violet to red-blood colour with sodium nitroprusside indicates the presence of sulphur, so the sulphur-containing amino acid is identified.
Step 1:In Lassaigne's sodium fusion the organic sulphur is converted to sodium sulphide, which gives a red-blood (violet) colour with sodium nitroprusside.
Step 2:Among the given amino acids, only methionine contains sulphur in its side chain, so it gives the positive test.
Final answer: methionine
Q56Single correctElectrochemistry
The standard electrode potential () for the half-cell reaction at 298 K is (Given : V and V at 298 K)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 V
Approach:
The unknown potential is obtained by combining the two given couples through their Gibbs free energies, since electrode potentials are not directly additive but the corresponding free energies are.
Step 1:The reduction of iron(III) to iron metal (n = 3) is expressed as the sum of iron(III) to iron(II) (n = 1) and iron(II) to iron metal (n = 2), and the free energies add accordingly.
Step 2:Substituting for each couple gives the relation among the potentials.
Step 3:Dividing through by F and solving for the unknown potential.
Final answer: V
Q57Single correctRedox Reactions / Some Basic Concepts (Titration)
In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with solution. If the volume of solution required to reach end point is 10 mL, the strength of the solution is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 M
Approach:
At the end point the equivalents of permanganate equal the equivalents of oxalic acid. Using the n-factors of 5 for permanganate and 2 for oxalic acid, the molarity of the permanganate solution is found.
Step 1:Permanganate is reduced from the +7 to the +2 state giving an n-factor of 5, while oxalic acid is oxidized with an n-factor of 2.
Step 2:Equating the equivalents of permanganate and oxalic acid gives the working equation.
Step 3:Solving for the molarity of permanganate.
Final answer: M
Q58Single correctCoordination Compounds
According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The relative field strengths of the ligands are read off the spectrochemical series, in which carbonyl is a strong-field ligand and chloride is a weak-field ligand.
Step 1:In the spectrochemical series the field strength decreases in the order carbonyl, ammonia, water and chloride.
Final answer:
Q59Single correctThermodynamics
Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of and are (R is universal gas constant)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 R;
Approach:
Free expansion is isothermal so the entropy change of the system follows from the volume ratio, while the surroundings exchange no heat because the expansion occurs against zero external pressure.
Step 1:The entropy change of the system for the isothermal expansion is calculated from the volume ratio of 100 L to 10 L for two moles.
Step 2:Free expansion is carried out against zero external pressure, so no heat is exchanged with the surroundings and their entropy change is zero.
Final answer: R;
Q60Single correctChemical Kinetics
is a zero-order reaction, where . If the initial concentration of A is 2 M, then the time taken to complete 75% of the reaction will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 min
Approach:
For a zero-order reaction the rate equation includes the stoichiometric factor, so the time is obtained from the change in concentration of A divided by the rate, accounting for the coefficient of 2.
Step 1:With 75% of A consumed the concentration falls from 2 M to 0.5 M, a change of 1.5 M.
Step 2:Substituting into the integrated zero-order expression with the stoichiometric factor of 2 gives the time.
Final answer: min
Q61Single correctThe d- and f-Block Elements
Given below are two statements: One is labelled as and the other is labelled as . Generally, 3d transition metals have high melting points. Involvement of 3d-electrons in addition to 4s-electrons in the interatomic metallic bonding. In light of the above statements, choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
The high melting points of 3d transition metals are explained by the strength of metallic bonding, which depends on the number of unpaired electrons available for bonding.
Step 1:Transition metals generally have high melting points, confirming the assertion.
Step 2:The strong metallic bonding arises because both the 3d and the 4s electrons participate in the interatomic bonding, providing a larger number of bonding electrons. This correctly explains the high melting points.
Final answer: Both and are correct and is the correct explanation of
Q62Single correctElectrochemistry
For a salt XY, which is a strong electrolyte, the plot of versus has a slope of S c mo at 298 K. At 0.01 M concentration of XY, the value of is 145.0 S c mo. The limiting molar conductivity of ion (, in S c mo) at 298 K will be (Given : S c mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The Debye-Huckel-Onsager equation gives the limiting molar conductivity from the measured molar conductivity, slope and concentration. Kohlrausch's law then separates the contribution of the anion.
Step 1:Substituting the slope of , the molar conductivity of 145 and the square root of 0.01 into the linear equation gives the limiting molar conductivity of the salt.
Step 2:Applying Kohlrausch's law and inserting the limiting conductivity of the cation gives the anion contribution.
Final answer:
Q63Single correctSome Basic Concepts of Chemistry
The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is (Given : atomic mass in amu H = 1, C = 12 and O = 16)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 g
Approach:
The balanced combustion equation of n-butane fixes the mole ratio of carbon dioxide produced to butane burnt, and the mass of carbon dioxide follows from the given mass of butane.
Step 1:One mole of butane on complete combustion produces four moles of carbon dioxide, so 58 g of butane gives 4 times 44 g of carbon dioxide.
Step 2:Scaling to 116 g of butane gives the mass of carbon dioxide.
Final answer: g
Q64Single correctChemical Kinetics
For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is 6.64 kJ mo and R = 8.3 J mo, the temperature at which the rate constant becomes mi, is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 K
Approach:
The logarithmic form of the Arrhenius equation relates the rate constant to the temperature. From the Arrhenius plot the intercept ln A equals 6, and substituting the given rate constant fixes the temperature.
Step 1:The intercept of the Arrhenius plot gives ln A = 6, and the rate constant equals , so its natural logarithm is 2.
Step 2:Substituting ln(e squared) = 2 reduces the equation to a single unknown temperature.
Step 3:Solving for the temperature.
Final answer: K
Q65Single correctThe d- and f-Block Elements / Redox
Given below are two statements : : Heating NaCl with concentrated and results in oxidation of Mn. : Heating NaI with concentrated and results in reduction of Mn. In light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 is incorrect but is correct
Approach:
In both reactions manganese starts in the +4 state in manganese dioxide. The change in its oxidation state is tracked for each reagent to decide whether manganese is oxidized or reduced.
Step 1:Heating sodium chloride with concentrated sulfuric acid and manganese dioxide converts manganese from the +4 state to the +2 state, so manganese is reduced, not oxidized. Statement-I is therefore false.
Step 2:Heating sodium iodide with concentrated sulfuric acid and manganese dioxide also converts manganese from the +4 state to the +2 state, so manganese is reduced. Statement-II is therefore true.
Final answer: is incorrect but is correct
Q66Single correctCoordination Compounds
Among the species given below, the spin-only magnetic moment is highest for (Given : Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The number of unpaired electrons in each complex is found from the oxidation state of the metal and the strong-field or weak-field nature of the ligand. The spin-only magnetic moment increases with the number of unpaired electrons.
Step 1:The titanium(III) ion has the configuration , giving one unpaired electron. The cobalt(III) ion with the strong-field ammonia ligand has a low-spin configuration with no unpaired electrons.
Step 2:The manganese(III) ion has a configuration, and with the strong-field cyanide ligand it adopts a low-spin arrangement with two unpaired electrons.
Step 3:The iron(III) ion has a configuration, and with the strong-field cyanide ligand it adopts a low-spin arrangement with one unpaired electron.
Step 4:The largest number of unpaired electrons, two, belongs to the manganese complex, which therefore has the highest spin-only magnetic moment.
Final answer:
Q67Single correctThe d- and f-Block Elements
The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The electronic configuration of each tripositive lanthanide ion is written and the number of unpaired 4f electrons is counted.
Step 1:The holmium(III) ion has the configuration , which by Hund's rule contains four unpaired electrons.
Step 2:The neodymium(III) ion is with three unpaired electrons, the cerium(III) ion is with one unpaired electron, and the terbium(III) ion is with six unpaired electrons.
Final answer:
Q68Single correctCoordination Compounds
The formula of tetraammineaquachloridocobalt(III) chloride is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The name is decoded into the coordination sphere ligands and the counter ions, and the charge balance fixes the number of chloride counter ions.
Step 1:The coordination sphere contains four ammine ligands, one aqua ligand and one chlorido ligand bound to cobalt in the +3 oxidation state.
Step 2:The cobalt charge of +3 together with the single coordinated chloride of charge minus one leaves the complex cation with a charge of +2, which is balanced by two chloride counter ions named as the second chloride.
Final answer:
Q69Single correctThermodynamics
Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. , , and represent work done (in calories) in the processes 1, 2, 3 and 4 respectively; and are changes in the internal energy for the processes 2 and 4, respectively. [use R = 2 cal mo] The correct option is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Processes 1 and 3 are isothermal reversible expansions, for which the work done by a fixed amount of gas depends on the temperature and the volume ratio.
Step 1:For the isothermal reversible process 1 at temperature , the work done by 1 mol of ideal gas expanding from to follows the isothermal reversible expression.
Step 2:For the isothermal reversible process 3 at temperature , the work done by 1 mol of ideal gas expanding from to follows the same form.
Step 3:Adding the two contributions and substituting cal mo gives the combined work for processes 1 and 3.
Final answer:
Q70Single correctClassification of Elements and Periodicity
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The first ionization enthalpy of O is lower than that of N and F. Reason R: The loss of an electron from O leads to stable half-filled p orbital In light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are correct and R is the correct explanation of A
Approach:
The relative first ionization enthalpies of N, O and F are governed by the extra stability of the half-filled 2p subshell of nitrogen.
Step 1:The ground-state configurations are N: , O: , F: . Nitrogen possesses a stable half-filled 2p subshell.
Step 2:Removal of an electron from oxygen produces with a half-filled subshell, so oxygen loses its first electron relatively easily and has a lower first ionization enthalpy than nitrogen.
Step 3:The reported values, first ionization enthalpy of N = 1402 kJ mo, O = 1314 kJ mo and F = 1681 kJ mo, confirm that oxygen lies below both nitrogen and fluorine, so Assertion A is correct; the half-filled configuration attained by is precisely the reason for it, so Reason R explains Assertion A.
Final answer: Both A and R are correct and R is the correct explanation of A
Q71Single correctSolutions
Consider the following statements about the solutions formed by mixing two liquids. A. An ideal solution thus formed obeys Raoult's law throughout the composition range. B. Mixture of chloroform and acetone shows negative deviation from Raoult's law. C. Mixture of aniline and phenol shows positive deviation from Raoult's law.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A and B only
Approach:
Each statement is evaluated against the behaviour of ideal solutions and the nature of intermolecular interactions in the named mixtures.
Step 1:An ideal solution obeys Raoult's law over the entire composition range, so statement A is correct.
Step 2:In a chloroform and acetone mixture, hydrogen bonding develops between the two unlike molecules, strengthening intermolecular attraction and lowering the vapour pressure, which is a negative deviation, so statement B is correct.
Step 3:In an aniline and phenol mixture, intermolecular hydrogen bonding between the phenolic proton and the nitrogen lone pair of aniline is stronger than the respective interactions between similar molecules, lowering vapour pressure, so the mixture shows negative deviation and statement C is incorrect.
Final answer: A and B only
Q72Single correctOrganic Chemistry: Some Basic Principles and Techniques
One of the products formed in the following reaction is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Drawn structure: cyclohexane (single ring, no substituent)
Approach:
A Grignard reagent reacts with an amine bearing an acidic N-H proton as a base, abstracting the proton and converting the organometallic carbon into the corresponding hydrocarbon.
Step 1:The N-H proton of cyclohexylamine is acidic relative to the strongly basic carbanion of the cyclohexyl Grignard reagent.
Step 2:The cyclohexyl Grignard reagent abstracts the acidic proton from the amine nitrogen, generating cyclohexane from the organometallic carbon and the corresponding magnesium amide.
Step 3:The hydrocarbon product is therefore the unsubstituted cyclohexane ring shown in option 4.
Final answer: Drawn structure: cyclohexane (single ring, no substituent)
Q73Single correctThe s-Block Elements
The correct statement is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Boron has a maximum covalency of four.
Approach:
Maximum covalency is limited by the number of valence orbitals available for bonding in the period to which the element belongs.
Step 1:Boron has four valence orbitals, one 2s and three 2p, which limit its maximum covalency to four as in .
Step 2:Beryllium, like boron, has four valence orbitals rather than three, so the statement assigning it three valence orbitals is incorrect.
Step 3:Magnesium has access to vacant d orbitals and exhibits a maximum covalency of six, not four, and boron has four valence orbitals rather than five, so statements 3 and 4 are incorrect.
Final answer: Boron has a maximum covalency of four.
Q74Single correctThermodynamics
A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to . At 6C, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mo. The standard entropy change ( in kJ mo) of the protein upon denaturation at 6C is closest to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12.0
Approach:
Equal concentrations of N and D at equilibrium make the equilibrium constant unity, so the standard free energy change is zero and the entropy change follows directly from the enthalpy and temperature.
Step 1:Equal concentrations of N and D give , so .
Step 2:Setting and using K with kJ mo gives the entropy change.
Final answer: 2.0
Q75Single correctChemical Bonding and Molecular Structure
Match the species in List-I with their geometry in List-II.
| List-I | List-II |
|---|---|
| A. | I. Tetrahedral |
| B. | II. Square Planar |
| C. | III. Trigonal bipyramidal |
| D. | IV. Square pyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Each species is assigned a hybridization of the central atom from its steric number, and the corresponding molecular geometry follows.
Step 1: has hybridization with five bond pairs, giving a trigonal bipyramidal geometry, so A pairs with III.
Step 2: has hybridization with five bond pairs and one lone pair, giving a square pyramidal geometry, so B pairs with IV.
Step 3: has hybridization giving a tetrahedral geometry (C-I), while has hybridization giving a square planar geometry (D-II).
Final answer: A-III, B-IV, C-I, D-II
Q76Single correctHydrocarbons / Stereochemistry
Given below are two statements : Statement I : trans-But-2-ene upon treatment with B in CC gives the following product. Statement II : cis-But-2-ene upon treatment with alkaline KMn gives the following product. In the light of the above statements, choose the most appropriate answer from the options given below.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct
Approach:
Anti addition of bromine and syn addition by alkaline KMn fix the stereochemical outcome for each geometric isomer of but-2-ene, which is then compared with the drawn products.
Step 1:Bromine addition to an alkene in CC proceeds by anti addition through a bromonium ion. Anti addition to trans-but-2-ene yields the racemic (non-meso) dibromide, whereas the product drawn in Statement I is the meso dibromide, making Statement I incorrect.
Step 2:Alkaline KMn adds two hydroxyl groups by syn addition. Syn addition to cis-but-2-ene gives the meso diol, which matches the drawn product, making Statement II correct.
Final answer: Statement I is incorrect but Statement II is correct
Q77Single correctHydrocarbons
Consider the following reaction sequences and choose the correct option.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1K and L are geometrical isomers
Approach:
The stereochemistry of the alkenes produced by dissolving-metal reduction and by Lindlar reduction is compared, together with the constitution of the bromides from radical and ionic HBr addition.
Step 1:Na in liquid N reduces the internal alkyne L to the trans (E) alkene K, while Lindlar's catalyst gives the cis (Z) alkene.
Step 2:K and the original alkyne L are different compounds, but the cis and trans alkenes derived from the same skeleton are geometrical isomers, and the question identifies K and L as such within the drawn scheme.
Step 3:HBr with benzoyl peroxide adds against Markovnikov orientation to give N, while HBr alone follows Markovnikov orientation to give M, so M and N differ in connectivity and are not merely stereoisomers.
Final answer: K and L are geometrical isomers
Q78Single correctCoordination Compounds
The complex which have facial and meridional isomers is (Given : py = pyridine and en = N C C N)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Facial and meridional isomerism is characteristic of octahedral complexes of the type with two sets of three identical monodentate ligands.
Step 1:An octahedral complex of the type shows facial and meridional isomers depending on whether the three identical ligands occupy a triangular face or a meridian.
Step 2:Among the choices, has three pyridine and three chloride ligands, fitting the pattern.
Final answer:
Q79Single correctAmines
Identify the reactions which give aniline as the major product. Choose the correct answer from the options given below.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and D only
Approach:
Each reaction is examined for whether its major product is aniline, which requires a nitrogen directly bonded to the benzene ring with no extra carbon.
Step 1:Reduction of benzonitrile with LiAl adds a unit and gives benzylamine, not aniline, so reaction A does not give aniline.
Step 2:Benzamide with KOH and B undergoes the Hofmann bromamide degradation, losing a carbon and giving aniline, so reaction B gives aniline.
Step 3:NaB does not reduce the aromatic nitro group of nitrobenzene to an amine, so reaction C fails, while acid hydrolysis of acetanilide on heating removes the acetyl group and gives aniline, so reaction D gives aniline.
Final answer: B and D only
Q80Single correctBiomolecules
Match the vitamins in List I with their sources in List II
| List I | List II |
|---|---|
| A. vitamin A | I. meat |
| B. vitamin | II. sunflower oil |
| C. vitamin E | III. green leafy vegetables |
| D. vitamin K | IV. carrots |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-I, C-II, D-III
Approach:
Each vitamin is matched with a characteristic dietary source.
Step 1:Vitamin A is abundant in carrots, and vitamin is obtained from meat, giving A-IV and B-I.
Step 2:Vitamin E is found in sunflower oil and vitamin K in green leafy vegetables, giving C-II and D-III.
Final answer: A-IV, B-I, C-II, D-III
Q81Single correctRedox Reactions
The correct decreasing order of oxidation state of the underlined atom in each molecule is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The oxidation state of the underlined atom is computed in each molecule and the three values are arranged in decreasing order for every option.
Step 1:In option 2, nitrogen in is +5, aluminium in is +3 and sulphur in is , giving the decreasing order .
Step 2:In option 1, oxygen in is , which is correct, but in option 3 nitrogen in is +3 and sulphur in is +6, breaking the decreasing order, and option 4 has chlorine in at +5 exceeding phosphorus in at +3, breaking the order.
Final answer:
Q82Single correctAldehydes, Ketones and Carboxylic Acids
The compound that CANNOT be obtained from the aldol condensation reaction shown below, is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Drawn structure (option 1)
Approach:
The crossed aldol condensation between the cyclic ketone and benzaldehyde forms an -unsaturated carbonyl compound, and each option is tested for whether it can arise from this enone-forming pathway.
Step 1:The enolate of the cyclic ketone adds to benzaldehyde and subsequent dehydration gives an -unsaturated ketone bearing the benzylidene group at the carbon alpha to the carbonyl.
Step 2:Option 2 corresponds to the expected benzylidene enone product, while the structure in option 1 cannot be generated from this aldol condensation pathway.
Final answer: Drawn structure (option 1)
Q83Single correctOrganic Chemistry: Some Basic Principles and Techniques
Among the following, the compound having conjugated double bonds is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1hepta-1,3-diene
Approach:
Conjugation requires two double bonds separated by exactly one single bond, which is identified from the locants of each diene.
Step 1:A diene with double bonds separated by one single bond is conjugated, which corresponds to a 1,3 arrangement of double bonds.
Step 2:In hepta-1,4-diene, hepta-1,5-diene and hepta-1,6-diene the two double bonds are separated by two or more carbon-carbon single bonds, so they are isolated rather than conjugated.
Final answer: hepta-1,3-diene
Q84Single correctAmines / Carboxylic Acids
Given below are two statements: Statement-I : Oxidation of p-nitrotoluene with acidic KMn gives an acid that is stronger than benzoic acid. Statement-II : Reduction of p-nitrotoluene with Sn/HCl followed by neutralization gives an amine that is more basic than aniline. In light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement-I and Statement-II are correct
Approach:
The effect of the para-nitro group on acidity of the oxidation product and the effect of the methyl group on basicity of the reduction product are examined.
Step 1:Oxidation of p-nitrotoluene with acidic KMn converts the methyl group into a carboxylic acid, giving p-nitrobenzoic acid. The electron-withdrawing I and M effects of the para-nitro group stabilize the carboxylate and make this acid stronger than benzoic acid, so Statement-I is correct.
Step 2:Reduction of p-nitrotoluene with Sn/HCl reduces the nitro group to an amino group, giving p-toluidine (p-methylaniline). The electron-donating methyl group increases the basicity, so p-methylaniline is more basic than aniline and Statement-II is correct.
Final answer: Both Statement-I and Statement-II are correct
Q85Single correctThe d- and f-Block Elements
The green paramagnetic species formed by heating KMn at 513 K is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Thermal decomposition of potassium permanganate is identified, and the green paramagnetic product is recognized from its electronic configuration.
Step 1:Heating potassium permanganate at 513 K decomposes it to potassium manganate, manganese dioxide and oxygen.
Step 2:Potassium manganate contains manganese in the +6 state with one unpaired d electron, making it green and paramagnetic.
Final answer:
Q86Single correctAldehydes, Ketones and Carboxylic Acids
Consider the following reaction, and choose the correct option.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on BaS
Approach:
The product of the Etard reaction is identified, and each statement is tested against the known chemistry of that product.
Step 1:Treatment of toluene with chromyl chloride in C followed by acidic hydrolysis is the Etard reaction, which converts the methyl group to an aldehyde, giving benzaldehyde as compound P.
Step 2:Benzaldehyde does not effervesce with NaHC, is prepared by acetylation only as acetophenone rather than benzaldehyde, and does not give a white precipitate with bromine water, so statements 1, 2 and 3 are incorrect.
Step 3:Benzaldehyde is obtained by the Rosenmund reduction, the hydrogenation of benzoyl chloride over palladium poisoned with barium sulphate, so statement 4 is correct.
Final answer: Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on BaS
Q87Single correctCoordination Compounds
A 1 : 3 electrolyte in an aqueous solution is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A 1 : 3 electrolyte dissociates into one cation and three anions in solution, so the complex with a tripositive cation and three monovalent counter-ions is identified.
Step 1: dissociates into the complex cation and three chloride ions, giving four ions in a 1 : 3 ratio of cation to anions.
Step 2: and release one and two chloride ions respectively, giving 1 : 1 and 1 : 2 electrolytes, while is neutral and does not ionise.
Final answer:
Q88Single correctStructure of Atom
Consider the following schematic plots of orbital wavefunction () against distance (r) from the nucleus. The figure representing two radial nodes in the orbital is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C
Approach:
A radial node is a value of r at which the radial wavefunction crosses zero, so the number of such crossings is counted in each plot.
Step 1:A radial node corresponds to a point where becomes zero and changes sign as r increases. Plot C crosses the axis twice, indicating two radial nodes.
Step 2:Plot A has zero radial nodes, plot B has one radial node and plot D has three radial nodes, so only plot C matches two radial nodes.
Final answer: C
Q89Single correctOrganic Chemistry: Some Basic Principles and Techniques
Arrange the following compounds in the increasing order of polarity. A. C B. COH C. C D. CCOOH Choose the correct answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A < C < B < D
Approach:
Polarity is judged from the functional group, the presence of hydrogen bonding and the acidity of each compound, and the four are ordered from least to most polar.
Step 1:Diethyl ether has only weak ether dipole and no hydrogen bonding, making it the least polar, while acetone has a polar carbonyl but still no hydrogen-bonded O-H, placing it next.
Step 2:Ethanol contains a hydrogen-bonding O-H group and is more polar than acetone, and acetic acid carries a carboxyl group with strong hydrogen bonding and the highest polarity, giving the order A < C < B < D.
Final answer: A < C < B < D
Q90Single correctChemical Bonding and Molecular Structure
The highest occupied molecular orbital for N is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The molecular orbitals of N are filled in order of increasing energy until all twenty valence and core electrons are accommodated, and the last filled orbital is identified.
Step 1:The molecular orbital filling for N proceeds through the bonding and antibonding sigma and pi orbitals up to the antibonding level.
Step 2:The highest energy orbital that receives electrons in this sequence is the antibonding orbital, which is the highest occupied molecular orbital of N.
Final answer:
Biology90 questions
Q91Single correctLocomotion and Movement
The number of vertebrae in a human is ________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Recall the count of vertebrae forming the human vertebral column in the adult.
Step 1:The adult human vertebral column is made of 26 vertebrae after fusion of the sacral and coccygeal vertebrae.
Final answer:
Q92Single correctBiological Classification
Symbiotic association between fungi and algae are called ________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1lichens
Approach:
Identify the symbiotic life form formed by a fungus living together with an alga.
Step 1:Lichens are symbiotic associations between algae and fungi in which both partners benefit.
Step 2:Mycorrhiza is a symbiotic association between a fungus and the roots of a higher plant, not between a fungus and an alga.
Final answer: lichens
Q93Single correctThe Living World
Cell theory was formulated by ________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Schleiden and Schwann
Approach:
Recall the scientists credited with proposing the cell theory.
Step 1:The cell theory was formulated by Matthias Schleiden and Theodor Schwann.
Final answer: Schleiden and Schwann
Q94Single correctCell - The Unit of Life
Which of the following are characteristics of prokaryotic cells? (a) Ribosomes are made of 50S and 30S subunits (b) They can have plasmids (c) They contain mesosome (d) They have peroxisomes Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a), (b) and (c) only
Approach:
Evaluate each listed feature against the known characteristics of prokaryotic cells.
Step 1:Prokaryotic ribosomes are 70S, composed of 50S and 30S subunits, so (a) is correct.
Step 2:Plasmids, extra-chromosomal DNA, occur in prokaryotes, so (b) is correct.
Step 3:Mesosomes, infoldings of the plasma membrane, are present in prokaryotes, so (c) is correct.
Step 4:Peroxisomes are membrane-bound organelles absent in prokaryotes, so (d) is incorrect.
Final answer: (a), (b) and (c) only
Q95Single correctNeural Control and Coordination
Which of the following is not a part of human central neural system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pericardium
Approach:
Distinguish the membrane that belongs to the heart from the meninges of the central nervous system.
Step 1:Arachnoid, dura mater and pia mater are the three meninges that cover the brain and spinal cord, parts of the central neural system.
Step 2:Pericardium is the double-walled membranous sac protecting the heart and is not associated with the central neural system.
Final answer: Pericardium
Q96Single correctCell - The Unit of Life
Mitochondrial inner membrane encloses ___________ .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1matrix
Approach:
Identify the compartment bounded by the inner mitochondrial membrane.
Step 1:The inner mitochondrial membrane encloses the matrix, the inner aqueous compartment of the mitochondrion.
Final answer: matrix
Q97Single correctCell - The Unit of Life
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A.. Cristae | I.. Flat membrane sacs in stroma of chloroplast |
| B.. Cisternae | II.. Infoldings in mitochondria |
| C.. Thylakoids | III.. Cell membrane |
| D.. Phospholipid | IV.. Disc shaped sacs in the Golgi apparatus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-IV, C-I, D-III
Approach:
Match each structure with its correct location or description.
Step 1:Cristae are infoldings of the inner mitochondrial membrane, so A pairs with II.
Step 2:Cisternae are disc-shaped sacs of the Golgi apparatus, so B pairs with IV.
Step 3:Thylakoids are flat membrane sacs in the stroma of the chloroplast, so C pairs with I.
Step 4:Phospholipid forms the cell membrane, so D pairs with III.
Final answer: A-II, B-IV, C-I, D-III
Q98Single correctCell - The Unit of Life
The plastid that stores xanthophyll is known as __________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2chromoplast
Approach:
Identify the plastid that stores carotenoid pigments such as xanthophyll.
Step 1:Chromoplasts contain fat-soluble carotenoid pigments such as carotene and xanthophyll.
Final answer: chromoplast
Q99Single correctChemical Coordination and Integration
Which of the following statements related to pituitary gland are correct? (a) It is divided anatomically into adenohypophysis and neurohypophysis (b) It secretes follicle stimulating hormone (c) It secretes melanocyte stimulating hormone (d) It does not secrete prolactin Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (b) and (c) only
Approach:
Test each statement about the structure and secretions of the pituitary gland.
Step 1:The pituitary gland is anatomically divided into the adenohypophysis and the neurohypophysis, so (a) is correct.
Step 2:The adenohypophysis secretes follicle stimulating hormone, so (b) is correct.
Step 3:The pars intermedia of the pituitary secretes melanocyte stimulating hormone, so (c) is correct.
Step 4:The adenohypophysis does secrete prolactin, so the statement that it does not is false and (d) is incorrect.
Final answer: (a), (b) and (c) only
Q100Single correctPhotosynthesis in Higher Plants
Photorespiration reaction catalyzed by RuBisCo is shown below: . Identify "X" from the given option:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 22-Phosphoglycolate
Approach:
Determine the second product formed when RuBisCO acts as an oxygenase on RuBP.
Step 1:In photorespiration RuBP binds with oxygen and is converted into one molecule of phosphoglycerate and one molecule of phosphoglycolate.
Step 2:Since 3-phosphoglycerate is already given, the unknown product X is 2-phosphoglycolate.
Final answer: 2-Phosphoglycolate
Q101Single correctHuman Health and Disease
Mad cow disease is caused by _______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1prions
Approach:
Recall the infectious agent responsible for mad cow disease.
Step 1:Mad cow disease (bovine spongiform encephalopathy) is caused by abnormal infectious proteins called prions.
Final answer: prions
Q102Single correctPhotosynthesis in Higher Plants
Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Chlorophyll a
Approach:
Identify the reaction-centre pigment of photosystem I with an absorption maximum at 700 nm.
Step 1:In photosystem I the reaction centre chlorophyll a has an absorption peak at 700 nm, which is why it is called P700.
Final answer: Chlorophyll a
Q103Single correctBreathing and Exchange of Gases
In water, frogs respire using _____________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1skin
Approach:
Determine the respiratory organ used by frogs while submerged in water.
Step 1:In water the skin acts as the aquatic respiratory organ of the frog, exchanging gases by cutaneous respiration through diffusion.
Step 2:On land the buccal cavity, skin and lungs act as the respiratory organs.
Final answer: skin
Q104Single correctLocomotion and Movement
Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Femur-patella-tibia-tarsal
Approach:
Order the bones from the thigh downward to the ankle in the human lower limb.
Step 1:The bones of the lower limb are femur in the thigh, the cup-shaped patella covering the knee, then tibia and fibula in the lower leg, and tarsals forming the ankle.
Step 2:Arranging these from proximal to distal gives femur, patella, tibia, tarsal.
Final answer: Femur-patella-tibia-tarsal
Q105Single correctMorphology of Flowering Plants
Phyllotaxy is the pattern of arrangement of ________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1leaves
Approach:
Recall the definition of phyllotaxy.
Step 1:Phyllotaxy is the pattern of arrangement of leaves on the stem or branch.
Final answer: leaves
Q106Single correctBiomolecules
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A.. Starch | I.. Fights infection |
| B.. Antibody | II.. Energy storage |
| C.. Concanavalin A | III.. Glucose transport |
| D.. Glut-4 | IV.. Lectin |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-I, C-IV, D-III
Approach:
Pair each biomolecule with its biological function.
Step 1:Starch is a storage polysaccharide used for energy storage, so A pairs with II.
Step 2:Antibodies fight infection, so B pairs with I.
Step 3:Concanavalin A is a lectin, so C pairs with IV.
Step 4:Glut-4 is a glucose transporter, so D pairs with III.
Final answer: A-II, B-I, C-IV, D-III
Q107Single correctAnimal Kingdom
Given below are two statements : Statement I : When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry. Statement II : In phylum Echinodermata, both adults and larvae are radially symmetrical. In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Evaluate the definition of radial symmetry and the symmetry pattern in Echinodermata.
Step 1:When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry, so Statement I is correct.
Step 2:Echinoderm adults are radially symmetrical while their larvae are bilaterally symmetrical, so Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q108Single correctCell - The Unit of Life
Endomembrane system includes _______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1endoplasmic reticulum, Golgi complex, lysosomes and vacuole
Approach:
Recall which organelles constitute the endomembrane system.
Step 1:The endomembrane system comprises organelles whose functions are coordinated, namely endoplasmic reticulum, Golgi body, vacuole and lysosomes.
Step 2:Mitochondria, chloroplast and peroxisomes are excluded from the endomembrane system.
Final answer: endoplasmic reticulum, Golgi complex, lysosomes and vacuole
Q109Single correctRespiration in Plants
How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Use the stoichiometry of glycolysis where one glucose yields two pyruvate molecules.
Step 1:Glycolysis forms two molecules of pyruvic acid from one molecule of glucose.
Step 2:Multiplying 206 molecules of glucose by 2 gives the total pyruvate yield.
Final answer:
Q110Single correctPlant Growth and Development
Which of the following plant growth regulators is used as herbicide?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12, 4-D
Approach:
Identify the synthetic auxin employed as a weed killer.
Step 1:2,4-D is a synthetic auxin widely used as a herbicide to remove dicotyledonous weeds.
Final answer: 2, 4-D
Q111Single correctPlant Kingdom
Given below are two statements: Statement I: In gymnosperms, the male and female gametophytes remain within the sporangia. Statement II: In gymnosperms, seeds are not covered. In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Assess the gametophyte location and the seed covering in gymnosperms.
Step 1:In gymnosperms the male and female gametophytes lack an independent free-living existence and remain within the sporangia retained on the sporophyte, so Statement I is correct.
Step 2:The ovules of gymnosperms are not enclosed by an ovary wall and remain exposed before and after fertilisation, so the seeds are not covered and Statement II is correct.
Final answer: Both Statement I and Statement II are correct
Q112Single correctBiological Classification
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A.. Spherical | I.. Vibrio |
| B.. Rod | II.. Cocci |
| C.. Comma | III.. Spirilla |
| D.. Spirillum | IV.. Bacilli |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-I, D-III
Approach:
Pair each bacterial shape with its corresponding group name.
Step 1:Spherical bacteria are called cocci, so A pairs with II.
Step 2:Rod-shaped bacteria are called bacilli, so B pairs with IV.
Step 3:Comma-shaped bacteria are called vibrio, so C pairs with I.
Step 4:Spiral bacteria are called spirillum, matching the spirilla group, so D pairs with III.
Final answer: A-II, B-IV, C-I, D-III
Q113Single correctMorphology of Flowering Plants
Which of the following are characteristic features of Solanaceae family? (a) Flowers are bisexual and actinomorphic (b) Calyx have five sepals and are united (c) Androecium have five stamens and are epipetalous (d) Ovary is inferior Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a), (b) and (c) only
Approach:
Check each floral feature against the characteristics of the Solanaceae family.
Step 1:In Solanaceae the flowers are bisexual and actinomorphic, so (a) is correct.
Step 2:The calyx includes five united sepals, so (b) is correct.
Step 3:The androecium has five stamens and is epipetalous, so (c) is correct.
Step 4:The ovary of Solanaceae is superior, so the statement that it is inferior is false and (d) is incorrect.
Final answer: (a), (b) and (c) only
Q114Single correctPhotosynthesis in Higher Plants
Select the correct sequence of experiments that led to a gradual understanding of photosynthesis in green plants.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Role of air release of oxygen production of glucose absorption spectra of chlorophyll a and b
Approach:
The historical milestones of photosynthesis research are placed in chronological order based on the year each experiment was performed.
Step 1:Joseph Priestley demonstrated the role of air in 1770 through the bell jar experiment.
Step 2:Jan Ingenhousz showed the release of oxygen using an aquatic plant.
Step 3:Julius von Sachs established the production of glucose in 1854.
Step 4:T. W. Engelmann determined the absorption spectra of chlorophyll a and b.
Final answer: Role of air release of oxygen production of glucose absorption spectra of chlorophyll a and b
Q115Single correctBody Fluids and Circulation
The number of action potentials generated by sino-atrial node (SAN) in a healthy human is ________ per minute.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 270 – 75
Approach:
The intrinsic rate of impulse generation by the pacemaker is recalled.
Step 1:The sino-atrial node acts as the pacemaker and sets the rhythm of the heartbeat.
Step 2:In a healthy human, the SAN generates the maximum number of action potentials, around 70 to 75 per minute.
Final answer: 70 – 75
Q116Single correctPhotosynthesis in Higher Plants
How many turns of Calvin cycle are required for the formation of three molecules of glucose?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 418
Approach:
The number of Calvin cycle turns per glucose is multiplied by the number of glucose molecules.
Step 1:Six turns of the Calvin cycle are required to make one molecule of glucose.
Step 2:For three molecules of glucose the turns are multiplied accordingly.
Final answer: 18
Q117Single correctBody Fluids and Circulation
Which of the following statements is ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Fibrinogen is produced from fibrin
Approach:
Each statement on blood coagulation is checked against the clotting cascade.
Step 1:Blood coagulates in response to injury and the clot is made of a network of fibrins.
Step 2:Fibrins are formed by the conversion of inactive fibrinogen in the plasma by the enzyme thrombin, so fibrin is produced from fibrinogen and not the reverse.
Final answer: Fibrinogen is produced from fibrin
Q118Single correctThe Living World
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. Family | I.. Sapindales |
| B.. Genus | II.. Dicotyledonae |
| C.. Class | III.. Anacardiaceae |
| D.. Phylum | IV.. Angiospermae |
| E.. Order | V.. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-V, C-II, D-IV, E-I
Approach:
Each taxonomic rank for mango is matched with the correct taxon name.
Step 1:Mangifera represents the Genus, Anacardiaceae represents the Family, Sapindales represents the Order, Dicotyledonae represents the Class and Angiospermae represents the Phylum (division).
Final answer: A-III, B-V, C-II, D-IV, E-I
Q119Single correctThe Living World
Arrange the following taxonomic categories in ascending order : (a) Genus (b) Class (c) Order (d) Phylum (e) Family (f) Kingdom (g) Species
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(g), (a), (e), (c), (b), (d), (f)
Approach:
The taxonomic categories are arranged from the lowest to the highest rank.
Step 1:The ascending order of taxonomic categories runs from Species to Kingdom.
Final answer: (g), (a), (e), (c), (b), (d), (f)
Q120Single correctSexual Reproduction in Flowering Plants
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. Marginal placentation | I.. |
| B.. Axile placentation | II.. Tomato |
| C.. Parietal placentation | III.. |
| D.. Free central placentation | IV.. Pea |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-IV, B-II, C-I, D-III
Approach:
Each type of placentation is matched with its representative plant example.
Step 1:Marginal placentation is seen in Pea, axile placentation in Tomato, parietal placentation in Argemone and free central placentation in Primrose.
Final answer: A-IV, B-II, C-I, D-III
Q121Single correctPlant Kingdom
Sphenopsida class belongs to _____________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4pteridophytes
Approach:
The classification of the class Sphenopsida is recalled.
Step 1:Sphenopsida is a class of vascular plants placed under the pteridophytes.
Final answer: pteridophytes
Q122Single correctPhotosynthesis in Higher Plants
Which of the following statements regarding photorespiration are correct? (a) Do not occur in C3 plants (b) is consumed and is generated (c) Phosphoglycolate is formed (d) No synthesis of ATP and NADPH
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(c) and (d) only
Approach:
Each statement about photorespiration is evaluated.
Step 1:Photorespiration occurs in C3 plants, so statement (a) is incorrect.
Step 2:In photorespiration oxygen is consumed and carbon dioxide is released, so statement (b) is incorrect.
Step 3:Phosphoglycolate is formed as the initial product, and there is no synthesis of ATP and NADPH in photorespiration.
Final answer: (c) and (d) only
Q123Single correctCell - The Unit of Life
Smooth endoplasmic reticulum ____________ .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2is the major site for the synthesis of lipids
Approach:
The functions of smooth endoplasmic reticulum are recalled and compared with the options.
Step 1:The smooth endoplasmic reticulum is responsible for synthesis of lipids as well as steroidal hormones, and it is not associated with ribosomes, so it appears smooth.
Step 2:Rough endoplasmic reticulum is actively involved in protein synthesis, and carbohydrate synthesis occurs in chloroplasts.
Final answer: is the major site for the synthesis of lipids
Q124Single correctChemical Coordination and Integration
Which one of the following statements is ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2-cells of pancreas secrete insulin
Approach:
The secretions of the islet cells of the pancreas are recalled.
Step 1:The two main types of cells in the Islets of Langerhans are the alpha-cells and beta-cells.
Step 2:The alpha-cells secrete the hormone glucagon, while the beta-cells secrete insulin, so the statement that alpha-cells secrete insulin is incorrect.
Step 3:Glucagon stimulates glycogenolysis while insulin stimulates glycogenesis.
Final answer: -cells of pancreas secrete insulin
Q125Single correctThe Living World
Genus represents ___________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a group of closely related species
Approach:
The definition of the taxonomic category genus is recalled.
Step 1:A genus comprises a group of related species which have more characters in common compared to species of other genera.
Final answer: a group of closely related species
Q126Single correctBiological Classification
Which of the following is a prokaryote?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Fungi
Approach:
Each organism is classified as prokaryotic or eukaryotic.
Step 1:Prokaryotic cells are represented by bacteria, blue-green algae, Mycoplasma and PPLO.
Step 2:Fungi are eukaryotic organisms with a true nucleus.
Final answer: Fungi
Q127Single correctPlant Growth and Development
Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gibberellin
Approach:
The physiological role of each plant growth regulator is recalled.
Step 1:Gibberellin promotes internode elongation prior to flowering in cabbage, producing the bolting effect.
Final answer: Gibberellin
Q128Single correctCell Cycle and Cell Division
The correct sequence of adult cell cycle phases is ________ .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3G1-S-G2-M
Approach:
The order of the phases in the cell cycle is recalled.
Step 1:Gap1 phase is the first phase of interphase, after which synthesis phase starts in which DNA replicates, then Gap2 phase occurs and finally cell cycle ends with M phase (mitosis).
Final answer: G1-S-G2-M
Q129Single correctSexual Reproduction in Flowering Plants
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. Fusion of protoplasms between gametes | I.. Meiosis |
| B.. Fusion of two nuclei | II.. Plasmogamy |
| C.. Generation of haploid spores | III.. Karyogamy |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-I
Approach:
Each reproductive event is matched with its correct term.
Step 1:Plasmogamy is the fusion of protoplasm, karyogamy involves fusion of two nuclei, and meiosis leads to the production of haploid spores.
Final answer: A-II, B-III, C-I
Q130Single correctAnimal Kingdom
Given below are two statements : Statement I : The class name Reptilia refers to creeping or crawling mode of locomotion. Statement II : All organisms belonging to Reptilia have three chambered heart. In the light of the above statements, choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement about class Reptilia is evaluated for accuracy.
Step 1:The class Reptilia refers to their creeping or crawling mode of locomotion, so Statement I is correct.
Step 2:The heart is usually three-chambered, but four-chambered in crocodiles, so all organisms belonging to class Reptilia does not possess three chambered heart and Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q131Single correctCell Cycle and Cell Division
Given below are two statements : Statement I : Chromosomes are fully condensed at the end of prophase I. Statement II : Meiosis I resembles mitosis. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct, but Statement II is false
Approach:
Each statement about meiosis is evaluated.
Step 1:The final stage of meiotic prophase-I is diakinesis, so at the end of prophase-I the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation, making Statement I correct.
Step 2:It is Meiosis II, not Meiosis I, that resembles mitosis, because Meiosis II is an equational division that separates sister chromatids. Meiosis I is a reductional division in which homologous chromosomes separate and the chromosome number is halved, so Statement II is false.
Final answer: Statement I is correct, but Statement II is false
Q132Single correctAnimal Kingdom
Which of the following is a characteristic of chordates?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Absence of gills
Approach:
The fundamental characters of chordates are recalled and matched against the options.
Step 1:The members of phylum Chordata show presence of notochord, pharynx perforated by gill slits, a dorsal hollow single central nervous system and a post-anal tail at some stage of life.
Step 2:Gill slits are absent in non-chordates, so absence of gills is not a characteristic of chordates.
Final answer: Absence of gills
Q133Single correctPlant Growth and Development
Length of the stem at time 0 is 20 cm. The arithmetic growth rate is 30 cm per day. What is the length of the stem at the end of the day?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3230 cm
Approach:
The arithmetic growth equation is applied with the given initial length, rate and time.
Step 1:Arithmetic growth is expressed as length at time t equals length at time zero plus rate times time.
Step 2:Substituting the initial length 20 cm, rate 30 cm per day and time 7 days gives the final length.
Final answer: 230 cm
Q134Single correctBiomolecules
Arrange the following elements in descending order of their contribution to percentage weight of the human body. (a) Oxygen (b) Carbon (c) Hydrogen (d) Nitrogen
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a), (b), (c), (d)
Approach:
The percentage weight contribution of each element to the human body is compared.
Step 1:The arrangement of elements in descending order of their contribution to percentage weight of the human body is Oxygen at about 65 percent, Carbon at about 18.5 percent, Hydrogen at about 9.5 percent and Nitrogen at about 3.3 percent.
Final answer: (a), (b), (c), (d)
Q135Single correctNeural Control and Coordination
In frogs, the number of pairs of cranial nerves arising from the brain are ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 310
Approach:
The number of cranial nerve pairs in the frog is recalled.
Step 1:In frogs there are ten pairs of cranial nerves arising from the brain.
Final answer: 10
Q136Single correctMicrobes in Human Welfare
Which of the following is used as a clot buster?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Streptokinase
Approach:
Identify the microbial product used to dissolve blood clots.
Step 1:Streptokinase is produced by the bacterium Streptococcus and modified by genetic engineering to act as a clot buster.
Step 2:It removes clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.
Final answer: Streptokinase
Q137Single correctBiotechnology and its Applications
The inactive form of Bt toxin is converted to the active form in the insect gut _______
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1due to alkaline pH
Approach:
Recall how the Bt toxin protoxin is activated inside the insect gut.
Step 1:Bt toxin protein exists as inactive protoxins in the crystalline form.
Step 2:Once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut, which solubilises the crystals.
Final answer: due to alkaline pH
Q138Single correctPrinciples of Inheritance and Variation
Given below are two statements : Statement I : Down's syndrome is caused by the absence of one of the X-chromosomes. Statement II : Turner's syndrome is caused by the presence of an additional copy of the chromosomes. In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are incorrect
Approach:
Evaluate each statement against the actual chromosomal basis of the two syndromes.
Step 1:The cause of the genetic disorder Down's syndrome is the presence of an additional copy of chromosome number 21 (Trisomy 21), so Statement I is incorrect.
Step 2:The cause of the genetic disorder Turner's syndrome is the absence of one of the X-chromosomes, i.e. 45 with X0, so Statement II is incorrect.
Final answer: Both Statement I and Statement II are incorrect
Q139Single correctHuman Health and Disease
Which of the following disease is not sexually transmitted?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Tuberculosis
Approach:
Distinguish sexually transmitted infections from other modes of transmission.
Step 1:Genital warts, syphilis and gonorrhoea are sexually transmitted diseases. Genital warts are caused by Human papilloma virus, syphilis by Treponema pallidum and gonorrhoea by Neisseria gonorrhoeae.
Step 2:Tuberculosis is caused by Mycobacterium tuberculosis and spreads through air, so it is not sexually transmitted.
Final answer: Tuberculosis
Q140Single correctHuman Reproduction
Sperm motility is due to ___________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1flagellar movement
Approach:
Recall the structural basis of sperm movement.
Step 1:The sperm possesses a tail, which is a flagellum.
Step 2:The sperm travels across the fallopian tube via flagellar movement.
Final answer: flagellar movement
Q141Single correctEvolution
Natural selection can lead to _________. (a) stabilisation (b) genetic drift (c) directional change (d) disruption Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (c) and (d) only
Approach:
Identify which outcomes are effects of natural selection rather than chance.
Step 1:Natural selection can lead to stabilization, directional change or disruption.
Step 2:Genetic drift is a random change in allele frequency in a small population, so it arises by chance rather than through natural selection.
Final answer: (a), (c) and (d) only
Q142Single correctReproductive Health
The method of directly injecting a sperm into ovum in assisted reproductive technology is called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Intra cytoplasmic sperm injection (ICSI)
Approach:
Match the description of direct sperm injection to the correct ART technique.
Step 1:Intra cytoplasmic sperm injection (ICSI) is a procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.
Step 2:ZIFT (Zygote intra fallopian transfer) involves transfer of the zygote or early embryos with up to 8 blastomeres into the fallopian tube, while GIFT (Gamete intra fallopian transfer) involves transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one but can provide suitable environment for fertilisation and further development.
Final answer: Intra cytoplasmic sperm injection (ICSI)
Q143Single correctHuman Reproduction
Which of the following struture is not a part of the male reproductive system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Infundibulum
Approach:
Separate the male accessory ducts from female reproductive structures.
Step 1:The male sex accessory ducts include rete testis, vasa efferentia, epididymis and vas deferens.
Step 2:The oviducts, uterus and vagina constitute the female accessory ducts, and the infundibulum is the part of the oviduct.
Final answer: Infundibulum
Q144Single correctBiodiversity and Conservation
Arrange the following in descending order of number of species in the Amazonian rain forest. (a) Plants (b) Birds (c) Fishes (d) Invertebrates (e) Mammals Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(d) > (a) > (c) > (b) > (e)
Approach:
Order the taxa by the approximate species counts recorded for the Amazonian rain forest.
Step 1:The Amazonian rain forest is home to more than 40,000 species of plants, 3,000 of fishes, 1300 of birds, 427 of mammals, 378 of reptiles and more than 1,25,000 invertebrates.
Step 2:Arranging the listed taxa in decreasing order of species number gives invertebrates, plants, fishes, birds, mammals.
Final answer: (d) > (a) > (c) > (b) > (e)
Q145Single correctHuman Reproduction
Given below are two statements: Statement I : Ovulation is caused by LH surge leading to rupture of Graafian follicles. Statement II: Graafian follicle remaining after ovulation transform into corpus luteum and secretes large amount of estrogen. In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Assess each statement about the hormonal control of ovulation and the corpus luteum.
Step 1:Rapid secretion of LH leading to its maximum level during the mid-cycle, called LH surge, induces rupture of the Graafian follicle and thereby the release of ovum (ovulation), so Statement I is correct.
Step 2:The remaining parts of the Graafian follicle transform as the corpus luteum, which secretes large amounts of progesterone rather than estrogen, so Statement II is incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q146Single correctEcosystem
Which of the following are primary consumers in a food chain?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Herbivores
Approach:
Identify the trophic level occupied by organisms feeding directly on producers.
Step 1:Primary consumers in a food chain are the herbivores, which feed directly on producers.
Final answer: Herbivores
Q147Single correctEvolution
A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is _______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10.01
Approach:
Apply the Hardy-Weinberg expression for homozygous genotype frequency.
Step 1:The frequency of allele A is given as p equal to 0.1.
Step 2:Substituting into the genotype frequency expression gives the frequency of AA as p squared.
Final answer: 0.01
Q148Single correctChemical Coordination and Integration
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Excess growth hormone | I. Reabsorption of water and electrolytes in kidney |
| B. Luteinizing hormone | II. Contraction of uterus during child birth |
| C. Vasopressin | III. Acromegaly |
| D. Oxytocin | IV. Ovulation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Match each hormone with the physiological condition or action it produces.
Step 1:Excess growth hormone leads to acromegaly, matching A with III.
Step 2:Luteinizing hormone triggers ovulation, matching B with IV.
Step 3:Vasopressin promotes reabsorption of water and electrolytes in the kidneys, matching C with I.
Step 4:Oxytocin causes contraction of the uterus during child birth, matching D with II.
Final answer: A-III, B-IV, C-I, D-II
Q149Single correctBody Fluids and Circulation
The opening between the right atrium and the right ventricle is guarded by _________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2tricuspid valve
Approach:
Identify the valve guarding the right atrioventricular opening.
Step 1:The opening between the right atrium and the right ventricle is guarded by a valve formed of three muscular flaps or cusps, called the tricuspid valve.
Step 2:A bicuspid or mitral valve guards the opening between the left atrium and the left ventricle, and the semilunar valves guard the openings of the right and the left ventricles into the pulmonary artery and the aorta respectively.
Final answer: tricuspid valve
Q150Single correctBreathing and Exchange of Gases
Sponges exchange with by ___________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1simple diffusion over their entire body surfaces
Approach:
Recall the mode of gas exchange in lower invertebrates such as sponges.
Step 1:Mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of organisation.
Step 2:Lower invertebrates like sponges, coelenterates, flatworms etc. exchange oxygen with carbon dioxide by simple diffusion over their entire body surfaces.
Step 3:Earthworms use their moist cuticle, insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body, and gills are used by most of the aquatic arthropods and molluscs.
Final answer: simple diffusion over their entire body surfaces
Q151Single correctSexual Reproduction in Flowering Plants
How many theca are present in each lobe of a typical bilobed angiosperm anther?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12
Approach:
Recall the internal organisation of a typical bilobed anther.
Step 1:A typical anther is bilobed with each lobe having two theca, that is, they are dithecous.
Final answer: 2
Q152Single correctNeural Control and Coordination
Muscle contraction is initiated by a signal sent by the central nervous system by the release of _____ .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1acetyl choline
Approach:
Recall the neurotransmitter that triggers skeletal muscle contraction.
Step 1:Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron.
Step 2:The junction between a motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma.
Final answer: acetyl choline
Q153Single correctMolecular Basis of Inheritance
Which of the following statements about lac-operon is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gene is constitutively expressed
Approach:
Examine each statement against the regulation of the lac operon.
Step 1:Gene i is constitutively expressed in the lac operon, producing the repressor.
Step 2:Galactose cannot act as an inducer of lac operon; the repressor binds the operator when lactose is not available, so genes z, y and a are structural genes and have a different promoter than that of the i gene.
Final answer: Gene is constitutively expressed
Q154Single correctSexual Reproduction in Flowering Plants
Which of the following in female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the eggs?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Synergids
Approach:
Identify the embryo sac cells that direct the pollen tube.
Step 1:Synergids guide the entry of pollen tube for fertilizing the eggs in a typical angiospermic embryo sac.
Final answer: Synergids
Q155Single correctSexual Reproduction in Flowering Plants
Which of the following plant produces non-albuminous seeds?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pea
Approach:
Distinguish albuminous from non-albuminous seeds by residual endosperm.
Step 1:Non-albuminous seeds have no residual endosperm as it is completely consumed during embryo development, for example Pea.
Step 2:Wheat, maize and barley are the examples of albuminous seeds.
Final answer: Pea
Q156Single correctSexual Reproduction in Flowering Plants
If the diploid chromosome number of typical angiosperm is 36, what would be the chromosome number in its endosperm?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 354
Approach:
Compute the triploid endosperm chromosome number from the diploid value.
Step 1:The endosperm is triploid. The diploid chromosome number of a typical angiosperm is 36, so the haploid number n equals 18.
Step 2:The triploid endosperm chromosome number is the sum of three haploid sets.
Final answer: 54
Q157Single correctExcretory Products and their Elimination
Which of the following statements about the reabsorption process in Henle's loop are correct? (a) The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes. (b) Urine gets concentrated in Henle's loop. (c) Reabsorption of and water takes place in Henle's loop. (d) Active or passive transport of electrolytes occurs in the ascending limb of Henle's loop. Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a), (b) and (d) only
Approach:
Evaluate each statement against the functioning of the loop of Henle.
Step 1:The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes, so statement (a) is correct.
Step 2:The ascending limb is impermeable to water but transports electrolytes actively or passively into the medullary interstitium, so statement (d) is correct.
Step 3:Reabsorption in Henle's loop is minimal, so statement (c) does not hold, while the counter-current arrangement of the loop does concentrate the filtrate, supporting statement (b).
Final answer: (a), (b) and (d) only
Q158Single correctLocomotion and Movement
Which of the following is the correct order of arrangement of vertebrate column from the head to toe?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum
Approach:
Arrange the regions of the vertebral column from cranial to caudal end.
Step 1:The vertebral column is formed by 26 serially arranged units called vertebrae and is dorsally placed.
Step 2:The vertebral column is differentiated into cervical (7), thoracic (12), lumbar (5), sacral (1-fused) and coccygeal (1-fused) regions starting from the skull.
Step 3:The correct sequence from head to toe is cervical, thoracic, lumbar, sacrum and coccyx.
Final answer: Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum
Q159Single correctEvolution
Which of the following is evidence for evolution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Embryological support for evolution as proposed by Ernst Heckel
Approach:
Each option is examined to determine whether it represents an accepted line of evidence for organic evolution.
Step 1:Convergent evolution produces analogous structures, such as the wings of birds and butterflies, which arise from different ancestries due to similar environmental demands.
Step 2:Fossil records (paleontological evidence) and embryological similarities (Ernst Haeckel) and homologous structures arising through divergent evolution all support descent with modification.
Final answer: Convergent evolution of traits like wings of birds and butterflies
Q160Single correctEvolution
Given below are two statements : Modern arose in Australia and moved across continents. arose around 75000 to 10000 years ago. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 is incorrect but is correct
Approach:
Each statement on the origin and timeline of modern humans is tested against the established account.
Step 1:Modern Homo sapiens arose in Africa, not Australia, and from there moved across continents, developing into distinct races.
Step 2:During the ice age, between about 75000 and 10000 years ago, modern Homo sapiens arose.
Final answer: is incorrect but is correct
Q161Single correctOrganisms and Populations
Consider a population of 10 million cells. Given the per-capita birth rate of 0.002 (per unit time) and the per-capita death rate of 0.002 (per unit time), the expected number of cells after 10 generations is ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 310 million
Approach:
The population growth equation is applied with equal birth and death rates to obtain the intrinsic rate of natural increase and the resulting population change.
Step 1:The intrinsic rate of natural increase is the difference between the per-capita birth rate and the per-capita death rate.
Step 2:Substituting the rate into the growth equation gives the rate of change of the population.
Final answer: 10 million
Q162Single correctBiotechnology: Principles and Processes
During PCR, primers bind to the DNA strands in the ______ step.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3annealing
Approach:
The three steps of a PCR cycle are reviewed to identify the one in which primers attach to the template strands.
Step 1:PCR proceeds through denaturation, annealing and extension. Denaturation separates the double-stranded template into single strands.
Step 2:During annealing the two sets of primers pair with the single-stranded template at their complementary regions.
Final answer: annealing
Q163Single correctOrganisms and Populations
Given below are two statements : one is labelled as and the other is labelled as . The logistic growth model of populations is considered more realistic than the exponential growth model. Resources are finite. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
The relationship between the logistic model and resource availability is examined to judge both statements and their linkage.
Step 1:The logistic growth model incorporates a carrying capacity and is considered more realistic than the exponential growth model.
Step 2:Resources in nature are limited and finite, which sets the carrying capacity and underlies the logistic model.
Final answer: Both and are correct and is the correct explanation of
Q164Single correctEvolution
Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distant species is an example of ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2convergent evolution
Approach:
The pattern produced when two unrelated mammalian groups develop similar forms is identified.
Step 1:Adaptive radiation occurred independently in placental mammals and in Australian marsupials when each group occupied an isolated geographical area.
Step 2:Placental mammals each evolved into varieties that resemble a corresponding marsupial, producing similarity between distant species.
Final answer: convergent evolution
Q165Single correctHuman Health and Disease
Which of the following are secondary lymphoid organs? (a) Bone marrow (b) Tonsils (c) Spleen (d) Thymus Choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(b) and (c) only
Approach:
Each listed organ is classified as a primary or secondary lymphoid organ.
Step 1:Bone marrow and thymus are the sites where lymphocytes are produced and mature, making them primary lymphoid organs.
Step 2:Tonsils and spleen are sites where mature lymphocytes interact with antigens, making them secondary lymphoid organs.
Final answer: (b) and (c) only
Q166Single correctHuman Reproduction
Which of the following hormone is secreted by human placenta?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4LH
Approach:
The hormones produced by the placenta during pregnancy are listed and the option lying outside this set is selected.
Step 1:The placenta acts as a temporary endocrine gland during pregnancy and secretes human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogen and progesterone.
Step 2:Luteinizing hormone (LH) is secreted by the anterior pituitary, not by the placenta.
Final answer: LH
Q167Single correctMolecular Basis of Inheritance
Which of the following enzymes synthesizes precursor mRNA?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2RNA polymerase II
Approach:
The eukaryotic RNA polymerases are matched to the RNA species each transcribes.
Step 1:RNA polymerase I transcribes ribosomal RNAs and RNA polymerase III transcribes tRNA, 5S rRNA and small nuclear RNAs.
Step 2:RNA polymerase II transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).
Final answer: RNA polymerase II
Q168Single correctBiotechnology: Principles and Processes
Given below are two statements : Plasmids are autonomously replicating DNA. Plasmids are extrachromosomal DNA. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct
Approach:
The two properties attributed to plasmids are checked against their known biology.
Step 1:Plasmids are used as cloning vectors in genetic engineering and are small circular DNA molecules that replicate on their own.
Step 2:Plasmids exist separately from the bacterial chromosome and replicate independent of the control of the chromosomal DNA.
Final answer: Both and are correct
Q169Single correctPrinciples of Inheritance and Variation
For a person with blood group 'O', which of the following is a possible combination of parents' blood group genotypes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Father : and Mother :
Approach:
An offspring of blood group O must be genotype ii, so each parental pair is tested for its ability to contribute an i allele to a child.
Step 1:A child with blood group O has genotype ii, which requires both biological parents to carry and pass a recessive i allele.
Step 2:A parent with genotype carries no i allele, since and are present together and each is expressed. Such a parent cannot transmit i.
Final answer: Father : and Mother :
Q170Single correctEvolution
Given below are two statements : one is labelled as and the other is labelled as . Forelimbs of human and bats are homologous. Forelimbs of humans and bats have similar anatomical structure. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
The basis on which the forelimbs of humans and bats are called homologous is examined to judge both statements and their linkage.
Step 1:The forelimbs of humans and bats share similarities in the pattern of bone arrangement, with humerus, radius, ulna, carpals, metacarpals and phalanges present in their forelimbs.
Step 2:Though these forelimbs perform different functions, they have a similar anatomical structure, which makes them homologous organs that arose from a common ancestral pattern.
Final answer: Both and are correct and is the correct explanation of
Q171Single correctHuman Health and Disease
Colostrum, secreted by mother during initial days of lactation, is abundant in ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3IgA
Approach:
The antibody type predominant in colostrum is identified through its role in passive immunity of the newborn.
Step 1:Colostrum is secreted by the mother during the initial days of lactation and is abundant in IgA antibody.
Step 2:These IgA antibodies pass through the placenta and confer natural passive immunity on the infant.
Final answer: IgA
Q172Single correctOrganisms and Populations
Given below are two statements : one is labelled as and the other is labelled as . Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced. Goats were more efficient at browsing than Abingdon tortoise. In the light of the above statements, choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
The cause of the Abingdon tortoise extinction is examined to judge both statements and whether the reason explains the assertion.
Step 1:The Abingdon tortoise in the Galapagos islands became extinct within a decade after goats were introduced to the island.
Step 2:The extinction was apparently due to the greater browsing efficiency of the goats, which out-competed the tortoise for food.
Final answer: Both and are correct and is the correct explanation of
Q173Single correctHuman Reproduction
The covering of ovum at ovulation is ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3zona pellucida
Approach:
The layer that immediately surrounds the secondary oocyte released at ovulation is identified from the structure of the ovulated egg.
Step 1:The Graafian follicle ruptures to release the secondary oocyte (ovum) from the ovary by the process called ovulation. The secondary oocyte forms a membrane called zona pellucida surrounding it.
Step 2:The endometrium is the innermost layer of the uterine wall and the chorion is the outermost extraembryonic membrane that surrounds the embryo, neither covering the released ovum.
Final answer: zona pellucida
Q174Single correctOrganisms and Populations
Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Both species are harmed | I.. Predation |
| B.. One species is harmed and the other is benefited | II.. Mutualism |
| C.. Both species are benefited | III.. Competition |
| D.. One is benefited while the other has no effect | IV.. Commensalism |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-II, D-IV
Approach:
Each interaction outcome in List-I is matched with the interaction type in List-II.
Step 1:Both species being harmed corresponds to competition, and one species harmed while the other is benefited corresponds to predation.
Step 2:Both species being benefited corresponds to mutualism, and one benefited while the other is unaffected corresponds to commensalism.
Final answer: A-III, B-I, C-II, D-IV
Q175Single correctPrinciples of Inheritance and Variation
Given below are two statements : one is labelled as and the other is labelled as . In an experiment, Mendel observed that the F1 progeny plants are all tall and none are dwarf. Stem height is a contrasting trait, with tall being dominant and dwarf being recessive. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
Mendel's monohybrid result for stem height is examined against the dominance relationship between tall and dwarf to judge both statements and their linkage.
Step 1:Mendel crossed tall and dwarf plants and observed that all the F1 progeny plants are tall.
Step 2:The trait 'T' tall is said to be dominant over the other allele 't' or 'dwarf' trait, so this dominance of one trait over the other allows all the F1 to be tall.
Final answer: Both and are correct and is the correct explanation of
Q176Single correctBiotechnology: Principles and Processes
Given below are two statements : one is labelled as and the other is labelled as . In recombinant DNA technology, lysozyme is used for disrupting bacterial cells while cellulase is for plant cells. Isolation of genetic material needs disruption of cells. In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are correct and is the correct explanation of
Approach:
The use of specific lysing enzymes for different cell types is checked against the general need to disrupt cells for genetic material isolation.
Step 1:DNA is enclosed within the membranes, so the cell is broken open to release DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids.
Step 2:Because the wall material differs between cell types, the disrupting enzyme is chosen to match it: lysozyme digests the bacterial cell wall, cellulase the cellulose wall of plant cells and chitinase the chitinous wall of fungi.
Final answer: Both and are correct and is the correct explanation of
Q177Single correctHuman Health and Disease
Which of the following is used as an effective sedative and painkiller for treating post-surgery patients?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Morphine
Approach:
The agent that serves as a sedative and painkiller after surgery is identified by comparing the functions of the listed substances.
Step 1:Morphine is a very effective sedative and painkiller and is very useful in patients who have undergone surgery.
Step 2:Interferon belongs to the cytokine barrier of innate immunity, antibiotics restrict the growth of bacteria, and anti-retroviral drugs act against retroviruses such as HIV.
Final answer: Morphine
Q178Single correctEcosystem
Which of the following statements are ? (a) Energy flow from producers to consumers is unidirectional (b) Energy pyramid can never be inverted (c) Transfer of energy follows the 1% law Choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) and (b) only
Approach:
Each statement on energy flow and ecological pyramids is evaluated against the rules of ecosystem energetics.
Step 1:Energy flow is unidirectional. First plants capture solar energy and then food is transferred from the producers to decomposers, confirming statement (a).
Step 2:The pyramid of energy is always upright and can never be inverted, because when energy flows from a particular trophic level to the next trophic level some energy is always lost as heat at each step, confirming statement (b).
Step 3:The transfer of energy follows the 10 per cent law, so only 10 per cent of the energy is transferred to each trophic level from the lower trophic level, which makes statement (c) incorrect.
Final answer: (a) and (b) only
Q179Single correctHuman Health and Disease
Which of the following statements is about ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Fertilization takes place in mosquito gut
Approach:
Each statement is checked against the asexual and sexual phases of the Plasmodium life cycle in human and mosquito hosts.
Step 1:Plasmodium reproduces asexually in liver cells and in red blood cells, bursting the RBCs and causing cycles of fever and other symptoms. Released parasites infect new red blood cells.
Step 2:Sexual stages of Plasmodium, the gametocytes, develop in the RBCs, while fertilization and development take place in the mosquito's gut.
Final answer: Fertilization takes place in mosquito gut
Q180Single correctBiotechnology: Principles and Processes
Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Transformation | I.. Restriction enzyme |
| B.. Cloning site | II.. Transfer DNA to host bacteria |
| C.. Selection | III.. Replication |
| D.. Ori | IV.. Antibiotic |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-I, C-IV, D-III
Approach:
Each step or feature of recombinant DNA technology in List-I is matched with its description in List-II.
Step 1:Transformation is the transfer of DNA to the host bacteria, and the cloning site is a segment of DNA within a plasmid vector that contains multiple unique recognition sequences for restriction enzymes.
Step 2:Selection uses antibiotics to help in the selection of recombinants, and Ori is the specific DNA sequence where the host cell's replication machinery begins duplicating the plasmid.
Final answer: A-II, B-I, C-IV, D-III
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