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NEET 2021 Sep 12 Question Paper with Solutions
All 200 questions from the NEET 2021 (Sep 12) paper — Physics (50), Chemistry (50) and Biology (100) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2021Chemistry PYQs 2021Biology PYQs 2021
- Questions
- 200
- Physics
- 50
- Chemistry
- 50
- Biology
- 100
Physics50 questions
Q1Single correctElectrostatic Potential and Capacitance
Two charged spherical conductors of radius and are connected by a wire. Then the ratio of surface charge densities of the spheres is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
When two conductors are joined by a wire they reach a common potential. Equating the potentials and using surface charge density gives the required ratio.
Step 1:Connecting the spheres by a conducting wire brings them to the same potential.
Step 2:Writing each charge in terms of its surface charge density yields the relation between densities.
Step 3:Rearranging gives the ratio of the surface charge densities.
Final answer:
Q2Single correctMechanical Properties of Fluids
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is , then the viscous force acting on the ball will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
At terminal velocity the net force on the ball is zero, so the viscous force equals the weight minus the buoyant force.
Step 1:The mass of the ball relates to its volume and density.
Step 2:The buoyant force equals the weight of the displaced glycerine of density .
Step 3:At constant velocity the viscous force balances the difference between weight and buoyancy.
Final answer:
Q3Single correctMoving Charges and Magnetism
An infinitely long straight conductor carries a current of as shown. An electron is moving with a speed of parallel to the conductor. The perpendicular distance between the electron and the conductor is at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The straight conductor sets up a magnetic field at the electron's location, and the moving electron experiences a magnetic Lorentz force. The field magnitude follows from the long-straight-wire formula, and the force follows from the charge, speed and field with the velocity perpendicular to the field.
Step 1:The magnetic field at perpendicular distance r from a long straight wire carrying current I is computed.
Step 2:The electron travels parallel to the wire, so its velocity is perpendicular to the magnetic field and the sine factor equals one.
Step 3:Substituting the electron charge magnitude, speed and field gives the force magnitude.
Final answer:
Q4Single correctElectric Charges and Fields
A dipole is placed in an electric field as shown. In which direction will it move?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Towards the right as its potential energy will decrease.
Approach:
A dipole tends to move so that its potential energy decreases. The orientation between the dipole moment and the field fixes the sign of the energy and the direction of motion toward the stronger field.
Step 1:The angle between the dipole moment and the field is for the given orientation.
Step 2:Moving into the region of stronger field where the orientation reduces this energy lowers the potential energy.
Step 3:The net force directs the dipole toward the right, where the field is stronger and the energy falls.
Final answer: Towards the right as its potential energy will decrease.
Q5Single correctRay Optics and Optical Instruments
A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4All of the above
Approach:
The merits of a large-aperture objective combine improved light gathering, finer resolution and clearer images, so each listed statement is correct.
Step 1:A larger aperture admits more light, raising the brightness and visibility of the image.
Step 2:The resolving power increases in proportion to the objective diameter, so a larger aperture separates closer objects.
Step 3:Greater brightness and resolution together improve the overall quality of the image, so every statement holds.
Final answer: All of the above
Q6Single correctUnits and Measurements
A screw gauge gives the following readings when used to measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on the main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on the main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 40.052 cm
Approach:
The least count of the screw gauge is the pitch divided by the number of circular-scale divisions, and the diameter combines the main-scale and circular-scale readings.
Step 1:With a pitch of 1 mm and 100 circular divisions, the least count follows directly.
Step 2:The diameter equals the main-scale reading plus the product of circular-scale reading and least count.
Final answer: 0.052 cm
Q7Single correctNuclei
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4216 MeV
Approach:
The total binding energy of each species equals its binding energy per nucleon times its mass number, and the gain is the difference between products and reactant.
Step 1:The total binding energy of the two product fragments uses 8.5 MeV per nucleon over a combined mass number of 240.
Step 2:The total binding energy of the parent nucleus uses 7.6 MeV per nucleon over mass number 240.
Step 3:The gain in binding energy is the difference between the product and reactant totals.
Final answer: 216 MeV
Q8Single correctMotion in a Straight Line
A small block slides down on a smooth inclined plane, starting from rest at time . Let be the distance travelled by the block in the interval to . Then, the ratio is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The block undergoes uniformly accelerated motion from rest, so the distance covered in the th second follows the standard relation, and the required ratio is formed for the th and th second.
Step 1:For motion from rest with acceleration , the distance in the nth second is proportional to .
Step 2:The distance in the th second is proportional to .
Step 3:Dividing the two distances gives the ratio.
Final answer:
Q9Single correctRay Optics and Optical Instruments
Find the value of the angle of emergence from the prism. Refractive index of the glass is .
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The ray refracts at the second face of the prism. Applying Snell's law at that face with the internal angle of incidence and the refractive index of glass gives the angle of emergence.
Step 1:At the emergent face the internal angle is , and Snell's law relates it to the emergent angle through the index .
Step 2:Solving for the sine of the emergent angle gives .
Step 3:The emergent angle therefore corresponds to .
Final answer:
Q10Single correctOscillations
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 40.628 s
Approach:
The spring constant follows from the given force and extension, and the time period of the suspended mass uses the spring-mass relation.
Step 1:The spring constant is the ratio of the applied force to the extension.
Step 2:The time period uses the suspended mass of 2 kg and the spring constant.
Step 3:Evaluating the expression gives the time period.
Final answer: 0.628 s
Q11Single correctRay Optics and Optical Instruments
A convex lens 'A' of focal length 20 cm and a concave lens 'B' of focal length 5 cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 215
Approach:
A parallel incident beam converges toward the focus of the convex lens. For the beam to leave the concave lens parallel, that converging point must coincide with the focus of the concave lens, fixing the separation.
Step 1:The parallel beam refracts through the convex lens and converges toward its focus at 20 cm.
Step 2:For the concave lens to render the beam parallel again, this converging point must lie at its virtual focus at 5 cm beyond the lens.
Final answer: 15
Q12Single correctCurrent Electricity
Column-I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.
| Column-I | Column-II |
|---|---|
| (A). Drift Velocity | (P). |
| (B). Electrical Resistivity | (Q). |
| (C). Relaxation Period | (R). |
| (D). Current Density | (S). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(A) - (R), (B) - (S), (C) - (P), (D) - (Q)
Approach:
Each electrical quantity is matched to its defining expression for conduction in a metallic conductor.
Step 1:Drift velocity equals the acceleration times the relaxation time, matching (R).
Step 2:Electrical resistivity equals the ratio of field to current density, matching (S).
Step 3:Relaxation period equals , matching (P).
Step 4:Current density equals , matching (Q).
Final answer: (A) - (R), (B) - (S), (C) - (P), (D) - (Q)
Q13Single correctSemiconductor Electronics
The electron concentration in an -type semiconductor is the same as hole concentration in a -type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Current in -type > current in -type
Approach:
With equal carrier concentrations, the conduction current depends on the carrier mobility. Comparing electron and hole mobilities fixes which current is larger.
Step 1:The current in each sample depends on the carrier concentration, charge, area and drift velocity, which itself depends on mobility.
Step 2:With identical concentrations, the ratio of the two currents reduces to the ratio of electron to hole mobilities.
Step 3:Electron mobility exceeds hole mobility, so the -type current is larger.
Final answer: Current in -type > current in -type
Q14Single correctNuclei
A radioactive nucleus undergoes spontaneous decay in the sequence , where Z is the atomic number of element X. The possible decay particles in the sequence are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Each decay step changes the atomic number in a characteristic way. Tracking the change in at every stage identifies the emitted particle.
Step 1:The first step lowers the atomic number from Z to , which corresponds to decay.
Step 2:The second step lowers the atomic number by two, from to , which is decay.
Step 3:The third step raises the atomic number from to , which corresponds to decay.
Final answer:
Q15Single correctWork, Energy and Power
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 28.1 kW
Approach:
The incident power equals the rate of fall of gravitational potential energy. Removing the 10% frictional loss gives the power delivered to the turbine.
Step 1:The incident power equals the product of , the height and the mass flow rate.
Step 2:With 10% of the input lost to friction, the generated power is 90% of the incident power.
Step 3:Expressing the result in kilowatts gives the generated power.
Final answer: 8.1 kW
Q16Single correctDual Nature of Radiation and Matter
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of watt will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The power of the source equals the number of photons emitted per second times the energy of each photon. Solving for that number gives the photon emission rate.
Step 1:The photon emission rate equals the power divided by the energy of a single photon.
Step 2:Substituting the power, wavelength, Planck constant and speed of light gives the rate.
Final answer:
Q17Single correctElectromagnetic Waves
For a plane electromagnetic wave propagating in -direction, which one of the following combination gives the correct possible directions for electric field and magnetic field respectively?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a plane electromagnetic wave the direction of propagation lies along . The pair whose cross product points along is the correct combination.
Step 1:The cross product of the fields must point along the -axis for propagation in that direction.
Step 2:Evaluating the cross product for the second combination gives a vector along .
Step 3:Every other combination gives a cross product that is either zero or not directed along , so only with is admissible.
Final answer:
Q18Single correctCurrent Electricity
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 . What will be the effective resistance if they are connected in series?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 44
Approach:
Four identical wires each have the same resistance . The parallel value fixes , and the series value is then four times .
Step 1:For four identical resistors in parallel the equivalent resistance is , which equals 0.25 .
Step 2:The same four resistors in series add directly to give four times .
Final answer: 4
Q19Single correctCurrent Electricity
In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 160 cm
Approach:
In a potentiometer the EMF balanced is proportional to the balancing length. The ratio of EMFs equals the ratio of lengths, which yields the new balance length.
Step 1:The balancing length is proportional to the EMF of the cell being measured.
Step 2:Substituting the two EMFs and the first length gives the new balance length.
Final answer: 60 cm
Q20Single correctElectrostatic Potential and Capacitance
Polar molecules are the molecules
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Having a permanent electric dipole moment
Approach:
Polar molecules are defined by the separation of their positive and negative charge centres, which gives them a fixed dipole moment.
Step 1:In polar molecules the centre of positive charges does not coincide with the centre of negative charges.
Step 2:This permanent separation gives the molecule a fixed electric dipole moment even without any external field.
Final answer: Having a permanent electric dipole moment
Q21Single correctElectrostatic Potential and Capacitance
A parallel plate capacitor has a uniform electric field 'E' in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The energy density of the field multiplied by the volume between the plates gives the total energy stored in the capacitor.
Step 1:The energy stored per unit volume of a uniform field equals .
Step 2:Integrating the constant energy density over the volume gives the total stored energy.
Final answer:
Q22Single correctElectrostatic Potential and Capacitance
The equivalent capacitance of the combination shown in the figure is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Points at the same potential allow the middle capacitor to be removed, after which the remaining capacitors form a simple parallel combination.
Step 1:Points 1, 2 and 3 are at the same potential, so the capacitor connected directly across them is short-circuited and stores no charge.
Step 2:The circuit redraws with the two remaining capacitors connected in parallel between the terminals.
Final answer:
Q23Single correctSemiconductor Electronics
Consider the following statements (A) and (B) and identify the correct answer.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of - junction lies between 0.1 V to 0.3 V.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of - junction lies between 0.1 V to 0.3 V.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A) is correct and (B) is incorrect.
Approach:
Each statement is checked against the standard behaviour of a zener diode and the typical potential barrier of a silicon junction.
Step 1:In reverse bias beyond the breakdown voltage, the zener voltage across the diode remains constant, so a zener diode used as a voltage regulator operates in reverse bias, making statement (A) correct.
Step 2:The potential barrier of a silicon - junction diode is about 0.7 V, so statement (B) giving 0.1 V to 0.3 V is incorrect.
Final answer: (A) is correct and (B) is incorrect.
Q24Single correctGravitation
A particle is released from height from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The particle falls under constant acceleration over a short range near the surface. Energy conservation relates the kinetic and potential energies, and the condition that kinetic energy is three times the potential energy fixes the height and speed.
Step 1:Falling a distance from rest gives the speed at height above the ground.
Step 2:The condition that kinetic energy equals three times the potential energy at height fixes the height.
Step 3:Substituting this height into the speed relation gives the speed at that instant.
Final answer:
Q25Single correctUnits and Measurements
If E and G respectively denote energy and gravitational constant, then has the dimensions of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The dimensions of the ratio follow by dividing the dimensional formula of energy by that of the gravitational constant.
Step 1:Energy has dimensions and the gravitational constant has dimensions .
Step 2:Dividing the dimensions of energy by those of the gravitational constant gives the dimensions of the ratio.
Final answer:
Q26Single correctKinetic Theory of Gases
Match Column - I and Column - II and choose the correct match from the given choices.
| Column - I | Column - II |
|---|---|
| (A). Root mean square speed of gas molecules | (P). |
| (B). Pressure exerted by ideal gas | (Q). |
| (C). Average kinetic energy of a molecule | (R). |
| (D). Total internal energy of 1 mole of a diatomic gas | (S). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Each kinetic-theory quantity is matched with its standard expression from the molecular model of an ideal gas.
Step 1:The root mean square speed of gas molecules is given by the square root of three times the gas constant times temperature divided by molar mass, so (A) pairs with (Q).
Step 2:The pressure exerted by an ideal gas equals one third of the number density times molecular mass times mean square speed, so (B) pairs with (P).
Step 3:The average kinetic energy of a single molecule is three halves of the Boltzmann constant times temperature, so (C) pairs with (S).
Step 4:A diatomic gas has five degrees of freedom, so the internal energy of one mole equals five halves of the gas constant times temperature, so (D) pairs with (R).
Final answer:
Q27Single correctOscillations
A body is executing simple harmonic motion with frequency '', the frequency of its potential energy is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The potential energy of a particle in simple harmonic motion depends on the square of the displacement, which doubles the oscillation frequency.
Step 1:The displacement of a particle executing simple harmonic motion of frequency n varies sinusoidally with time.
Step 2:Substituting the displacement into the potential energy expression gives a sine-squared dependence, which can be written using a double-angle term.
Step 3:The sine-squared term oscillates at twice the angular frequency, so the potential energy completes two cycles for every one cycle of displacement.
Final answer:
Q28Single correctThermal Properties of Matter
A cup of coffee cools from to in t minutes, when the room temperature is . The time taken by a similar cup of coffee to cool from to at a room temperature same at is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Newton's law of cooling relates the rate of temperature drop to the difference between the average body temperature and the surroundings.
Step 1:Applying the law to the first cooling from ninety to eighty degrees gives one relation for the constant K.
Step 2:Applying the same law to the second cooling from eighty to sixty degrees over time t-one gives a second relation.
Step 3:Substituting the value of K from the first relation into the second relation yields the required time.
Final answer:
Q29Single correctElectromagnetic Waves
A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by
The displacement current between the plates of the capacitor, would then be given by
The displacement current between the plates of the capacitor, would then be given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The displacement current equals the rate of change of charge on the capacitor, found by differentiating the charge expression.
Step 1:The charge on the capacitor at any instant is the capacitance times the applied alternating voltage.
Step 2:Differentiating the charge with respect to time gives the displacement current.
Final answer:
Q30Single correctNuclei
The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The remaining activity after a given time is found by raising one half to the number of half-lives elapsed.
Step 1:The number of half-lives elapsed in 150 hours with a half-life of 100 hours is the ratio of the two times.
Step 2:The fraction of activity remaining is one half raised to the power three halves.
Final answer:
Q31Single correctAlternating Current
An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is A. The impedance of the circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The source voltage is the phasor sum of the resistive and net reactive voltages, and impedance is the ratio of rms voltage to rms current.
Step 1:The resultant source voltage is the phasor combination of the resistor voltage and the difference of the inductor and capacitor voltages.
Step 2:The rms current is the amplitude of the current divided by root two.
Step 3:The impedance is the rms source voltage divided by the rms current.
Final answer:
Q32Single correctMoving Charges and Magnetism
A thick current carrying cable of radius '' carries current '' uniformly distributed across its cross-section. The variation of magnetic field due to the cable with the distance '' from the axis of the cable is represented by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(graph) magnetic field rises linearly with r inside the cable up to r = R then falls hyperbolically as 1/r outside
Approach:
Ampere's circuital law gives a magnetic field that grows linearly inside the conductor and falls inversely with distance outside it.
Step 1:For points inside the cable, Ampere's law gives a field directly proportional to the distance from the axis.
Step 2:For points outside the cable, the enclosed current is constant and the field varies inversely with distance.
Step 3:The field therefore rises along a straight line up to the surface and then decreases along a hyperbola, matching the third graph.
Final answer: (graph) linear rise to r = R then 1/r decay
Q33Single correctDual Nature of Radiation and Matter
An electromagnetic wave of wavelength '' is incident on a photosensitive surface of negligible work function. If 'm' mass of photoelectron emitted from the surface has de-Broglie wavelength , then :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
With negligible work function the entire photon energy becomes electron kinetic energy, which links the incident wavelength to the de Broglie wavelength.
Step 1:With the work function taken as negligible, the photon energy equals the maximum kinetic energy of the photoelectron.
Step 2:The de Broglie wavelength is Planck's constant divided by the momentum, so the momentum is expressed in terms of the de Broglie wavelength.
Step 3:Squaring the relation gives the incident wavelength in terms of the de Broglie wavelength.
Final answer:
Q34Single correctGravitation
The escape velocity from the Earth's surface is . The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Expressing escape velocity in terms of density and radius shows it is directly proportional to radius for equal densities.
Step 1:Writing the mass in terms of density and radius converts the escape velocity into a function of density and radius.
Step 2:For equal densities the escape velocity scales linearly with radius, so a planet of four times the radius has four times the escape velocity.
Final answer:
Q35Single correctUnits and Measurements
If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Energy is written as a product of powers of force, acceleration and time, and the exponents are found by comparing dimensions on both sides.
Step 1:Energy is assumed to depend on powers of force, acceleration and time, and each is replaced by its dimensions in mass, length and time.
Step 2:Comparing powers of mass gives a equal to one.
Step 3:Comparing powers of length gives a plus b equal to two, so b equals one.
Step 4:Comparing powers of time gives minus two a minus two b plus c equal to minus two, and substituting a and b gives c equal to two.
Final answer:
Q36Single correctElectromagnetic Induction
Two conducting circular loops of radii and are placed in the same plane with their centres coinciding. If , the mutual inductance M between them will be directly proportional to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Mutual inductance is found by passing a current through the large loop, evaluating the magnetic field it produces over the region of the small loop, and computing the flux linked with the small loop per unit current. Because the small loop lies near the centre of the large one, the central field of the large loop is taken as uniform across the small loop.
Step 1:A current I in the large loop of radius produces a magnetic field at its centre.
Step 2:Since is much larger than , this central field is essentially uniform over the small loop, so the flux through the small loop is the field times its area.
Step 3:Dividing the flux linkage by the current gives the mutual inductance.
Final answer:
Q37Single correctRay Optics and Optical Instruments
A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 420 cm from the plane mirror, it would be a virtual image
Approach:
The lens forms an image that acts as object for the plane mirror, and the mirror produces an equally distant virtual image behind it.
Step 1:Applying the lens formula with object distance sixty centimetres and focal length thirty centimetres gives a real image sixty centimetres from the lens.
Step 2:The mirror is forty centimetres from the lens, so the lens image lies twenty centimetres behind the mirror and serves as a virtual object for the mirror.
Step 3:A plane mirror forms an image at an equal distance on its other side, so the final image is twenty centimetres in front of the mirror and is virtual.
Final answer: 20 cm from the plane mirror, it would be a virtual image
Q38Single correctCurrent Electricity
Three resistors having resistances , and are connected as shown in the given circuit. The ratio of currents in terms of resistances used in the circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The two parallel resistors carry the same potential difference, and Kirchhoff's current law splits the incoming current between them.
Step 1:Resistors r-two and r-three are in parallel, so the potential difference across them is equal, relating their branch currents.
Step 2:Applying the junction rule, the total current splits into the two parallel branch currents.
Step 3:Taking the ratio of the branch current to the total current gives the required result.
Final answer:
Q39Single correctSemiconductor Electronics
For the given circuit, the input digital signals are applied at the terminals , and . What would be the output at the terminal ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Output stays at 5 V throughout the interval to
Approach:
The gate combination is reduced to a Boolean expression and evaluated across each time interval to obtain the output waveform.
Step 1:The combination of logic gates reduces to the sum of the AND of A and B with the AND of the complements of B and C.
Step 2:Evaluating the expression across each successive time interval gives a high output wherever either product term is true.
Step 3:The truth table gives for every one of the six time intervals, so the output holds a constant high level of 5 V.
Final answer: Output stays at 5 V throughout the interval to
Q40Single correctAlternating Current
A series LCR circuit containing 5.0 H inductor, 80 F capacitor and 40 resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at resonant angular frequency are likely to be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 346 rad/s and 54 rad/s
Approach:
The half-power frequencies are located symmetrically about the resonant frequency at intervals of the resistance divided by twice the inductance.
Step 1:The resonant angular frequency is the reciprocal of the square root of the product of inductance and capacitance.
Step 2:The lower half-power frequency is the resonant frequency minus the resistance over twice the inductance.
Step 3:The upper half-power frequency is the resonant frequency plus the resistance over twice the inductance.
Final answer: 46 rad/s and 54 rad/s
Q41Single correctMotion in a Plane
A car starts from rest and accelerates at 5 m/. At s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at s?
(Take m/)
(Take m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 420 m/s, 10 m/
Approach:
The ball keeps the car's horizontal velocity at release and gains vertical velocity under gravity, while its acceleration after release is g.
Step 1:The horizontal velocity of the ball at release equals the car's velocity after four seconds of acceleration from rest.
Step 2:After release the ball falls freely for two seconds, gaining vertical velocity under gravity.
Step 3:The resultant speed is the magnitude of the horizontal and vertical velocity components.
Step 4:Once released, the only acceleration acting on the ball is gravitational, of magnitude ten metres per second squared.
Final answer: 20 m/s, 10 m/
Q42Single correctGravitation
A particle of mass 'm' is projected with a velocity () from the surface of the earth. ( escape velocity) The maximum height above the surface reached by the particle is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Conservation of mechanical energy between the surface and the highest point gives the maximum height in terms of the escape-velocity fraction.
Step 1:The particle is launched at k times the escape velocity, so its speed is less than escape velocity since k is less than one.
Step 2:Equating total mechanical energy at the surface and at the maximum height and substituting the escape velocity relation gives a relation for the height.
Step 3:Solving for the height yields the maximum height in terms of the radius and the fraction k.
Final answer:
Q43Single correctMotion in a Plane
A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle '' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, , is then given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The speed of circular motion sets the projection speed, and the maximum-height formula equated to four times the radius gives the angle.
Step 1:The uniform speed equals the circumference divided by the period of revolution.
Step 2:The projectile launched at this speed attains a maximum height equal to four times the radius.
Step 3:Substituting the circular speed and solving for the sine of the angle gives the projection angle.
Final answer:
Q44Single correctMoving Charges and Magnetism
For and and
What will be the complete expression for ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The magnetic force expression is expanded as a cross product and the components are compared to solve for the field components.
Step 1:Expanding the cross product of velocity and field as a determinant yields the three force components.
Step 2:Comparing the i and k components gives two relations that fix the field components.
Step 3:Solving the relations gives B equal to minus six and B-zero equal to minus eight, completing the field vector.
Final answer:
Q45Single correctLaws of Motion
A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is ( m/) nearly :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 24.2 kg m/s
Approach:
Impulse equals the change in linear momentum, with the ball arriving and leaving at the same speed in opposite directions.
Step 1:The speed of the ball just before and just after impact follows from free fall through ten metres.
Step 2:The impulse is the difference between the upward final momentum and the downward initial momentum.
Step 3:Evaluating the magnitude gives the impulse imparted to the ball.
Final answer: 4.2 kg m/s
Q46Single correctMoving Charges and Magnetism
A uniform conducting wire of length and resistance '' is wound up as a current carrying coil in the shape of,
(i) an equilateral triangle of side ''.
(ii) a square of side ''.
The magnetic dipole moments of the coil in each case respectively are :
(i) an equilateral triangle of side ''.
(ii) a square of side ''.
The magnetic dipole moments of the coil in each case respectively are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
The magnetic moment of each coil is the product of the number of turns, the current and the area of one loop.
Step 1:A length of twelve a wound into an equilateral triangle of side a forms four turns, and the moment is the number of turns times current times triangle area.
Step 2:The same length wound into a square of side a forms three turns, and the moment is the number of turns times current times square area.
Final answer: and
Q47Single correctElectrostatic Potential and Capacitance
Twenty seven drops of same size are charged at 200 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 41980 V
Approach:
Conservation of volume relates the radii of the small and big drops, and conservation of charge then gives the potential of the big drop.
Step 1:Conservation of volume for twenty-seven small drops merging into one big drop gives the big-drop radius as three times the small-drop radius.
Step 2:The total charge of the big drop is twenty-seven times the charge of a small drop.
Step 3:The potential of the big drop is the total charge over the big radius, which equals nine times the small-drop potential.
Final answer: 1980 V
Q48Single correctSystem of Particles and Rotational Motion
A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass 'm' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. ( m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 kg
Approach:
Taking torques about the wedge support and setting their sum to zero gives the unknown suspended mass.
Step 1:Taking moments about the wedge at the forty centimetre mark, the two kilogram mass at the twenty centimetre mark provides an anticlockwise torque of four newton metres.
Step 2:The rod's weight acts at its centre at the hundred centimetre mark, sixty centimetres from the wedge, giving a clockwise torque of three newton metres.
Step 3:The unknown mass at the one hundred sixty centimetre mark, one hundred twenty centimetres from the wedge, gives a clockwise torque of twelve m newton metres.
Step 4:Setting the net torque about the wedge to zero and solving gives the unknown mass.
Final answer: kg
Q49Single correctSystem of Particles and Rotational Motion
From a circular ring of mass 'M' and radius 'R' an arc corresponding to a sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times ''. Then the value of 'K' is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Removing a quarter of the ring leaves three quarters of the mass, and the moment of inertia of every ring element is at the same radius.
Step 1:Removing the ninety degree sector removes one quarter of the mass, leaving three quarters of the original mass.
Step 2:Every element of the remaining ring lies at the same radius from the central axis, so the moment of inertia is the remaining mass times radius squared.
Step 3:Comparing with K times M R squared gives the value of K.
Final answer:
Q50Single correctAlternating Current
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10.2 A
Approach:
For an ideal transformer the input power equals the output power, which fixes the primary current.
Step 1:For an ideal transformer the input power on the primary equals the output power delivered to the lamp.
Step 2:Dividing the output power by the primary voltage gives the primary current.
Final answer: 0.2 A
Chemistry50 questions
Q51Single correctHydrocarbons
Dihedral angle of least stable conformer of ethane is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Ethane has two extreme conformations: the eclipsed and the staggered. Their relative stability is governed by torsional strain, which depends on the dihedral angle between the C-H bonds on the two carbons.
Step 1:In the staggered conformer the C-H bonds on adjacent carbons are as far apart as possible, giving a dihedral angle of 60 degrees and minimum torsional strain.
Step 2:In the eclipsed conformer the C-H bonds on adjacent carbons overlap directly, giving a dihedral angle of 0 degrees and maximum torsional strain.
Final answer:
Q52Single correctThe p-Block Elements
Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Noble gases have very high melting and boiling points
Approach:
Each statement is tested against the known properties of group 18 elements, which exist as monatomic gases held together only by weak van der Waals forces.
Step 1:Noble gases are non-polar monatomic species, so they dissolve only slightly in water; this statement is correct.
Step 2:The only interatomic attraction in noble gases is weak dispersion (London) force, so their melting and boiling points are very low, not high; this statement is incorrect.
Step 3:Because the atoms have completely filled shells, the induced-dipole dispersion forces between them are weak; this statement is correct.
Step 4:Noble gases resist gaining an electron, giving large positive electron gain enthalpies; this statement is correct.
Final answer: Noble gases have very high melting and boiling points
Q53Single correctHydrocarbons / Haloalkanes
The major product of the following chemical reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Addition of HBr to an unsymmetrical alkene in the presence of an organic peroxide proceeds by a free-radical (anti-Markovnikov) mechanism, so the bromine adds to the terminal, less substituted carbon.
Step 1:The peroxide generates a bromine radical, which adds to the terminal carbon of the double bond to form the more stable secondary carbon radical.
Step 2:Abstraction of hydrogen from HBr by this radical completes the chain, placing bromine on the end carbon and hydrogen on the inner carbon.
Final answer:
Q54Single correctStructure of Atom
A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The wavelength of electromagnetic radiation is found from the relation between the speed of light, frequency and wavelength.
Step 1:The frequency is converted from kilohertz to hertz.
Step 2:Rearranging the wave relation gives the wavelength as the speed of light divided by the frequency.
Final answer:
Q55Single correctThe d- and f-Block Elements
The incorrect statement among the following is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Most of the trivalent Lanthanoid ions are colourless in the solid state
Approach:
Each statement is checked against the characteristic properties of the f-block elements.
Step 1:Actinoid contraction from element to element is greater than lanthanoid contraction because the 5f electrons shield the nuclear charge poorly; this statement is correct.
Step 2:Many trivalent lanthanoid ions have partially filled 4f orbitals and absorb in the visible region, so they are coloured both in the solid state and in aqueous solution; the statement claiming most are colourless is incorrect.
Step 3:Lanthanoids show typical metallic character and conduct heat and electricity well; this statement is correct.
Step 4:Finely divided actinoids are highly reactive metals; this statement is correct.
Final answer: Most of the trivalent Lanthanoid ions are colourless in the solid state
Q56Single correctThe Solid State
Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The number of voids in a close-packed structure is fixed relative to the number of atoms present in the unit cell.
Step 1:A hexagonal primitive (hcp) unit cell contains 6 atoms.
Step 2:The number of octahedral voids equals the number of atoms, while the number of tetrahedral voids is twice the number of atoms.
Final answer:
Q57Single correctAmines
Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Hinsberg's reagent (benzenesulphonyl chloride) reacts differently with primary, secondary and tertiary amines; the product that bears an N-H bond is acidic and dissolves in alkali.
Step 1:A primary amine reacts with benzenesulphonyl chloride to give an N-alkylbenzene sulphonamide that retains an acidic N-H hydrogen.
Step 2:This N-H hydrogen, being acidic, is removed by alkali, so the sulphonamide of a primary amine dissolves in aqueous potassium hydroxide.
Step 3:A secondary amine gives a sulphonamide with no N-H hydrogen and therefore stays insoluble in alkali, and a tertiary amine does not react at all, so only the primary amine meets the condition.
Final answer:
Q58Single correctRedox Reactions
Which of the following reactions is the metal displacement reaction? Choose the right option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
A metal displacement reaction is one in which a metal in a compound is displaced from it by another, more reactive metal acting as the reducing agent.
Step 1:Reactions 1 and 4 are thermal decompositions, in which a single compound breaks into simpler substances.
Step 2:In the third reaction iron displaces hydrogen from hydrochloric acid, making it a hydrogen displacement, not a metal displacement.
Step 3:In the second reaction aluminium reduces chromic oxide and displaces chromium metal, the defining feature of a metal displacement reaction.
Final answer:
Q59Single correctChemistry in Everyday Life
Given below are two statements :
Statement I :
Aspirin and Paracetamol belong to the class of narcotic analgesics.
Statement II :
Morphine and Heroin are non-narcotic analgesics.
In the light of the above statements, choose the correct answer from the options given below.
Statement I :
Aspirin and Paracetamol belong to the class of narcotic analgesics.
Statement II :
Morphine and Heroin are non-narcotic analgesics.
In the light of the above statements, choose the correct answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are false
Approach:
The two statements are tested against the standard classification of analgesics into narcotic and non-narcotic types.
Step 1:Aspirin and paracetamol are non-narcotic analgesics, so the claim that they are narcotic analgesics is false.
Step 2:Morphine and heroin are narcotic analgesics, so the claim that they are non-narcotic is false.
Final answer: Both Statement I and Statement II are false
Q60Single correctAlcohols, Phenols and Ethers
What is the IUPAC name of the organic compound formed in the following chemical reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 42-methylbutan-2-ol
Approach:
A Grignard reagent adds to the carbonyl carbon of a ketone, and aqueous work-up gives a tertiary alcohol whose structure follows from combining the two fragments.
Step 1:Ethylmagnesium bromide delivers an ethyl group to the carbonyl carbon of acetone, forming an alkoxide intermediate.
Step 2:Hydrolysis of the alkoxide yields a tertiary alcohol bearing two methyl groups and one ethyl group on the carbinol carbon.
Step 3:Numbering the longest chain of four carbons with the hydroxyl on carbon two and a methyl branch on the same carbon gives the IUPAC name.
Final answer: 2-methylbutan-2-ol
Q61Single correctThe p-Block Elements (Group 17)
Statement I : Acid strength increases in the order given as .
Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.
Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both statement I and Statement II are true
Approach:
The acidic strength of hydrogen halides is governed by the H-X bond dissociation enthalpy, which depends on the size of the halogen atom.
Step 1:Within the periodic table, moving down the group the halogen size increases from fluorine to iodine, so the H-X bond becomes longer and weaker.
Step 2:A weaker H-X bond releases the proton more readily, increasing enthalpy of dissociation in solution, so acid strength rises in the order HF, HCl, HBr, HI.
Step 3:Statement II correctly explains Statement I by linking increasing atomic size to decreasing bond strength and rising acidity, so both statements are true.
Final answer: Both statement I and Statement II are true
Q62Single correctThermodynamics
Which one among the following is the correct option for right relationship between and for one mole of ideal gas?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The difference between the molar heat capacities at constant pressure and constant volume is obtained by comparing the heat supplied in each process for one mole of an ideal gas.
Step 1:At constant volume the heat supplied raises the internal energy, so the molar heat capacity at constant volume equals the internal energy change per degree.
Step 2:At constant pressure the heat supplied equals the enthalpy change, and enthalpy relates to internal energy through the pressure-volume term.
Step 3:Dividing through by the temperature change gives the difference of the molar heat capacities.
Final answer:
Q63Single correctChemical Bonding and Molecular Structure
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). | (i). Square pyramidal |
| (b). | (ii). Trigonal planar |
| (c). | (iii). Octahedral |
| (d). | (iv). Trigonal bipyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Approach:
The shape of each molecule follows from VSEPR theory by counting the bond pairs and lone pairs around the central atom.
Step 1:Phosphorus pentachloride has five bond pairs and no lone pair, giving a trigonal bipyramidal shape.
Step 2:Sulphur hexafluoride has six bond pairs and no lone pair, giving an octahedral shape.
Step 3:Bromine pentafluoride has five bond pairs and one lone pair, giving a square pyramidal shape.
Step 4:Boron trifluoride has three bond pairs and no lone pair, giving a trigonal planar shape.
Final answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Q64Single correctSolutions
The following solutions were prepared by dissolving 10 g of glucose in 250 ml of water , 10 g of urea in 250 ml of water and 10 g of sucrose in 250 ml of water . The right option for the decreasing order of osmotic pressure of these solutions is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Osmotic pressure is a colligative property proportional to the molar concentration of solute, so for equal masses in equal volumes the solute of lowest molar mass gives the highest osmotic pressure.
Step 1:Since the same mass of solute is dissolved in the same volume, the smaller the molar mass the larger the molar concentration and the greater the osmotic pressure.
Step 2:The molar masses follow the order sucrose greater than glucose greater than urea, so the concentration order reverses to urea greater than glucose greater than sucrose.
Step 3:Therefore the osmotic pressures rank with urea highest, glucose next and sucrose lowest.
Final answer:
Q65Single correctGeneral Principles and Processes of Isolation of Elements
Which one of the following methods can be used to obtain highly pure metal which is liquid at room temperature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Distillation
Approach:
A metal that is liquid at room temperature and has a low boiling point can be purified by vaporising it and condensing the pure vapour, a process suited to volatile metals.
Step 1:Distillation purifies low boiling metals such as mercury and zinc by boiling the impure metal and condensing the vapour, leaving non-volatile impurities behind.
Step 2:Mercury is a metal that is liquid at room temperature and low boiling, so distillation gives it in highly pure form.
Final answer: Distillation
Q66Single correctHydrogen
Tritium, a radioactive isotope of hydrogen, emits which of the following particles?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Beta
Approach:
The decay particle of tritium is identified from its nuclear instability, which arises from an excess of neutrons relative to protons.
Step 1:Tritium has one proton and two neutrons, an unstable neutron-rich nucleus.
Step 2:A neutron in the nucleus converts into a proton and an electron, and the emitted electron is a beta particle.
Final answer: Beta
Q67Single correctThe Solid State
The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The number of body centred lattices among the 14 Bravais lattices is found by listing which crystal systems admit a body centred arrangement.
Step 1:Among the 14 Bravais lattices the body centred unit cell occurs only in the cubic, tetragonal and orthorhombic crystal systems.
Step 2:Counting these three systems gives the total number of body centred unit cells.
Final answer:
Q68Single correctHaloalkanes and Haloarenes
The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Saytzeff's Rule
Approach:
The preferred alkene formed in dehydrohalogenation is determined by the relative stability of the possible products, governed by an elimination rule.
Step 1:Removal of hydrogen bromide from 2-bromopentane can give either pent-2-ene or pent-1-ene depending on which beta hydrogen is lost.
Step 2:The more substituted alkene pent-2-ene is more stable and forms as the major product, which is the prediction of Saytzeff's rule.
Final answer: Saytzeff's Rule
Q69Single correctGeneral Principles and Processes of Isolation of Elements
The maximum temperature that can be achieved in blast furnace is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Upto 2200 K
Approach:
The blast furnace used to extract iron reaches its highest temperature in the combustion zone near the tuyeres, and that ceiling fixes which reduction steps can occur.
Step 1:In the blast furnace the combustion of coke with the hot air blast raises the temperature to its highest value near the bottom of the furnace.
Step 2:The maximum temperature attained in the blast furnace is about 2200 K.
Final answer: Upto 2200 K
Q70Single correctElectrochemistry
The molar conductance of NaCl, HCl and at infinite dilution are 126.45, 426.16 and 91.0 S c mo respectively. The molar conductance of at infinite dilution is. Choose the right option for your answer.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 S c mo
Approach:
By Kohlrausch law of independent migration of ions, the limiting molar conductance of acetic acid is obtained by combining those of acetate sodium, hydrogen chloride and sodium chloride.
Step 1:The limiting molar conductance of acetic acid equals that of sodium acetate plus that of hydrochloric acid minus that of sodium chloride.
Step 2:Adding and subtracting the given values gives the limiting molar conductance of acetic acid.
Final answer: S c mo
Q71Single correctCoordination Compounds
Ethylene diaminetetraacetate (EDTA) ion is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hexadentate ligand with four "O" and two "N" donor atoms
Approach:
The denticity of EDTA is determined by counting the donor atoms that can simultaneously coordinate to a central metal ion.
Step 1:The ethylenediaminetetraacetate ion carries two amino nitrogen atoms and four carboxylate oxygen atoms capable of coordinating.
Step 2:All six of these donor atoms can bind a single metal ion at the same time, making EDTA a hexadentate ligand.
Final answer: Hexadentate ligand with four "O" and two "N" donor atoms
Q72Single correctThe s-Block Elements
The structures of beryllium chloride in solid state and vapour phase, are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Chain and dimer, respectively
Approach:
The structure adopted by beryllium chloride depends on its physical state, which determines whether it polymerises or exists as discrete molecules.
Step 1:In the solid state beryllium chloride has a polymeric chain structure in which each beryllium is bridged to neighbouring chlorine atoms.
Step 2:In the vapour phase at moderate temperature beryllium chloride exists as a chloro-bridged dimer.
Final answer: Chain and dimer, respectively
Q73Single correctHydrocarbons
The correct structure of 2,6-Dimethyl-dec-4-ene is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(structure) ten-carbon chain carrying methyl groups on C2 and C6 with the double bond between C4 and C5, i.e. 2,6-dimethyldec-4-ene
Approach:
The correct structure is built by drawing the ten-carbon parent chain, placing the double bond between carbons four and five and adding methyl groups at carbons two and six.
Step 1:The parent name dec indicates a ten-carbon main chain.
Step 2:The suffix dec-4-ene places the carbon-carbon double bond between the fourth and fifth carbons of the chain.
Step 3:Two methyl substituents are placed on the second and sixth carbons of the chain.
Final answer: (structure) ten-carbon chain carrying methyl groups on C2 and C6 with the double bond between C4 and C5, i.e. 2,6-dimethyldec-4-ene
Q74Single correctSurface Chemistry
The right option for the statement "Tyndall effect is exhibited by", is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Starch solution
Approach:
The Tyndall effect, the scattering of light by suspended particles, is shown only by colloidal dispersions whose particle size lies in the colloidal range, so the colloid has to be identified.
Step 1:Sodium chloride, glucose and urea form true solutions whose particles are too small to scatter light.
Step 2:Starch dispersed in water forms a colloidal solution whose larger particles scatter light and exhibit the Tyndall effect.
Final answer: Starch solution
Q75Single correctPolymers
Which one of the following polymers is prepared by addition polymerisation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Teflon
Approach:
Each polymer is classified by its mode of formation, distinguishing those built by repeated addition of monomers from those formed by condensation with loss of small molecules.
Step 1:Nylon-66, Novolac and Dacron are prepared by condensation polymerisation, in which monomers join with elimination of small molecules such as water.
Step 2:Teflon is formed by the repeated addition of tetrafluoroethene units without loss of any molecule, making it an addition polymer.
Final answer: Teflon
Q76Single correctBiomolecules
The RBC deficiency is deficiency disease of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Vitamin
Approach:
Each B-complex vitamin is linked to a characteristic deficiency disorder, and the disease tied to red blood cell formation identifies the required vitamin.
Step 1:Deficiency of vitamin B12 leads to pernicious anaemia, a condition in which red blood cell formation is impaired.
Step 2:Vitamin B2 (riboflavin) deficiency causes cheilosis and skin disorders, vitamin B6 (pyridoxine) deficiency causes convulsions, and vitamin B1 (thiamine) deficiency causes beri-beri, none of which is the RBC deficiency disease.
Final answer: Vitamin
Q77Single correctHaloalkanes and Haloarenes
The correct sequence of bond enthalpy of 'C—X' bond is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Bond enthalpy of the carbon-halogen bond depends on the bond length, which increases down the halogen group; a longer bond is weaker.
Step 1:The size of the halogen atom increases from F to I, so the C—X bond length increases in the order C—F < C—Cl < C—Br < C—I.
Step 2:A shorter bond corresponds to a higher bond enthalpy. The reported dissociation enthalpies are C—F = 452, C—Cl = 351, C—Br = 293 and C—I = 234 kJ mol⁻¹.
Step 3:Arranging in decreasing bond enthalpy gives the required sequence.
Final answer:
Q78Single correctSome Basic Concepts of Chemistry
An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The mole ratio of the elements is obtained by dividing each mass percentage by its atomic mass, and the simplest whole-number ratio gives the empirical formula.
Step 1:Carbon is 78% and hydrogen is the remaining 22%. The relative moles of carbon are 78/12 = 6.5 and of hydrogen are 22/1 = 22.
Step 2:Dividing each value by the smaller (6.5) gives carbon as 1 and hydrogen as 22/6.5 = 3.38, which rounds to 3.
Step 3:The simplest whole-number ratio corresponds to the empirical formula CH3.
Final answer:
Q79Single correctThe s-Block Elements
Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Beryllium chloride
Approach:
Covalent character of an alkaline earth halide is governed by the polarising power of the cation, which is largest for the smallest cation.
Step 1:Beryllium has the smallest size and the highest charge density among the alkaline earth metals, giving it strong polarising power according to Fajan's rules.
Step 2:This strong polarisation makes beryllium chloride covalent, so it dissolves in organic solvents, whereas the chlorides of calcium, strontium and the larger metals are predominantly ionic.
Final answer: Beryllium chloride
Q80Single correctChemical Bonding and Molecular Structure
B is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3s and 6
Approach:
The hybridisation of boron follows from the three sigma bonds it forms, and the electron count around it follows from those three shared pairs.
Step 1:Boron in BF3 forms three B—F sigma bonds with no lone pair, giving three electron domains and sp2 hybridisation with a trigonal planar shape.
Step 2:Three bonding pairs surround the boron atom, contributing 3 × 2 = 6 electrons, which falls short of an octet and makes the molecule electron deficient.
Final answer: s and 6
Q81Single correctIonic Equilibrium
The p of dimethylamine and p of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 37.75
Approach:
Dimethylammonium acetate is a salt of a weak acid and a weak base, for which the pH depends on both pKa and pKb through a standard relation.
Step 1:For a salt of a weak acid and a weak base the pH is given by 7 + (pKa - pKb)/2.
Step 2:Substituting the values gives 7 + (4.77 - 3.27)/2.
Final answer: 7.75
Q82Single correctStates of Matter
Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(graph) P falls hyperbolically with V, with the 600 K isotherm highest and the 200 K isotherm lowest
Approach:
Boyle's law fixes the shape of each P-V curve as a rectangular hyperbola, and the constant PV product rises with temperature, ordering the curves.
Step 1:Boyle's law states that at constant temperature pressure is inversely proportional to volume, so each isotherm is a rectangular hyperbola in the P-V plane.
Step 2:The constant k equals nRT, so it increases with temperature and the 600 K hyperbola lies above the 400 K and 200 K hyperbolae.
Final answer: (graph) P falls hyperbolically with V, with the 600 K isotherm highest and the 200 K isotherm lowest
Q83Single correctChemical Kinetics
For a reaction A B, enthalpy of reaction is kJ mo and enthalpy of activation is 9.6 kJ mo. The correct potential energy profile for the reaction is shown in option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(graph) PE profile in which A rises over a small barrier and settles at a product level B below A
Approach:
The sign of the reaction enthalpy fixes whether products lie above or below reactants, and the activation enthalpy fixes the barrier height relative to reactants.
Step 1:The reaction enthalpy is negative, so the products B lie below the reactants A and the reaction is exothermic.
Step 2:Using the relation between enthalpies, the backward activation enthalpy is the forward activation enthalpy minus the reaction enthalpy, giving 9.6 - (-4.2) = 13.8 kJ mol⁻¹.
Step 3:An exothermic profile starts at A, crosses the 9.6 kJ mo forward barrier and settles at B below A, with the reverse barrier of 13.8 kJ mo larger than the forward one.
Final answer: (graph) PE profile in which A rises over a small barrier and settles at a product level B below A
Q84Single correctOrganic Chemistry - Basic Principles
The compound which shows metamerism is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Metamerism requires a functional group, typically an ether, flanked by alkyl chains that can be distributed in different ways while keeping the same molecular formula.
Step 1:Metamers differ in the alkyl groups attached on either side of the same functional group, so an ether of sufficient carbon count is needed.
Step 2:The formula C4H10O corresponds to ethers such as CH3-CH2-O-CH2-CH3 and CH3-O-CH(CH3)-CH3, which carry different alkyl chains around the oxygen.
Final answer:
Q85Single correctThe d- and f-Block Elements
Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Lanthanoid contraction
Approach:
The unusual closeness in size of a second and third transition series pair is explained by the contraction across the intervening lanthanoid series.
Step 1:Across the lanthanoid series the poor shielding by 4f electrons causes a steady decrease in size known as the lanthanoid contraction.
Step 2:This contraction offsets the expected increase from Zr to Hf, leaving their atomic radii almost identical at about 160 pm and 159 pm.
Final answer: Lanthanoid contraction
Q86Single correctCoordination Compounds
Match List-I with List-II.
Magnetic moment, BM (where n = number of unpaired electrons)
Choose the correct answer from the options given below.
Magnetic moment, BM (where n = number of unpaired electrons)
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). | (i). 5.92 BM |
| (b). | (ii). 0 BM |
| (c). | (iii). 4.90 BM |
| (d). | (iv). 1.73 BM |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
The number of unpaired electrons in each iron complex is fixed by the oxidation state and the strong- or weak-field nature of the ligand, and the spin-only formula then gives the magnetic moment.
Step 1:In [Fe(CN)6]3- iron is +3 (d5) with the strong-field cyanide ligand, leaving one unpaired electron and a moment of 1.73 BM.
Step 2:In [Fe(H2O)6]3+ iron is +3 (d5) with the weak-field water ligand, giving five unpaired electrons and a moment of 5.92 BM.
Step 3:In [Fe(CN)6]4- iron is +2 (d6) with strong-field cyanide, pairing all electrons to give zero unpaired electrons and 0 BM.
Step 4:In [Fe(H2O)6]2+ iron is +2 (d6) with weak-field water, leaving four unpaired electrons and a moment of 4.90 BM.
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q87Single correctAmines
The reagent '' in the given sequence of chemical reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Replacing the diazonium group by hydrogen requires a mild reducing reagent, and ethanol serves this role.
Step 1:The aromatic amine is converted to its diazonium salt with nitrous acid generated from sodium nitrite and hydrochloric acid at low temperature.
Step 2:Treatment of the diazonium salt with ethanol replaces the diazonium group by hydrogen, giving 1,3,5-tribromobenzene, so reagent R is ethanol.
Final answer:
Q88Single correctClassification of Elements and Periodicity
From the following pairs of ions which one is not an iso-electronic pair?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Isoelectronic species carry the same number of electrons, so the electron count of each ion in a pair is compared.
Step 1:O2- and F- both have 10 electrons, Na+ and Mg2+ both have 10 electrons, and Mn2+ and Fe3+ both have 23 electrons, so these three pairs are isoelectronic.
Step 2:Fe2+ has 26 - 2 = 24 electrons while Mn2+ has 25 - 2 = 23 electrons, so this pair is not isoelectronic.
Final answer:
Q89Single correctChemical Kinetics
The slope of Arrhenius plot of first order reaction is K. The value of of the reaction is. Choose the correct option for your answer.
[Given R = 8.314 J]
[Given R = 8.314 J]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 141.5 kJ mo
Approach:
The Arrhenius equation in logarithmic form makes the slope of ln k against 1/T equal to -Ea/R, from which Ea follows.
Step 1:Writing the Arrhenius equation logarithmically shows that the slope of the ln k versus 1/T plot equals -Ea/R.
Step 2:Setting the given slope equal to -Ea/R gives -5 × 10³ = -Ea/8.314, so Ea = 5 × 10³ × 8.314 J/mol.
Step 3:Converting to kilojoules gives an activation energy of about 41.5 kJ mol⁻¹.
Final answer: 41.5 kJ mo
Q90Single correctThermodynamics
For irreversible expansion of an ideal gas under isothermal condition, the correct option is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For an ideal gas the internal energy depends only on temperature, and the total entropy change is governed by the spontaneity of the irreversible process.
Step 1:Under isothermal conditions the temperature change is zero, so for an ideal gas the internal energy change is zero.
Step 2:An irreversible expansion is a spontaneous process, for which the total entropy change is positive and therefore nonzero.
Final answer:
Q91Single correctThe p-Block Elements
In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2O < S < Se < Te : Increasing p values
Approach:
Each sequence is tested against the genuine periodic trend, and the one whose ordering contradicts the indicated property is selected.
Step 1:Down group 16 the hydrides become stronger acids, so pKa decreases from H2O to H2Te rather than increasing.
Step 2:Hydrohalic acid strength does increase from HF to HI, the acidic character of the group 15 hydrides does increase down the group, and the oxidising power of the group 14 dioxides does increase towards , so those three sequences match their stated trends.
Final answer: O < S < Se < Te : Increasing p values
Q92Single correctAldehydes, Ketones and Carboxylic Acids
The intermediate compound '' in the following chemical reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(structure) benzene ring bearing a group, the Etard intermediate
Approach:
The conversion of toluene to benzaldehyde by chromyl chloride is the Etard reaction, which proceeds through a characteristic chromium complex intermediate.
Step 1:In the Etard reaction, chromyl chloride attacks the methyl group of toluene to form a brown chromium complex of the type C6H5-CH(OCrOHCl2)2.
Step 2:Acidic hydrolysis of this complex releases benzaldehyde.
Final answer: (structure) benzene ring bearing a group, the Etard intermediate
Q93Single correctSolutions
The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is :
[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3336 mm of Hg
Approach:
For an ideal solution the total vapour pressure is the sum of the partial pressures, each obtained from Raoult's law using the mole fractions.
Step 1:With benzene to octane in a 3 : 2 molar ratio, the mole fraction of benzene is 3/5 and that of octane is 2/5.
Step 2:Applying Raoult's law gives the total pressure as 280 × (3/5) + 420 × (2/5).
Step 3:Adding the partial pressures gives the total vapour pressure.
Final answer: 336 mm of Hg
Q94Single correctElectrochemistry
The molar conductivity of 0.007 M acetic acid is 20 S c mo. What is the dissociation constant of acetic acid? Choose the correct option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 mol
Approach:
The degree of dissociation is the ratio of molar conductivity to limiting molar conductivity, and Ostwald's dilution law then yields the dissociation constant.
Step 1:The limiting molar conductivity of acetic acid is the sum of the ionic contributions, 350 + 50 = 400 S cm² mol⁻¹.
Step 2:The degree of dissociation is the ratio of the measured to the limiting molar conductivity, 20/400 = 1/20.
Step 3:Applying Ostwald's dilution law with the small alpha approximation gives Ka = Cα² = 0.007 × (1/20)².
Final answer: mol
Q95Single correctHydrocarbons
Consider the above reaction and identify the missing reagent/chemical.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Decarboxylation of a sodium carboxylate to the alkane uses soda lime, a mixture of sodium hydroxide and calcium oxide.
Step 1:Heating the sodium salt of a carboxylic acid with soda lime removes carbon dioxide and produces the alkane with one fewer carbon.
Step 2:Soda lime is sodium hydroxide combined with calcium oxide in a 3 : 1 ratio, so the missing reagent alongside NaOH is calcium oxide.
Final answer:
Q96Single correctAldehydes, Ketones and Carboxylic Acids
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
(a).  | (i). Hell-Volhard-Zelinsky reaction |
(b).  | (ii). Gattermann-Koch reaction |
(c).  | (iii). Haloform reaction |
(d).  | (iv). Esterification |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Approach:
Each transformation is identified by its reagents and substrate, which names the reaction in List-II.
Step 1:The formylation of benzene with carbon monoxide and hydrogen chloride in presence of anhydrous aluminium chloride and copper chloride is the Gattermann-Koch reaction.
Step 2:A methyl ketone treated with a sodium hypohalite undergoes the haloform reaction.
Step 3:An alcohol reacting with a carboxylic acid in presence of concentrated sulphuric acid undergoes esterification.
Step 4:Alpha halogenation of a carboxylic acid using halogen and red phosphorus followed by hydrolysis is the Hell-Volhard-Zelinsky reaction.
Final answer: (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
Q97Single correctAldehydes, Ketones and Carboxylic Acids
The product formed in the following chemical reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(structure) the ring ketone reduced to with the ester side chain left intact
Approach:
Sodium borohydride is a mild reducing agent whose selectivity toward different carbonyl groups determines which functional group changes.
Step 1:Sodium borohydride reduces the ketone carbonyl of the cyclohexanone ring to a secondary alcohol.
Step 2:Sodium borohydride does not reduce esters, so the side-chain ester group remains unchanged in the product.
Final answer: (structure) the ring ketone reduced to with the ester side chain left intact
Q98Single correctChemical Bonding and Molecular Structure
Which of the following molecules is non-polar in nature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A molecule is non-polar when its symmetric geometry causes the individual bond dipoles to cancel to a net zero dipole moment.
Step 1:SbCl5 has a trigonal bipyramidal shape in which the bond moments are arranged symmetrically.
Step 2:The vector sum of the bond moments in this symmetric arrangement is zero, so SbCl5 is non-polar, whereas POCl3, CH2O and NO2 have unsymmetrical geometries and net dipole moments.
Final answer:
Q99Single correctEnvironmental Chemistry
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). | (i). Acid rain |
| (b). | (ii). Smog |
| (c). | (iii). Ozone depletion |
| (d). | (iv). Tropospheric pollution |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Approach:
Each chemical process is associated with the pollution phenomenon it drives in the atmosphere.
Step 1:Oxidation of sulphur dioxide to sulphur trioxide occurs in the troposphere, linking process (a) to tropospheric pollution.
Step 2:Photolysis of hypochlorous acid releases chlorine radicals that destroy stratospheric ozone, linking process (b) to ozone depletion.
Step 3:The attack of sulphuric acid on calcium carbonate, as in the corrosion of marble, is a consequence of acid rain, linking process (c) to acid rain.
Step 4:Photodissociation of nitrogen dioxide produces atomic oxygen that initiates photochemical smog, linking process (d) to smog.
Final answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Q100Single correctStates of Matter
Choose the correct option for the total pressure (in atm.) in a mixture of 4 g and 2 g confined in a total volume of one litre at 0°C is :
[Given R = 0.082 L atm mo, T = 273 K]
[Given R = 0.082 L atm mo, T = 273 K]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 325.18
Approach:
The total number of moles of the gas mixture is found from the masses, and the ideal gas equation then gives the total pressure.
Step 1:The moles of oxygen are 4/32 = 1/8 and the moles of hydrogen are 2/2 = 1.
Step 2:The total number of moles is the sum, 1/8 + 1 = 9/8.
Step 3:Applying the ideal gas equation gives the total pressure as (9/8 × 0.082 × 273)/1.
Final answer: 25.18
Biology100 questions
Q101Single correctMorphology of Flowering Plants
Which of the following plants is monoecious?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
A monoecious plant bears both male and female sex organs on the same individual.
Step 1:In a monoecious plant, male and female reproductive structures occur on the same plant body.
Step 2:Most species of the green alga Chara are monoecious, bearing antheridia and oogonia on the same thallus.
Step 3:Cycas circinalis, Carica papaya and Marchantia polymorpha are dioecious, with male and female organs on separate individuals.
Final answer:
Q102Single correctSexual Reproduction in Flowering Plants
A typical angiosperm embryo sac at maturity is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 18-nucleate and 7-celled
Approach:
The structure of the mature angiosperm embryo sac follows the typical Polygonum type.
Step 1:A typical angiosperm embryo sac contains three antipodals, one central cell, one egg cell and two synergids.
Step 2:The central cell carries two polar nuclei, so the embryo sac is eight nucleate while remaining seven celled.
Final answer: 8-nucleate and 7-celled
Q103Single correctPlant Kingdom
Gemmae are present in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Some Liverworts
Approach:
Gemmae are green, multicellular asexual buds of bryophytes.
Step 1:Gemmae are green, multicellular asexual buds produced by some liverworts such as Marchantia in cup-like gemma cups.
Step 2:Mosses reproduce vegetatively by fragmentation and budding of protonema, not by gemmae.
Step 3:Pteridophytes and gymnosperms normally do not reproduce asexually by gemmae.
Final answer: Some Liverworts
Q104Single correctCell Cycle and Cell Division
When the centromere is situated in the middle of two equal arms of chromosomes, the chromosome is referred as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Metacentric
Approach:
Chromosome type is named from the position of the centromere along the chromosome.
Step 1:When the centromere lies at the middle of two equal arms, the chromosome is metacentric.
Step 2:A centromere slightly away from the middle gives a sub-metacentric chromosome; one very close to the end gives an acrocentric chromosome; one at the terminal position gives a telocentric chromosome.
Final answer: Metacentric
Q105Single correctCell Cycle and Cell Division
Which of the following stages of meiosis involves division of centromere?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Anaphase II
Approach:
Centromere splitting separates sister chromatids and occurs only at anaphase II in meiosis.
Step 1:Telophase I marks the last stage of meiosis I; during this phase the chromatids reach the poles and start uncoiling, with no centromere division.
Step 2:Chromosomes form two parallel plates in metaphase I and one plate in metaphase II, but the centromere stays undivided.
Step 3:Division of the centromere occurs in anaphase II, separating the sister chromatids.
Final answer: Anaphase II
Q106Single correctPhotosynthesis in Higher Plants
The first stable product of fixation in Sorghum is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Oxaloacetic acid
Approach:
Sorghum is a C4 plant, so the first stable product of carbon fixation is a four-carbon acid.
Step 1:In a C4 plant such as Sorghum, the first stable product of CO2 fixation is the four-carbon oxaloacetic acid.
Step 2:Phosphoglyceric acid is the first stable product in the C3 cycle, and pyruvic acid is the three-carbon product of glycolysis; succinic acid is an intermediate of the Krebs cycle.
Final answer: Oxaloacetic acid
Q107Single correctEvolution
The factor that leads to Founder effect in a population is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Genetic drift
Approach:
The founder effect is a special case of genetic drift seen in small founding populations.
Step 1:A change in gene frequency in a small population by chance is known as genetic drift; the founder effect is its outcome.
Step 2:When a few individuals are dispersed and act as founders of a new isolated population, the founder's allele frequencies differ from the original population.
Step 3:Natural selection, recombination and mutation act differently and have direction, unlike the random sampling of genetic drift.
Final answer: Genetic drift
Q108Single correctCell - The Unit of Life
Which of the following is an statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles
Approach:
The incorrect statement is found by checking each against established cell biology.
Step 1:Mature sieve tube elements possess a peripheral cytoplasm and a large central vacuole but lack a nucleus, so the first statement is incorrect.
Step 2:Microbodies occur in both plant and animal cells, the perinuclear space of the nuclear envelope separates nuclear and cytoplasmic materials, and nuclear pores allow two-way movement of proteins and RNA; these statements are correct.
Final answer: Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles
Q109Single correctOrganisms and Populations
Amensalism can be represented as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Species A (–); Species B (0)
Approach:
Amensalism is the interaction where one species is harmed while the other is unaffected.
Step 1:In amensalism one organism of different species is harmed while the other neither benefits nor is harmed, so one species is marked negative and the other zero.
Step 2:A (+) (+) pair denotes mutualism and a (+) (–) pair denotes interactions like predation or parasitism, while a (–) (–) pair denotes competition.
Final answer: Species A (–); Species B (0)
Q110Single correctEcosystem
The amount of nutrients, such as carbon, nitrogen, phosphorus and calcium present in the soil at any given time, is referred as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Standing state
Approach:
The amount of inorganic nutrients present in the soil at a given time defines a specific ecological term.
Step 1:The amount of inorganic substances such as carbon, nitrogen, phosphorus and calcium present in soil at a given time is the standing state.
Step 2:The amount of living material present in different trophic levels at a given time is the standing crop, while a climax community is the final stable community in equilibrium with the environment of an area.
Final answer: Standing state
Q111Single correctPlant Kingdom
Which of the following algae contains mannitol as reserve food material?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Reserve food material identifies the algal class to which each genus belongs.
Step 1:Ectocarpus is a brown alga of class Phaeophyceae, whose members store food as mannitol and laminarin.
Step 2:Ulothrix and Volvox belong to Chlorophyceae and store starch as reserve food; Gracilaria is a member of red algae Rhodophyceae and stores floridean starch.
Final answer:
Q112Single correctOrganisms and Populations
Inspite of interspecific competition in nature, which mechanism the competing species might have evolved for their survival?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Resource partitioning
Approach:
Competing species coexist by dividing the contested resource rather than eliminating one another.
Step 1:Competing species coexist through resource partitioning, by which they avoid competition and one species is not driven to extinction.
Step 2:In mutualism two organisms are equally benefited, in predation one organism is the predator and the other the prey, and competitive release describes the expansion of distribution when a strong competitor is removed.
Final answer: Resource partitioning
Q113Single correctPlant Kingdom
Genera like and produce two kinds of spores. Such plants are known as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Heterosporous
Approach:
Plants producing two distinct kinds of spores are described by a specific term.
Step 1:Selaginella and Salvinia produce microspores and megaspores, so they are known as heterosporous.
Step 2:Most pteridophytes produce a single type of spore and are homosporous; sorus terms refer to spore clusters of ferns and do not apply here.
Final answer: Heterosporous
Q114Single correctPlant Growth and Development
The site of perception of light in plants during photoperiodism is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Leaf
Approach:
Photoperiodic light is perceived at a specific plant organ before flowering is induced.
Step 1:The site of perception of light in plants during photoperiodism is the leaf.
Step 2:The site of perception of low temperature stimulus during vernalisation is the shoot apex and embryo, which differs from photoperiodic perception.
Final answer: Leaf
Q115Single correctPlant Growth and Development
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Plasticity
Approach:
The ability of plants to follow different developmental pathways has a defined name.
Step 1:The ability of a plant to follow different pathways in response to environment or phases of life to form different structures is called plasticity.
Step 2:Heterophylly in cotton, coriander and larkspur, where leaves of juvenile and mature plants differ, illustrates plasticity.
Final answer: Plasticity
Q116Single correctCell - The Unit of Life
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). Cristae | (i). Primary constriction in chromosome |
| (b). Thylakoids | (ii). Disc-shaped sacs in Golgi apparatus |
| (c). Centromere | (iii). Infoldings in mitochondria |
| (d). Cisternae | (iv). Flattened membranous sacs in stroma of plastids |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Each structure is matched to the organelle feature it describes.
Step 1:The inner membrane of mitochondria forms infoldings called cristae, so (a) matches (iii).
Step 2:Thylakoids are flattened membranous sacs in the stroma of plastids, so (b) matches (iv).
Step 3:The centromere is the primary constriction that holds two chromatids together in a chromosome, so (c) matches (i); cisternae are the disc-shaped sacs of the Golgi apparatus, so (d) matches (ii).
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q117Single correctTransport in Plants
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). Cohesion | (i). More attraction in liquid phase |
| (b). Adhesion | (ii). Mutual attraction among water molecules |
| (c). Surface tension | (iii). Water loss in liquid phase |
| (d). Guttation | (iv). Attraction towards polar surfaces |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Approach:
Each property of water is matched to its definition.
Step 1:Cohesion is the mutual attraction among water molecules, so (a) matches (ii), and adhesion is the attraction of water molecules to polar surfaces, so (b) matches (iv).
Step 2:Surface tension expresses the greater attraction of water molecules in the liquid phase than in the gaseous phase, so (c) matches (i), and guttation is the loss of water in liquid form from the leaf margins, so (d) matches (iii).
Final answer: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Q118Single correctBiotechnology - Principles and Processes
DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Bright orange bands
Approach:
The colour of DNA bands depends on the stain and the radiation used for viewing.
Step 1:After the bands are stained on a gel, they are viewed under ultraviolet light.
Step 2:Ethidium bromide intercalates into the DNA and the bands appear bright orange in colour under UV radiation.
Final answer: Bright orange bands
Q119Single correctSexual Reproduction in Flowering Plants
The term used for transfer of pollen grains from anthers of one plant to stigma of a different plant which, during pollination, brings genetically different types of pollen grains to stigma, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Xenogamy
Approach:
The pollination term is identified from the source of pollen and its genetic relation to the stigma.
Step 1:Xenogamy refers to the transfer of pollen grains from the anthers of one plant to the stigma of a different plant, which during pollination brings genetically different pollen grains to the stigma.
Step 2:Geitonogamy is the transfer of pollen grains from the anther to the stigma of another flower of the same plant; chasmogamy refers to flowers that open, and cleistogamy to flowers that do not open.
Final answer: Xenogamy
Q120Single correctPlant Kingdom
Which of the following algae produce Carrageen?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Red algae
Approach:
Carrageen is a hydrocolloid obtained commercially from a particular algal group.
Step 1:The cell wall of red algae is composed of agar, carrageen and funori along with cellulose.
Step 2:Brown algae have cell walls of cellulose and algin in a pectin frame, green algae have cellulose and pectin walls, and blue-green algae have mucopeptide walls.
Final answer: Red algae
Q121Single correctEcosystem
Which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pyramid of biomass in sea is generally upright.
Approach:
Each statement on ecological pyramids is tested for correctness.
Step 1:The pyramid of biomass in the sea is generally inverted because the biomass of zooplanktons is higher than that of phytoplanktons, as the smaller phytoplanktons multiply faster yet support a much larger standing crop of zooplankton.
Step 2:The statement that the pyramid of biomass in the sea is generally upright contradicts this and is therefore not correct.
Step 3:The pyramid of energy is always upright and the pyramid of numbers in a grassland ecosystem is upright, so those statements are correct.
Final answer: Pyramid of biomass in sea is generally upright.
Q122Single correctEcosystem
In the equation GPP – R = NPP
R represents :
R represents :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Respiration losses
Approach:
The productivity equation links gross primary productivity, respiration and net primary productivity.
Step 1:In the equation GPP – R = NPP, the term R stands for respiration losses.
Step 2:GPP is the gross primary productivity and NPP is the net primary productivity left after respiration losses.
Final answer: Respiration losses
Q123Single correctMorphology of Flowering Plants
Diadelphous stamens are found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pea
Approach:
Diadelphous condition is named from how the stamens unite into two bundles.
Step 1:Stamens are said to be diadelphous when they are united into two bundles, as in pea.
Step 2:China rose has monoadelphous stamens and Citrus has polyadelphous stamens; monoadelphous stamens are grouped into one bundle while polyadelphous stamens occur in more than two bundles.
Final answer: Pea
Q124Single correctBiotechnology and its Applications
When gene targeting involving gene amplification is attempted in an individual's tissue to treat disease, it is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gene therapy
Approach:
The method that corrects a gene defect within a patient's tissue has a defined name.
Step 1:Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child or embryo.
Step 2:Biopiracy is the use of bioresources by multinational companies and other organisations without proper authorisation from the countries and people concerned, while molecular diagnosis refers to the act or process of determining the nature or cause of a disease.
Final answer: Gene therapy
Q125Single correctAnatomy of Flowering Plants
Match List-I with List-II. Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| a. Lenticels | i. Phellogen |
| b. Cork cambium | ii. Suberin deposition |
| c. Secondary cortex | iii. Exchange of gases |
| d. Cork | iv. Phelloderm |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Each structure of the periderm and its associated function or synonym is paired using the standard NCERT account of secondary growth in dicot stems.
Step 1:Lenticels are the lens-shaped pores in the periderm through which gaseous exchange between internal tissues and the atmosphere takes place.
Step 2:Cork cambium, the lateral meristem that gives rise to the periderm, is also termed phellogen.
Step 3:The phellogen cuts off cells towards the inside that mature into the secondary cortex, also called phelloderm.
Step 4:The phellogen cuts off cells towards the outside that become cork (phellem); their walls are heavily impregnated with suberin.
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q126Single correctPrinciples of Inheritance and Variation
The production of gametes by the parents, formation of zygotes by the and plants, can be understood from a diagram called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Punnett square
Approach:
The graphical checkerboard used in Mendelian genetics to enumerate the gamete combinations and resulting genotypes of a cross is identified from its standard name.
Step 1:The grid devised by Reginald C. Punnett tabulates all possible gametes of one parent against those of the other along the two axes.
Step 2:Filling each cell with the union of the corresponding row and column gametes yields the genotypes of the offspring, allowing the genotypic and phenotypic ratios of the F2 generation to be read directly.
Final answer: Punnett square
Q127Single correctBiotechnology - Principles and Processes
Which of the following is a correct sequence of steps in a PCR (Polymerase Chain Reaction)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Denaturation, Annealing, Extension
Approach:
Order the three repeated phases of a PCR cycle by temperature and function.
Step 1:Denaturation at about 94 degrees Celsius separates the double-stranded template into single strands.
Step 2:Annealing follows as the two oligonucleotide primers bind to their complementary sequences on the single strands.
Step 3:Extension by a thermostable DNA polymerase builds the new strands from the primers.
Final answer: Denaturation, Annealing, Extension
Q128Single correctPrinciples of Inheritance and Variation
Mutations in plant cells can be induced by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Gamma rays
Approach:
Identify the agent capable of altering DNA and inducing mutation.
Step 1:Several kinds of radiation such as gamma rays, X-rays and UV-rays cause mutation by acting as physical mutagens.
Step 2:Such induced mutation in plants develops improved varieties; the first natural mutant variety of maize grain known as zeatin originated from a tobacco cytokinin. Kinetin and zeatin are cytokinins, not mutagens.
Final answer: Gamma rays
Q129Single correctBiotechnology: Principles and Processes / Plant Breeding
Match List-I with List-II. Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| a. Protoplast fusion | i. Totipotency |
| b. Plant tissue culture | ii. Pomato |
| c. Meristem culture | iii. Somaclones |
| d. Micropropagation | iv. Virus free plants |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Approach:
Each plant tissue-culture technique is paired with its characteristic product or underlying principle following the NCERT account of in-vitro plant biotechnology.
Step 1:Fusion of protoplasts from two different species followed by culture of the hybrid produces somatic hybrids; the tomato-potato somatic hybrid is the classic Pomato.
Step 2:The capacity of a cultured plant cell to regenerate a whole plant is totipotency, the basis of all plant tissue culture.
Step 3:Culture of the apical meristem, which the virus is unable to invade, yields virus-free plants.
Step 4:Plants raised through micropropagation arise from somatic tissue and are termed somaclones, genetically identical to the parent.
Final answer: (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Q130Single correctBiotechnology - Principles and Processes
During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2DNA
Approach:
Recall which macromolecule is precipitated by chilled ethanol after enzymatic removal of others.
Step 1:Various enzymes such as ribonuclease, protease and others are added to break down substances like proteins, RNA and others, so once all these substances are broken down, DNA is left which is precipitated out by adding chilled ethanol.
Step 2:Purified DNA appears as collected threads when chilled ethanol is added to the preparation.
Final answer: DNA
Q131Single correctMolecular Basis of Inheritance
Complete the flow chart on central dogma.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein
Approach:
Trace the flow of genetic information from DNA to protein.
Step 1:Formation of DNA from DNA is replication, the first step labelled (a).
Step 2:Formation of mRNA from DNA is transcription, the step labelled (b).
Step 3:Formation of protein from mRNA is translation, the step labelled (c).
Step 4:The final product (d) is protein.
Final answer: (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein
Q132Single correctBiotechnology - Principles and Processes
Which of the following is not an application of PCR (Polymerase Chain Reaction)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Purification of isolated protein
Approach:
PCR amplifies a chosen DNA segment, so the listed application that has nothing to do with nucleic acids is the exception.
Step 1:PCR is the polymerase chain reaction, used to make multiple copies of a gene, hence PCR is used for gene amplification.
Step 2:PCR-based assays have been developed that detect the presence of gene sequences of infectious agents, making PCR useful in molecular diagnosis and in detecting mutations.
Step 3:Protein is not the target of PCR; hence PCR has no role in its purification.
Final answer: Purification of isolated protein
Q133Single correctPlant Growth and Development
The plant hormone used to destroy weeds in a field
(1)
(2)
(3)
(4)
SolutionAnswer: Option 32, 4-D
Approach:
Identify the synthetic auxin applied as a herbicide.
Step 1:Some synthetic auxins are used as weedicides.
Step 2:2,4-D is widely used to remove broad-leaved weeds or dicotyledonous weeds in cereal crops or monocotyledonous plants.
Step 3:IAA and IBA are natural auxins, and NAA is a synthetic auxin used for other purposes.
Final answer: 2, 4-D
Q134Single correctAnatomy of Flowering Plants
Match List-I with List-II.
Select the correct answer from the options given below.
Select the correct answer from the options given below.
| List - I | List - II |
|---|---|
| (a). Cells with active cell division capacity | (i). Vascular tissues |
| (b). Tissue having all cells similar in structure and function | (ii). Meristematic tissue |
| (c). Tissue having different types of cells | (iii). Sclereids |
| (d). Dead cells with highly thickened walls and narrow lumen | (iv). Simple tissue |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Approach:
Match each anatomical description with the correct plant tissue type.
Step 1:Meristematic tissues are those tissues which have cells with active cell division capacity.
Step 2:Simple tissues have cells which have all the cells similar in structure and function.
Step 3:Vascular tissues are complex permanent tissues hence they have different types of cells.
Step 4:Sclereids are sclerenchymatous cells which are dead with highly thickened walls and narrow lumen.
Final answer: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Q135Single correctBiomolecules
Which of the following are not secondary metabolites in plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Amino acids, glucose
Approach:
Distinguish primary metabolites from secondary metabolites among the listed compounds.
Step 1:Amino acids and glucose are included under the category of primary metabolites as they have identifiable functions and play known roles in normal physiological processes.
Step 2:Morphine, codeine, vinblastin, curcumin, rubber and gums are included under the category of secondary metabolites as their role or functions in host organisms is not known yet; many of them are useful to human welfare.
Final answer: Amino acids, glucose
Q136Single correctRespiration in Plants
Which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2In ETC (Electron Transport Chain), one molecule of NADH + gives rise to 2 ATP molecules, and one gives rise to 3 ATP molecules
Approach:
Compare each statement against established facts of aerobic respiration to find the false one.
Step 1:In the electron transport chain one NADH + gives rise to 3 ATP molecules and one gives rise to 2 ATP molecules, so the statement quoting 2 and 3 has the two figures the wrong way round.
Step 2:Oxygen acts only at the terminal stage as the final electron acceptor, ATP is synthesised through complex V, and oxidation-reduction reactions establish the proton gradient, so the remaining statements are correct.
Final answer: In ETC (Electron Transport Chain), one molecule of NADH + gives rise to 2 ATP molecules, and one gives rise to 3 ATP molecules
Q137Single correctMineral Nutrition
Match Column-I with Column-II.
Choose the correct answer from options given below.
Choose the correct answer from options given below.
| Column-I | Column-II |
|---|---|
| (a). Nitrococcus | (i). Denitrification |
| (b). Rhizobium | (ii). Conversion of ammonia to nitrite |
| (c). Thiobacillus | (iii). Conversion of nitrite to nitrate |
| (d). Nitrobacter | (iv). Conversion of atmospheric nitrogen to ammonia |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Approach:
Assign each bacterium to its specific role in the nitrogen cycle.
Step 1:Conversion of atmospheric nitrogen N to N (ammonia) is carried out by fixers such as Rhizobium.
Step 2:N is converted to NO (nitrite) by nitrifying bacteria such as Nitrococcus.
Step 3:NO is converted to NO (nitrate) by nitrifying bacteria called Nitrobacter.
Step 4:Thiobacillus carries out denitrification, a process which converts NO/NO to .
Final answer: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Q138Single correctMorphology of Flowering Plants
Match Column-I with Column-II
Select the correct answer from the options given below.
Select the correct answer from the options given below.
| Column-I | Column-II |
|---|---|
| (a). | (i). Brassicaceae |
| (b). | (ii). Liliaceae |
| (c). | (iii). Fabaceae |
| (d). | (iv). Solanaceae |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Approach:
Identify each plant family from its characteristic floral formula.
Step 1:The Solanaceae family bears a floral formula with gamosepalous calyx, gamopetalous corolla, five stamens and a bicarpellary syncarpous gynoecium with superior ovary.
Step 2:The Liliaceae family shows a perianth-based formula with six tepals, six stamens and a tricarpellary superior ovary.
Step 3:The Fabaceae family shows numerous stamens with a polycarpellary feature in its formula and a monocarpellary superior ovary pattern matching the apocarpous gynoecium with many ovules.
Step 4:The Brassicaceae family shows a tetradynamous androecium with stamens in the pattern 1+2 and a bicarpellary ovary.
Final answer: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Q139Single correctMolecular Basis of Inheritance
Identify the correct statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria
Approach:
Each statement on transcription and RNA processing is tested against the NCERT account, and the single correct one is retained.
Step 1:Capping adds an unusual nucleotide, methyl guanosine triphosphate, to the 5' end of the hnRNA, not the 3' end, so statement 1 is incorrect.
Step 2:It is the template strand, not the coding strand, that is copied into mRNA, so statement 3 is incorrect.
Step 3:Split gene arrangement with introns and exons is a feature of eukaryotes, not prokaryotes, so statement 4 is incorrect.
Step 4:In bacteria the termination factor Rho associates with RNA polymerase and alters its activity so that transcription stops, making statement 2 correct.
Final answer: RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria
Q140Single correctBiological Classification / Reproduction
Which of the following statements is correct ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Fusion of protoplasms between two motile or non-motile gametes is called plasmogamy
Approach:
Each definition is checked against the NCERT terminology for fungal sexual reproduction, nutrition and cyanobacterial nitrogen fixation.
Step 1:Karyogamy is the fusion of two nuclei, not the fusion of two cells, so statement 1 is incorrect.
Step 2:Organisms that depend on living plants or animals are parasites; saprophytes obtain nutrition from dead and decaying organic matter, so statement 3 is incorrect.
Step 4:Cyanobacteria fix atmospheric nitrogen in specialized thick-walled cells called heterocysts, not sheath cells, so statement 4 is incorrect.
Step 5:Plasmogamy is the fusion of protoplasms between two motile or non-motile gametes, which is the correct definition.
Final answer: Fusion of protoplasms between two motile or non-motile gametes is called plasmogamy
Q141Single correctAnatomy of Flowering Plants
Select the correct pair.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cells of medullary rays that form part of cambial ring - Interfascicular cambium
Approach:
Match each cell description with the correct anatomical term to find the valid pair.
Step 1:When the cells of the medullary rays differentiate they give rise to the new cambium called interfascicular cambium, so that pairing is the correct one.
Step 2:Large colourless empty cells in the epidermis of grass leaves are bulliform cells, not subsidiary cells, so that pairing is wrong.
Step 3:In dicot leaves the vascular bundles are surrounded by large thick-walled bundle sheath cells, not conjunctive tissue, so that pairing is wrong.
Step 4:Loose parenchyma cells rupturing the epidermis and forming a lens-shaped opening in bark form a lenticel, not spongy parenchyma, so that pairing is wrong.
Final answer: Cells of medullary rays that form part of cambial ring - Interfascicular cambium
Q142Single correctMolecular Basis of Inheritance
DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Repetitive DNA
Approach:
The class of DNA sequence whose variation forms the basis of DNA fingerprinting is identified from the NCERT account.
Step 1:A large portion of the eukaryotic genome consists of repetitive DNA, the specific regions whose differences DNA fingerprinting identifies.
Step 2:The technique relies on Variable Number of Tandem Repeats (VNTR), a category of satellite (repetitive) DNA showing very high polymorphism, which serves as the probe.
Final answer: Repetitive DNA
Q143Single correctSexual Reproduction in Flowering Plants
In some members of which of the following pairs of families, pollen grains retain their viability for months after release?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Rosaceae ; Leguminosae
Approach:
Recall the families in which pollen retains long viability versus those with short viability.
Step 1:In members of some plant families like Solanaceae, Rosaceae and Leguminosae the pollen grains retain their viability for months after release.
Step 2:In cereals (Poaceae) pollen grains retain viability for around 30 minutes, so any pair containing Poaceae is excluded.
Final answer: Rosaceae ; Leguminosae
Q144Single correctPhotosynthesis in Higher Plants
Which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cyclic photophosphorylation involves both PS I and PS II
Approach:
Check each statement about photophosphorylation to find the false one.
Step 1:Cyclic photophosphorylation involves only PS I, whereas both PS I and PS II operate in non-cyclic photophosphorylation where ATP and NADPH + are both formed, so the statement crediting cyclic photophosphorylation with NADPH synthesis is wrong.
Step 2:Both PS I and PS II are found on grana lamellae whereas stroma lamellae have PS I only and lack NADP reductase, supporting statements 2 and 3.
Final answer: Cyclic photophosphorylation involves both PS I and PS II
Q145Single correctMolecular Basis of Inheritance
What is the role of RNA polymerase III in the process of transcription in eukaryotes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Transcribes tRNA, 5s rRNA and snRNA
Approach:
Recall the transcripts produced by each eukaryotic RNA polymerase.
Step 1:RNA polymerase III transcribes tRNA, 5S rRNA and snRNA.
Step 2:RNA polymerase I transcribes 5.8S, 18S and 28S rRNA, and RNA polymerase II transcribes hnRNA which is the precursor of mRNA.
Final answer: Transcribes tRNA, 5s rRNA and snRNA
Q146Single correctOrganisms and Populations
In the exponential growth equation , e represents
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The base of natural logarithms
Approach:
Interpret the symbol e in the exponential population growth equation.
Step 1:In the exponential growth equation , e represents the base of natural logarithms.
Step 2:Here is population density after time t, is population density at time zero, and r is the intrinsic rate of natural increase called biotic potential.
Final answer: The base of natural logarithms
Q147Single correctBiotechnology - Principles and Processes
Now a days it is possible to detect the mutated gene causing cancer by allowing radioactive probe to hybridise its complementary DNA in a clone of cells, followed by its detection using autoradiography because :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3mutated gene does not appear on a photographic film as the probe has no complementarity with it
Approach:
Reason out why a mutated gene fails to register in autoradiography with a normal probe.
Step 1:Autoradiography allows the detection or localisation of radioactive isotope within a biological sample; a probe is a radiolabelled single-stranded RNA or DNA depending on the technique.
Step 2:To identify the mutated gene a probe is allowed to hybridise to its complementary DNA in a clone of cells, but if the gene has mutated, hybridisation does not take place and the probe will not appear on the photographic film.
Step 3:Because the mutated gene lacks complementarity with the normal probe, it does not appear on the photographic film.
Final answer: mutated gene does not appear on a photographic film as the probe has no complementarity with it
Q148Single correctBiomolecules
Match List-I with List-II.
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). Protein | (i). C = C double bonds |
| (b). Unsaturated fatty acid | (ii). Phosphodiester bonds |
| (c). Nucleic acid | (iii). Glycosidic bonds |
| (d). Polysaccharide | (iv). Peptide bonds |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Pair each biomolecule with the characteristic bond that links its monomers or structure.
Step 1:In a polypeptide or protein, amino acids are linked by a peptide bond which is formed when the carboxyl (-COOH) group of one amino acid reacts with the amino (-N) group of the next amino acid with the elimination of a water moiety.
Step 2:Unsaturated fatty acids contain one or more C = C double bonds.
Step 3:In nucleic acids, a phosphate moiety links the 3'-carbon of one sugar of the nucleotide to the 5'-carbon of the sugar of the succeeding nucleotide, the bond between the phosphate and hydroxyl group being an ester bond and giving a phosphodiester bond.
Step 4:In a polysaccharide, the individual monosaccharides are linked by a glycosidic bond.
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q149Single correctCell Cycle and Cell Division
Match List-I with List-II. Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| a. S phase | i. Proteins are synthesized |
| b. phase | ii. Inactive phase |
| c. Quiescent stage | iii. Interval between mitosis and initiation of DNA replication |
| d. phase | iv. DNA replication |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Each phase of the cell cycle is paired with its defining event following the NCERT description of interphase.
Step 1:During the S (synthesis) phase the DNA content is doubled by replication.
Step 2:In the G2 phase the cell grows and proteins are synthesized in preparation for mitosis.
Step 3:Cells that exit the cycle into the quiescent stage (G0) remain metabolically active but do not proliferate, an inactive phase with respect to division.
Step 4:The G1 phase is the interval between the end of mitosis and the initiation of DNA replication.
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q150Single correctBiotechnology - Principles and Processes
Plasmid pBR322 has PstI restriction enzyme site within gene am that confers ampicillin resistance. If this enzyme is used for inserting a gene for -galactoside production and the recombinant plasmid is inserted in an E.coli strain
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1It will not be able to confer ampicillin resistance to the host cell
Approach:
Apply insertional inactivation to the ampicillin resistance gene of pBR322.
Step 1:pBR322 is a commonly used cloning vector; when the gene for -galactoside is inserted into the site of the am gene by using PstI, the recombinant E.coli will lose ampicillin resistance due to insertional inactivation of the antibiotic resistance gene.
Step 2:The host recombinant cell will produce -galactoside which is not a novel protein but does have ampicillin resistance.
Step 3:A recombinant E.coli is produced and the host cell will not undergo lysis due to insertion of -galactoside gene.
Final answer: It will not be able to confer ampicillin resistance to the host cell
Q151Single correctBiotechnology and its Applications
Identify the pair
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Drugs Ricin
Approach:
Each pairing links a class of biomolecule with an example; the task is to find the mismatched pair.
Step 1:Ricin is a toxin obtained from the seeds of the Ricinus plant, alongside vinblastin and curcumin.
Step 2:Codeine and morphine are alkaloids, abrin is a toxin obtained from the plant Abrus, and concanavalin A is a lectin, so the other three pairs are correctly matched.
Final answer: Drugs Ricin
Q152Single correctCell Cycle and Cell Division
The fruit fly has 8 chromosomes (2n) in each cell. During interphase of Mitosis if the number of chromosomes at phase is 8, what would be the number of chromosomes after S phase?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 18
Approach:
S phase replicates the DNA but does not change the chromosome count, so the number remains the same across the phase.
Step 1:During S phase the duplication of DNA occurs, so the amount of DNA increases but the chromosome number does not change.
Step 2:Since the number of chromosomes at phase is 8, the number after S phase is also 8.
Final answer: 8
Q153Single correctAnimal Kingdom
Match List - I with List - II
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). Metamerism | (i). Coelenterata |
| (b). Canal system | (ii). Ctenophora |
| (c). Comb plates | (iii). Annelida |
| (d). Cnidoblasts | (iv). Porifera |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Approach:
Each diagnostic feature is assigned to the phylum where it is characteristic.
Step 1:Metamerism, the segmentation of the body into serially repeated units, is commonly seen in members of phylum Annelida.
Step 2:A water canal system is present in members of phylum Porifera, and comb plates (ciliated plates) used in locomotion characterise phylum Ctenophora.
Step 3:Cnidoblasts or cnidocytes are characteristic of cnidarians (Coelenterata).
Final answer: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Q154Single correctReproductive Health
Match List-I with List-II.
Choose the correct answer from the options given below
Choose the correct answer from the options given below
| List-I | List-II |
|---|---|
| (a). Vaults | (i). Entry of sperm through Cervix is blocked |
| (b). IUDs | (ii). Removal of Vas deferens |
| (c). Vasectomy | (iii). Phagocytosis of sperms within the Uterus |
| (d). Tubectomy | (iv). Removal of fallopian tube |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Approach:
Each contraceptive method is matched with its mechanism of action.
Step 1:Diaphragms, cervical caps and vaults are barrier methods for females that work by blocking the entry of sperms through the cervix.
Step 2:IUDs increase phagocytosis of sperms within the uterus.
Step 3:Vasectomy involves cutting or removal of a part of the vas deferens, and tubectomy involves removal of a part of the fallopian tube.
Final answer: (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
Q155Single correctMolecular Basis of Inheritance
If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3T : 30 ; G : 20 ; C : 20
Approach:
Chargaff's rule gives equal amounts of complementary bases, allowing the remaining percentages to be derived.
Step 1:By Chargaff's rule for double-stranded DNA, adenine equals thymine, so thymine is also 30%.
Step 2:Adenine and thymine together account for 60%, leaving 40% to be shared equally between guanine and cytosine.
Step 3:Since guanine equals cytosine, each makes up 20%.
Final answer: T : 30 ; G : 20 ; C : 20
Q156Single correctMolecular Basis of Inheritance
Which of the following RNAs is not required for the synthesis of protein?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4siRNA
Approach:
Three RNA types carry out translation, while the remaining type has a regulatory role unrelated to protein synthesis.
Step 1:siRNAs are small interfering RNAs, also called silencing RNAs, that are double-stranded and non-coding.
Step 2:mRNA is the messenger that carries genetic information from DNA, tRNA carries amino acids to the mRNA during translation, and rRNA forms ribosomes which are involved in translation.
Final answer: siRNA
Q157Single correctReproductive Health
Which one of the following is an example of Hormone releasing IUD?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2LNG-20
Approach:
Intrauterine devices are classified as non-medicated, copper-releasing, or hormone-releasing, and the example must fall in the hormone-releasing group.
Step 1:LNG-20 is a hormone releasing IUD which makes the uterus unsuitable for implantation and the cervix hostile to sperms.
Step 2:Multiload 375, CuT and Cu7 are copper releasing IUDs which suppress sperm motility and the fertilizing capacity of sperms.
Final answer: LNG-20
Q158Single correctDigestion and Absorption
Succus entericus is referred to as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Intestinal juice
Approach:
The term succus entericus is matched with the correct digestive secretion by its site of production.
Step 1:Succus entericus is the name given to the intestinal juice, the acidic fluid found in the stomach is chyme, pancreatic acinar cells produce pancreatic juice, and gastric glands in the stomach secrete gastric juice.
Final answer: Intestinal juice
Q159Single correctHuman Health and Disease
Chronic auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscle is called as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Myasthenia gravis
Approach:
The described chronic autoimmune attack on the neuromuscular junction identifies a specific muscle disorder.
Step 1:Myasthenia gravis is a chronic auto immune disorder affecting neuromuscular junction, leading to fatigue, weakening and paralysis of skeletal muscle.
Step 2:Gout is due to deposition of uric acid crystals in joints, arthritis is inflammation of joints, and muscular dystrophy is a genetic disorder causing progressive degeneration of skeletal muscle.
Final answer: Myasthenia gravis
Q160Single correctBiotechnology and its Applications
With regard to insulin choose options.
(a) C-peptide is not present in mature insulin.
(b) The insulin produced by rDNA technology has C-peptide.
(c) The pro-insulin has C-peptide
(d) A-peptide and B-peptide of insulin are interconnected by disulphide bridges.
Choose the answer from the options given below
(a) C-peptide is not present in mature insulin.
(b) The insulin produced by rDNA technology has C-peptide.
(c) The pro-insulin has C-peptide
(d) A-peptide and B-peptide of insulin are interconnected by disulphide bridges.
Choose the answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a), (c) and (d) only
Approach:
Each statement about insulin structure and its synthesis is evaluated for correctness.
Step 1:Insulin is synthesized as a pro-hormone which contains an A-chain, a B-chain and an extra stretch called the C-peptide.
Step 2:The C-peptide is removed during maturation, so it is not present in mature insulin, making statement (a) correct.
Step 3:Chains A and B are interconnected by disulphide bridges, so statement (d) is correct, whereas insulin produced by rDNA technology lacks the C-peptide, making statement (b) incorrect.
Final answer: (a), (c) and (d) only
Q161Single correctAnimal Kingdom
Which one of the following belongs to the family Muscidae?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4House fly
Approach:
Each insect is placed in its family to identify the member of Muscidae.
Step 1:Housefly belongs to the family Muscidae, class Insecta and phylum Arthropoda.
Step 2:Fire flies are placed in family Lampyridae, grasshopper in family Acrididae, and cockroach in family Blattidae.
Final answer: House fly
Q162Single correctMolecular Basis of Inheritance
Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2DNA dependent RNA polymerase
Approach:
The enzyme that single-handedly drives all three steps of prokaryotic transcription is identified by its holoenzyme structure and function.
Step 1:In prokaryotes, the DNA dependent RNA polymerase is a holoenzyme made of polypeptides () and , and it is responsible for initiation, elongation and termination during transcription.
Step 2:DNase degrades DNA, DNA dependent DNA polymerase acts in DNA replication, and DNA ligase joins the discontinuously synthesised fragments of DNA.
Final answer: DNA dependent RNA polymerase
Q163Single correctHuman Reproduction
Receptors for sperm binding in mammals are present on:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Zona pellucida
Approach:
The egg layer carrying sperm-binding receptors in mammals is identified among the egg coverings.
Step 1:Zona pellucida has receptors for sperm binding (ZP3 receptors) in mammals.
Step 2:Corona radiata is a layer of radially arranged cells of membrana granulosa, and the perivitelline space is the gap between the vitelline membrane and the zona pellucida.
Final answer: Zona pellucida
Q164Single correctCell Cycle and Cell Division
The centriole undergoes duplication during:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1S-phase
Approach:
The phase in which the centriole replicates is identified within interphase events.
Step 1:During S phase of cell cycle replication of DNA takes place, and in animal cells the centriole also duplicates in the cytoplasm.
Step 2:In phase there is duplication of mitochondria, chloroplast and Golgi bodies, and the tubulin protein is also synthesized during this phase.
Final answer: S-phase
Q165Single correctAnimal Kingdom
Which one of the following organisms bears hollow and pneumatic long bones?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Neophron
Approach:
Hollow pneumatic long bones are a flight adaptation, so the organism must be a bird.
Step 1:Hollow and pneumatic long bones are present in animals that belong to class Aves, and Neophron belongs to this class.
Step 2:Ornithorhynchus (Platypus) and Macropus (Kangaroo) belong to class Mammalia, while Hemidactylus (Wall lizard) is a member of class Reptilia.
Final answer: Neophron
Q166Single correctChemical Coordination and Integration
Erythropoietin hormone which stimulates R.B.C. formation is produced by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Juxtaglomerular cells of the kidney
Approach:
The site of erythropoietin production is identified among endocrine cell groups.
Step 1:Juxtaglomerular cells of the kidney secrete erythropoietin hormone which stimulates RBC formation.
Step 2:Alpha cells of pancreas produce hormone glucagon, the cells of rostral adenohypophysis synthesize hormones of anterior lobe of pituitary, and the cells of bone marrow are responsible for formation of formed elements.
Final answer: Juxtaglomerular cells of the kidney
Q167Single correctAnimal Kingdom
Match the following:
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List-I | List-II |
|---|---|
| (a). Physalia | (i). Pearl oyster |
| (b). Limulus | (ii). Portuguese Man of War |
| (c). Ancylostoma | (iii). Living fossil |
| (d). Pinctada | (iv). Hookworm |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each genus is matched with its common name or distinguishing description.
Step 1:Physalia is commonly known as Portuguese Man of War, and Limulus is considered a living fossil and commonly known as king crab.
Step 2:Ancylostoma is a roundworm commonly known as hookworm, and Pinctada is commonly known as pearl oyster, included in phylum Mollusca.
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q168Single correctBiotechnology - Principles and Processes
During the process of gene amplification using PCR, if very high temperature is not maintained in the beginning, then which of the following steps of PCR will be affected first?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Denaturation
Approach:
The PCR step that depends on the initial high temperature is identified from the cycle sequence.
Step 1:High temperature about 94 degrees Celsius is required for the process of denaturation whic is the first step of PCR.
Step 2:Annealing is performed at 50 to 60 degrees Celsius which is the second step that can get affected, extension follows annealing, and ligation is not a step of PCR.
Final answer: Denaturation
Q169Single correctStructural Organisation in Animals
Which of the following statements wrongly represents the nature of smooth muscle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Communication among the cells is performed by intercalated discs
Approach:
Each statement is checked against the properties of smooth muscle to find the incorrect one.
Step 1:Intercalated discs are found in cardiac muscle cells, not smooth muscle, so the statement attributing them to smooth muscle is incorrect.
Step 2:Smooth muscle fibres are non-striated and involuntary in nature and are present in the wall of blood vessels, uterus, gall bladder and alimentary canal.
Final answer: Communication among the cells is performed by intercalated discs
Q170Single correctCell - The Unit of Life
The organelles that are included in the endomembrane system are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Endoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
Approach:
The endomembrane system groups organelles whose functions are coordinated and continuous, excluding semi-autonomous and non-membranous organelles.
Step 1:The endomembrane system consist of endoplasmic reticulum, Golgi complex, vacuoles and lysosomes.
Step 2:Mitochondria is a semi-autonomous cell organelle, and ribosomes are non-membranous organelles, so both are excluded from the endomembrane system.
Final answer: Endoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
Q171Single correctMicrobes in Human Welfare
Match List-I with List-II
Choose the correct answer from the options given below
Choose the correct answer from the options given below
| List-I | List-II |
|---|---|
| (a). Aspergillus niger | (i). Acetic Acid |
| (b). Acetobacter aceti | (ii). Lactic Acid |
| (c). Clostridium butylicum | (iii). Citric Acid |
| (d). Lactobacillus | (iv). Butyric Acid |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Each microbe is matched with the organic acid it is used to produce industrially.
Step 1:Aspergillus niger is involved in production of citric acid, and Acetobacter aceti is involved in production of acetic acid.
Step 2:Clostridium butylicum is involved in production of butyric acid, and Lactobacillus is involved in the production of lactic acid.
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q172Single correctBiotechnology and its Applications
For effective treatment of the disease, early diagnosis and understanding its pathophysiology is very important. Which of the following molecular diagnostic techniques is very useful for early detection?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3ELISA Technique
Approach:
An early-detection molecular diagnostic technique works either by detecting a pathogen-specific nucleic acid sequence or by detecting the antigen or antibodies.
Step 1:ELISA can be used for early detection of an infection either by detecting the presence of pathogenic antigen or by detecting the antibodies synthesized against the pathogen.
Step 2:Southern blotting detects a specific DNA sequence in a given sample and can detect the presence of pathogenic DNA or RNA prior to antibody formation, while hybridization with a labelled ssDNA or ssRNA probe is used to find a mutated gene, and Western blotting detects a specific protein.
Final answer: ELISA Technique
Q173Single correctStructural Organisation in Animals
Which of the following characteristics is with respect to cockroach?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A ring of gastric caeca is present at the junction of midgut and hind gut
Approach:
Each anatomical statement about the cockroach is checked against its morphology to find the incorrect one.
Step 1:A ring of gastric caecae is present at the junction of foregut and midgut, at the junction of midgut and hindgut malpighian tubules are present, so the statement placing gastric caeca between midgut and hindgut is incorrect.
Step 2:Hypopharynx lies within the cavity enclosed by mouthparts, in females the 7th sternum is boat shaped and together with the 8th and 9th sterna forms a genital pouch, and the 10th abdominal segment in both sexes bears a pair of anal cerci.
Final answer: A ring of gastric caeca is present at the junction of midgut and hind gut
Q174Single correctBody Fluids and Circulation
Persons with 'AB' blood group are called as "Universal recipients". This is due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Absence of antibodies, anti-A and anti-B, in plasma
Approach:
Universal recipient status depends on which antibodies are absent from the plasma so that donor red cells are not agglutinated.
Step 1:Persons with AB blood group have both antigens A and B on the surface of RBCs and lack antibodies anti-A and anti-B in their plasma.
Step 2:Because no antibodies are present, persons with AB blood group can accept blood from persons with AB as well as the other groups of blood without agglutination, so they are called Universal recipients.
Final answer: Absence of antibodies, anti-A and anti-B, in plasma
Q175Single correctEnvironmental Issues
Dobson units are used to measure thickness of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ozone
Approach:
The Dobson unit is matched to the atmospheric quantity it measures.
Step 1:The thickness of the ozone in a column of air from the ground to the top of atmosphere is measured in terms of Dobson units.
Step 2:The lowermost layer of atmosphere is the troposphere, CFCs are ozone depleting substances, and ozone found in the upper part of atmosphere, the stratosphere, is called good ozone.
Final answer: Ozone
Q176Single correctAnimal Kingdom
Read the following statements
(a) Metagenesis is observed in Helminths.
(b) Echinoderms are triploblastic and coelomate animals.
(c) Round worms have organ-system level of body organization.
(d) Comb plates present in ctenophores help in digestion.
(e) Water vascular system is characteristic of Echinoderms.
Choose the answer from the options given below.
(a) Metagenesis is observed in Helminths.
(b) Echinoderms are triploblastic and coelomate animals.
(c) Round worms have organ-system level of body organization.
(d) Comb plates present in ctenophores help in digestion.
(e) Water vascular system is characteristic of Echinoderms.
Choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(b), (c) and (e) are correct
Approach:
Each statement is evaluated against the diagnostic features of the respective animal phyla.
Step 1:Metagenesis, the alternation of asexual polyp and sexual medusa generations, occurs in members of phylum Coelenterata (Cnidaria), not in Helminths, so statement (a) is incorrect.
Step 2:Echinoderms are triploblastic and coelomate animals, so statement (b) is correct.
Step 3:Roundworms (Aschelminthes) show organ-system level of body organization, so statement (c) is correct.
Step 4:Comb plates in ctenophores function in locomotion, not in digestion, so statement (d) is incorrect.
Step 5:Water vascular system, used in locomotion, capture and transport of food and respiration, is a characteristic feature of Echinoderms, so statement (e) is correct.
Final answer: (b), (c) and (e) are correct
Q177Single correctPrinciples of Inheritance and Variation
In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 325%
Approach:
Sickle cell anaemia is an autosomal recessive disorder, so the diseased phenotype appears only in homozygous recessive progeny.
Step 1:Both parents are heterozygous carriers, represented as .
Step 2:The monohybrid cross between two heterozygotes yields a genotypic ratio of one homozygous dominant, two heterozygous and one homozygous recessive.
Step 3:Only the homozygous recessive individuals express sickle cell anaemia, which is one out of four progeny.
Final answer: 25%
Q178Single correctBody Fluids and Circulation
Which enzyme is responsible for the conversion of inactive fibrinogens to fibrins?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Thrombin
Approach:
The terminal step of blood coagulation is identified along with the enzyme catalysing fibrin formation.
Step 1:During coagulation of blood, an enzyme complex thrombokinase helps in the conversion of prothrombin (present in plasma) into thrombin.
Step 2:Thrombin further catalyses the conversion of inactive fibrinogens into fibrin monomers that form a network of threads.
Step 3:Renin is secreted by JG cells of the kidney, and epinephrine is an adrenal hormone, so neither acts on fibrinogen.
Final answer: Thrombin
Q179Single correctBreathing and Exchange of Gases
Select the favourable conditions required for the formation of oxyhaemoglobin at the alveoli.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1High , low , less , lower temperature
Approach:
The conditions promoting association of oxygen with haemoglobin in the alveoli are derived from the oxygen dissociation curve.
Step 1:Binding of oxygen to haemoglobin is favoured by a high partial pressure of oxygen, which prevails in the alveoli.
Step 2:Low partial pressure of carbon dioxide, low hydrogen ion concentration and lower temperature shift the dissociation curve to favour formation of oxyhaemoglobin.
Step 3:The alveolar conditions of high pO2, low pCO2, lesser hydrogen ion concentration and lower temperature together favour oxyhaemoglobin formation.
Final answer: High , low , less , lower temperature
Q180Single correctDigestion and Absorption
Sphincter of oddi is present at:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Junction of hepato-pancreatic duct and duodenum
Approach:
The anatomical location of the sphincter of Oddi within the digestive tract is identified.
Step 1:The bile duct and the pancreatic duct open together into the duodenum as the common hepato-pancreatic duct.
Step 2:This common duct is guarded by a sphincter called the sphincter of Oddi at its junction with the duodenum.
Step 3:The ileo-caecal valve lies at the ileum-caecum junction and the gastro-oesophageal sphincter regulates the opening of oesophagus into the stomach, so neither corresponds to the sphincter of Oddi.
Final answer: Junction of hepato-pancreatic duct and duodenum
Q181Single correctCell Cycle and Cell Division
Which stage of meiotic prophase shows terminalisation of chiasmata as its distinctive feature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Diakinesis
Approach:
The substages of prophase I are recalled to locate where chiasmata terminalisation occurs.
Step 1:Chiasmata, the X-shaped structures, are formed in diplotene stage while it terminalise in diakinesis stage.
Step 2:Bivalents are formed in zygotene stage and crossing over takes place in pachytene stage, so these stages do not show terminalisation.
Step 3:Compaction of chromosomal material occurs in leptotene stage, which precedes chiasma formation.
Final answer: Diakinesis
Q182Single correctBiotechnology and its Applications
Which of the following is an objective of Biofortification in crops?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Improve resistance to diseases
Approach:
The defined aims of biofortification are compared with each listed objective to find the one that falls outside them.
Step 1:Biofortification aims to improve vitamin content, protein content and micronutrient and mineral content of crops.
Step 2:Improvement of resistance to diseases is not a stated objective of biofortification, which targets nutritional quality rather than disease resistance.
Final answer: Improve resistance to diseases
Q183Single correctBreathing and Exchange of Gases
The partial pressures (in mm Hg) of oxygen and carbon dioxide at alveoli (the site of diffusion) are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
The standard partial pressures of respiratory gases in alveolar air are recalled.
Step 1:In atmospheric air the partial pressure of oxygen is 159 mm Hg and that of carbon dioxide is 0.3 mm Hg.
Step 2:In the alveoli the partial pressure of oxygen is 104 mm Hg and the partial pressure of carbon dioxide is 40 mm Hg.
Step 3:In deoxygenated blood the partial pressure of oxygen is 40 mm Hg and that of carbon dioxide is 45 mm Hg, while in oxygenated blood the partial pressure of oxygen is 95 mm Hg and that of carbon dioxide is 40 mm Hg.
Final answer: and
Q184Single correctHuman Health and Disease
Venereal diseases can spread through :
(a) Using sterile needles
(b) Transfusion of blood from infected person
(c) Infected mother to foetus
(d) Kissing
(e) Inheritance
Choose the answer from the options given below
(a) Using sterile needles
(b) Transfusion of blood from infected person
(c) Infected mother to foetus
(d) Kissing
(e) Inheritance
Choose the answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(b) and (c) only
Approach:
Each listed route is checked against the recognised modes of transmission of venereal (sexually transmitted) diseases.
Step 1:Venereal diseases are transmitted by sharing of infected needles, surgical instruments with infected person, transfusion of blood or from an infected mother to foetus.
Step 2:Use of sterile needles does not transmit infection, so statement (a) is incorrect.
Step 3:Venereal diseases are not transmitted through kissing or inheritance, so statements (d) and (e) are incorrect.
Final answer: (b) and (c) only
Q185Single correctBiotechnology - Principles and Processes
A specific recognition sequence identified by endonucleases to make cuts at specific positions within the DNA is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Palindromic Nucleotide sequences
Approach:
The nature of the sequence recognised by restriction endonucleases is identified.
Step 1:Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA and once it finds its specific recognition sequence it bind to DNA and cuts each of the two strands of DNA.
Step 2:Okazaki fragments arise during discontinuous lagging-strand replication and poly(A) tail is a post-transcriptional modification in eukaryotes, so neither is a restriction recognition site.
Step 3:A PCR primer sequence is termed degenerate if some of its position have several possible bases, which is unrelated to endonuclease recognition.
Final answer: Palindromic Nucleotide sequences
Q186Single correctMolecular Basis of Inheritance
Which one of the following statements about Histones is ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The pH of histones is slightly acidic
Approach:
Each statement is evaluated against the known structural and chemical properties of histone proteins.
Step 1:Histones are rich in the basic amino acids lysine and arginine, which carry positively charged side chains, so statements (3) and (4) are correct.
Step 2:There are five types of histone proteins, namely H1, H2A, H2B, H3 and H4, and four of them occur in pairs to form a unit of 8 molecules called the histone octamer, so statement (1) is correct.
Step 3:The pH of histones is basic, not acidic, owing to the abundance of basic amino acids, so statement (2) is the wrong statement.
Final answer: The pH of histones is slightly acidic
Q187Single correctLocomotion and Movement
During muscular contraction which of the following events occur?
(a) 'H' zone disappears
(b) 'A' band widens
(c) 'I' band reduces in width
(d) Myosine hydrolyzes ATP, releasing the ADP and Pi.
(e) Z-lines attached to actins are pulled inwards.
Choose the answer from the options given below:
(a) 'H' zone disappears
(b) 'A' band widens
(c) 'I' band reduces in width
(d) Myosine hydrolyzes ATP, releasing the ADP and Pi.
(e) Z-lines attached to actins are pulled inwards.
Choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a), (c), (d), (e) only
Approach:
Each event is verified against the sliding filament mechanism of muscle contraction.
Step 1:During contraction the globular head of myosin acts as ATPase and hydrolyses ATP molecule and eventually leads to the formation of cross bridge, so statement (d) is correct.
Step 2:The myosin head pulls the actin filament towards the centre of 'A-band', so the Z-line attached to these actins are also pulled inwards thereby causing a shortening of the sarcomere, so statement (e) is correct.
Step 3:The thin myofilaments move past the thick myofilaments due to which the H-zone narrows, which reduces the length of I-band but retains the length of A-band, so statements (a) and (c) are correct while (b) is incorrect.
Final answer: (a), (c), (d), (e) only
Q188Single correctLocomotion and Movement
Match List-I with List-II
Choose the correct answer from the options given below
Choose the correct answer from the options given below
| List -I | List -II |
|---|---|
| (a). Scapula | (i). Cartilaginous joints |
| (b). Cranium | (ii). Flat bone |
| (c). Sternum | (iii). Fibrous joints |
| (d). Vertebral column | (iv). Triangular flat bone |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Approach:
Each skeletal structure in List-I is matched to its descriptive feature or joint type in List-II.
Step 1:Scapula is a large triangular flat bone situated in the dorsal part of the thorax, so (a) matches (iv).
Step 2:Fibrous joint is shown by the flat skull bones which fuse end-to-end with the help of dense fibrous connective tissues in the form of sutures to form the cranium, so (b) matches (iii).
Step 3:Sternum is a flat bone on the ventral midline of thorax, so (c) matches (ii).
Step 4:Cartilaginous joints between the adjacent vertebrae in the vertebral column permit limited movements, so (d) matches (i).
Final answer: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Q189Single correctHuman Reproduction
Which of these is an important component of initiation of parturition in humans?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Release of Prolactin
Approach:
The hormonal events that trigger the onset of parturition are reviewed to identify the exception.
Step 1:At the end of gestation, the completely developed foetus is expelled out, a process called parturition controlled by a complex neuroendocrine mechanism.
Step 2:The estrogen and progesterone ratio increases as the estrogen levels rise significantly, and prostaglandins stimulate uterine contractions, so these contribute to initiation.
Step 3:Oxytocin, the main hormone also called birth hormone, is released by maternal pituitary and brings about strong uterine contractions, whereas prolactin is a lactation hormone that has no role in initiation of parturition.
Final answer: Release of Prolactin
Q190Single correctBreathing and Exchange of Gases
A person goes to high altitude and experiences 'altitude sickness' with symptoms like breathing difficulty and heart palpitations.
Due to low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
In the light of the above statements, choose the correct answer from the options given below
Due to low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
In the light of the above statements, choose the correct answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both (A) and (R) are true and (R) is the correct explanation of (A)
Approach:
The truth of the assertion and reason and their causal relationship are assessed for altitude sickness.
Step 1:Altitude sickness can be experienced at high altitude where the body does not get enough oxygen due to low atmospheric pressure and causes nausea, fatigue and heart palpitations, so the assertion is true.
Step 2:The low atmospheric pressure at high altitude lowers the partial pressure of oxygen and reduces oxygen supply to the body, so the reason is true.
Step 3:The reduced oxygen availability stated in the reason directly accounts for the breathing difficulty and palpitations stated in the assertion, so the reason is the correct explanation of the assertion.
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q191Single correctBiomolecules
Following are the statements with reference to 'lipids'.
(a) Lipids having only single bonds are called unsaturated fatty acids.
(b) Lecithin is a phospholipid.
(c) Trihydroxy propane is glycerol.
(d) Palmitic acid has 20 carbon atoms including carboxyl carbon.
(e) Arachidonic acid has 16 carbon atoms.
Choose the answer from the options given below.
(a) Lipids having only single bonds are called unsaturated fatty acids.
(b) Lecithin is a phospholipid.
(c) Trihydroxy propane is glycerol.
(d) Palmitic acid has 20 carbon atoms including carboxyl carbon.
(e) Arachidonic acid has 16 carbon atoms.
Choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(b) and (c) only
Approach:
Each statement about lipids is checked against the definitions and carbon counts of the named molecules.
Step 1:Lipids having only single bonds between carbon atoms are saturated fatty acids, not unsaturated, so statement (a) is incorrect.
Step 2:Lecithin is a phospholipid found in cell membrane and glycerol is trihydroxy propane, so statements (b) and (c) are correct.
Step 3:Palmitic acid has 16 carbon atoms including carboxyl carbon and arachidonic acid has 20 carbon atoms, so statements (d) and (e) are incorrect.
Final answer: (b) and (c) only
Q192Single correctHuman Health and Disease
The Adenosine deaminase deficiency results into
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Dysfunction of Immune system
Approach:
The physiological consequence of adenosine deaminase deficiency is identified.
Step 1:Adenosine deaminase (ADA) enzyme is crucial for the immune system to function, hence its deficiency results in the dysfunction of immune system.
Step 2:Hyposecretion of hormones of the adrenal cortex causes Addison's disease, which is unrelated to this enzyme.
Step 3:Parkinson's disease is a long-term degenerative disorder of the central nervous system, and disorders affecting GIT and associated glands are called digestive disorders, so neither results from ADA deficiency.
Final answer: Dysfunction of Immune system
Q193Single correctHuman Reproduction
Which of the following secretes the hormone, relaxin, during the later phase of pregnancy?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Corpus luteum
Approach:
The source of relaxin during the later phase of pregnancy is identified among the listed structures.
Step 1:The hormone relaxin is produced in the later phase of pregnancy by the ovary.
Step 2:A Graafian follicle is not formed during pregnancy, and neither the uterus nor the foetus produces relaxin, so none of those three can be the source.
Step 3:Relaxin is produced by the corpus luteum present in the ovary, since the ruptured Graafian follicle is called corpus luteum and it retains its endocrine function.
Final answer: Corpus luteum
Q194Single correctStructural Organisation in Animals
Identify the types of cell junctions that help to stop the leakage of the substances across a tissue and facilitation of communication with neighbouring cells via rapid transfer of ions and molecules.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Tight junctions and Gap junctions, respectively
Approach:
The two described functions are matched to the corresponding types of intercellular junctions in order.
Step 1:Tight junctions help to stop leakage of substances from leaking across a tissue, so the first function corresponds to tight junctions.
Step 2:Gap junctions facilitate communication between cells by connecting the cytoplasm of adjoining cells for rapid transfer of ions and molecules, so the second function corresponds to gap junctions.
Step 3:Adhering junctions cement neighbouring cells together but neither block leakage nor allow communication, so any pairing that uses them fails.
Final answer: Tight junctions and Gap junctions, respectively
Q195Single correctReproductive Health
Which of the following is a step in Multiple Ovulation Embryo Transfer Technology (MOET)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cow is administered hormone having LH like activity for super ovulation
Approach:
Each statement is compared with the actual steps of the MOET protocol used for cattle improvement.
Step 1:Multiple Ovulation Embryo Transfer Technology is used for herd improvement.
Step 2:Cows are administered hormones with FSH-like activity for superovulation, so administering a hormone with LH-like activity for superovulation is not a correct step.
Step 3:In MOET 6-8 eggs are produced per cycle, cows are fertilised by artificial insemination and the fertilised embryos are transferred to surrogate mothers at the 8-32 cell stage, so these are valid steps.
Final answer: Cow is administered hormone having LH like activity for super ovulation
Q196Single correctHuman Health and Disease
Match List-I with List-II
Choose the correct answer from the options given below
Choose the correct answer from the options given below
| List -I | List -II |
|---|---|
| (a). Filariasis | (i). |
| (b). Amoebiasis | (ii). |
| (c). Pneumonia | (iii). |
| (d). Ringworm | (iv). |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Each disease in List-I is matched to its causative organism in List-II.
Step 1:Filariasis is the disease caused by Wuchereria bancrofti, the filarial worm, so (a) matches (iii).
Step 2:Amoebiasis or amoebic dysentery is caused by the protozoan parasite Entamoeba histolytica in the large intestine of human, so (b) matches (iv).
Step 3:Pneumonia is caused by bacteria like Streptococcus pneumoniae and Haemophilus influenzae, so (c) matches (i).
Step 4:Ringworm is caused by fungi belonging to genera Microsporum, Trichophyton and Epidermophyton, so (d) matches (ii).
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q197Single correctOrganisms and Populations
Match List-I with List - II
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List - I | List - II |
|---|---|
| (a). Allen's Rule | (i). Kangaroo rat |
| (b). Physiological adaptation | (ii). Desert lizard |
| (c). Behavioural adaptation | (iii). Marine fish at depth |
| (d). Biochemical adaptation | (iv). Polar seal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Each adaptation type in List-I is matched to the organism that exemplifies it in List-II.
Step 1:Polar seal generally has shorter ears and limbs to minimise heat loss, illustrating Allen's rule, so (a) matches (iv).
Step 2:Kangaroo rat exhibits physiological adaptation, so (b) matches (i).
Step 3:Desert lizard shows behavioural adaptation, as it lacks the physiological ability to cope up with extreme temperature but manages the body temperature by behavioural means, so (c) matches (ii).
Step 4:Marine fishes at depth are adapted biochemically to survive in great depths in ocean, so (d) matches (iii).
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q198Single correctEvolution
Match List - I with List - II
Choose the correct answer from the options given below.
Choose the correct answer from the options given below.
| List - I | List - II |
|---|---|
| (a). Adaptive radiation | (i). Selection of resistant varieties due to excessive use of herbicides and pesticides |
| (b). Convergent evolution | (ii). Bones of forelimbs in Man and Whale |
| (c). Divergent evolution | (iii). Wings of Butterfly and Bird |
| (d). Evolution by anthropogenic action | (iv). Darwin Finches |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Approach:
Each evolutionary pattern in List-I is matched to its illustrative example in List-II.
Step 1:Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography, for example Darwin's finches, so (a) matches (iv).
Step 2:Analogous organs which are not anatomically similar structures though they perform similar functions are a result of convergent evolution, for example wings of butterfly and of bird, so (b) matches (iii).
Step 3:Homologous organs which are anatomically similar structures but perform different functions according to their needs are a result of divergent evolution, for example bones of forelimbs in man and whale, so (c) matches (ii).
Step 4:Evolution by anthropogenic action means evolution due to human interference, for example antibiotic resistant microbes and herbicide resistant varieties and pesticide resistant varieties, so (d) matches (i).
Final answer: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Q199Single correctMolecular Basis of Inheritance
The codon 'AUG' codes for methionine and phenylalanine.
'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the answer from the options given below.
'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is true
Approach:
Each statement about codon assignments is checked against the standard genetic code.
Step 1:AUG has dual functions, it codes for methionine and it also acts as initiator codon, but it does not code for phenylalanine, so Statement I is incorrect.
Step 2:The codons AAA and AAG both code for the amino acid lysine, so Statement II is true.
Final answer: Statement I is incorrect but Statement II is true
Q200Single correctStructural Organisation in Animals
Following are the statements about prostomium of earthworm.
(a) It serves as a covering for mouth.
(b) It helps to open cracks in the soil into which it can crawl.
(c) It is one of the sensory structures.
(d) It is the first body segment.
Choose the answer from the options given below.
(a) It serves as a covering for mouth.
(b) It helps to open cracks in the soil into which it can crawl.
(c) It is one of the sensory structures.
(d) It is the first body segment.
Choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a), (b) and (c) are correct
Approach:
Each statement about the prostomium is evaluated against earthworm morphology.
Step 1:The anterior end of the earthworm has a mouth which has a covering called prostomium, so statement (a) is correct.
Step 2:Prostomium acts as a wedge to force open cracks in the soil, so statement (b) is correct.
Step 3:Prostomium has receptors, so it is sensory in function, so statement (c) is correct.
Step 4:The first body segment of earthworm is the peristomium, not the prostomium, so statement (d) is incorrect.
Final answer: (a), (b) and (c) are correct
Frequently Asked Questions
How many questions are in the NEET 2021 Sep 12 paper?
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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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