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NEET 2020 Sep 13 Question Paper with Solutions
All 180 questions from the NEET 2020 (Sep 13) 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 2020Chemistry PYQs 2020Biology PYQs 2020
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q136Single correctAtoms
For which one of the following, Bohr model is not valid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Singly ionised neon atom (N)
Approach:
The Bohr model applies only to single-electron (hydrogen-like) species, so the system with more than one electron is identified.
Step 1:The Bohr model treats a single electron orbiting a nucleus and ignores electron-electron interaction.
Step 2:Hydrogen, singly ionised helium and a deuteron atom each retain a single electron, so each is hydrogen-like.
Step 3:Singly ionised neon N retains nine electrons, introducing strong electron-electron repulsion that the Bohr model cannot describe.
Final answer: Singly ionised neon atom (N)
Q137Single correctElectromagnetic Waves
The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is :
(c = speed of electromagnetic waves)
(c = speed of electromagnetic waves)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The energy densities stored in the electric and magnetic fields of an electromagnetic wave are compared.
Step 1:The intensity of an electromagnetic wave is shared by the electric and magnetic field energy densities.
Step 2:Substituting E = cB and = 1/( ) into the magnetic energy density gives the magnetic term in terms of E.
Step 3:The two contributions are equal, so their ratio is unity.
Final answer:
Q138Single correctWave Optics
The Brewsters angle for an interface should be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Brewster's law is applied for light passing from a rarer to a denser medium to bound the polarising angle.
Step 1:Brewster's angle satisfies tan() equal to the refractive index of the interface.
Step 2:For an interface from a rarer to a denser medium the refractive index exceeds one, so the tangent of the angle exceeds one.
Step 3:The angle of incidence cannot reach the grazing value, bounding it below ninety degrees.
Final answer:
Q139Single correctKinetic Theory
A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 2C.
Its density is : (R = 8.3 J mo )
Its density is : (R = 8.3 J mo )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The ideal gas law is rewritten in terms of density using the molar mass of hydrogen.
Step 1:The ideal gas law expressed through density relates pressure, molar mass and temperature.
Step 2:The molar mass of hydrogen gas is two grams per mole and the temperature is 300 kelvin.
Step 3:Substituting the pressure 249000 pascal gives the density.
Final answer:
Q140Single correctRay Optics
A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is , then the angle of incidence is nearly equal to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The prism geometry is used with the emergence-normal condition to relate the incidence angle to the prism angle.
Step 1:The ray emerges normally from the second surface, so the angle of refraction at that surface is zero.
Step 2:The prism angle equals the sum of the two refraction angles, fixing the first refraction angle as A.
Step 3:For a small angle prism Snell's law at the first surface reduces to i equal to mu times .
Final answer:
Q141Single correctThermodynamics
Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2adiabatic
Approach:
The conditions of the free-expansion experiment are matched to the defining feature of a thermodynamic process.
Step 1:The system is thermally insulated, so no heat is exchanged with the surroundings.
Step 2:A process carried out with no heat exchange is by definition adiabatic.
Step 3:The gas expands into vacuum against zero external pressure, so no work is done and the internal energy stays constant.
Final answer: adiabatic
Q142Single correctDual Nature of Radiation and Matter
The energy equivalent of 0.5 g of a substance is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Einstein's mass-energy relation converts the given mass into its energy equivalent.
Step 1:The mass is expressed in kilograms and the speed of light is taken as three times ten to the eighth metres per second.
Step 2:Substituting into Einstein's relation gives the energy.
Final answer:
Q143Single correctGravitation
A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 232 N
Approach:
The inverse-square dependence of gravitational force on distance from the centre is applied.
Step 1:At a height equal to half the radius the distance from the centre becomes three halves of the earth's radius.
Step 2:The force scales as the inverse square of distance, so the new force is the surface weight multiplied by the square of the ratio of radii.
Step 3:Evaluating the product gives the gravitational force at that height.
Final answer: 32 N
Q144Single correctSemiconductor Electronics
The solids which have the negative temperature coefficient of resistance are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4insulators and semiconductors
Approach:
The temperature behaviour of resistance is examined for each class of solid.
Step 1:In metals a rise in temperature increases lattice vibrations and resistance, giving a positive temperature coefficient.
Step 2:In semiconductors and insulators a rise in temperature releases more charge carriers, decreasing resistance.
Step 3:Both insulators and semiconductors therefore show a negative temperature coefficient of resistance.
Final answer: insulators and semiconductors
Q145Single correctOscillations
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 rad
Approach:
The displacement and acceleration of a simple harmonic oscillator are compared as functions of time.
Step 1:The acceleration in simple harmonic motion is proportional to the negative of the displacement.
Step 2:Writing acceleration as a sine function introduces a phase shift of pi relative to the displacement.
Step 3:The displacement and acceleration are exactly out of phase by pi radians.
Final answer: rad
Q146Single correctUnits and Measurements
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitch of the screw gauge is :
The pitch of the screw gauge is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.5 mm
Approach:
The relation between least count, pitch and the number of circular scale divisions is rearranged for the pitch.
Step 1:The least count equals the pitch divided by the number of divisions on the circular scale.
Step 2:Rearranging gives the pitch as the product of the least count and the number of divisions.
Final answer: 0.5 mm
Q147Single correctWaves
In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2524 Hz
Approach:
The beat frequency equals the magnitude of the difference between the two string frequencies. The two candidate values for B differ from A by 6 Hz; the direction in which the beat frequency changes when the tension of B is lowered selects the correct candidate.
Step 1:An initial beat frequency of 6 Hz with A at 530 Hz places B at either 524 Hz or 536 Hz.
Step 2:Lowering the tension in B reduces its frequency. Taking the 524 Hz candidate, the frequency drops below 524 Hz, increasing the gap from A to 7 Hz, which matches the observation.
Step 3:Taking the 536 Hz candidate, lowering the tension drops B toward 530 Hz, which would decrease the beat frequency to about 5 Hz, contradicting the observed increase; this candidate is rejected.
Final answer: 524 Hz
Q148Single correctSystem of Particles and Rotational Motion
Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass.
The centre of mass of the system from the 5 kg particle is nearly at a distance of :
The centre of mass of the system from the 5 kg particle is nearly at a distance of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 367 cm
Approach:
The centre of mass position is computed by the weighted mean of the two particle positions.
Step 1:The 5 kg particle is placed at the origin and the 10 kg particle at one metre along the rod.
Step 2:Substituting the masses and positions into the centre of mass formula gives the distance from the 5 kg particle.
Step 3:Converting to centimetres gives about sixty-seven centimetres.
Final answer: 67 cm
Q149Single correctSystem of Particles and Rotational Motion
Find the torque about the origin when a force of 3 N acts on a particle whose position vector is 2 m.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 Nm
Approach:
The torque is found as the cross product of the position vector and the force vector.
Step 1:The position vector lies along the z-axis and the force lies along the y-axis.
Step 2:The cross product of the z and y unit vectors gives the negative x direction.
Step 3:Multiplying the magnitudes gives the torque vector.
Final answer: Nm
Q150Single correctDual Nature of Radiation and Matter
Light with an average flux of 20 W/c falls on a non-reflecting surface at normal incidence having surface area 20 c. The energy received by the surface during time span of 1 minute is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The total energy is the product of flux, surface area and the time of exposure.
Step 1:The incident power equals the flux multiplied by the surface area.
Step 2:The energy received over one minute equals the power multiplied by sixty seconds.
Step 3:Expressing the energy in scientific notation gives the received energy.
Final answer:
Q151Single correctElectric Charges and Fields
A spherical conductor of radius 10 cm has a charge of C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Outside a uniformly charged conductor the field equals that of a point charge at the centre.
Step 1:The point at 15 cm lies outside the 10 cm sphere, so the entire charge behaves as if concentrated at the centre.
Step 2:Substituting the constant, charge and distance into the point-charge field expression gives the field.
Step 3:Evaluating the quotient gives the magnitude of the electric field.
Final answer:
Q152Single correctElectrostatic Potential and Capacitance
In a certain region of space with volume 0.2 , the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1zero
Approach:
The electric field is obtained from the spatial rate of change of the potential.
Step 1:The potential is constant at five volts throughout the region, so its spatial derivative vanishes.
Step 2:The electric field is the negative gradient of the potential, which is zero for a constant potential.
Final answer: zero
Q153Single correctSemiconductor Electronics
The increase in the width of the depletion region in a p-n junction diode is due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2reverse bias only
Approach:
The effect of bias polarity on the depletion region of a p-n junction is examined.
Step 1:Under reverse bias the external field pulls majority carriers away from the junction, widening the depletion region.
Step 2:Under forward bias majority carriers are pushed toward the junction, narrowing the depletion region.
Step 3:The increase in width therefore arises only under reverse bias.
Final answer: reverse bias only
Q154Single correctAlternating Current
A 40 F capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 32.5 A
Approach:
The capacitive reactance is computed and used with Ohm's law for the rms current.
Step 1:The capacitive reactance is found from the frequency and capacitance.
Step 2:Dividing the rms voltage by the reactance gives the rms current.
Final answer: 2.5 A
Q155Single correctKinetic Theory
The mean free path for a gas, with molecular diameter d and number density n can be expressed as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The standard kinetic-theory expression for the mean free path is identified.
Step 1:The mean free path is inversely proportional to the number density and to the square of the molecular diameter.
Step 2:Including the geometric factor from relative motion gives the full expression.
Final answer:
Q156Single correctSemiconductor Electronics
For transistor action, which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The base region must be very thin and lightly doped.
Approach:
The structural and biasing requirements for transistor action are recalled and applied to each statement.
Step 1:For transistor action the emitter junction is forward biased while the collector junction is reverse biased, so a claim that both junctions are forward biased is wrong.
Step 2:The base must be very thin and lightly doped so that most injected carriers reach the collector.
Final answer: The base region must be very thin and lightly doped.
Q157Single correctDual Nature of Radiation and Matter
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4zero
Approach:
The photoelectric threshold condition is checked after the frequency is halved.
Step 1:The incident frequency is initially 1.5 times the threshold frequency.
Step 2:Halving the frequency brings it below the threshold frequency.
Step 3:Below the threshold frequency no electrons are emitted regardless of intensity, so the photoelectric current is zero.
Final answer: zero
Q158Single correctNuclei
When a uranium isotope is bombarded with a neutron, it generates , three neutrons and :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The missing fragment is found by conserving mass number and atomic number in the fission reaction.
Step 1:The total mass number before the reaction is 235 plus 1 from the neutron, equal to 236.
Step 2:Subtracting the krypton mass number and the three emitted neutrons leaves the fragment mass number.
Step 3:The total atomic number is 92, and subtracting 36 for krypton leaves 56, identifying barium.
Final answer:
Q159Single correctAtoms and Nuclei
The energy required to break one bond in DNA is J. This value in eV is nearly:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Energy in joule is converted to electron volt by dividing by the charge of one electron, since .
Step 1:One electron volt equals J, so the joule value is divided by this factor.
Step 2:Carrying out the division gives the energy in electron volt.
Final answer:
Q160Single correctLaws of Motion
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For two masses over a frictionless pulley, the acceleration follows from the net driving weight divided by the total mass.
Step 1:The heavier mass is 6 kg and the lighter mass is 4 kg, so the difference of the weights drives the motion.
Step 2:Dividing the net driving weight by the total mass of the system gives the acceleration.
Final answer:
Q161Single correctMechanical Properties of Solids
A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to when mass M is suspended from its free end. The expression for Young's modulus is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Young's modulus is the ratio of stress to strain, where stress is the applied force per unit area and strain is the fractional change in length.
Step 1:The suspended mass produces a stretching force equal to its weight, and the extension is the increase in length.
Step 2:Substituting the force, area, original length, and extension into the definition of Young's modulus gives the required expression.
Final answer:
Q162Single correctKinetic Theory of Gases
The average thermal energy for a mono-atomic gas is : ( is Boltzmann constant and T, absolute temperature)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By the equipartition theorem, each degree of freedom contributes of average energy, and a monatomic gas has three translational degrees of freedom.
Step 1:A monatomic gas molecule has only three translational degrees of freedom.
Step 2:Multiplying the number of degrees of freedom by gives the average thermal energy.
Final answer:
Q163Single correctCurrent Electricity
Which of the following graph represents the variation of resistivity () with temperature (T) for copper?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Graph: rising from near the origin, curving upward (concave up) as T increases
Approach:
Copper is a metallic conductor. Its resistivity rises with temperature because lattice vibrations scatter conduction electrons more strongly as temperature increases. The correct graph is the one in which resistivity grows from a small value near absolute zero and increases with temperature.
Step 1:For metals the temperature coefficient of resistivity is positive, so resistivity must increase with temperature; a falling curve cannot describe copper.
Step 2:Near absolute zero the resistivity of copper tends to a small residual value and rises with temperature, becoming linear at higher temperatures; the curve therefore starts low and bends upward rather than being a straight line through a high intercept.
Step 3:The graph that increases from near the origin and curves upward as temperature rises corresponds to copper.
Final answer: Graph: rising from near the origin, curving upward (concave up) as T increases
Q164Single correctCurrent Electricity
The color code of a resistance is given below
(see figure: bands coloured Yellow, Violet, Brown, Gold)
The values of resistance and tolerance, respectively, are
(see figure: bands coloured Yellow, Violet, Brown, Gold)
The values of resistance and tolerance, respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The first two colour bands give the significant figures, the third band gives the decimal multiplier, and the fourth band gives the tolerance.
Step 1:Yellow corresponds to 4 and Violet to 7, giving the first two significant digits as 47.
Step 2:Brown is the multiplier with value , and Gold gives a tolerance of five percent.
Final answer:
Q165Single correctWave Optics
In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3four times
Approach:
Fringe width is directly proportional to the screen distance and inversely proportional to the slit separation, so the two changes are combined multiplicatively.
Step 1:Halving the slit separation doubles the fringe width through the inverse dependence on .
Step 2:Doubling the screen distance doubles the fringe width again through the direct dependence on , so the two factors combine.
Final answer: four times
Q166Single correctElectrostatic Potential and Capacitance
The capacitance of a parallel plate capacitor with air as medium is . With the introduction of a dielectric medium, the capacitance becomes . The permittivity of the medium is :
()
()
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The dielectric constant is the ratio of the filled capacitance to the air capacitance, and the permittivity of the medium is this dielectric constant times the permittivity of free space.
Step 1:Dividing the dielectric-filled capacitance by the air capacitance gives the dielectric constant.
Step 2:Multiplying the dielectric constant by the permittivity of free space gives the permittivity of the medium.
Final answer:
Q167Single correctUnits and Measurements
Dimensions of stress are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Stress is defined as force per unit area, so its dimensions follow from the dimensions of force divided by those of area.
Step 1:Force has dimensions and area has dimensions .
Step 2:Dividing the dimensions of force by those of area gives the dimensions of stress.
Final answer:
Q168Single correctRay Optics and Optical Instruments
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 rad
Approach:
The limit of resolution of a telescope is given by the diffraction formula relating the wavelength of light to the diameter of the objective.
Step 1:The wavelength is 600 nm and the objective diameter is 2 m.
Step 2:Substituting the wavelength and diameter into the diffraction formula gives the angular limit of resolution.
Final answer: rad
Q169Single correctAlternating Current
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is . If instead C is removed from the circuit, the phase difference is again between current and voltage. The power factor of the circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Equal phase differences when either reactive element is removed imply equal capacitive and inductive reactances, which is the resonance condition for the full circuit.
Step 1:Removing L leaves R and C with phase angle , and removing C leaves R and L with the same phase angle, so the two reactances are equal.
Step 2:With equal reactances the net reactance of the full LCR circuit vanishes, leaving a purely resistive circuit at resonance with unit power factor.
Final answer:
Q170Single correctElectrostatic Potential and Capacitance
A short electric dipole has a dipole moment of C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of with the dipole axis is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The potential of a short dipole at a point depends on the dipole moment, the cosine of the angle from the axis, and the inverse square of the distance.
Step 1:The dipole moment is C m, the angle is with , and the distance is 0.6 m.
Step 2:Substituting these values along with the Coulomb constant into the dipole potential formula gives the potential.
Final answer:
Q171Single correctMagnetism and Matter
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A . The permeability of the material of the rod is:
( T m )
( T m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 T m
Approach:
The permeability of a magnetic material relates to its susceptibility through the relative permeability. Substituting the susceptibility into the relation gives the permeability directly; the magnetising field value is not required for this relation.
Step 1:The relative permeability follows from the susceptibility.
Step 2:Multiplying the permeability of free space by the relative permeability gives the permeability of the rod.
Step 3:Expressing the result in standard form yields the permeability of the material.
Final answer: T m
Q172Single correctMoving Charges and Magnetism
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is :
( T m )
( T m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 T
Approach:
The magnetic field at the centre of a long solenoid is the permeability of free space times the number of turns per unit length times the current.
Step 1:The number of turns per unit length is the total turns divided by the length in metres.
Step 2:Substituting the permeability, turn density, and current gives the magnetic field.
Final answer: T
Q173Single correctCurrent Electricity
A charged particle having drift velocity of m in an electric field of V, has a mobility in of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Mobility is defined as the magnitude of drift velocity per unit electric field, so the drift velocity is divided by the field.
Step 1:The drift velocity is m and the electric field is V .
Step 2:Dividing the drift velocity by the electric field gives the mobility.
Final answer:
Q174Single correctThermal Properties of Matter
The quantities of heat required to raise the temperature of two solid copper spheres of radii and () through 1 K are in the ratio :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
For the same temperature rise and identical material, the heat required is proportional to the mass and hence to the volume, which scales as the cube of the radius.
Step 1:With the same material, specific heat, and temperature change, the heat required is proportional to the cube of the radius.
Step 2:Substituting and cubing gives the ratio of heats.
Final answer:
Q175Single correctDual Nature of Radiation and Matter
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is nm, the potential difference is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The de Broglie wavelength of an electron accelerated through a potential difference follows the standard relation in which the wavelength in nanometres equals .
Step 1:Setting the standard expression equal to the given wavelength relates the potential to the wavelength.
Step 2:Squaring the square root of the potential gives the potential difference.
Final answer:
Q176Single correctUnits and Measurements
Taking into account of the significant figures, what is the value of 9.99 m 0.0099 m?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
In subtraction the result is rounded to the same number of decimal places as the quantity with the fewest decimal places.
Step 1:Direct subtraction of the two lengths gives an unrounded value.
Step 2:The quantity 9.99 m has two decimal places, the fewest, so the result is rounded to two decimal places.
Final answer: m
Q177Single correctMotion in a Straight Line
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 m
Approach:
The height is found from the kinematic relation connecting the final and initial velocities, the acceleration due to gravity, and the displacement.
Step 1:The initial downward velocity is 20 m/s and the impact velocity is 80 m/s, with gravity acting downward at 10 m/.
Step 2:Rearranging the kinematic relation and substituting gives the height of the tower.
Final answer: m
Q178Single correctMechanical Properties of Fluids
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 g
Approach:
Capillary rise is inversely proportional to the radius, while the mass of risen water equals the product of density, cross-sectional area, and height, so its dependence on radius is examined directly.
Step 1:Substituting the capillary rise into the mass expression shows that mass is proportional to the first power of the radius, since the area contributes while the height contributes .
Step 2:Doubling the radius doubles the risen mass, so the new mass is twice the original 5 g.
Final answer: g
Q179Single correctCurrent Electricity
A resistance wire connected in the left gap of a metre bridge balances a 10 resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 of the resistance wire is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
The metre bridge balance gives the unknown resistance from the ratio of the two segments, and dividing the total wire length by this resistance gives the length per ohm.
Step 1:The balance point divides the wire in the ratio 3 : 2, so the unknown resistance relates to the 10 ohm resistance through this ratio.
Step 2:Dividing the total wire length of 1.5 m by the total resistance of 15 ohm gives the length corresponding to one ohm.
Final answer: m
Q180Single correctSemiconductor Electronics
For the logic circuit shown, the truth table is :
(see figure: inputs A and B feed a gate combination producing output Y)
(see figure: inputs A and B feed a gate combination producing output Y)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A | B | Y
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
Approach:
The logic circuit shown is a combination of gates whose net behaviour is determined by tracing the inputs through to the output for each input combination.
Step 1:Tracing the inputs through the gate combination, the output is high only when both inputs are high, which is the AND behaviour.
Step 2:Listing the output for every input pair produces high output only for the (1,1) row.
Final answer: A | B | Y
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
Chemistry45 questions
Q91Single correctChemical Bonding and Molecular Structure
Identify a molecule which does not exist.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Molecular orbital theory predicts stability from the bond order, which equals one half of the difference between bonding and antibonding electrons.
Step 1:For the four electrons fill the bonding and antibonding orbitals equally.
Step 2:Substituting into the bond order expression gives a value of zero, so no net bond forms.
Step 3:A bond order of zero corresponds to a molecule that does not exist, whereas , and all have positive bond orders.
Final answer:
Q92Single correctEquilibrium
Find the solubility of in 0.1 M NaOH. Given that the ionic product of is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
In the presence of a strong base the hydroxide ion concentration is fixed by NaOH, and the solubility of the sparingly soluble hydroxide follows from its solubility product.
Step 1:The 0.1 M NaOH supplies the hydroxide ions, so the common ion effect sets the hydroxide concentration.
Step 2:The solubility equals the dissolved nickel ion concentration, obtained by rearranging the solubility product expression.
Step 3:Substituting the values gives the solubility.
Final answer:
Q93Single correctThe p-Block Elements and Allotropes of Carbon
Identify the correct statements from the following :
(a) is used as refrigerant for ice-cream and frozen food.
(b) The structure of contains twelve six carbon rings and twenty five carbon rings.
(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.
(d) CO is colorless and odourless gas.
(a) is used as refrigerant for ice-cream and frozen food.
(b) The structure of contains twelve six carbon rings and twenty five carbon rings.
(c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline.
(d) CO is colorless and odourless gas.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(c) and (d) only
Approach:
Each of the four printed statements is judged on its own chemistry, and the statements that survive are collected.
Step 1:The refrigerant used for ice-cream and frozen food is dry ice, i.e. SOLID carbon dioxide; gaseous has no such use, so statement (a) as printed is wrong.
Step 2:Buckminsterfullerene is built from twenty six-membered and twelve five-membered carbon rings, so the printed counts in statement (b) are wrong.
Step 3:ZSM-5 is the shape-selective zeolite used industrially to convert alcohols to gasoline, so statement (c) is correct.
Step 4:Carbon monoxide is a colourless, odourless and highly poisonous gas, so statement (d) is correct.
Step 5:Only statements (c) and (d) survive.
Final answer: (c) and (d) only
Q94Single correctThermodynamics
Hydrolysis of sucrose is given by the following reaction.
If the equilibrium constant is at 300 K, the value of at the same temperature will be :
If the equilibrium constant is at 300 K, the value of at the same temperature will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The standard reaction free energy relates to the equilibrium constant through the standard thermodynamic relation.
Step 1:The standard reaction free energy is the negative product of the gas constant, temperature and the natural logarithm of the equilibrium constant.
Step 2:Substituting , and gives the expression directly.
Final answer:
Q95Single correctAldehydes, Ketones and Carboxylic Acids
Identify compound X in the following sequence of reactions
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Benzal chloride (benzene ring bearing )
Approach:
Free-radical chlorination of the methyl group and subsequent hydrolysis are tracked to identify the intermediate that yields an aldehyde.
Step 1:Photochemical chlorination of toluene replaces hydrogen on the side chain; the degree of substitution determines whether hydrolysis gives an alcohol, aldehyde or acid.
Step 2:Hydrolysis at 373 K converts a geminal dichloride on the side chain into a carbonyl group, producing benzaldehyde.
Step 3:Therefore X is the benzal chloride bearing the group on the ring.
Final answer: Benzal chloride (benzene ring bearing )
Q96Single correctClassification of Elements and Periodicity in Properties
Identify the incorrect match.
| Name | IUPAC Official Name |
|---|---|
| (a). Unnilunium | (i). Mendelevium |
| (b). Unniltrium | (ii). Lawrencium |
| (c). Unnilhexium | (iii). Seaborgium |
| (d). Unununnium | (iv). Darmstadtium |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(d), (iv)
Approach:
The IUPAC systematic name encodes the atomic number digit by digit using nil=0, un=1, bi=2, tri=3, hex=6, with the suffix -ium. Decoding each name gives its atomic number, which fixes the official element name; the pairing that decodes to a different element is the incorrect match.
Step 1:Unnilunium decodes as un-nil-un = 1-0-1, atomic number 101, whose official name is Mendelevium, so the pairing (a)-(i) is a correct match.
Step 2:Unnilhexium decodes as un-nil-hex = 1-0-6, atomic number 106, whose official name is Seaborgium, so the pairing (c)-(iii) is a correct match.
Step 3:Unununnium decodes as un-un-un = 1-1-1, atomic number 111, whose official name is Roentgenium, whereas Darmstadtium is element 110; the pairing (d)-(iv) is therefore the incorrect match.
Final answer: (d), (iv)
Q97Single correctThe Solid State
An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
In a body-centred cubic cell the atoms touch along the body diagonal, which fixes the ratio between atomic radius and edge length.
Step 1:Along the body diagonal of a bcc unit cell, four atomic radii span the diagonal length .
Step 2:Solving for the radius gives a factor of times the edge length.
Step 3:Substituting the edge length of 288 pm gives the required radius.
Final answer:
Q98Single correctChemical Bonding and Molecular Structure
Which of the following set of molecules will have zero dipole moment ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene
Approach:
A molecule has zero net dipole moment when its symmetric geometry causes the individual bond dipoles to cancel.
Step 1:Trigonal planar boron trifluoride, linear beryllium difluoride and linear carbon dioxide are all symmetric, so their bond dipoles cancel to zero.
Step 2:1,4-dichlorobenzene places the two chlorine atoms diametrically opposite, so the two C-Cl dipoles cancel, whereas the 1,3 isomer leaves a net dipole.
Step 3:Ammonia, water and nitrogen trifluoride are bent or pyramidal and retain a net dipole, eliminating the other sets.
Final answer: Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene
Q99Single correctElectrochemistry
On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Oxygen gas
Approach:
The species liberated at the anode is decided by comparing the oxidation potentials of water and the sulphate ion.
Step 1:During electrolysis of dilute sulphuric acid, oxidation occurs at the anode and reduction at the cathode.
Step 2:Water is preferentially oxidised over the sulphate ion at the platinum anode, releasing oxygen gas.
Step 3:Hydrogen appears at the cathode, so the anode product is oxygen gas.
Final answer: Oxygen gas
Q100Single correctAlcohols, Phenols and Ethers
Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Tert. butyl alcohol
Approach:
Grignard addition to a ketone followed by hydrolysis raises the degree of substitution at the carbonyl carbon, defining the class of alcohol formed.
Step 1:The methyl carbanion of the Grignard reagent adds to the carbonyl carbon of acetone, which already bears two methyl groups.
Step 2:Hydrolysis of the alkoxide gives a carbon bonded to three methyl groups and one hydroxyl, a tertiary alcohol.
Step 3:The product is 2-methyl-2-propanol, commonly named tert. butyl alcohol.
Final answer: Tert. butyl alcohol
Q101Single correctThe p-Block Elements (Group 16)
Which of the following oxoacid of sulphur has linkage?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3, peroxodisulphuric acid
Approach:
The presence of a peroxide linkage is identified from the structure of each sulphur oxoacid.
Step 1:Sulphurous and sulphuric acids contain only S-O and S-OH bonds with no oxygen-oxygen bonds.
Step 2:Pyrosulphuric acid links two sulphur atoms through a bridging oxygen, again without a peroxide linkage.
Step 3:Peroxodisulphuric acid carries a bridge between the two sulphate units.
Final answer: , peroxodisulphuric acid
Q102Single correctAmines
Which of the following amine will give the carbylamine test?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Aniline (benzene ring bearing )
Approach:
The carbylamine (isocyanide) test is specific to primary amines, so the structure with a free group is selected.
Step 1:The carbylamine test produces a foul-smelling isocyanide only when a primary amine reacts with chloroform and alcoholic potassium hydroxide.
Step 2:Aniline carries a primary group, while the N-methyl, N,N-dimethyl and N-ethyl anilines are secondary or tertiary amines.
Step 3:Therefore aniline alone responds to the carbylamine test.
Final answer: Aniline (benzene ring bearing )
Q103Single correctThe d- and f-Block Elements / Coordination Compounds
The calculated spin only magnetic moment of ion is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The spin-only magnetic moment follows from the number of unpaired electrons in the ion's electronic configuration.
Step 1:Chromium has the configuration ; removing two electrons gives with a configuration.
Step 2:Substituting four unpaired electrons into the spin-only formula gives the moment.
Final answer:
Q104Single correctThermodynamics
The correct option for free expansion of an ideal gas under adiabatic condition is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
The conditions of an adiabatic free expansion of an ideal gas are applied to heat, work and temperature change.
Step 1:An adiabatic process exchanges no heat with the surroundings.
Step 2:Free expansion occurs against zero external pressure, so the work done is zero.
Step 3:With both heat and work zero, the internal energy stays constant, so for an ideal gas the temperature does not change.
Final answer: and
Q105Single correctSolutions
The freezing point depression constant of benzene is 5.12 K kg . The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The depression in freezing point of a dilute solution is proportional to the molality through the cryoscopic constant.
Step 1:The depression equals the product of the cryoscopic constant and the molality for a non-electrolyte solute.
Step 2:Substituting the cryoscopic constant and molality gives the depression.
Final answer:
Q106Single correctElectrochemistry
The number of Faradays(F) required to produce 20 g of calcium from molten (Atomic mass of Ca = 40 g ) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11
Approach:
Faraday's law relates the moles of metal deposited to the charge passed through the number of electrons in the reduction.
Step 1:The number of moles of calcium produced is the mass divided by the atomic mass.
Step 2:Reduction of one calcium ion requires two electrons, so two Faradays are needed per mole of calcium.
Step 3:Multiplying the moles of calcium by two gives the Faradays required.
Final answer: 1
Q107Single correctAldehydes, Ketones and Carboxylic Acids
Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cross Aldol condensation
Approach:
The reaction type is decided by whether the carbonyl partners carry alpha-hydrogens and whether they differ.
Step 1:Acetophenone has alpha-hydrogens and forms a carbanion under dilute base, while benzaldehyde lacks alpha-hydrogens and acts as the electrophile.
Step 2:Condensation between two different carbonyl compounds, one of which provides the enolate, is a cross aldol condensation.
Step 3:The base-promoted reaction between benzaldehyde and acetophenone is therefore a cross aldol condensation.
Final answer: Cross Aldol condensation
Q108Single correctSurface Chemistry / Purification Techniques
Paper chromatography is an example of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Partition chromatography
Approach:
The classification of paper chromatography is based on the mechanism by which components are separated.
Step 1:In paper chromatography the moisture held by the cellulose acts as the stationary liquid phase while the developing solvent is the mobile phase.
Step 2:Separation arises from the differing distribution of solutes between the two liquid phases, which is partition.
Step 3:Paper chromatography is therefore classed as partition chromatography.
Final answer: Partition chromatography
Q109Single correctChemical Kinetics
An increase in the concentration of the reactants of a reaction leads to change in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4collision frequency
Approach:
The quantity affected by reactant concentration is identified from collision theory.
Step 1:Activation energy, heat of reaction and threshold energy are fixed by the nature of the reaction and do not depend on concentration.
Step 2:Raising the reactant concentration increases the number of molecules per unit volume, so collisions occur more often.
Step 3:The quantity that changes with concentration is therefore the collision frequency.
Final answer: collision frequency
Q110Single correctStates of Matter / Solutions
A mixture of and Ar gases in a cylinder contains 7 g of and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of is :
[Use atomic masses (in g ) : N = 14, Ar = 40]
[Use atomic masses (in g ) : N = 14, Ar = 40]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The partial pressure of a gas in a mixture equals its mole fraction times the total pressure, by Dalton's law.
Step 1:The moles of each gas follow from mass divided by molar mass.
Step 2:The mole fraction of nitrogen is its moles divided by the total moles.
Step 3:Multiplying the mole fraction by the total pressure gives the partial pressure of nitrogen.
Final answer:
Q111Single correctThe d- and f-Block Elements / Metallurgy
Identify the correct statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pig iron can be moulded into a variety of shapes.
Approach:
Each statement on iron and copper metallurgy is checked against standard facts.
Step 1:Wrought iron is the purest form of iron with very low carbon content, so statement (1) is wrong, and blister copper owes its appearance to escaping sulphur dioxide, not carbon dioxide, so statement (2) is wrong.
Step 2:Vapour phase refining of nickel uses the Mond process; the Van Arkel method is applied to titanium and zirconium, so statement (3) is wrong.
Step 3:Pig iron has a high carbon content and can be cast and moulded into various shapes, so statement (4) is correct.
Final answer: Pig iron can be moulded into a variety of shapes.
Q112Single correctGeneral Organic Chemistry
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Hyperconjugation
Approach:
The relative stability of the two carbocations is explained by the electronic effects available to the methyl groups attached to the cationic carbon.
Step 1:A tertiary butyl cation carries three methyl groups on the positive carbon, compared with fewer for the secondary cation.
Step 2:Each C-H bond of the adjacent methyl groups donates electron density into the empty orbital through hyperconjugation, with more such bonds available in the tertiary cation.
Step 3:The larger number of hyperconjugative interactions stabilises the tertiary cation more than the secondary one.
Final answer: Hyperconjugation
Q113Single correctSurface Chemistry
Which of the following is a cationic detergent?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cetyltrimethyl ammonium bromide
Approach:
A cationic detergent carries a positively charged hydrophilic head, so the surfactant with a quaternary ammonium cation is selected.
Step 1:Sodium lauryl sulphate, sodium stearate and sodium dodecylbenzene sulphonate all carry negatively charged heads, making them anionic detergents.
Step 2:Cetyltrimethyl ammonium bromide contains a quaternary ammonium cation as the active surface-active group.
Step 3:Therefore cetyltrimethyl ammonium bromide is the cationic detergent.
Final answer: Cetyltrimethyl ammonium bromide
Q114Single correctHydrocarbons
Elimination reaction of 2-Bromo-pentane to form pent-2-ene is
(a) -Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction
(a) -Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a), (b), (c)
Approach:
Each labelled statement is evaluated against the mechanism of dehydrohalogenation of a secondary alkyl halide.
Step 1:Loss of HBr from 2-bromopentane removes a hydrogen from the carbon adjacent (beta) to the halogen-bearing carbon, so the process is a beta-elimination.
Step 2:Two beta-hydrogens are available; removal toward C-3 gives the more substituted pent-2-ene, which is the major product predicted by Zaitsev's rule.
Step 3:Removal of a hydrogen and a halogen from adjacent carbons is, by definition, a dehydrohalogenation.
Step 4:Only statements (a), (b) and (c) describe this elimination correctly.
Final answer: (a), (b), (c)
Q115Single correctSolutions
The mixture which shows positive deviation from Raoult's law is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ethanol + Acetone
Approach:
Positive deviation arises when intermolecular forces in the mixture are weaker than those in the pure components.
Step 1:Ethanol is strongly hydrogen bonded in the pure state; adding acetone breaks these hydrogen bonds, weakening the average interaction and raising the vapour pressure above the ideal value.
Step 2:Benzene and toluene are structurally similar and form a nearly ideal mixture; chloroethane with bromoethane is also close to ideal.
Step 3:Acetone and chloroform form a hydrogen bond between the components, strengthening interactions and producing negative deviation.
Final answer: Ethanol + Acetone
Q116Single correctCoordination Compounds
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The relative field strength of ligands is set by their position in the spectrochemical series.
Step 1:In the spectrochemical series the ordering of these donors by increasing field strength is thiocyanate (S-bonded), fluoride, oxalate, then cyanide.
Step 2:Cyanide is a strong-field ligand at the high end, while S-bonded thiocyanate is the weakest of the four.
Final answer:
Q117Single correctBiomolecules
Which of the following is a basic amino acid ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Lysine
Approach:
An amino acid is basic when its side chain carries an additional basic group such as an amine.
Step 1:Lysine bears a second amino group on its side chain, giving it a net basic character.
Step 2:Serine and tyrosine carry neutral hydroxyl-containing side chains and alanine carries a neutral methyl group, so none of these is basic.
Final answer: Lysine
Q118Single correctSolutions
HCl was passed through a solution of , and NaCl. Which compound(s) crystallise(s)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Only NaCl
Approach:
Passing HCl gas increases the chloride ion concentration; the salt with the lowest solubility under this common-ion effect crystallises out.
Step 1:Dissolved HCl raises the chloride ion concentration sharply, shifting the dissolution equilibria of the chloride salts.
Step 2:Among the three salts sodium chloride is the least soluble under this high chloride concentration, so it precipitates while the more soluble magnesium and calcium chlorides remain in solution.
Final answer: Only NaCl
Q119Single correctPolymers
Which of the following is a natural polymer?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1cis-1, 4-polyisoprene
Approach:
A natural polymer is one produced by living organisms rather than synthesised industrially.
Step 1:Natural rubber is cis-1,4-polyisoprene, obtained from the latex of rubber trees.
Step 2:Butadiene-styrene copolymer, polybutadiene and butadiene-acrylonitrile copolymer are all synthetic rubbers.
Final answer: cis-1, 4-polyisoprene
Q120Single correctp-Block Elements
Which of the following is not correct about carbon monoxide?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
Approach:
Each statement is tested against the known toxicology and formation of carbon monoxide.
Step 1:Carbon monoxide binds haemoglobin to give carboxyhaemoglobin, which is far more stable than oxyhaemoglobin; the statement calling it less stable is therefore incorrect.
Step 2:Carbon monoxide does form carboxyhaemoglobin, lowers the oxygen-carrying capacity of blood and arises from incomplete combustion, so those statements are correct.
Final answer: The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
Q121Single correctBiomolecules
Sucrose on hydrolysis gives
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3-D-Glucose + -D-Fructose
Approach:
Sucrose is a disaccharide whose hydrolysis releases its two constituent monosaccharide units.
Step 1:Acidic or enzymatic hydrolysis of sucrose cleaves the glycosidic linkage between its glucose and fructose units.
Step 2:The products are alpha-D-glucose and beta-D-fructose.
Final answer: -D-Glucose + -D-Fructose
Q122Single correctThe d- and f-Block Elements
The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with is responsible for the transmission of nerve signals.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Potassium
Approach:
The metal ion described is identified from its enzymatic and nerve-signalling roles alongside sodium.
Step 1:Potassium ions activate numerous enzymes and take part in glucose oxidation that yields ATP.
Step 2:Together with sodium ions, potassium ions govern the transmission of nerve impulses across membranes.
Final answer: Potassium
Q123Single correctSome Basic Concepts of Chemistry
Which one of the followings has maximum number of atoms ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 41 g of Li(s) [Atomic mass of Li = 7]
Approach:
The number of atoms in 1 g of each sample is found from moles times Avogadro's number, accounting for the atomicity of oxygen.
Step 1:For silver 1 g corresponds to 1/108 mol of atoms, and for magnesium 1 g corresponds to 1/24 mol of atoms.
Step 2:For oxygen gas 1 g is 1/32 mol of molecules, giving 2/32 = 1/16 mol of atoms; for lithium 1 g is 1/7 mol of atoms.
Step 3:Comparing the molar amounts of atoms, 1/7 is the largest, so lithium contains the most atoms.
Final answer: 1 g of Li(s) [Atomic mass of Li = 7]
Q124Single correctStructure of Atom
The number of protons, neutrons and electrons in , respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 171, 104 and 71
Approach:
Proton, neutron and electron counts follow from the atomic number and mass number of the nuclide.
Step 1:The atomic number 71 gives 71 protons, and the neutral atom carries an equal number of electrons.
Step 2:Subtracting the atomic number from the mass number gives the neutron count.
Step 3:The ordered values are 71 protons, 104 neutrons and 71 electrons.
Final answer: 71, 104 and 71
Q125Single correctRedox Reactions
What is the change in oxidation number of carbon in the following reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 to
Approach:
The oxidation number of carbon is determined in the reactant methane and in the product carbon tetrachloride.
Step 1:In methane carbon is bonded to four less electronegative hydrogens, giving carbon an oxidation number of minus four.
Step 2:In carbon tetrachloride carbon is bonded to four more electronegative chlorines, giving carbon an oxidation number of plus four.
Step 3:Carbon therefore changes from minus four to plus four.
Final answer: to
Q126Single correctThe d- and f-Block Elements
Identify the incorrect statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The oxidation states of chromium in and are not the same.
Approach:
Each statement is tested against the established chemistry of transition elements.
Step 1:Chromium is in the plus six state in both chromate and dichromate, so the claim that the oxidation states differ is false.
Step 2:The chromous ion is a stronger reducing agent than the ferrous ion, transition metals are catalytic through variable oxidation states and complex formation, and interstitial compounds trap small atoms in metal lattices, so those statements are correct.
Final answer: The oxidation states of chromium in and are not the same.
Q127Single correctThermodynamics
For the reaction, , the correct option is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 and
Approach:
The signs of the enthalpy and entropy changes are deduced from bond formation and the change in the number of gaseous species.
Step 1:Two chlorine atoms combine to form a chlorine molecule; bond formation releases energy, so the reaction enthalpy is negative.
Step 2:Two gaseous atoms are converted into one gaseous molecule, lowering the number of gas particles and hence the disorder, so the entropy change is negative.
Step 3:Both the enthalpy and entropy changes are negative.
Final answer: and
Q128Single correctSurface Chemistry
Measuring Zeta potential is useful in determining which property of colloidal solution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Stability of the colloidal particles
Approach:
The physical meaning of zeta potential is linked to the property it characterises in a colloid.
Step 1:Zeta potential measures the electric potential at the slipping plane of the charged colloidal particles.
Step 2:A larger zeta potential means stronger mutual repulsion between particles, so it serves as a measure of colloidal stability.
Final answer: Stability of the colloidal particles
Q129Single correctp-Block Elements
Urea reacts with water to form A which will decompose to form B. B when passed through (aq), deep blue colour solution C is formed. What is the formula of C from the following ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The reaction sequence is traced from urea through its hydrolysis products to the final copper complex.
Step 1:Urea reacts with water to give ammonium carbamate as A, which decomposes to release ammonia as B.
Step 2:Passing ammonia into aqueous copper ions forms the deep blue tetraamminecopper(II) complex.
Step 3:The deep blue solution C is the tetraamminecopper(II) ion.
Final answer:
Q130Single correctThe s-Block Elements
Match the following and identify the correct option.
| List I | List II |
|---|---|
| (a). | (i). |
| (b). Temporary hardness of water | (ii). An electron deficient hydride |
| (c). | (iii). Synthesis gas |
| (d). | (iv). Non-planar structure |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Approach:
Each species in List I is paired with its correct description in List II.
Step 1:A mixture of carbon monoxide and hydrogen is known as synthesis gas, pairing (a) with (iii).
Step 2:Temporary hardness arises from dissolved magnesium and calcium bicarbonates, pairing (b) with (i).
Step 3:Diborane is an electron deficient hydride, pairing (c) with (ii), while hydrogen peroxide has a non-planar open-book structure, pairing (d) with (iv).
Step 4:The complete matching.
Final answer: (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Q131Single correctThe p-Block Elements
Which is the correct option?
| Oxide | Nature |
|---|---|
| (a). CO | (i). Basic |
| (b). BaO | (ii). Neutral |
| (c). | (iii). Acidic |
| (d). | (iv). Amphoteric |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Approach:
Each oxide is classified by its acid-base character.
Step 1:Carbon monoxide is a neutral oxide, pairing (a) with (ii).
Step 2:Barium oxide is a basic oxide, pairing (b) with (i).
Step 3:Aluminium oxide is amphoteric, pairing (c) with (iv), and dichlorine heptoxide is acidic, pairing (d) with (iii).
Step 4:The complete matching.
Final answer: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Q132Single correctChemical Kinetics
The rate constant for a first order reaction is . The time required to reduce 2.0 g of the reactant to 0.2 g is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3500 s
Approach:
The time is obtained from the integrated first order rate law using the initial and final amounts.
Step 1:The ratio of initial to final reactant amount is ten, whose logarithm is one.
Step 2:Substituting the rate constant and the logarithm into the integrated rate law gives the time.
Final answer: 500 s
Q133Single correctHydrocarbons
An alkene on ozonolysis gives methanal as one of the product. Its structure is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A cyclohexane ring bearing a (allyl) substituent
Approach:
Ozonolysis cleaves the carbon-carbon double bond into two carbonyl fragments; methanal forms only from a terminal CH2= carbon.
Step 1:Methanal arises when the alkene carries a terminal methylene group, since cleavage of the terminal carbon gives formaldehyde.
Step 2:The allyl-substituted cyclohexane carries a terminal group, and ozonolysis cleaves that double bond to give methanal as one product.
Final answer: A cyclohexane ring bearing a (allyl) substituent
Q134Single correctHydrocarbons
Which of the following alkane cannot be made in good yield by Wurtz reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3n-Heptane
Approach:
The Wurtz reaction couples two alkyl halides and gives good yields only for symmetrical alkanes with an even number of carbon atoms.
Step 1:The Wurtz coupling joins two identical alkyl groups, so a symmetrical alkane with an even carbon count is obtained cleanly.
Step 2:n-Heptane has seven carbons, an odd number, and cannot be formed from two equal halves; making it would require a mixture of halides giving a mixture of products in poor yield.
Final answer: n-Heptane
Q135Single correctAlcohols, Phenols and Ethers
Anisole on cleavage with HI gives
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Phenol (benzene ring with )
Approach:
Cleavage of an aryl alkyl ether by hydroiodic acid breaks the alkyl-oxygen bond, giving the alkyl iodide and the phenol.
Step 1:Anisole is methyl phenyl ether; with HI the bond between the methyl carbon and oxygen is cleaved because aryl-oxygen bonds are not broken.
Step 2:Cleavage of the aryl-alkyl ether gives phenol and methyl iodide, since the iodide attacks the methyl carbon and never the aromatic ring.
Final answer: Phenol (benzene ring with )
Biology90 questions
Q1Single correctEcology
Which of the following is not an attribute of a population?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Species interaction
Approach:
Population attributes are statistical properties measured for a group of individuals of the same species, distinguishing them from properties of a single organism.
Step 1:A population possesses characteristics such as birth rate (natality), death rate (mortality), sex ratio, age distribution and population density.
Step 2:Species interaction describes the relationship between two different species within a community and operates at the community level, not the population level.
Final answer: Species interaction
Q2Single correctCell Structure and Function
The process of growth is maximum during
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Log phase
Approach:
Growth in living organisms follows a sigmoid pattern in which the rate of growth changes across distinct phases.
Step 1:The sigmoid growth curve consists of a lag phase, a log (exponential) phase and a stationary phase.
Step 2:During the log phase, cells divide and enlarge at the fastest rate, making growth maximum.
Final answer: Log phase
Q3Single correctMorphology of Flowering Plants
The roots that originate from the base of the stem are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Fibrous roots
Approach:
Roots are classified by their origin into the tap root system and the fibrous root system, which arise from different parts of the plant.
Step 1:In monocotyledonous plants the primary root is short lived and is replaced by a large number of roots that originate from the base of the stem.
Step 2:The primary root and its lateral branches form the tap root system, which develops from the radicle and not from the stem base.
Final answer: Fibrous roots
Q4Single correctHuman Health and Disease
Match the following diseases with the causative organism and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Typhoid | (i). |
| (b). Pneumonia | (ii). |
| (c). Filariasis | (iii). |
| (d). Malaria | (iv). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Each disease is paired with its specific causative organism using standard NCERT associations from Human Health and Disease.
Step 1:Typhoid is a bacterial disease caused by Salmonella typhi, which spreads through contaminated food and water and is detected by the Widal test.
Step 2:Pneumonia of the alveoli is caused by the bacteria Haemophilus influenzae and Streptococcus pneumoniae, filling the alveoli with fluid.
Step 3:Filariasis (elephantiasis) is caused by the filarial worms Wuchereria bancrofti and Wuchereria malayi, transmitted by mosquitoes.
Step 4:Malaria is caused by the protozoan parasite Plasmodium, transmitted through the bite of the female Anopheles mosquito.
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q5Single correctReproductive Health
In which of the following techniques, the embryos are transferred to assist those females who cannot conceive?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1ZIFT and IUT
Approach:
Assisted reproductive technologies are distinguished by whether gametes, zygotes or early embryos are transferred to the female reproductive tract.
Step 1:In Zygote Intra Fallopian Transfer the zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.
Step 2:In Intra Uterine Transfer an embryo with more than 8 blastomeres is transferred into the uterus.
Step 3:GIFT transfers an ovum and ICSI directly injects a sperm into the ovum, so these methods do not transfer embryos.
Final answer: ZIFT and IUT
Q6Single correctPrinciples of Inheritance and Variation
Identify the wrong statement with reference to the gene 'I' that controls ABO blood groups.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3When and are present together, they express same type of sugar.
Approach:
The gene I controlling ABO blood groups shows multiple allelism and codominance, and each statement is judged against these properties.
Step 1:The gene I has three alleles, the I-A, I-B and i, and a diploid individual carries only two of them, so these statements are correct.
Step 2:The I-A allele produces one form of sugar and the I-B allele produces a different form of sugar, and the i allele produces no sugar.
Step 3:When the I-A and I-B alleles are present together they express two different types of sugar, not the same type, so this statement is wrong.
Final answer: When and are present together, they express same type of sugar.
Q7Single correctBiotechnology - Principles and Processes
Choose the correct pair from the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ligases - Join the two DNA molecules
Approach:
Each enzyme is matched with its true function in DNA metabolism and recombinant DNA technology.
Step 1:Ligases catalyse the formation of phosphodiester bonds and join two DNA molecules, so this pair is correct.
Step 2:Polymerases synthesise new DNA strands, nucleases cut the phosphodiester bonds rather than separate strands, and exonucleases remove nucleotides from the ends rather than at specific internal sites.
Final answer: Ligases - Join the two DNA molecules
Q8Single correctPrinciples of Inheritance and Variation
Select the correct match
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Sickle cell anaemia Autosomal recessive trait, chromosome-11
Approach:
Each genetic disorder is matched with its correct mode of inheritance and chromosomal basis.
Step 1:Sickle cell anaemia is an autosomal recessive trait whose gene is located on chromosome 11, so this match is correct.
Step 2:Haemophilia is X linked and not Y linked, phenylketonuria is an autosomal recessive trait and not dominant, and thalassemia is autosomal and not X linked.
Final answer: Sickle cell anaemia Autosomal recessive trait, chromosome-11
Q9Single correctAnimal Kingdom
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Gregarious, polyphagous pest | (i). |
| (b). Adult with radial symmetry and larva with bilateral symmetry | (ii). Scorpion |
| (c). Book lungs | (iii). |
| (d). Bioluminescence | (iv). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Each diagnostic feature is matched with the representative animal that shows it.
Step 1:Locusta is a gregarious polyphagous pest and Asterias shows radial symmetry as an adult with a bilaterally symmetrical larva.
Step 2:Scorpion respires through book lungs and Ctenoplana exhibits bioluminescence.
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q10Single correctHuman Health and Disease
The infectious stage of that enters the human body is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Sporozoites
Approach:
The life cycle of Plasmodium is traced to identify the stage transmitted by the mosquito to the human host.
Step 1:When an infected female Anopheles mosquito bites a person, it injects sporozoites present in its salivary glands into the human blood.
Step 2:Trophozoites develop later inside human liver and red blood cells, and gametocytes are taken up by the mosquito, so these stages do not initiate the human infection.
Final answer: Sporozoites
Q11Single correctBiomolecules
Identify the substances having glycosidic bond and peptide bond, respectively in their structure
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Inulin, insulin
Approach:
Glycosidic bonds link monosaccharide units in polysaccharides, while peptide bonds link amino acids in proteins, and each pair is tested against this.
Step 1:Inulin is a polymer of fructose units joined by glycosidic bonds.
Step 2:Insulin is a protein hormone whose amino acids are joined by peptide bonds.
Step 3:Cholesterol and lecithin are lipids and glycerol is an alcohol, so the other pairs fail to satisfy both bond requirements.
Final answer: Inulin, insulin
Q12Single correctSexual Reproduction in Flowering Plants
The plant parts which consist of two generations - one within the other
(a) Pollen grains inside the anther
(b) Germinated pollen grain with two male gametes
(c) Seed inside the fruit
(d) Embryo sac inside the ovule
(a) Pollen grains inside the anther
(b) Germinated pollen grain with two male gametes
(c) Seed inside the fruit
(d) Embryo sac inside the ovule
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) and (d)
Approach:
A structure shows two generations one within the other when a haploid gametophyte develops inside the diploid sporophytic tissue.
Step 1:Pollen grains are the male gametophyte generation lying inside the diploid anther wall, representing two generations one within the other.
Step 2:The embryo sac is the female gametophyte generation present inside the diploid ovule, again showing two generations one within the other.
Step 3:A germinated pollen grain with male gametes and a seed inside a fruit do not represent one gametophyte enclosed within sporophytic tissue in this sense.
Final answer: (a) and (d)
Q13Single correctMineral Nutrition
The product(s) of reaction catalyzed by nitrogenase in root nodules of leguminous plants is/are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ammonia and hydrogen
Approach:
The nitrogenase reaction is written out to identify all products formed during biological nitrogen fixation.
Step 1:Nitrogenase reduces dinitrogen to ammonia, and the same reaction also evolves hydrogen gas as a product.
Step 2:Nitrate is produced by nitrifying bacteria, not by nitrogenase, and oxygen inactivates the enzyme, so neither nitrate nor oxygen can be a nitrogenase product.
Final answer: Ammonia and hydrogen
Q14Single correctCell Cycle and Cell Division
Identify the correct statement with regard to phase (Gap 1) of interphase.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cell is metabolically active, grows but does not replicate its DNA.
Approach:
The events of each phase of the cell cycle are recalled to characterise the G1 phase.
Step 1:During the G1 phase the cell is metabolically active and grows continuously but does not replicate its DNA.
Step 2:DNA replication occurs in the S phase and nuclear division occurs in the M phase, so the other statements describe different phases.
Final answer: Cell is metabolically active, grows but does not replicate its DNA.
Q15Single correctStructural Organisation in Animals
Cuboidal epithelium with brush border of microvilli is found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Proximal convoluted tubule of nephron
Approach:
The location of cuboidal epithelium bearing a brush border of microvilli is identified from its role in absorption.
Step 1:The proximal convoluted tubule of the nephron is lined by cuboidal epithelium whose microvilli form a brush border that increases the surface area for reabsorption.
Step 2:The intestinal lining is columnar, while the salivary gland ducts and Eustachian tube lack this characteristic cuboidal brush border.
Final answer: Proximal convoluted tubule of nephron
Q16Single correctCell - The Unit of Life
Which of the following statements about inclusion bodies is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2These are involved in ingestion of food particles
Approach:
The properties of cytoplasmic inclusion bodies are recalled to find the statement that does not apply to them.
Step 1:Inclusion bodies are non membrane bound structures that lie free in the cytoplasm and represent reserve materials such as glycogen, oil and protein.
Step 2:Inclusion bodies store reserve substances and are not involved in the ingestion of food particles, so this statement is incorrect.
Final answer: These are involved in ingestion of food particles
Q17Single correctCell - The Unit of Life
Which is the important site of formation of glycoproteins and glycolipids in eukaryotic cells?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Golgi bodies
Approach:
The cell organelle responsible for synthesising glycoproteins and glycolipids is identified from its function in glycosylation.
Step 1:Golgi bodies carry out glycosylation, attaching sugar residues to proteins and lipids to form glycoproteins and glycolipids.
Step 2:The endoplasmic reticulum, peroxisomes and polysomes perform other functions and are not the principal site of this synthesis.
Final answer: Golgi bodies
Q18Single correctBiotechnology - Principles and Processes
In gel electrophoresis, separated DNA fragments can be visualized with the help of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ethidium bromide in UV radiation
Approach:
The staining agent and the radiation needed to view DNA fragments after gel electrophoresis are identified.
Step 1:Separated DNA fragments are stained with ethidium bromide, which intercalates into the DNA.
Step 2:Exposure to ultraviolet radiation makes the ethidium bromide bound DNA fluoresce as bright orange bands.
Final answer: Ethidium bromide in UV radiation
Q19Single correctBreathing and Exchange of Gases
Identify the wrong statement with reference to transport of oxygen
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Higher conc. in alveoli favours the formation of oxyhaemoglobin
Approach:
The factors governing the binding of oxygen to haemoglobin are examined to find the statement that contradicts the oxygen dissociation behaviour.
Step 1:Oxygen binding depends mainly on the partial pressure of oxygen, while a high partial pressure of carbon dioxide and a low partial pressure of carbon dioxide influence binding as stated.
Step 2:A high hydrogen ion concentration lowers the affinity of haemoglobin for oxygen and promotes the dissociation rather than the formation of oxyhaemoglobin, so this statement is wrong.
Final answer: Higher conc. in alveoli favours the formation of oxyhaemoglobin
Q20Single correctMorphology of Flowering Plants
Ray florets have
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Inferior ovary
Approach:
The position of the ovary in ray florets of the family Asteraceae is determined from the arrangement of floral parts.
Step 1:In ray florets the calyx, corolla and stamens arise above the ovary, making the flower epigynous.
Step 2:In an epigynous flower the ovary lies below the other floral whorls and is therefore an inferior ovary.
Final answer: Inferior ovary
Q21Single correctBiotechnology - Principles and Processes
The specific palindromic sequence which is recognized by EcoRI is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The recognition site of the restriction enzyme EcoRI is recalled and confirmed to be a palindrome.
Step 1:EcoRI recognises the six base pair sequence GAATTC, whose complementary strand read in the same five prime to three prime direction is also GAATTC, confirming a palindrome.
Step 2:GGAACC, CTTAAG read 5' to 3' and GGATCC are either not palindromic in the EcoRI sense or are the recognition sequences of other restriction enzymes.
Final answer:
Q22Single correctBiotechnology - Principles and Processes
Identify the wrong statement with regard to Restriction Enzymes.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Sticky ends can be joined by using DNA ligases.
Approach:
Each statement about restriction enzymes is checked against the recognition and cutting mechanism described in the NCERT text.
Step 1:A restriction enzyme inspects the length of a DNA sequence and binds when it finds its specific recognition sequence, and it cuts DNA at palindromic sites and is widely used in genetic engineering.
Step 2:Sticky ends produced by restriction enzymes are joined by DNA ligases, which is a true property of the technique.
Final answer: Sticky ends can be joined by using DNA ligases.
Q23Single correctMicrobes in Human Welfare
Which of the following is put into Anaerobic sludge digester for further sewage treatment?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Activated sludge
Approach:
The sequence of secondary sewage treatment is traced to identify the material fed into the anaerobic sludge digester.
Step 1:During secondary treatment a part of the activated sludge is used as inoculum, and the remaining major part is pumped into large anaerobic sludge digesters.
Step 2:Inside the digester anaerobic bacteria digest the bacteria and fungi of the sludge and produce a mixture of biogas.
Final answer: Activated sludge
Q24Single correctBreathing and Exchange of Gases
Select the correct events that occur during inspiration.
(a) Contraction of diaphragm
(b) Contraction of external inter-costal muscles
(c) Pulmonary volume decreases
(d) Intra pulmonary pressure increases
(a) Contraction of diaphragm
(b) Contraction of external inter-costal muscles
(c) Pulmonary volume decreases
(d) Intra pulmonary pressure increases
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a) and (b)
Approach:
Each of the four printed events is tested against what happens to the thoracic cavity during inspiration.
Step 1:Contraction of the diaphragm flattens it and increases the antero-posterior volume of the thoracic chamber, so event (a) occurs during inspiration.
Step 2:Contraction of the external inter-costal muscles lifts the ribs and sternum, increasing the dorso-ventral volume, so event (b) also occurs during inspiration.
Step 3:The increase in thoracic volume enlarges the pulmonary volume and therefore lowers the intra-pulmonary pressure below atmospheric pressure, which is what draws air in; events (c) and (d) describe the opposite changes and belong to expiration.
Step 4:The events that occur during inspiration are therefore (a) and (b).
Final answer: (a) and (b)
Q25Single correctStructural Organisation in Animals
If the head of cockroach is removed, it may live for few days because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.
Approach:
The cockroach has a segmentally arranged nervous system in which most of the nervous tissue lies along the ventral side of the body, so removal of the head leaves much of it functional.
Step 1:The cockroach nervous system consists of paired ganglia connected by a double ventral nerve cord running along the ventral surface, with only a fraction of the nervous tissue contained in the head.
Step 2:Because the head holds only a small portion of the nervous system, decapitation does not immediately stop the segmental ganglia from controlling body functions, allowing survival for a few days.
Final answer: the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.
Q26Single correctAnimal Kingdom
Which of the following statements are true for the phylum-Chordata?
(a) In Urochordata notochord extends from head to tail and it is present throughout their life.
(b) In Vertebrata notochord is present during the embryonic period only.
(c) Central nervous system is dorsal and hollow.
(d) Chordata is divided into 3 subphyla : Hemichordata, Tunicata and Cephalochordata.
(a) In Urochordata notochord extends from head to tail and it is present throughout their life.
(b) In Vertebrata notochord is present during the embryonic period only.
(c) Central nervous system is dorsal and hollow.
(d) Chordata is divided into 3 subphyla : Hemichordata, Tunicata and Cephalochordata.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(b) and (c)
Approach:
Each statement is tested against the defining features of phylum Chordata to identify the two that are correct.
Step 1:In Urochordata the notochord is present only in the larval tail and not throughout life, so statement (a) is false; in Vertebrata the notochord is replaced by a vertebral column and persists only in the embryonic stage, so statement (b) is true.
Step 2:A dorsal, hollow central nervous system is a fundamental chordate character, so statement (c) is true; Chordata is divided into three subphyla Urochordata, Cephalochordata and Vertebrata, not the set listed, so statement (d) is false.
Final answer: (b) and (c)
Q27Single correctBiotechnology: Principles and Processes
Match the organism with its use in biotechnology.
Select the correct option from the following:
Select the correct option from the following:
| Column-I | Column-II |
|---|---|
| (a). | (i). Cloning vector |
| (b). | (ii). Construction of first rDNA molecule |
| (c). | (iii). DNA polymerase |
| (d). | (iv). Cry proteins |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Approach:
Each microorganism is paired with the biotechnological tool or role it provides, following NCERT Biotechnology: Principles and Processes.
Step 1:Bacillus thuringiensis produces insecticidal Cry proteins (Bt toxin), whose genes are introduced into crops to confer pest resistance.
Step 2:Thermus aquaticus supplies the thermostable Taq DNA polymerase used to synthesise DNA strands during the polymerase chain reaction.
Step 3:Agrobacterium tumefaciens carries the Ti plasmid that is disarmed and used as a cloning vector to deliver genes into plant cells.
Step 4:The antibiotic resistance gene of Salmonella typhimurium was linked to the native plasmid of Escherichia coli by Cohen and Boyer to construct the first recombinant DNA molecule.
Final answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Q28Single correctMineral Nutrition
Select the correct option.
| List - I | List - II |
|---|---|
| (a). Iron | (i). Photolysis of water |
| (b). Zinc | (ii). Pollen germination |
| (c). Boron | (iii). Required for chlorophyll biosynthesis |
| (d). Manganese | (iv). IAA biosynthesis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Approach:
Each essential mineral element is paired with its characteristic physiological role in the plant.
Step 1:Iron is required for the formation of chlorophyll, so (a) matches (iii); zinc is needed for the biosynthesis of the auxin IAA, so (b) matches (iv).
Step 2:Boron is essential for pollen germination and pollen tube growth, so (c) matches (ii); manganese activates the splitting of water during the light reaction, so (d) matches (i) photolysis of water.
Final answer: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Q29Single correctAnatomy of Flowering Plants
Identify the incorrect statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Sapwood is the innermost secondary xylem and is lighter in colour
Approach:
Each statement about heartwood and sapwood is checked against the structure of secondary xylem to find the false one.
Step 1:Heartwood is the dead, non-conducting central wood that provides mechanical support and is dark due to deposition of tannins, resins and oils, making statements 1 and 4 correct; sapwood is the outer functional wood that conducts water and minerals, making statement 2 correct.
Step 2:Sapwood is the peripheral (outer), not the innermost, region of secondary xylem and is lighter in colour, so the claim that it is the innermost secondary xylem is false.
Final answer: Sapwood is the innermost secondary xylem and is lighter in colour
Q30Single correctBiomolecules
Match the following.
Choose the correct option from the following.
Choose the correct option from the following.
| List - I | List - II |
|---|---|
| (a). Inhibitor of catalytic activity | (i). Ricin |
| (b). Possess peptide bonds | (ii). Malonate |
| (c). Cell wall material in fungi | (iii). Chitin |
| (d). Secondary metabolite | (iv). Collagen |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
Approach:
Each biomolecule is matched to its defining biochemical role or location.
Step 1:Malonate is a competitive inhibitor of succinate dehydrogenase, so (a) matches (ii); collagen is a protein built from peptide bonds, so (b) matches (iv).
Step 2:Chitin forms the fungal cell wall, so (c) matches (iii); ricin is a toxic secondary metabolite obtained from castor, so (d) matches (i).
Final answer: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
Q31Single correctHuman Reproduction
Meiotic division of the secondary oocyte is completed
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4At the time of fusion of a sperm with an ovum
Approach:
The timing of completion of the second meiotic division of the secondary oocyte during oogenesis is identified.
Step 1:The secondary oocyte released at ovulation is arrested in metaphase of the second meiotic division and does not complete it before fertilisation.
Step 2:The second meiotic division is completed only after a sperm enters the secondary oocyte, producing the ovum and the second polar body, so completion coincides with fusion of sperm with the ovum.
Final answer: At the time of fusion of a sperm with an ovum
Q32Single correctBiodiversity and Conservation
According to Robert May, the global species diversity is about
(1)
(2)
(3)
(4)
SolutionAnswer: Option 47 million
Approach:
The estimate of total global species diversity attributed to Robert May is recalled.
Step 1:About 1.5 million species have actually been described and recorded by taxonomists, which is a count of known species rather than the total estimate.
Step 2:Robert May's conservative statistical estimate places the global species diversity at about 7 million.
Final answer: 7 million
Q33Single correctMolecular Basis of Inheritance
The first phase of translation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Aminoacylation of tRNA
Approach:
The earliest molecular event in the process of translation is identified.
Step 1:Translation begins with the charging of transfer RNA, in which an amino acid is attached to its specific tRNA in an energy-dependent step called aminoacylation or amino acid activation.
Step 2:Binding of mRNA to the ribosome and recognition of anti-codons occur after the tRNAs are charged, so those events are subsequent steps.
Final answer: Aminoacylation of tRNA
Q34Single correctBiodiversity and Conservation
Which of the following regions of the globe exhibits highest species diversity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Amazon forests
Approach:
Species diversity increases towards the equator, so the most equatorial tropical rainforest is identified as the richest region.
Step 1:Tropical regions near the equator support the greatest species richness because of stable climate, abundant solar energy and long evolutionary time.
Step 2:Among the listed regions the Amazonian tropical rainforest holds the greatest biological diversity on Earth, exceeding the Western Ghats, Madagascar and the Himalayas.
Final answer: Amazon forests
Q35Single correctBiotechnology and its Applications
Which of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The functional insulin has A and B chains linked together by hydrogen bonds.
Approach:
Each statement about insulin synthesis and structure is verified to identify the false one.
Step 1:In humans insulin is first made as the precursor proinsulin that carries an extra stretch called the C-peptide, and genetically engineered human insulin (humulin) is produced in Escherichia coli, making statements 1, 2 and 4 correct.
Step 2:The mature functional insulin has its A and B chains held together by disulphide bonds, not hydrogen bonds, so statement 3 is incorrect.
Final answer: The functional insulin has A and B chains linked together by hydrogen bonds.
Q36Single correctAnatomy of Flowering Plants
The transverse section of a plant shows following anatomical features :
(a) Large number of scattered vascular bundles surrounded by bundle sheath
(b) Large conspicuous parenchymatous ground tissue
(c) Vascular bundles conjoint and closed
(d) Phloem parenchyma absent
Identify the category of plant and its part :
(a) Large number of scattered vascular bundles surrounded by bundle sheath
(b) Large conspicuous parenchymatous ground tissue
(c) Vascular bundles conjoint and closed
(d) Phloem parenchyma absent
Identify the category of plant and its part :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Monocotyledonous stem
Approach:
The listed anatomical features are matched to the characteristic transverse-section pattern of a particular plant organ.
Step 1:Numerous scattered vascular bundles each surrounded by a sclerenchymatous bundle sheath, with conjoint and closed bundles and absent phloem parenchyma, are diagnostic of a monocot stem.
Step 2:Conspicuous parenchymatous ground tissue without differentiation into cortex and pith further supports the monocot stem, distinguishing it from roots, which have radial bundles.
Final answer: Monocotyledonous stem
Q37Single correctAnimal Kingdom
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). 6-15 pairs of gill slits | (i). Trygon |
| (b). Heterocercal caudal fin | (ii). Cyclostomes |
| (c). Air Bladder | (iii). Chondrichthyes |
| (d). Poison sting | (iv). Osteichthyes |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each chordate feature is matched with the taxon or organism that characteristically possesses it.
Step 1:Cyclostomes bear 6 to 15 pairs of gill slits, so (a) matches (ii); the cartilaginous fishes (Chondrichthyes) have a heterocercal caudal fin, so (b) matches (iii).
Step 2:The air bladder for buoyancy is a feature of the bony fishes (Osteichthyes), so (c) matches (iv); the sting ray Trygon carries a poison sting, so (d) matches (i).
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q38Single correctEvolution
From his experiments, S.L. Miller produced amino acids by mixing the following in a closed flask
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1, , and water vapor at
Approach:
The composition and conditions of the Miller-Urey experiment that yielded amino acids are recalled.
Step 1:Stanley Miller created a closed apparatus containing the reducing-atmosphere gases methane, hydrogen and ammonia together with water vapour.
Step 2:Electric sparks were passed at a temperature of about 800 degrees Celsius, simulating lightning, which led to the formation of amino acids.
Final answer: , , and water vapor at
Q39Single correctEvolution
Embryological support for evolution was disapproved by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Karl Ernst von Baer
Approach:
The scientist who rejected the use of embryological similarity as proof of evolution is identified.
Step 1:Embryological support for evolution, based on Ernst Haeckel's proposal that embryos repeat adult ancestral stages, was disproved by Karl Ernst von Baer.
Step 2:Von Baer showed that embryos never pass through the adult stages of other animals, so the recapitulation interpretation was invalid.
Final answer: Karl Ernst von Baer
Q40Single correctTransport in Plants
The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and in early morning is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Root pressure
Approach:
The driving force behind the appearance of liquid water at leaf tips during the night is identified.
Step 1:Loss of liquid water from the tips or margins of leaves is called guttation, and it occurs through specialised pores called hydathodes.
Step 2:At night, when transpiration is minimal, the positive hydrostatic pressure developed in the xylem by active absorption at the roots, that is root pressure, forces water out as droplets.
Final answer: Root pressure
Q41Single correctBiomolecules
Secondary metabolites such as nicotine, strychnine and caffeine are produced by plants for their
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Defence action
Approach:
The biological function of plant alkaloids such as nicotine, strychnine and caffeine is identified.
Step 1:Nicotine, strychnine and caffeine are nitrogen-containing alkaloids classed as secondary metabolites that are not involved in primary growth or nutrition.
Step 2:These toxic alkaloids deter herbivores and pathogens, so they serve a defensive role for the plant.
Final answer: Defence action
Q42Single correctPhotosynthesis in Higher Plants
The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 21 molecule of 3-C compound
Approach:
The products of the oxygenase activity of RuBisCO acting on RuBP during photorespiration are determined.
Step 1:When RuBisCO binds oxygen instead of carbon dioxide, the five-carbon substrate RuBP is split into one molecule of the three-carbon 3-phosphoglycerate and one molecule of the two-carbon 2-phosphoglycolate.
Step 2:In terms of a 3-carbon compound the oxygenation yields only one molecule of the 3-carbon phosphoglycerate, distinguishing it from carboxylation, which gives two such molecules.
Final answer: 1 molecule of 3-C compound
Q43Single correctBiotechnology and its Applications
Bt cotton variety that was developed by the introduction of toxin gene of Bacillus thuringiensis (Bt) is resistant to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Insect pests
Approach:
The target of resistance conferred by the Bt toxin gene introduced into cotton is identified.
Step 1:The cry gene from Bacillus thuringiensis encodes Cry proteins that form insecticidal crystalline toxins.
Step 2:When expressed in cotton, the Cry toxin is lethal to chewing insect larvae such as the cotton bollworm, making the plant resistant to insect pests.
Final answer: Insect pests
Q44Single correctEvolution
Which of the following refer to correct example(s) of organisms which have evolved due to changes in environment brought about by anthropogenic action?
(a) Darwin's Finches of Galapagos islands.
(b) Herbicide resistant weeds.
(c) Drug resistant eukaryotes.
(d) Man-created breeds of domesticated animals like dogs.
(a) Darwin's Finches of Galapagos islands.
(b) Herbicide resistant weeds.
(c) Drug resistant eukaryotes.
(d) Man-created breeds of domesticated animals like dogs.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(b), (c) and (d)
Approach:
Each example is tested for whether the evolutionary change was driven by human-induced (anthropogenic) environmental change.
Step 1:Darwin's finches diversified through natural adaptive radiation on the Galapagos islands without human influence, so statement (a) is not an example of anthropogenic action.
Step 2:Herbicide-resistant weeds, drug-resistant eukaryotes and the many breeds of domesticated dogs all arose because of human activities such as herbicide use, drug use and artificial selection, so (b), (c) and (d) are correct.
Final answer: (b), (c) and (d)
Q45Single correctHuman Health and Disease
Identify the wrong statement with reference to immunity.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Active immunity is quick and gives full response.
Approach:
Each statement about active and passive immunity is checked against their definitions and properties to find the wrong one.
Step 1:Active immunity develops when exposure to a living or dead antigen induces the host to make its own antibodies, while passive immunity results from directly receiving ready-made antibodies, as when antibodies pass from mother to foetus, so statements 1, 2 and 4 are correct.
Step 2:Active immunity is slow because it takes time for the host to mount an antibody response, so describing it as quick is wrong.
Final answer: Active immunity is quick and gives full response.
Q46Single correctStrategies for Enhancement in Food Production
By which method was a new breed 'Hisardale' of sheep formed by using Bikaneri ewes and Marino rams?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cross breeding
Approach:
Crossing animals of two different breeds to combine desirable traits defines the breeding method that produced Hisardale.
Step 1:Hisardale is a hybrid sheep breed developed by mating Bikaneri ewes with Marino rams, two distinct breeds.
Step 2:Mating between two different breeds is termed cross breeding, which combines desirable qualities of both parents.
Final answer: Cross breeding
Q47Single correctDigestion and Absorption
Identify the correct statement with reference to human digestive system.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ileum is a highly coiled part
Approach:
Each statement is checked against the known anatomy of the human alimentary canal.
Step 1:The ileum is the highly coiled distal part of the small intestine, so the statement describing it as highly coiled is correct.
Step 2:The ileum is itself a part of the small intestine and opens into the large intestine, not into the small intestine; serosa is the outermost layer; the vermiform appendix arises from the caecum.
Final answer: Ileum is a highly coiled part
Q48Single correctMicrobes in Human Welfare
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). | (i). Cyclosporin-A |
| (b). | (ii). Butyric Acid |
| (c). | (iii). Citric Acid |
| (d). | (iv). Blood cholesterol lowering agent |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Approach:
Each microbe is paired with the metabolic product it commercially yields.
Step 1:Clostridium butylicum is a fermenting bacterium that produces butyric acid, giving (a)-(ii).
Step 2:Trichoderma polysporum yields the immunosuppressant cyclosporin-A, giving (b)-(i); Monascus purpureus is a yeast producing a blood-cholesterol-lowering statin, giving (c)-(iv); Aspergillus niger produces citric acid, giving (d)-(iii).
Final answer: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Q49Single correctExcretory Products and their Elimination
Presence of which of the following conditions in urine are indicative of Diabetes Mellitus?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ketonuria and Glycosuria
Approach:
The urinary indicators specific to Diabetes Mellitus are identified.
Step 1:In Diabetes Mellitus, raised blood glucose causes glucose to appear in urine, a condition called glycosuria.
Step 2:Impaired glucose utilisation increases fat breakdown, producing ketone bodies that are excreted in urine as ketonuria; together glycosuria and ketonuria signal the disorder.
Final answer: Ketonuria and Glycosuria
Q50Single correctBiological Classification
Floridean starch has structure similar to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Amylopectin and glycogen
Approach:
The reserve food of red algae is compared structurally with known polysaccharides.
Step 1:Floridean starch is the stored carbohydrate of Rhodophyceae (red algae).
Step 2:Its branched-chain structure resembles that of amylopectin and glycogen.
Final answer: Amylopectin and glycogen
Q51Single correctHuman Health and Disease
Select the option including all sexually transmitted diseases.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gonorrhoea, Syphilis, Genital herpes
Approach:
Each listed set of diseases is checked to see whether every disease in it is transmitted by sexual contact.
Step 1:Gonorrhoea, syphilis and genital herpes are all spread by sexual contact.
Step 2:Malaria and filaria are transmitted by mosquito vectors and cancer is non-communicable, so any set containing them is not an all-STD set.
Final answer: Gonorrhoea, Syphilis, Genital herpes
Q52Single correctCell Cycle and Cell Division
Match the following with respect to meiosis.
Select the correct option from the following.
Select the correct option from the following.
| Column-I | Column-II |
|---|---|
| (a). Zygotene | (i). Terminalization |
| (b). Pachytene | (ii). Chiasmata |
| (c). Diplotene | (iii). Crossing over |
| (d). Diakinesis | (iv). Synapsis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Approach:
Each sub-stage of meiotic prophase I is paired with the event that characterises it.
Step 1:Synapsis, the pairing of homologous chromosomes, begins at zygotene, giving (a)-(iv).
Step 2:Crossing over occurs at pachytene, giving (b)-(iii); chiasmata become visible at diplotene, giving (c)-(ii); terminalization of chiasmata occurs at diakinesis, giving (d)-(i).
Final answer: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Q53Single correctBiological Classification
Which of the following pairs is of unicellular algae?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 and
Approach:
The pair in which both genera are single-celled forms is identified.
Step 1:Chlorella is a unicellular green alga and Spirulina is a unicellular (filamentous single-celled) cyanobacterium, both used as protein-rich food.
Step 2:Laminaria and Sargassum are large multicellular brown algae, Gelidium and Gracilaria are multicellular red algae, and Volvox is a colonial form, so those pairs are excluded.
Final answer: and
Q54Single correctHuman Reproduction
Which of the following hormone levels will cause release of ovum (ovulation) from the graffian follicle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1High concentration of Estrogen
Approach:
The hormonal condition that triggers ovulation from the Graafian follicle is determined.
Step 1:During the follicular phase the growing Graafian follicle secretes a high concentration of estrogen.
Step 2:The peak estrogen level induces the LH surge that causes rupture of the follicle and release of the ovum.
Final answer: High concentration of Estrogen
Q55Single correctBiotechnology and its Applications
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Bt cotton | (i). Gene therapy |
| (b). Adenosine deaminase deficiency | (ii). Cellular defence |
| (c). RNAi | (iii). Detection of HIV infection |
| (d). PCR | (iv). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Each biotechnology term is matched with the concept it is associated with.
Step 1:Bt cotton carries genes from Bacillus thuringiensis, giving (a)-(iv); adenosine deaminase deficiency is treated by gene therapy, giving (b)-(i).
Step 2:RNA interference acts as a cellular defence mechanism, giving (c)-(ii); the polymerase chain reaction is used to detect HIV infection, giving (d)-(iii).
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q56Single correctEnvironmental Issues
Montreal protocol was signed in 1987 for control of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Emission of ozone depleting substances
Approach:
The objective of the Montreal Protocol is recalled.
Step 1:The Montreal Protocol of 1987 aimed to limit the release of substances that deplete the stratospheric ozone layer.
Step 2:Greenhouse gas emission is addressed by the Kyoto Protocol, while transboundary movement of genetically modified organisms and e-waste fall under other agreements.
Final answer: Emission of ozone depleting substances
Q57Single correctBiological Classification
Which of the following is correct about viroids?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2They have free RNA without protein coat
Approach:
The molecular nature of viroids is recalled.
Step 1:A viroid is an infectious agent smaller than a virus, discovered by T. O. Diener.
Step 2:It consists of a free, low-molecular-weight RNA that lacks the protein coat present in viruses.
Final answer: They have free RNA without protein coat
Q58Single correctMorphology of Flowering Plants
The ovary is half inferior in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Plum
Approach:
The plant with a half-inferior ovary (perigynous flower) is identified.
Step 1:A half-inferior ovary occurs in a perigynous flower, where the gynoecium is situated on a rim with other floral parts at its margin.
Step 2:Among the given plants, plum (a rose family member) shows the perigynous arrangement with a half-inferior ovary.
Final answer: Plum
Q59Single correctDigestion and Absorption
The enzyme enterokinase helps in conversion of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2trypsinogen into trypsin
Approach:
The substrate and product of enterokinase action are recalled.
Step 1:Enterokinase is secreted by the intestinal mucosa and acts on the inactive pancreatic zymogen trypsinogen.
Step 2:It converts trypsinogen into the active enzyme trypsin, which then activates other pancreatic zymogens.
Final answer: trypsinogen into trypsin
Q60Single correctEcosystem
Match the trophic levels with their correct species examples in grassland ecosystem.
Select the correct option.
Select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Fourth trophic level | (i). Crow |
| (b). Second trophic level | (ii). Vulture |
| (c). First trophic level | (iii). Rabbit |
| (d). Third trophic level | (iv). Grass |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each trophic level in a grassland food chain is assigned its representative organism.
Step 1:Grass is the producer at the first trophic level, giving (c)-(iv); the rabbit is the primary consumer at the second trophic level, giving (b)-(iii).
Step 2:The crow as a secondary consumer occupies the third trophic level, giving (d)-(i); the vulture as a top consumer occupies the fourth trophic level, giving (a)-(ii).
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q61Single correctPrinciples of Inheritance and Variation
How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in one character with contrasting traits?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 314
Approach:
The number of true-breeding pea varieties chosen by Mendel is recalled.
Step 1:Mendel studied seven pairs of contrasting characters in the garden pea, each pair differing in a single character.
Step 2:Each pair consists of two true-breeding varieties, so seven pairs amount to fourteen varieties.
Final answer: 14
Q62Single correctNeural Control and Coordination
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Organ of Corti | (i). Connects middle ear and pharynx |
| (b). Cochlea | (ii). Coiled part of the labyrinth |
| (c). Eustachian tube | (iii). Attached to the oval window |
| (d). Stapes | (iv). Located on the basilar membrane |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
Approach:
Each part of the ear is matched with its correct anatomical description.
Step 1:The organ of Corti rests on the basilar membrane of the cochlea, giving (a)-(iv); the cochlea is the coiled portion of the inner-ear labyrinth, giving (b)-(ii).
Step 2:The Eustachian tube connects the middle ear cavity with the pharynx, giving (c)-(i); the stapes is the ossicle attached to the oval window, giving (d)-(iii).
Final answer: (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
Q63Single correctSexual Reproduction in Flowering Plants
In water hyacinth and water lily, pollination takes place by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Insects or wind
Approach:
The pollinating agents of water hyacinth and water lily are recalled.
Step 1:Although these are aquatic plants, their flowers emerge above the water surface and are exposed to air.
Step 2:Such emergent flowers are pollinated by insects or by wind, as in most land plants.
Final answer: Insects or wind
Q64Single correctPlant Growth and Development
Name the plant growth regulator which upon spraying on sugarcane crop, increases the length of stem, thus increasing the yield of sugarcane crop.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gibberellin
Approach:
The growth regulator that lengthens sugarcane stems is identified.
Step 1:Gibberellins promote internodal elongation of the stem.
Step 2:Spraying gibberellin on sugarcane increases the length of the stem and thereby the cane yield.
Final answer: Gibberellin
Q65Single correctPhotosynthesis in Higher Plants
In light reaction, plastoquinone facilitates the transfer of electrons from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1PS-II to Cytf complex
Approach:
The position of plastoquinone in the photosynthetic electron transport chain is recalled.
Step 1:In the light reaction, electrons leave excited photosystem II and pass to the mobile carrier plastoquinone.
Step 2:Plastoquinone then delivers these electrons to the cytochrome b6f complex, pumping protons across the membrane.
Final answer: PS-II to Cytf complex
Q66Single correctPlant Growth and Development
Which of the following is not an inhibitory substance governing seed dormancy?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gibberellic acid
Approach:
The substance that does not promote seed dormancy is identified.
Step 1:Abscisic acid, phenolic acids and para-ascorbic acid are inhibitory substances that maintain seed dormancy.
Step 2:Gibberellic acid breaks dormancy and promotes germination, so it is not an inhibitory substance.
Final answer: Gibberellic acid
Q67Single correctMolecular Basis of Inheritance
Name the enzyme that facilitates opening of DNA helix during transcription.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4RNA polymerase
Approach:
The enzyme responsible for unwinding DNA during transcription is identified.
Step 1:Transcription is catalysed by RNA polymerase, which binds the promoter and moves along the template.
Step 2:RNA polymerase itself unwinds and opens the DNA helix as it synthesises RNA, so a separate helicase is not required during transcription.
Final answer: RNA polymerase
Q68Single correctExcretory Products and their Elimination
Which of the following would help in prevention of diuresis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Reabsorption of Na and water from renal tubules due to aldosterone
Approach:
Diuresis means increased urine output; the process that reduces it by conserving water is identified.
Step 1:Aldosterone promotes reabsorption of sodium ions and water from the renal tubules, reducing the volume of urine formed.
Step 2:Reduced urine volume opposes diuresis, so aldosterone-driven reabsorption helps prevent it.
Final answer: Reabsorption of Na and water from renal tubules due to aldosterone
Q69Single correctEcosystem
In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gross primary productivity is always more than net primary productivity
Approach:
Gross primary productivity (GPP) is the total rate of photosynthetic organic matter production, while net primary productivity (NPP) is the biomass remaining after subtracting respiratory losses. Comparing the two definitions fixes their relationship.
Step 1:Gross primary productivity is the rate of production of organic matter during photosynthesis by producers.
Step 2:A fraction of GPP is consumed by producers in respiration, and the remainder is net primary productivity available to consumers.
Step 3:Since respiratory loss is positive, GPP exceeds NPP for any living producer.
Final answer: Gross primary productivity is always more than net primary productivity
Q70Single correctHuman Reproduction
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Placenta | (i). Androgens |
| (b). Zona pellucida | (ii). Human Chorionic Gonadotropin (hCG) |
| (c). Bulbo-urethral glands | (iii). Layer of the ovum |
| (d). Leydig cells | (iv). Lubrication of the Penis |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each structure is matched to its specific secretion or function. The placenta secretes hCG, the zona pellucida is a layer of the ovum, bulbo-urethral glands lubricate the penis, and Leydig cells secrete androgens.
Step 1:The placenta acts as an endocrine tissue and secretes human Chorionic Gonadotropin.
Step 2:The zona pellucida is the transparent glycoprotein layer surrounding the ovum.
Step 3:Bulbo-urethral glands secrete a fluid that lubricates the penis.
Step 4:Leydig cells of the testis synthesise and secrete androgens.
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q71Single correctPlant Kingdom
Strobili or cones are found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Equisetum
Approach:
Strobili are compact aggregations of sporophylls bearing sporangia. Among the listed plants, the one that produces distinct strobili is identified.
Step 1:A strobilus is a cone-like cluster of sporophylls that carries sporangia.
Step 2:Equisetum, a pteridophyte, bears terminal strobili composed of sporangiophores.
Step 3:Salvinia and Pteris bear sori, and Marchantia is a liverwort lacking strobili.
Final answer: Equisetum
Q72Single correctCell Cycle and Cell Division
Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (). This process occurs at the end of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 phase
Approach:
The point at which non-dividing cells leave the cell cycle to enter the quiescent G0 stage is identified from the phases of interphase described in NCERT Cell Cycle and Cell Division.
Step 1:Interphase is divided into the G1 phase, the S phase of DNA replication, and the G2 phase that precedes mitosis.
Step 2:Cells that do not divide further leave the cycle at the end of the G1 phase, before DNA replication begins, and pass into the quiescent G0 stage while remaining metabolically active.
Step 3:The quiescent G0 cells retain the ability to re-enter the cycle at G1 when stimulated to divide again.
Final answer: phase
Q73Single correctEvolution
Flippers of Penguins and Dolphins are examples of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Convergent evolution
Approach:
Flippers in penguins (birds) and dolphins (mammals) perform the same function but arise from different ancestral lineages, defining the type of evolutionary pattern.
Step 1:Penguins and dolphins belong to unrelated classes yet both possess flippers adapted to swimming.
Step 2:When different lineages independently evolve similar features under similar selection pressures, the process is convergent evolution.
Final answer: Convergent evolution
Q74Single correctMolecular Basis of Inheritance
If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is bp, then the length of the DNA is approximately
(1)
(2)
(3)
(4)
SolutionAnswer: Option 32.2 meters
Approach:
The total length of the DNA equals the number of base pairs multiplied by the rise per base pair. The product is converted from nanometres to metres.
Step 1:The length is the product of the number of base pairs and the distance per base pair.
Step 2:Converting nanometres to metres uses the factor .
Step 3:Expressed to two significant figures the total length is about 2.2 metres.
Final answer: 2.2 meters
Q75Single correctBody Fluids and Circulation
The QRS complex in a standard ECG represents
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Depolarisation of ventricles
Approach:
Each wave of an electrocardiogram corresponds to a specific electrical event in the cardiac cycle. The QRS complex is identified with one such event.
Step 1:The P wave represents atrial depolarisation and the T wave represents ventricular repolarisation.
Step 2:The QRS complex marks the depolarisation of the ventricles, which precedes their contraction.
Final answer: Depolarisation of ventricles
Q76Single correctBody Fluids and Circulation
Match the following columns and select the correct option.
| Column - I | Column - II |
|---|---|
| (a). Eosinophils | (i). Immune response |
| (b). Basophils | (ii). Phagocytosis |
| (c). Neutrophils | (iii). Release histaminase, destructive enzymes |
| (d). Lymphocytes | (iv). Release granules containing histamine |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Approach:
Each leucocyte type is paired with its characteristic function. Eosinophils release histaminase, basophils release histamine, neutrophils carry out phagocytosis, and lymphocytes mediate the immune response.
Step 1:Eosinophils resist infections and release histaminase and destructive enzymes during allergic reactions.
Step 2:Basophils release granules containing histamine, serotonin and heparin.
Step 3:Neutrophils are the most abundant phagocytic cells that engulf pathogens.
Step 4:Lymphocytes are responsible for the body's immune response.
Final answer: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Q77Single correctMolecular Basis of Inheritance
Which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Adenine pairs with thymine through two H-bonds
Approach:
Complementary base pairing in DNA is governed by a fixed number of hydrogen bonds. The pairing between adenine and thymine is identified by counting these bonds.
Step 1:Adenine forms a complementary pair with thymine in double-stranded DNA.
Step 2:The adenine-thymine pair is held by two hydrogen bonds, whereas guanine-cytosine is held by three.
Final answer: Adenine pairs with thymine through two H-bonds
Q78Single correctBiotechnology - Principles and Processes
The sequence that controls the copy number of the linked DNA in the vector, is termed
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ori site
Approach:
A cloning vector contains specific sequences with defined roles. The element governing how many copies of the linked DNA are produced is identified.
Step 1:The origin of replication is the sequence where replication of the vector begins.
Step 2:The number of copies of the linked DNA depends on the origin of replication, so the ori site controls copy number.
Final answer: Ori site
Q79Single correctBiomolecules
Identify the basic amino acid from the following.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Lysine
Approach:
Amino acids are classified as acidic, basic or neutral based on the nature of their side chain. The one bearing an additional amino group in its side chain is basic.
Step 1:Glutamic acid carries an extra carboxyl group, making it acidic, while tyrosine and valine are neutral.
Step 2:Lysine possesses an additional amino group in its side chain, which makes it a basic amino acid.
Final answer: Lysine
Q80Single correctChemical Coordination and Integration
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Pituitary gland | (i). Grave's disease |
| (b). Thyroid gland | (ii). Diabetes mellitus |
| (c). Adrenal gland | (iii). Diabetes insipidus |
| (d). Pancreas | (iv). Addison's disease |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Each endocrine gland is linked to a disorder arising from its dysfunction. Pituitary maps to diabetes insipidus, thyroid to Grave's disease, adrenal to Addison's disease, and pancreas to diabetes mellitus.
Step 1:Deficiency of antidiuretic hormone from the pituitary causes diabetes insipidus.
Step 2:Over-activity of the thyroid gland produces Grave's disease.
Step 3:Hypofunction of the adrenal cortex causes Addison's disease.
Step 4:Inadequate insulin secretion by the pancreas results in diabetes mellitus.
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q81Single correctChemical Coordination and Integration
Select the correct statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Glucocorticoids stimulate gluconeogenesis.
Approach:
Each statement concerns hormonal control of blood glucose. The roles of glucocorticoids, glucagon and insulin are checked to find the accurate statement.
Step 1:Glucocorticoids promote the formation of glucose from non-carbohydrate precursors, that is gluconeogenesis.
Step 2:Glucagon raises blood glucose and is linked with hyperglycemia, and insulin lowers blood glucose and is linked with hypoglycemia, so the other statements are incorrect.
Final answer: Glucocorticoids stimulate gluconeogenesis.
Q82Single correctStructural Organisation in Animals
Which one of the following is the most abundant protein in the animals?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Collagen
Approach:
The most abundant protein in the animal body is a structural protein of connective tissue; each listed protein is weighed on overall abundance.
Step 1:Collagen is the principal structural protein of skin, bone, tendon and other connective tissues throughout the body.
Step 2:Haemoglobin, lectin and insulin are restricted to specific cells or functions and are far less abundant than collagen.
Final answer: Collagen
Q83Single correctPrinciples of Inheritance and Variation
Experimental verification of the chromosomal theory of inheritance was done by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Morgan
Approach:
The chromosomal theory was proposed by Sutton and Boveri, but its experimental proof came from work on a model organism. The scientist responsible for that verification is identified.
Step 1:Sutton and Boveri proposed that chromosomes carry the hereditary factors, framing the chromosomal theory of inheritance.
Step 2:Thomas Hunt Morgan experimentally verified the theory through breeding experiments on Drosophila.
Final answer: Morgan
Q84Single correctLocomotion and Movement
Match the following columns and select the correct option.
| Column-I | Column-II |
|---|---|
| (a). Floating Ribs | (i). Located between second and seventh ribs |
| (b). Acromion | (ii). Head of the Humerus |
| (c). Scapula | (iii). Clavicle |
| (d). Glenoid cavity | (iv). Do not connect with the sternum |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Approach:
Each skeletal feature is matched with its correct anatomical description. Floating ribs do not reach the sternum, the acromion articulates with the clavicle, the scapula lies over the ribs, and the glenoid cavity holds the head of the humerus.
Step 1:Floating ribs are the last two pairs that remain unattached to the sternum.
Step 2:The acromion process of the scapula articulates with the clavicle.
Step 3:The scapula is a triangular bone situated dorsally between the second and seventh ribs.
Step 4:The glenoid cavity receives the head of the humerus to form the shoulder joint.
Final answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Q85Single correctRespiration in Plants
The number of substrate level phosphorylations in one turn of citric acid cycle is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2One
Approach:
Substrate level phosphorylation in the citric acid cycle produces a high-energy phosphate compound at a specific step. The number of such events per turn is counted.
Step 1:In the citric acid cycle, the conversion of succinyl-CoA to succinate generates one molecule of GTP (or ATP) by substrate level phosphorylation.
Step 2:No other step of a single turn produces a high-energy phosphate directly, so the count is one.
Final answer: One
Q86Single correctCell Cycle and Cell Division
Dissolution of the synaptonemal complex occurs during
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Diplotene
Approach:
The synaptonemal complex forms and breaks down at defined substages of prophase I. The substage marking its dissolution is identified.
Step 1:The synaptonemal complex assembles during zygotene as homologous chromosomes pair.
Step 2:The synaptonemal complex breaks down during diplotene, when homologues begin to separate and chiasmata become visible.
Final answer: Diplotene
Q87Single correctAnimal Kingdom
Bilaterally symmetrical and acoelomate animals are exemplified by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Platyhelminthes
Approach:
The required phylum must combine bilateral symmetry with the absence of a body cavity, and each listed phylum is checked for both features.
Step 1:Ctenophora shows radial symmetry, Aschelminthes are pseudocoelomate, and Annelida are coelomate.
Step 2:Platyhelminthes are bilaterally symmetrical and lack a coelom, making them acoelomate.
Final answer: Platyhelminthes
Q88Single correctSexual Reproduction in Flowering Plants
The body of the ovule is fused within the funicle at
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hilum
Approach:
The ovule is attached to the placenta through the funicle, and a particular region marks where the funicle joins the ovule body. That junction is identified.
Step 1:The funicle is the stalk that connects the ovule to the placenta of the ovary.
Step 2:The point where the body of the ovule is fused with the funicle is the hilum.
Final answer: Hilum
Q89Single correctStructural Organisation in Animals
Goblet cells of alimentary canal are modified from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Columnar epithelial cells
Approach:
Goblet cells are mucus-secreting cells of the gut lining. Their cell of origin among the epithelial types lining the alimentary canal is identified.
Step 1:The lining of the alimentary canal is largely composed of columnar epithelial cells.
Step 2:Some of these columnar cells become modified into mucus-secreting goblet cells.
Final answer: Columnar epithelial cells
Q90Single correctHuman Health and Disease
Snow-blindness in Antarctic region is due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Inflammation of cornea due to high dose of UV-B radiation
Approach:
Snow-blindness is a radiation injury linked to ozone depletion over the Antarctic. The specific cause is identified among the listed effects.
Step 1:Depletion of the ozone layer over Antarctica increases the dose of UV-B radiation reaching the surface.
Step 2:High doses of UV-B cause inflammation of the cornea, a condition known as snow-blindness.
Final answer: Inflammation of cornea due to high dose of UV-B radiation
Frequently Asked Questions
How many questions are in the NEET 2020 Sep 13 paper?
The NEET 2020 Sep 13 paper has 180 questions — Physics (45), Chemistry (45) and Biology (90). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for NEET?
NEET awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Biology across 180 questions for 720 marks.
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