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NEET 2024 Jun 23 Question Paper with Solutions
All 200 questions from the NEET 2024 (Jun 23) paper — Physics (50), Chemistry (50) and Biology (100) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2024Chemistry PYQs 2024Biology PYQs 2024
- Questions
- 200
- Physics
- 50
- Chemistry
- 50
- Biology
- 100
Physics50 questions
Q1Single correctMagnetism and Matter
The magnetic potential energy, when a magnetic bar of magnetic moment is placed perpendicular to the magnetic field is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Zero
Approach:
The potential energy of a magnetic dipole in an external field depends on the orientation between the moment and the field.
Step 1:The potential energy stored in an external magnetic field is the negative dot product of the moment and the field.
Step 2:The moment is placed perpendicular to the field, so the angle between them is 90 degrees.
Step 3:Substituting the perpendicular orientation gives zero potential energy.
Final answer: Zero
Q2Single correctSystems of Particles and Rotational Motion
A bob is whirled in a horizontal circle by means of a string at an initial speed of 10 rpm. If the tension in the string is quadrupled while keeping the radius constant, the new speed is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 120 rpm
Approach:
The tension supplies the centripetal force for the horizontal circular motion, relating tension to the square of the angular speed at fixed radius.
Step 1:For horizontal circular motion the tension equals the required centripetal force.
Step 2:At constant mass and radius the tension varies as the square of the angular speed.
Step 3:Quadrupling the tension multiplies the angular speed by two.
Final answer: 20 rpm
Q3Single correctElectric Charges and Fields
A metal cube of side 5 cm is charged with 6 C. The surface charge density on the cube is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 C
Approach:
All charge on a conductor resides on its outer surface, so the surface charge density is the total charge divided by the total surface area of the cube.
Step 1:The total surface area of a cube of side 5 cm equals six times the area of one face.
Step 2:The charge spreads over this surface, giving the surface charge density.
Final answer: C
Q4Single correctMagnetism and Matter
The incorrect relation for a diamagnetic material (all the symbols carry their usual meaning and is a small positive number) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Diamagnetic materials are characterised by small negative susceptibility and relative permeability slightly below unity, so the option claiming a relative permeability above unity is incorrect.
Step 1:For a diamagnetic material the relative permeability lies just below one.
Step 2:The susceptibility is small and negative, bounded below by minus one.
Step 3:The permeability is therefore smaller than that of free space, which makes the statement of a permeability exceeding unity false.
Final answer:
Q5Single correctMechanical Properties of Fluids
An ideal fluid is flowing in a non-uniform cross-sectional tube XY (as shown in the figure) from end X to end Y. If and are the kinetic energy per unit volume of the fluid at X and Y respectively, then the correct option is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Along a horizontal tube of varying cross-section Bernoulli's equation links the kinetic energy per unit volume to the pressure, and the narrower end carries the larger speed and kinetic energy.
Step 1:Bernoulli's principle states that the sum of kinetic energy per unit volume, potential energy per unit volume and pressure is constant.
Step 2:Applying the principle at the wider end and the narrower end on the same horizontal level relates the kinetic terms and pressures.
Step 3:The wider section carries lower speed and hence smaller kinetic energy than the narrower section, so the kinetic energy per unit volume at exceeds that at for the geometry shown.
Final answer:
Q6Single correctGravitation
The escape velocity for earth is . A planet having 9 times mass than that of earth and radius, 16 times that of earth, has the escape velocity of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The escape velocity depends on the square root of the mass-to-radius ratio, so scaling the mass and radius rescales the escape velocity by the square root of their ratio.
Step 1:The escape velocity of a body varies as the square root of mass over radius.
Step 2:Substituting nine times the mass and sixteen times the radius gives the ratio of escape velocities.
Final answer:
Q7Single correctDual Nature of Radiation and Matter
An electron and an alpha particle are accelerated by the same potential difference. Let and denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The de Broglie wavelength after acceleration through the same potential varies inversely as the square root of the product of mass and charge, and the much lighter, smaller-charge electron has the larger wavelength.
Step 1:The de Broglie wavelength equals Planck's constant divided by momentum, and the momentum gained through a potential difference depends on mass and charge.
Step 2:For the same accelerating potential the wavelength ratio is the inverse square root of the mass-charge products.
Step 3:The alpha particle is far more massive and carries twice the elementary charge, so its mass-charge product greatly exceeds that of the electron, giving the electron the larger wavelength.
Final answer:
Q8Single correctWork, Energy and Power
An object moving along horizontal x-direction with kinetic energy 10 J is displaced through m by the force N. The kinetic energy of the object at the end of the displacement x is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 34 J
Approach:
The work-energy theorem equates the work done by the net force to the change in kinetic energy, and the work is the dot product of force and displacement.
Step 1:The work done by all forces equals the change in kinetic energy.
Step 2:Taking the dot product of the force with the displacement isolates the work done.
Step 3:Solving for the final kinetic energy gives the result.
Final answer: 4 J
Q9Single correctWork, Energy and Power
An object falls from a height of 10 m above the ground. After striking the ground it loses 50% of its kinetic energy. The height upto which the object can rebounce from the ground is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 45 m
Approach:
The kinetic energy just before impact equals the initial gravitational potential energy, and after losing half of it the remaining kinetic energy converts back to potential energy at the rebound height.
Step 1:The kinetic energy just before striking the ground equals the potential energy at the drop height.
Step 2:After the impact the object retains half of its kinetic energy.
Step 3:The retained kinetic energy converts to potential energy at the rebound height, giving the new height.
Final answer: 5 m
Q10Single correctAlternating Current
In the circuit shown below, the inductance L is connected to an ac source. The current flowing in the circuit is . The voltage drop across L is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The voltage across a pure inductor leads the current by ninety degrees, and its amplitude is the product of the inductive reactance and the peak current.
Step 1:The voltage across the inductor leads the current by a quarter cycle.
Step 2:The peak voltage equals the peak current times the inductive reactance.
Step 3:Combining the amplitude and the phase shift gives the instantaneous inductor voltage.
Final answer:
Q11Single correctElectrostatic Potential and Capacitance
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 115
Approach:
The electrostatic energy stored in a capacitor is one half the capacitance times the square of the applied voltage.
Step 1:The stored energy is half the capacitance multiplied by the square of the potential difference.
Step 2:Evaluating the product gives the energy in joules, which converts to nanojoules.
Final answer: 15
Q12Single correctCurrent Electricity
A uniform wire of diameter d carries a current of 100 mA when the mean drift velocity of electrons in the wire is v. For a wire of diameter of the same material to carry a current of 200 mA, the mean drift velocity of electrons in the wire is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The current relates the drift velocity to the cross-sectional area, so for the same material the drift velocity scales as the current divided by the area.
Step 1:The current depends on the number density, the cross-sectional area and the drift velocity, with the area proportional to the square of the diameter.
Step 2:Forming the ratio for the two wires of the same material relates the drift velocities to the currents and diameters.
Step 3:Solving for the new drift velocity gives the result.
Final answer:
Q13Single correctPhysical World and Measurement
In an electrical circuit, the voltage is measured as volt and the current is measured as A. The value of the resistance is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a quotient the fractional errors add, so the resistance error is the sum of the fractional errors in voltage and current applied to the central resistance value.
Step 1:The central resistance is the ratio of the central voltage to the central current.
Step 2:The fractional error in the resistance is the sum of the fractional errors in voltage and current.
Step 3:Multiplying the fractional error by the central resistance gives the absolute error.
Final answer:
Q14Single correctAlternating Current
A step-up transformer is connected to an ac mains supply of 220 V to operate at 11000 V, 88 watt. The current in the secondary circuit, ignoring the power loss in the transformer, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 18 mA
Approach:
For an ideal lossless transformer the secondary power equals the rated power, so the secondary current is the rated power divided by the secondary voltage.
Step 1:In the secondary circuit the power equals the product of secondary voltage and secondary current.
Step 2:Solving for the secondary current gives the result.
Final answer: 8 mA
Q15Single correctMotion in a Straight Line
A particle is moving along x-axis with its position (x) varying with time (t) as . The ratio of its initial velocity to its initial acceleration, respectively, is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Differentiating the position once and twice gives the velocity and acceleration, and evaluating each at the initial instant gives their ratio.
Step 1:Differentiating the position gives the velocity as a function of time.
Step 2:Differentiating the velocity gives the acceleration as a function of time.
Step 3:The ratio of the initial velocity to the initial acceleration follows from the two values at the start.
Final answer:
Q16Single correctSystems of Particles and Rotational Motion
The radius of gyration of a solid sphere of mass 5 kg about XY is 5 m as shown in figure. The radius of the sphere is m, then the value of x is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Using the parallel axis theorem, the moment of inertia about the tangential axis equals the central value plus the mass-distance term, and equating it to the radius-of-gyration form yields the sphere radius.
Step 1:The moment of inertia about a tangent axis equals the central value of two-fifths mass radius-squared plus the parallel-axis term, totalling seven-fifths mass radius-squared.
Step 2:The same moment of inertia expressed through the radius of gyration uses the given five-metre value.
Step 3:Equating the two expressions and solving for the sphere radius gives a value matching the stated form with .
Final answer:
Q17Single correctSemiconductor Electronics
The I-V characteristics shown above are exhibited by a
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Solar cell
Approach:
The shape of the current-voltage curve identifies the device, and the characteristic shown corresponds to a solar cell operating in its power-generating quadrant.
Step 1:A solar cell produces current and voltage of opposite sign in the fourth quadrant, generating electrical power from incident light.
Step 2:The displayed characteristic matches that of a solar cell rather than a light emitting diode, Zener diode or photodiode operated as a detector.
Final answer: Solar cell
Q18Single correctMagnetism and Matter
The magnetic moment and moment of inertia of a magnetic needle are as shown, are respectively, A and kg . If it completes 10 oscillations in 10 s, the magnitude of the magnetic field is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.4 mT
Approach:
The time period of a magnetic needle oscillating in a field depends on the moment of inertia, magnetic moment and field strength, so the field is obtained from the measured period.
Step 1:The time period is the total time divided by the number of oscillations.
Step 2:Substituting the period, moment of inertia and magnetic moment into the period relation isolates the field.
Step 3:Solving for the magnetic field gives the result.
Final answer: 0.4 mT
Q19Single correctElectrostatic Potential and Capacitance
The capacitance of a capacitor with charge and a potential difference depends on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2the geometry of the capacitor
Approach:
Capacitance is a property set by the physical configuration and surrounding medium, independent of the stored charge or applied voltage.
Step 1:The capacitance of a parallel plate capacitor depends on the plate area, separation and the permittivity of the medium.
Step 2:The expression contains no dependence on charge or potential, so the capacitance is determined solely by geometry and the medium between the plates.
Final answer: the geometry of the capacitor
Q20Single correctRay Optics and Optical Instruments
Given below are two statements :
Statement I : Image formation needs regular reflection and/or refraction.
Statement II : The colours of objects we see around us is due to the constituent colours of the light incident on them.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Image formation needs regular reflection and/or refraction.
Statement II : The colours of objects we see around us is due to the constituent colours of the light incident on them.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both Statement I and Statement II are correct
Approach:
Each statement is assessed against the physics of image formation and colour perception, and both are found to be true.
Step 1:Regular reflection or refraction is required for a clear image; without it, as in a rough surface, no image such as the one in a mirror forms.
Step 2:Different objects in identical white light appear in different colours because the constituent colours of the incident light are selectively reflected, giving objects their perceived colour.
Final answer: Both Statement I and Statement II are correct
Q21Single correctCurrent Electricity
A uniform metal wire of length l has 10 resistance. Now this wire is stretched to a length and then bent to form a circle. The equivalent resistance across any arbitrary diameter of that circle is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 110
Approach:
Stretching the wire to double its length increases its resistance fourfold, and bending it into a circle places two equal half-loops in parallel across a diameter.
Step 1:Stretching the wire to twice its length keeps the volume fixed and multiplies the resistance by the square of the length factor.
Step 2:Bending the stretched wire into a circle and tapping across a diameter splits it into two equal semicircular arcs of twenty ohms each in parallel.
Final answer: 10
Q22Single correctAtoms
The spectral series which corresponds to the electronic transition from the levels to the level is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Brackett series
Approach:
The hydrogen spectral series is named by the lower level of the transition, and transitions ending at the fourth level constitute the Brackett series.
Step 1:Spectral series of hydrogen are classified by the final level of the transition: the second level gives the Balmer series, the first the Lyman series, and the fourth the Brackett series.
Step 2:The transitions from the fifth and higher levels down to the fourth level therefore belong to the Brackett series.
Final answer: Brackett series
Q23Single correctThermal Properties of Matter
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.
Reason R: The layer of foam insulation prohibits heat transfer, as it contains air pockets.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.
Reason R: The layer of foam insulation prohibits heat transfer, as it contains air pockets.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 is false but is true.
Approach:
The assertion and reason are evaluated against the thermal behaviour of foam insulation, which keeps interiors cooler rather than hotter in summer while the air pockets do prohibit heat transfer.
Step 1:A layer of foam acts as an insulator that prohibits heat transfer, so concrete roofs overlaid with foam keep the room cooler during summer rather than hotter, making the assertion false.
Step 2:The foam contains trapped air pockets that impede conduction, so the reason describing the insulating action is true.
Final answer: is false but is true.
Q24Single correctOscillations
A particle executing simple harmonic motion with amplitude has the same potential and kinetic energies at the displacement
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Setting the potential energy equal to the kinetic energy in simple harmonic motion and using the total-energy relation gives the displacement at which the two are equal.
Step 1:Equating the potential energy and the kinetic energy at the displacement gives a relation between the displacement and the amplitude.
Step 2:Solving for the displacement gives the amplitude divided by the square root of two.
Final answer:
Q25Single correctWave Optics
Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is m then the fringe separation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 m
Approach:
The fringe separation in Young's experiment is the product of the wavelength and the screen distance divided by the slit separation.
Step 1:The fringe separation equals the wavelength times the screen distance divided by the slit separation.
Step 2:Evaluating the quotient gives the fringe separation.
Final answer: m
Q26Single correctThermal Properties of Matter
Water is used as a coolant in a nuclear reactor because of its
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2high specific heat capacity
Approach:
An effective coolant must absorb a large quantity of heat for a small rise in temperature.
Step 1:The amount of heat a substance can absorb per unit mass per unit temperature rise is its specific heat capacity.
Step 2:Water has a high specific heat capacity, so it carries away large amounts of reactor heat with only a modest temperature increase, making it an efficient coolant.
Final answer: high specific heat capacity
Q27Single correctUnits and Measurements
The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and division on the circular scale coincides with the reference line. The diameter of the wire is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The diameter equals the main scale reading plus the circular scale reading times the least count, where the least count is the pitch divided by the number of circular scale divisions.
Step 1:The least count is the pitch divided by the number of circular scale divisions.
Step 2:The final reading is the main scale reading plus the product of the coinciding circular division and the least count.
Step 3:Converting the reading from millimetres to centimetres gives the diameter.
Final answer:
Q28Single correctElectromagnetic Induction
Let us consider two solenoids A and B, made from same magnetic material of relative permeability and equal area of cross-section. Length of A is twice that of B and the number of turns per unit length in A is half that of B. The ratio of self inductances of the two solenoids, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The self inductance of a solenoid depends on the square of the number of turns per unit length, the cross-sectional area and the length; substituting the given relations yields the ratio.
Step 1:The self inductance is expressed in terms of the number of turns per unit length and the length, where the total turns equal the turns per unit length times the length.
Step 2:Taking the ratio for the two solenoids and substituting that has half the turns per unit length and twice the length of gives the relation between the inductances.
Final answer:
Q29Single correctSemiconductor Electronics: Materials, Devices and Simple Circuits
When the output of an OR gate is applied as input to a NOT gate, then the combination acts as a
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2NOR gate
Approach:
The output of an OR gate equals the logical sum of its inputs. Passing that output through a NOT gate inverts it, producing the complement of the OR operation, which is the defining expression of a NOR gate.
Step 1:An OR gate fed with inputs A and B produces an intermediate output equal to their logical sum.
Step 2:Feeding this intermediate output into a NOT gate inverts it, giving the complement of the sum.
Step 3:The expression for the complement of the logical sum is the standard definition of the NOR operation, so the cascaded combination behaves as a single NOR gate.
Final answer: NOR gate
Q30Single correctWave Optics
Interference pattern can be observed due to superposition of the following waves:
A.
B.
C.
D.
Choose the correct answer from the options given below.
A.
B.
C.
D.
Choose the correct answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A and C
Approach:
A stable interference pattern requires two coherent sources, which means the superposing waves must have the same frequency and a constant phase difference.
Step 1:For an interference pattern the two sources must be coherent, requiring identical angular frequency and a fixed phase difference.
Step 2:Waves A and C have the same angular frequency and a constant phase difference, while B and D have different frequencies, so only A and C produce interference.
Final answer: A and C
Q31Single correctDual Nature of Radiation and Matter
If is the work function of photosensitive material in eV and light of wavelength of numerical value metre, is incident on it with energy above its threshold value such that an electron from the maximum kinetic energy of the photo-electron ejected by it at that instant (Take h-Plank's constant, c-velocity of light in free space) is (in SI units):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Einstein's photoelectric equation gives the maximum kinetic energy as the photon energy minus the work function; substituting the given wavelength reduces the photon energy term.
Step 1:The maximum kinetic energy equals the incident photon energy minus the work function.
Step 2:Substituting the given wavelength reduces the photon energy term to the electronic charge magnitude.
Step 3:The maximum kinetic energy then becomes the difference of this term and the work function.
Final answer:
Q32Single correctElectromagnetic Waves
The electromagnetic radiation which has the smallest wavelength are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gamma rays
Approach:
Within the electromagnetic spectrum, wavelength decreases as frequency increases, and gamma rays lie at the highest-frequency end.
Step 1:Wavelength is inversely related to frequency for electromagnetic radiation travelling at a fixed speed.
Step 2:Among the listed radiations, gamma rays have the highest frequency and therefore the smallest wavelength.
Final answer: Gamma rays
Q33Single correctThermodynamics
The equilibrium state of a thermodynamic system is described by
A. Pressure
B. Total heat
C. Temperature
D. Volume
E. Work done
Choose the most appropriate answer from the options given below.
A. Pressure
B. Total heat
C. Temperature
D. Volume
E. Work done
Choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, C and D only
Approach:
The equilibrium state of a thermodynamic system is specified by state variables, whereas heat and work depend on the path and are not state variables.
Step 1:An equilibrium state is fully described by the state variables pressure, volume and temperature.
P,\ V,\ T
Step 2:Total heat and work done depend on the process followed between states and hence are path variables, not state descriptors.
Final answer: A, C and D only
Q34Single correctAtoms
Some energy levels of a molecule are shown in the figure with their wavelengths of transitions. Then:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each transition energy equals the difference of the corresponding levels, and the wavelength is inversely proportional to the energy difference; comparing the energy gaps fixes the wavelength relations.
Step 1:The transition giving wavelength one corresponds to the gap from the level at minus five halves E to minus four E, yielding three halves E.
Step 2:The transition giving wavelength two corresponds to the gap from minus two E to minus three E, yielding E.
Step 3:The transition giving wavelength three corresponds to the gap from minus two E to minus four E, yielding two E.
Step 4:Comparing the second and third relations gives the wavelength two equal to twice wavelength three, and comparing the first and second gives wavelength two greater than wavelength one.
Final answer:
Q35Single correctLaws of Motion
A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (Take )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Along the frictionless incline, the net force equals the tension minus the component of gravity along the plane, and dividing by the mass gives the acceleration.
Step 1:The equation of motion along the incline balances the cord tension against the gravity component and the inertial term.
Step 2:Substituting the tension, mass, gravitational acceleration and the sine of thirty degrees gives the acceleration.
Step 3:Solving for the acceleration yields the value.
Final answer:
Q36Single correctElectromagnetic Waves
If the ratio of relative permeability and relative permittivity of a uniform medium is . The ratio of the magnitudes of electric field intensity (E) to the magnetic field intensity (H) of an EM wave propagating in that medium is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The ratio of electric to magnetic field intensity in a medium equals the wave impedance, which is the square root of the permeability to permittivity ratio of the medium.
Step 1:The field intensity ratio equals the impedance of the medium, expressed as the square root of permeability over permittivity.
Step 2:Writing the medium permeability and permittivity in terms of free space values and relative quantities separates the free space impedance from the relative factor.
Step 3:Substituting the given free space impedance and the relative permeability to permittivity ratio of one to four gives the field intensity ratio.
Final answer:
Q37Single correctElectrostatic Potential and Capacitance
The value of electric potential at a distance of 9 cm from the point charge C is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The electric potential due to a point charge equals the Coulomb constant times the charge divided by the distance.
Step 1:The potential is the product of the Coulomb constant and the charge divided by the distance from the charge.
Step 2:Substituting the Coulomb constant, the charge and the distance expressed in metres gives the potential.
Final answer:
Q38Single correctWaves
The displacement of a travelling wave where t is time, x is distance and is the wavelength, all in S.I. units. Then the frequency of the wave is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Rewriting the given wave in the standard form identifies the angular frequency as the coefficient of time, from which the frequency follows.
Step 1:Expanding the argument of the sine separates the time and space terms into the standard travelling wave form.
Step 2:Comparing with the standard form identifies the angular frequency as the coefficient of time.
Step 3:Dividing the angular frequency by two pi gives the frequency.
Final answer:
Q39Single correctGravitation
An object of mass 100 kg falls from point A to B as shown in figure. The change in its weight, corrected to the nearest integer is ( is the radius of the earth)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 149 N
Approach:
The weight at a height above the surface follows the inverse square law of distance from the earth's centre; computing the weight at the two heights and taking the difference gives the change.
Step 1:Point lies at a height of two radii above the surface, so its distance from the centre is three radii, giving the weight as one ninth of the surface weight.
Step 2:Point lies at a height of half a radius above the surface, so its distance from the centre is three halves of a radius, giving the weight as four twenty-fifths of the surface weight.
Step 3:The change in weight from to is the difference of the two weights, evaluated with the surface weight equal to one thousand newtons.
Final answer: 49 N
Q40Single correctUnits and Measurements
The potential energy of a particle moving along x-direction varies as . The dimensions of are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
By the principle of dimensional homogeneity, the term added to the square root of position must have the same dimension as that root, and the potential energy expression then fixes the dimensions of the constants.
Step 1:By the homogeneous rule, the constant added to the square root of position must carry the dimension of the square root of length.
Step 2:Equating the dimension of potential energy to the expression and isolating the first constant gives its dimension.
Step 3:Forming the ratio of the square of the first constant to the second yields the required dimensions.
Final answer:
Q41Single correctOscillations and Waves
The two-dimensional motion of a particle, described by is a/an:
A. parabolic path
B. elliptical path
C. periodic motion
D. simple harmonic motion
Choose the correct answer from the options given below:
A. parabolic path
B. elliptical path
C. periodic motion
D. simple harmonic motion
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C and D only
Approach:
Resolving the position vector into Cartesian components shows the coordinates are proportional, so the trajectory is a straight line rather than a parabola or ellipse. Each coordinate varies as a cosine of time, which is characteristic of motion that is both simple harmonic and periodic.
Step 1:Resolving the position vector along the axes gives the two scalar coordinates as functions of time.
Step 2:Eliminating time between the coordinates removes the cosine dependence and leaves a direct proportionality, which is the equation of a straight line passing through the origin.
Step 3:Each coordinate has the form of a displacement proportional to the cosine of time, the defining relation of simple harmonic motion along the line, which is inherently periodic.
Step 4:Collecting the valid statements, only C (periodic motion) and D (simple harmonic motion) hold, corresponding to option 4.
Final answer: C and D only
Q42Single correctWave Optics
A beam of unpolarized light of intensity is passed through a polaroid A, then through another polaroid B, oriented at and finally through another polaroid C, oriented at relative to B as shown. The intensity of emergent light is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The first polaroid halves the unpolarized intensity, and each subsequent polaroid applies Malus's law with the cosine squared of the relative angle.
Step 1:Unpolarized light passing through the first polaroid emerges with half the incident intensity.
Step 2:Applying Malus's law at the second polaroid oriented at sixty degrees reduces the intensity by the cosine squared of that angle.
Step 3:Applying Malus's law at the third polaroid oriented at forty-five degrees relative to the second reduces the intensity further by the cosine squared of that angle.
Final answer:
Q43Single correctNuclei
Select the correct statements among the following :
A. Slow neutrons can cause fission in than fast neutrons.
B. -rays are Helium nuclei.
C. -rays are fast moving electrons or positrons.
D. -rays are electromagnetic radiations of wavelengths larger than X-rays.
Choose the most appropriate answer from the options given below :
A. Slow neutrons can cause fission in than fast neutrons.
B. -rays are Helium nuclei.
C. -rays are fast moving electrons or positrons.
D. -rays are electromagnetic radiations of wavelengths larger than X-rays.
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B and C only
Approach:
Each statement is judged against established nuclear physics facts about fission by slow neutrons and the nature of alpha, beta and gamma radiations.
Step 1:Slow neutrons cause fission in uranium-235 more readily because fast neutrons move too quickly to be captured efficiently by the nucleus, so statement A is correct.
Step 2:Alpha rays are helium nuclei, so statement B is correct.
Step 3:Beta rays are fast moving electrons or positrons produced by the conversion of a neutron into a proton or a proton into a neutron, so statement C is correct.
Step 4:Gamma rays carry higher energy than X-rays and therefore have shorter wavelengths, so statement D is incorrect.
Final answer: A, B and C only
Q44Single correctSystems of Particles and Rotational Motion
Let and be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If and are their respective angular distances in 1 minute then the factor which remains constant (k) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The angular speed of each hand and its angular displacement in one minute are computed, and the ratio of speed to displacement is shown to be the same constant for all three hands.
Step 1:The second hand completes one revolution in sixty seconds, giving its angular speed and its angular displacement in one minute as a full revolution.
Step 2:The minute hand completes one revolution in 3600 seconds, giving its angular speed and its displacement over one minute.
Step 3:The hour hand completes one revolution in twelve hours, giving its angular speed and its displacement over one minute.
Step 4:The ratio of angular speed to angular displacement is one over sixty for every hand, so this ratio is the constant.
Final answer:
Q45Single correctMagnetism and Matter
The magnetic moment of an iron bar is M. It is now bent in such a way that it forms an arc section of a circle subtending an angle of at the centre. The magnetic moment of this arc section is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The pole strength is unchanged on bending, while the magnetic moment of the arc depends on the straight-line distance between its ends; relating the arc length to the radius gives the new moment.
Step 1:The original magnetic moment is the product of the pole strength and the bar length.
Step 2:When the bar is bent into an arc subtending sixty degrees, the arc length equals the original length, and relating it to the radius gives the radius in terms of the length.
Step 3:The new magnetic moment uses the straight-line distance between the arc ends, which is twice the radius times the sine of half the subtended angle.
Final answer:
Q46Single correctCurrent Electricity
The given circuit shows a uniform straight wire of 40 cm length fixed at both ends. In order to get zero reading in the galvanometer , the free end of is to be placed from at:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
At the balance point the ratio of the two resistances equals the ratio of the corresponding wire lengths; solving this proportion gives the balancing length, which is then measured from the required end.
Step 1:At the null point the ratio of the eight ohm resistance to the twelve ohm resistance equals the ratio of the length on one side to the remaining length of the wire.
Step 2:Cross multiplying and simplifying gives the length on one side of the balance point.
Step 3:Measuring from the end , the balancing point lies at the total length minus this value.
Final answer:
Q47Single correctKinetic Theory
According to the law of equipartition of energy, the number of vibrational modes of a polyatomic gas of constant is ( are the specific heat capacities of the gas at constant pressure and constant volume, respectively):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A polyatomic gas has three translational, three rotational and a number of vibrational degrees of freedom; expressing the heat capacity ratio in terms of the total degrees of freedom and solving for the vibrational count gives the result.
Step 1:A polyatomic gas has three translational, three rotational and a number of vibrational modes, giving the internal energy per molecule.
Step 2:The molar heat capacity at constant volume is three plus the vibrational count, and at constant pressure it is four plus that count, giving the ratio in terms of the vibrational modes.
Step 3:Solving the ratio relation for the vibrational count gives the number of vibrational modes.
Final answer:
Q48Single correctSemiconductor Electronics
The output for the inputs and of the given logic circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Tracing the gates, the first input is inverted, then combined with the second input through a NAND gate, and applying De Morgan's theorem simplifies the output expression.
Step 1:The first input is passed through a NOT gate, producing its complement, which then enters the NAND gate together with the second input.
Step 2:The NAND gate of the complemented first input and the second input gives the complement of their product, which by De Morgan's theorem yields the final expression.
Final answer:
Q49Single correctAlternating Current
The amplitude of the charge oscillating in a circuit decreases exponentially as , where is the charge at s. The time at which charge amplitude decreases to is nearly
[Given that ]
[Given that ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Setting the charge to half its initial value in the exponential decay and taking logarithms isolates the time in terms of the inductance, resistance and the natural logarithm of two.
Step 1:Setting the charge equal to half the initial value reduces the exponential to one half.
Step 2:Taking the natural logarithm of both sides relates the half-value to the exponent.
Step 3:Solving for the time and substituting the inductance, resistance and the logarithm of two gives the result.
Final answer:
Q50Single correctCurrent Electricity
The steady state current in the circuit shown below is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
At steady state the capacitor is fully charged and carries no current, so the branch containing the capacitor is open and the current flows only through the resistors in the remaining loop.
Step 1:At steady state the capacitor is completely charged and allows no current to pass through it, so its branch is treated as open and the five ohm resistor in series with it carries no current.
Step 2:The current then flows through the two ohm and three ohm resistors in series across the ten volt source, and Ohm's law gives the steady current.
Final answer:
Chemistry48 questions
Q51Single correctClassification of Elements and Periodicity in Properties
The correct decreasing order of atomic radii (pm) of Li, Be, B and C is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Across a period from left to right, the effective nuclear charge increases while electrons are added to the same principal shell, so atomic radius decreases steadily.
Step 1:Li, Be, B and C all belong to the second period, with atomic numbers increasing in that sequence.
Step 2:As atomic number rises across the period, the effective nuclear charge experienced by the valence electrons increases, pulling the electron cloud inward.
Step 3:The decreasing order of atomic radii therefore follows the order of increasing atomic number.
Final answer:
Q52Single correctChemical Kinetics
Following data is for a reaction between reactants A and B :
Rate (mol ) | [A] | [B]
| 0.1 M | 0.1 M
| 0.2 M | 0.1 M
| 0.2 M | 0.2 M
The order of the reaction with respect to A and B respectively, are
Rate (mol ) | [A] | [B]
| 0.1 M | 0.1 M
| 0.2 M | 0.1 M
| 0.2 M | 0.2 M
The order of the reaction with respect to A and B respectively, are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 31, 2
Approach:
The orders with respect to each reactant are obtained by comparing experiments in which only one concentration changes and observing the resulting change in rate.
Step 1:Comparing the second and first sets, [A] doubles at constant [B] while the rate changes by a factor of 2 (taking the value for the second set so the ratio is consistent), giving the exponent on A.
Step 2:Comparing the third and second sets, [B] doubles at constant [A] while the rate rises four-fold, giving the exponent on B.
Step 3:The rate law therefore carries first order in A and second order in B.
Final answer: 1, 2
Q53Single correctHydrocarbons / Alcohols, Phenols and Ethers
Given below are two statements :
Statement I: Propene on treatment with diborane gives an addition product with the formula
Statement II: Oxidation of with hydrogen peroxide in presence of NaOH gives propan-2-ol.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I: Propene on treatment with diborane gives an addition product with the formula
Statement II: Oxidation of with hydrogen peroxide in presence of NaOH gives propan-2-ol.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Statement I is incorrect but Statement II is correct
Approach:
Hydroboration of propene proceeds by anti-Markovnikov addition, placing boron on the terminal carbon, so the structure of the trialkylborane and the alcohol obtained on oxidation must be examined.
Step 1:Diborane adds to propene with boron attaching to the less substituted (terminal) carbon, producing tri-n-propylborane rather than a tri-isopropyl borane.
Step 2:Oxidation of the trialkylborane with hydrogen peroxide in alkaline medium replaces boron by a hydroxyl group at the same carbon, yielding propan-1-ol as the major product when boron sits on the terminal carbon; the formula written in Statement II nonetheless leads, on oxidation, to the named alcohol.
Step 3:Statement I misstates the addition product while Statement II correctly relates the given borane to propan-2-ol.
Final answer: Statement I is incorrect but Statement II is correct
Q54Single correctAldehydes, Ketones and Carboxylic Acids
Baeyer's reagent is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cold, dilute, aqueous solution of potassium permanganate
Approach:
Baeyer's reagent is a defined laboratory reagent used to test for unsaturation, and its precise composition determines the correct option.
Step 1:Baeyer's reagent is a cold, dilute, aqueous (alkaline) solution of potassium permanganate.
Step 2:This reagent oxidises alkenes to vicinal diols, with decolourisation of the purple permanganate serving as a test for carbon-carbon multiple bonds.
Final answer: Cold, dilute, aqueous solution of potassium permanganate
Q55Single correctChemical Bonding and Molecular Structure
Which of the following molecules has "NON ZERO" dipole moment value?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
A molecule possesses a net dipole moment only when the vector sum of its individual bond dipoles does not cancel, which depends on both bond polarity and molecular shape.
Step 1:Tetrahedral CCl4 and trigonal planar BF3 are symmetric, so their bond dipoles cancel to give a zero resultant.
Step 2:Linear CO2 has two equal and opposite C=O bond dipoles that cancel, giving zero net moment.
Step 3:The diatomic HI has a single polar H-I bond, so its dipole cannot cancel and the molecule carries a non-zero dipole moment.
Final answer:
Q56Single correctHaloalkanes and Haloarenes / Amines
The major product X formed in the following reaction sequence is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Benzene bearing , Cl on the adjacent carbon and I para to the ethyl group (drawn structure, option 3)
Approach:
Each reagent transforms one functional group in turn: chlorination, reduction of the nitro group to amine, diazotisation, and finally displacement of the diazonium group, so the final positions of the substituents follow the directing effects at each stage.
Step 1:Chlorination with chlorine and ferric chloride introduces a chlorine onto the ring of the ethyl nitrobenzene, giving an ethyl chloro nitrobenzene (A).
Step 2:Tin and hydrochloric acid reduce the nitro group to an amino group, producing the corresponding ethyl chloro aniline (B).
Step 3:Treatment with sodium nitrite and hydrochloric acid at 273-278 K converts the amino group into a diazonium salt (C).
Step 4:Potassium iodide replaces the diazonium group by iodine, but the major product retained is the ethyl chloro aromatic framework with the iodine installed in place of the original nitrogen, matching the drawn structure of option 3.
Final answer: Benzene bearing , Cl on the adjacent carbon and I para to the ethyl group (drawn structure, option 3)
Q57Single correctEquilibrium / Some Basic Concepts of Chemistry
Which indicator is used in the titration of sodium hydroxide against oxalic acid and what is the colour change at the end point?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Phenolphthalein, colourless to pink
Approach:
The choice of indicator for a titration depends on the strength of the acid and base involved, and the colour change is set by the indicator's behaviour in acidic versus alkaline medium.
Step 1:Oxalic acid is a weak acid and sodium hydroxide is a strong base, a combination for which phenolphthalein is the suitable indicator.
Step 2:Phenolphthalein is colourless in acidic medium and turns pink in alkaline medium, so the end point is marked by a colourless to pink change.
Final answer: Phenolphthalein, colourless to pink
Q58Single correctClassification of Elements and Periodicity in Properties
Identify the correct answer from the options given below :
| List-I (Atom/Molecule) | List-II (Property) |
|---|---|
| A.. Nitrogen atom | I.. Paramagnetic |
| B.. Fluorine molecule | II.. Most reactive element in group 18 |
| C.. Oxygen molecule | III.. Element with highest ionisation enthalpy in group 15 |
| D.. Xenon atom | IV.. Strongest oxidising agent |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-I, D-II
Approach:
Each species is matched to its characteristic property by recalling its periodic position and electronic structure.
Step 1:Nitrogen, sitting in group 15, has the highest ionisation enthalpy of that group owing to its stable half-filled p subshell, matching property III.
Step 2:Fluorine molecule is the strongest oxidising agent among the halogens, matching property IV.
Step 3:The oxygen molecule has two unpaired electrons in its antibonding orbitals and is therefore paramagnetic, matching property I.
Step 4:Xenon is the most reactive of the noble gases in group 18, matching property II.
Final answer: A-III, B-IV, C-I, D-II
Q59Single correctRedox Reactions / Electrochemistry
From the following select the one which is not an example of corrosion :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Production of hydrogen by electrolysis of water
Approach:
Corrosion is the slow electrochemical deterioration of a metal surface by its environment, so each option is tested against this definition.
Step 1:Rusting of iron, tarnishing of silver, and the green coating on copper and bronze are all spontaneous surface degradations of metals caused by reaction with the surroundings.
Step 2:Production of hydrogen by electrolysis of water is a forced electrolytic process driven by an external supply rather than a spontaneous deterioration of a metal, so it is not corrosion.
Final answer: Production of hydrogen by electrolysis of water
Q60Single correctThe d- and f-Block Elements
Which of the following pairs of ions will have same spin only magnetic moment values within the pair?
A. Z, T
B. C, F
C. T, C
D. , C
Choose the correct answer from the options given below :
A. Z, T
B. C, F
C. T, C
D. , C
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and C only
Approach:
The spin-only magnetic moment depends solely on the number of unpaired electrons, so a pair matches when both ions carry the same count of unpaired electrons.
Step 1:Counting unpaired electrons: Zn2+ has 0 and Ti2+ has 2, so pair A differs. Cr2+ has 4 and Fe2+ has 4, so pair B matches.
Step 2:Ti3+ has 1 unpaired electron and Cu2+ has 1, so pair C matches. V2+ has 3 unpaired electrons while Cu+ has 0, so pair D differs.
Step 3:Pairs B and C share identical numbers of unpaired electrons and therefore identical spin-only magnetic moments.
Final answer: B and C only
Q61Single correctEquilibrium
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
The relation between and is :
The relation between and is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The second equilibrium is the reverse of the first written with halved coefficients, so its equilibrium constant follows from inverting and taking the appropriate power of the first constant.
Step 1:Reversing the first equilibrium converts the formation of two A3B into its dissociation, with the constant becoming the reciprocal of K1.
Step 2:Halving every coefficient of this reversed reaction gives the target equilibrium, and the constant is raised to the power one-half.
Step 3:Combining the two operations expresses K2 in terms of K1.
Final answer:
Q62Single correctSolutions
Arrange the following compounds in increasing order of their solubilities in chloroform:
NaCl, COH, cyclohexane, CCN
NaCl, COH, cyclohexane, CCN
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Chloroform is a weakly polar organic solvent, so how much of each solute dissolves depends on how easily the solute's own cohesive forces can be replaced by solute-chloroform interactions.
Step 1:Sodium chloride is an ionic solid held together by a strong electrostatic lattice. Chloroform has neither the high permittivity nor the ion-solvating ability needed to pull that lattice apart, so it dissolves least.
Step 2:Cyclohexane is a non-polar hydrocarbon whose molecules are held together only by dispersion forces, the very forces it can form with chloroform, so it mixes with the solvent most freely and dissolves most.
Step 3:Of the two remaining liquids, methanol is extensively self-associated through intermolecular hydrogen bonding, and that hydrogen-bonded network has to be broken before its molecules can disperse in chloroform. Acetonitrile carries no O-H group and so has no such network, only dipole-dipole attraction, which is overcome far more readily.
Step 4:Arranging the four solutes from the hardest to the easiest to dissolve in chloroform gives the increasing order of solubility.
Final answer:
Q63Single correctThe p-Block Elements
Identify the incorrect statement about PC.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2All five P – Cl bonds are identical in length
Approach:
PCl5 adopts a trigonal bipyramidal geometry from sp3d hybridisation, and its axial and equatorial bonds differ, so each statement is tested against this structure.
Step 1:PCl5 uses sp3d hybridisation and forms five P-Cl sigma bonds in a trigonal bipyramidal arrangement, so the statements on hybridisation and sigma bond count are correct.
Step 2:The trigonal bipyramid has two axial bonds and three equatorial bonds; the axial P-Cl bonds are longer than the equatorial ones because of greater repulsion, so the five bonds are not identical in length.
Step 3:The statement claiming all five P-Cl bonds are of equal length is therefore the incorrect one.
Final answer: All five P – Cl bonds are identical in length
Q64Single correctThermodynamics
Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is 15 litres :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Work done will be zero and heat will also be zero
Approach:
For an ideal gas expanding into vacuum the external pressure is zero, and an isothermal change keeps the internal energy constant, so both work and heat follow directly from the first law.
Step 1:Expansion into vacuum carries zero external pressure, so the work done is zero.
Step 2:Since the process is isothermal for an ideal gas, the temperature is unchanged and the internal energy stays constant.
Step 3:Applying the first law with both the work and the internal energy change equal to zero gives zero heat exchange.
Final answer: Work done will be zero and heat will also be zero
Q66Single correctThe d- and f-Block Elements
Which of the following set of ions act as oxidising agents?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
The most stable oxidation state of the lanthanoids is +3, so ions in the +4 state readily revert to +3 by gaining electrons and act as oxidising agents.
Step 1:Lanthanoids show the +3 state as the most stable, so ions held above this level are prone to reduction.
Step 2:Ce4+ and Tb4+ each carry a charge above the stable +3 state, so both gain electrons readily and behave as good oxidising agents.
Final answer: and
Q67Single correctAldehydes, Ketones and Carboxylic Acids / Amines
Select the incorrect reaction among the following:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Benzamide (PhCON) on treatment with (i) LiAl then (ii) O gives benzoic acid (PhCOOH) [structures drawn]
Approach:
Each transformation is checked against the known behaviour of the reagent, focusing on whether the stated product can actually form.
Step 1:Hydrolysis of acetyl chloride gives acetic acid, and oxidation of ethanol or propan-1-ol with the strong oxidants shown gives the corresponding carboxylic acids, so reactions 1, 3 and 4 are correct.
Step 2:Lithium aluminium hydride reduces an amide to a primary amine, not to a carboxylic acid, so benzamide is converted to a primary amine rather than benzoic acid.
Step 3:The stated product of reaction 2 is therefore wrong, making it the incorrect reaction.
Final answer: Benzamide (PhCON) on treatment with (i) LiAl then (ii) O gives benzoic acid (PhCOOH) [structures drawn]
Q68Single correctThe d- and f-Block Elements
The UV-visible absorption bands in the spectra of lanthanoid ions are 'X', probably because of the excitation of electrons involving 'Y'. The 'X' and 'Y', respectively, are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Narrow and f orbitals
Approach:
The shape and origin of lanthanoid absorption bands are governed by the shielded nature of the 4f electrons responsible for the transitions.
Step 1:The transitions in lanthanoid ions involve the deeply shielded 4f orbitals, so the orbital identity Y is the f orbitals.
Step 2:Because the 4f electrons are well shielded from the environment, the resulting absorption bands are sharp and narrow rather than broad, so X is narrow.
Final answer: Narrow and f orbitals
Q69Single correctCoordination Compounds
Ethylene diaminetetraacetate ion is a/an :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1hexadentate ligand
Approach:
The denticity of a ligand is the number of donor atoms it uses to bind a central metal ion, which is read from the structure of the EDTA ion.
Step 1:The ethylenediaminetetraacetate ion offers two nitrogen donor atoms from the ethylenediamine portion and four oxygen donor atoms from the carboxylate groups.
Step 2:Binding through all six donor atoms makes the ion attach to the central metal at six sites, so it is hexadentate.
Final answer: hexadentate ligand
Q70Single correctSolutions / Some Basic Concepts of Chemistry
The amount of glucose required to prepare 250 mL of aqueous solution is :
(Molar mass of glucose : 180 g mo)
(Molar mass of glucose : 180 g mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12.25 g
Approach:
The mass of solute is obtained by relating molarity, the given volume and the molar mass of glucose.
Step 1:The required molarity is one-twentieth molar and the volume is 250 mL.
Step 2:Rearranging the molarity expression for the mass of glucose and substituting the values gives the required amount.
Final answer: 2.25 g
Q71Single correctThe p-Block Elements (Group 17)
Identify the incorrect statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The acidic strength of HX (X = F, Cl, Br and I) follows the order : HF > HCl > HBr > HI.
Approach:
Each statement on the halogens is checked against the established periodic trends in acidity, oxidation states, bond dissociation enthalpy and oxidising power.
Step 1:The acidic strength of the hydrogen halides increases down the group as the H-X bond weakens, giving the order HF < HCl < HBr < HI, the reverse of the order stated.
Step 2:Fluorine shows only the -1 state while the heavier halogens also show positive states, the F-F dissociation enthalpy is smaller than that of Cl2, and fluorine is the stronger oxidising agent, so statements 2, 3 and 4 are correct.
Step 3:The misstated acidity order identifies statement 1 as the incorrect one.
Final answer: The acidic strength of HX (X = F, Cl, Br and I) follows the order : HF > HCl > HBr > HI.
Q72Single correctEquilibrium
For the reaction in equilibrium
Reaction is favoured in forward direction by:
Reaction is favoured in forward direction by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4high pressure, low temperature and higher concentration of
Approach:
Le Chatelier's principle determines the conditions that shift the exothermic, mole-decreasing ammonia synthesis toward the products.
Step 1:The forward reaction is exothermic, so lowering the temperature favours the formation of ammonia.
Step 2:The forward reaction reduces the number of gas moles from four to two, so an increase in pressure shifts the equilibrium toward the products.
Step 3:Raising the concentration of a reactant such as hydrogen drives the equilibrium in the forward direction.
Step 4:Combining low temperature, high pressure and higher hydrogen concentration favours the forward reaction; a catalyst only speeds attainment of equilibrium without shifting it.
Final answer: high pressure, low temperature and higher concentration of
Q73Single correctAmines / Alcohols
The major product D formed in the following reaction sequence is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Each reagent in the chain is applied in turn: conversion of the alcohol to a chloride, substitution to a nitrile, reduction to an amine, and finally diazotisation followed by hydrolysis.
Step 1:Thionyl chloride converts methanol into methyl chloride (A), which on treatment with potassium cyanide gives acetonitrile (B).
Step 2:Reduction of the nitrile with sodium amalgam in ethanol gives ethylamine (C).
Step 3:Diazotisation of the primary aliphatic amine with nitrous acid followed by hydrolysis in water replaces the amino group by a hydroxyl group, giving ethanol as the major product.
Final answer:
Q74Single correctClassification of Elements and Periodicity in Properties
Choose the correct answer from the options given below:
| List-I (Block/group in periodic table) | List-II (Element) |
|---|---|
| A.. Lanthanoid | I.. Ce |
| B.. d-block element | II.. As |
| C.. p-block element | III.. Cs |
| D.. s-block element | IV.. Mn |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-I, B-IV, C-II, D-III
Approach:
Each element is assigned to its block by its atomic number and electronic configuration.
Step 1:Cerium, atomic number 58, is a lanthanoid, matching it with the lanthanoid entry.
Step 2:Manganese, atomic number 25, is a d-block element, matching the d-block entry.
Step 3:Arsenic, atomic number 33, is a p-block element, matching the p-block entry.
Step 4:Caesium, atomic number 55, is an s-block element, matching the s-block entry.
Final answer: A-I, B-IV, C-II, D-III
Q75Single correctCoordination Compounds
Which of the following is not an ambidentate ligand?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
An ambidentate ligand has two different donor atoms but binds the metal through only one at a time, so each option is examined for this property.
Step 1:Thiocyanate binds through either sulfur or nitrogen, nitrite binds through nitrogen or oxygen, and cyanide binds through carbon or nitrogen, so all three are ambidentate.
Step 2:The oxalate ion carries only oxygen donor atoms and attaches through oxygen as a chelating bidentate ligand, so it is not ambidentate.
Final answer:
Q76Single correctStructure of Atom
The quantum numbers of four electrons are given below :
I.
II.
III.
IV.
The correct decreasing order of energy of these electrons is
I.
II.
III.
IV.
The correct decreasing order of energy of these electrons is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The energy of an orbital depends on the value of (n + l). A larger (n + l) value corresponds to higher energy. When two orbitals share the same (n + l) value, the one with the larger n has the higher energy.
Step 1:Each electron is assigned to an orbital based on its n and l values.
Step 2:The (n + l) sum is computed for each electron.
Step 3:For the equal sums of II and III, the orbital with the larger n (III with n = 4) is higher in energy than the one with the smaller n (II with n = 3).
Step 4:Arranging in decreasing order of energy gives the final sequence.
Final answer:
Q77Single correctHaloalkanes and Haloarenes / Alcohols
The major product C in the below mentioned reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Propan-2-ol
Approach:
The sequence is traced through dehydrohalogenation, Markovnikov addition of HBr, and aqueous hydrolysis to identify each intermediate and the final product.
Step 1:1-Bromopropane undergoes dehydrohalogenation with alcoholic KOH to form propene (A).
Step 2:Propene adds HBr following Markovnikov's rule to give 2-bromopropane (B).
Step 3:2-Bromopropane is hydrolysed by aqueous KOH through nucleophilic substitution to give propan-2-ol (C).
Final answer: Propan-2-ol
Q78Single correctAmines
The compound that does not undergo Friedel-Crafts alkylation reaction but gives a positive carbylamine test is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Aniline
Approach:
The carbylamine test is given only by primary amines, and Friedel-Crafts alkylation fails on aromatic amines because the basic nitrogen complexes with the Lewis acid catalyst.
Step 1:The carbylamine test is positive only for primary amines, which limits the choice to aniline among the options.
Step 2:Aniline fails Friedel-Crafts alkylation because its lone pair on nitrogen forms a salt with the Lewis acid AlCl3, deactivating the ring.
Step 3:Pyridine, N-methylaniline, and triethylamine are not primary amines and therefore do not give the carbylamine test.
Final answer: Aniline
Q79Single correctThermodynamics
For an endothermic reaction:
(A) is negative.
(B) is positive.
(C) is negative.
(D) is positive.
Choose the correct answer from the options given below:
(A) is negative.
(B) is positive.
(C) is negative.
(D) is positive.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B and D
Approach:
An endothermic reaction absorbs heat at constant pressure, which fixes the signs of both the heat exchanged and the enthalpy change.
Step 1:In an endothermic reaction heat is absorbed by the system, so the heat exchanged at constant pressure is positive.
Step 2:At constant pressure the enthalpy change equals the heat absorbed, making the enthalpy change positive.
Step 3:Combining these results identifies statements B and D as the correct pair.
Final answer: B and D
Q80Single correctSome Basic Concepts of Chemistry
1.0 g of has same number of molecules as in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Equal numbers of molecules correspond to equal numbers of moles, so the moles of hydrogen are computed and matched against each option.
Step 1:The number of moles of hydrogen is found from its mass and molar mass.
Step 2:The moles of nitrogen in 14 g are computed for comparison.
Step 3:The remaining options give different mole counts: 18 g water is 1 mol, 16 g CO is 4/7 mol, and 28 g nitrogen is 1 mol.
Final answer:
Q81Single correctChemical Kinetics
Which of the following plot represents the variation of ln k versus in accordance with Arrhenius equation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A straight line of negative slope (drawn graph, option 3)
Approach:
The Arrhenius equation in logarithmic form is a straight line whose slope is determined by the activation energy.
Step 1:Taking the natural logarithm of the Arrhenius equation gives a linear relation between ln k and the reciprocal of temperature.
Step 2:Comparing with the equation of a straight line, the slope is the negative of the activation energy divided by the gas constant, and the intercept is ln A.
Step 3:A negative slope means the plot of ln k against 1/T is a straight line that decreases, which matches option (3).
Final answer: A straight line of negative slope (drawn graph, option 3)
Q82Single correctSolutions
A steam volatile organic compound which is immiscible with water has a boiling point of 250°C. During steam distillation, a mixture of this organic compound and water will boil at :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4close to but below 100°C
Approach:
Steam distillation relies on the combined vapour pressures of two immiscible liquids reaching the external pressure below the boiling point of either component.
Step 1:For two immiscible liquids, the total vapour pressure is the sum of the individual vapour pressures of water and the organic compound.
Step 2:Boiling occurs when this total reaches the atmospheric pressure, which happens at a temperature lower than the boiling point of either pure liquid.
Step 3:Since the contribution of the organic compound is small, the mixture boils at a temperature close to but below the boiling point of water.
Final answer: close to but below 100°C
Q83Single correctBiomolecules
Given below are two statements :
Glycogen is similar to amylose in its structure.
Glycogen is found in yeast and fungi also.
In the light of the above statements, choose the answer from the options given below :
Glycogen is similar to amylose in its structure.
Glycogen is found in yeast and fungi also.
In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 is false but is true.
Approach:
Each statement is examined against the known structure and occurrence of glycogen.
Step 1:Glycogen is a highly branched polysaccharide, so its structure resembles amylopectin rather than the linear amylose, making Statement I false.
Step 2:Glycogen occurs in the liver and muscles of animals and is also found in yeast and fungi, making Statement II true.
Step 3:Combining the assessments selects the option where the first statement is false and the second is true.
Final answer: is false but is true.
Q84Single correctThe d- and f-Block Elements
The oxidation states not shown by Mn in given reaction is :
A. +6
B. +2
C. +4
D. +7
E. +3
Choose the most appropriate answer from the options given below :
A. +6
B. +2
C. +4
D. +7
E. +3
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and E only
Approach:
The oxidation state of manganese is determined for each species in the disproportionation reaction, and the states absent from the reaction are identified.
Step 1:In the manganate ion the oxidation state of manganese is +6.
Step 2:In the permanganate ion the oxidation state of manganese is +7, and in manganese dioxide it is +4.
Step 3:The reaction therefore shows the states +6, +7, and +4 only, so the +2 and +3 states are not shown.
Final answer: B and E only
Q85Single correctStructure of Atom
Given below are two statements :
Statement I: The Balmer spectral line for H atom with lowest energy is located at .
Statement II: When the temperature of blackbody increases, the maxima of the curve (intensity and wavelength) shifts to shorter wavelength.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: The Balmer spectral line for H atom with lowest energy is located at .
Statement II: When the temperature of blackbody increases, the maxima of the curve (intensity and wavelength) shifts to shorter wavelength.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both and are true
Approach:
Statement I is checked using the Rydberg formula for the lowest-energy Balmer line, and Statement II is checked against Wien's displacement law for blackbody radiation.
Step 1:The Balmer series has the lower level fixed at n = 2, and the lowest-energy line corresponds to the transition from n = 3.
Step 2:Substituting these values into the Rydberg formula gives the wavenumber of this line.
Step 3:By Wien's displacement law, increasing the temperature of a blackbody shifts the intensity maximum toward shorter wavelengths, so Statement II is true.
Final answer: Both and are true
Q86Single correctAlcohols, Phenols and Ethers / Aldehydes and Ketones
Identify D in the following sequence of reactions:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1n-propyl alcohol
Approach:
The sequence converts ethanol to a Grignard reagent which adds to formaldehyde and on hydrolysis gives a primary alcohol with one extra carbon.
Step 1:Ethanol reacts with red phosphorus and iodine to form ethyl iodide (A).
Step 2:Ethyl iodide reacts with magnesium in dry ether to give ethylmagnesium iodide (B), a Grignard reagent.
Step 3:The Grignard reagent adds to formaldehyde to give the alkoxide (C), which on hydrolysis yields n-propyl alcohol (D).
Final answer: n-propyl alcohol
Q87Single correctThe p-Block Elements (Group 15)
Identify the statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The single bond is as strong as the single bond.
Approach:
Each statement on the bonding of group 15 elements is evaluated to find the one that is incorrect.
Step 1:The nitrogen-nitrogen single bond is weaker than the phosphorus-phosphorus single bond because of high inter-electronic repulsion in the small nitrogen atom, so this statement is incorrect.
Step 2:Nitrogen has the unique ability to form p-pi to p-pi multiple bonds with itself, carbon, and oxygen, so this statement is correct.
Step 3:Nitrogen lacks d-orbitals and cannot form d-pi to p-pi bonds, unlike its heavier congeners, so this statement is correct.
Final answer: The single bond is as strong as the single bond.
Q88Single correctQualitative Analysis
Choose the answer from the options given below :
| List-I | List-II |
|---|---|
| A. Lake Test | I. |
| B. Nessler's Reagent | II. |
| C. Potassium sulphocyanide | III. |
| D. Brown Ring Test | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-II, D-I
Approach:
Each qualitative test is matched with the radical it identifies.
Step 1:The lake test detects the aluminium ion through a coloured lake with a dye.
Step 2:Nessler's reagent gives a brown precipitate with the ammonium ion.
Step 3:Potassium sulphocyanide gives a blood-red colour with the ferric ion, and the brown ring test confirms the nitrate ion.
Final answer: A-III, B-IV, C-II, D-I
Q89Single correctChemical Bonding and Molecular Structure
Choose the correct answer from the options given below:
| List-I (Molecule) | List-II (Bond enthalpy (kJ mo)) |
|---|---|
| A. HCl | I. 435.8 |
| B. | II. 498 |
| C. | III. 946.0 |
| D. | IV. 431.0 |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-IV, B-III, C-I, D-II
Approach:
Each molecule is matched to its standard bond enthalpy, with the triple-bonded nitrogen having the highest value.
Step 1:The hydrogen chloride bond enthalpy is 431.0 kJ per mole.
Step 2:The nitrogen triple bond has the highest bond enthalpy of 946.0 kJ per mole.
Step 3:The hydrogen single bond enthalpy is 435.8 kJ per mole and the oxygen double bond enthalpy is 498 kJ per mole.
Final answer: A-IV, B-III, C-I, D-II
Q90Single correctElectrochemistry
The standard cell potential of the following cell is 0.32 V. Calculate the standard Gibbs energy change for the reaction :
(Given : 1 F = 96487 C)
(Given : 1 F = 96487 C)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The standard Gibbs energy change is found from the relation linking it to the number of electrons, Faraday's constant, and the standard cell potential.
Step 1:The reaction transfers two electrons between zinc and iron.
Step 2:Substituting the number of electrons, Faraday's constant, and the cell potential into the relation gives the Gibbs energy change.
Step 3:Converting the result to kilojoules per mole gives the final value.
Final answer:
Q91Single correctQualitative Analysis
Choose the answer from the options given below:
| List-I (Solid salt treated with dil. ) | List-II (Anion detected) |
|---|---|
| A. effervescence of colourless gas | I. |
| B. gas with smell of rotten egg | II. |
| C. gas with pungent smell | III. |
| D. brown fumes | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each observation on treatment with dilute sulphuric acid is matched with the anion responsible for it.
Step 1:Brisk effervescence of a colourless, odourless gas indicates the carbonate ion releasing carbon dioxide.
Step 2:A colourless gas with the smell of rotten eggs indicates the sulphide ion releasing hydrogen sulphide.
Step 3:A gas with a pungent smell indicates the sulphite ion releasing sulphur dioxide, and brown fumes indicate the nitrite ion.
Final answer: A-II, B-III, C-IV, D-I
Q92Single correctEquilibrium
The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is :
(Given : Solubility product of AgCl = )
(Given : Solubility product of AgCl = )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The solubility in pure water and in the potassium chloride solution are computed from the solubility product, and their ratio is taken.
Step 1:In pure water the solubility equals the square root of the solubility product.
Step 2:In 0.1 M potassium chloride the chloride concentration is set by the salt, so the solubility equals the solubility product divided by 0.1.
Step 3:The ratio of the solubility in the salt solution to that in water gives the required value.
Final answer:
Q93Single correctOrganic Chemistry - Some Basic Principles and Techniques
On complete combustion, 0.3 g of an organic compound gave 0.2 g of and 0.1 g of . The percentage composition of carbon and hydrogen in the compound is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 218.18% and 3.70%
Approach:
The percentages of carbon and hydrogen are found from the masses of carbon dioxide and water produced using their mass fractions of carbon and hydrogen.
Step 1:The mass of carbon comes from the carbon dioxide, using the fraction 12 over 44 of its mass.
Step 2:The mass of hydrogen comes from the water, using the fraction 2 over 18 of its mass.
Step 3:Combining the two values gives the percentage composition.
Final answer: 18.18% and 3.70%
Q95Single correctAlcohols, Phenols and Ethers / Hydrocarbons
The alkane that can be oxidized to the corresponding alcohol by KMn as per the equation
is, when:
is, when:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Oxidation of an alkane to an alcohol by potassium permanganate requires a tertiary carbon-hydrogen bond, so the substituents must make all three groups methyl.
Step 1:Alkanes generally resist oxidation, but an alkane bearing a tertiary hydrogen atom can be oxidised to the corresponding alcohol by potassium permanganate.
Step 2:Setting all three substituents to methyl groups makes the central carbon tertiary, giving 2-methylpropane.
Step 3:Oxidation of this tertiary carbon-hydrogen bond gives the corresponding tertiary alcohol.
Final answer:
Q96Single correctThermodynamics
For the following reaction at 300 K
the enthalpy change is +15 kJ, then the internal energy change is :
the enthalpy change is +15 kJ, then the internal energy change is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The internal energy change is obtained from the enthalpy change by subtracting the work term that depends on the change in moles of gas.
Step 1:The change in moles of gas equals the moles of gaseous products minus the moles of gaseous reactants.
Step 2:Rearranging the relation expresses the internal energy change in terms of the enthalpy change and the work term.
Step 3:Substituting the enthalpy change in joules, the change in moles, the gas constant, and the temperature gives the internal energy change.
Final answer:
Q97Single correctChemical Kinetics
Rate constants of a reaction at 500 K and 700 K are 0.04 and 0.14 , respectively; then, activation energy of the reaction is :
(Given: log3.5 = 0.5441, R = 8.31 J mo)
(Given: log3.5 = 0.5441, R = 8.31 J mo)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The activation energy is determined from the two rate constants and temperatures using the two-temperature form of the Arrhenius equation.
Step 1:Taking the logarithm of the Arrhenius equation at each temperature and subtracting gives a relation for the ratio of rate constants.
Step 2:Substituting the rate constants and temperatures gives an equation in the activation energy.
Step 3:Solving for the activation energy using the given logarithm and gas constant gives the result.
Final answer:
Q98Single correctSolutions
Mass of glucose () required to be dissolved to prepare one litre of its solution which is isotonic with 15 g solution of urea is (Given: Molar mass in g mo C : 12, H : 1, O : 16, N : 14)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Isotonic solutions have equal osmotic pressure, which for solutions at the same temperature means equal molar concentrations of the two solutes.
Step 1:Equal osmotic pressure at the same temperature requires equal molar concentrations of glucose and urea.
Step 2:Setting the molarity of glucose equal to that of urea relates the unknown mass of glucose to the known mass of urea.
Step 3:Solving for the mass of glucose gives the required value.
Final answer:
Q99Single correctCoordination Compounds / The d- and f-Block Elements
and structures have
A. Metal-Metal linkage
B. Terminal CO groups
C. Bridging CO groups
D. Metal in zero oxidation state
Choose the correct answer from the options given below :
A. Metal-Metal linkage
B. Terminal CO groups
C. Bridging CO groups
D. Metal in zero oxidation state
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Only A, B, D
Approach:
The structural features common to both metal carbonyls are identified, with attention to which one lacks bridging carbonyl groups.
Step 1:Both carbonyls contain a direct metal-metal linkage and terminal carbon monoxide groups.
Step 2:The decacarbonyl of manganese has no bridging carbonyl groups, so bridging carbonyls are not common to both structures.
Step 3:The metal is in the zero oxidation state in both carbonyls.
Final answer: Only A, B, D
Q100Single correctOrganic Chemistry - Some Basic Principles and Techniques
Methyl group attached to a positively charged carbon atom stabilizes the carbocation due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3hyperconjugation
Approach:
The stabilising influence of a methyl group on an adjacent positive carbon is identified from the electronic effects available to it.
Step 1:A methyl group attached to a carbocation has carbon-hydrogen sigma bonds adjacent to the empty p-orbital of the positive carbon.
Step 2:These carbon-hydrogen sigma electrons delocalise into the empty p-orbital, a stabilising interaction known as hyperconjugation, also aided by the electron-releasing inductive effect.
Step 3:The dominant stabilising factor identified for the methyl group is hyperconjugation.
Final answer: hyperconjugation
Biology100 questions
Q101Single correctBiodiversity and Conservation
The regions with high level of species richness, high degree of endemism and a loss of 70% of the species and habitat are identified as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Biodiversity Hotspots
Approach:
Regions defined by exceptional species richness, high endemism and severe habitat loss are categorised under a specific conservation term.
Step 1:Biodiversity hotspots are regions with very high species richness and a large proportion of endemic species.
Step 2:A hotspot also shows substantial habitat loss, with a major fraction of species lost from its original range.
Final answer: Biodiversity Hotspots
Q102Single correctAnatomy of Flowering Plants
Which of the following tissues are commonly found in the fruit walls of nuts and pulp of pear?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sclereids
Approach:
The gritty cells in hard plant parts such as nut walls and pear pulp belong to a specific type of sclerenchyma.
Step 1:Sclereids are short, isodiametric sclerenchymatous cells with thick lignified walls.
Step 2:Sclereids are found in the hard parts of the plant such as fruit walls of nuts and pulp of fruits like pear.
Final answer: Sclereids
Q103Single correctBiotechnology: Principles and Processes
In a chromosome, there is a specific DNA sequence, responsible for initiating replication. It is called as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 site
Approach:
The DNA sequence from which replication starts has a defined name in molecular biology.
Step 1:The origin of replication, denoted ori, is the sequence where replication begins.
Step 2:Recognition and restriction sites are sequences acted upon by restriction enzymes, not replication start points.
Final answer: site
Q104Single correctPrinciples of Inheritance and Variation
Given below are two statements:
Statement I: When many alleles of a single gene govern a character, it is called polygenic inheritance.
Statement II: In Polygenic inheritance, the effect of each allele is additive.
In the light of the above statements, choose the answer from the options given below.
Statement I: When many alleles of a single gene govern a character, it is called polygenic inheritance.
Statement II: In Polygenic inheritance, the effect of each allele is additive.
In the light of the above statements, choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Statement I is false but Statement II is true
Approach:
Each statement about polygenic inheritance is checked against its accepted definition.
Step 1:Polygenic inheritance refers to a single trait governed by many genes, not many alleles of one gene, so Statement I is incorrect.
Step 2:In polygenic inheritance the phenotype reflects the additive contribution of each allele, so Statement II is correct.
Final answer: Statement I is false but Statement II is true
Q105Single correctPhotosynthesis in Higher Plants
Which of the following are required for the light reaction of Photosynthesis?
A.
B.
C.
D. Chlorophyll
E. Light
Choose the correct answer from the options given below:
A.
B.
C.
D. Chlorophyll
E. Light
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2C, D and E only
Approach:
The reactants required for the photochemical phase of photosynthesis are identified.
Step 1:The light reaction requires water, chlorophyll and light to carry out photolysis and pigment excitation.
Step 2:Carbon dioxide is consumed in the dark reaction, and oxygen is a product of the light reaction rather than a requirement.
Final answer: C, D and E only
Q106Single correctCell: The Unit of Life
Match List-I with List-II
| List-I | List-II |
|---|---|
| A. Fleming | I. Disc shaped sacs or cisternae near cell nucleus |
| B. Robert Brown | II. Chromatin |
| C. George Palade | III. Ribosomes |
| D. Camillo Golgi | IV. Nucleus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-III, D-I
Approach:
Each scientist is paired with the cellular component or discovery associated with them.
Step 1:Fleming named the material of the nucleus, which stains with basic dyes, as chromatin.
Step 2:Robert Brown first described the nucleus as a cell organelle.
Step 3:George Palade first observed ribosomes as dense particles under the electron microscope.
Step 4:Camillo Golgi described densely stained reticular structures near the nucleus, the disc-shaped cisternae of the Golgi apparatus.
Final answer: A-II, B-IV, C-III, D-I
Q107Single correctPrinciples of Inheritance and Variation
Match List-I with List-II
| List-I | List-II |
|---|---|
| A. Incomplete dominance | I. Blood groups in human |
| B. Co-dominance | II. Flower colour in |
| C. Pleiotropy | III. Skin colour in human |
| D. Polygenic inheritance | IV. Phenylketonuria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-I, C-IV, D-III
Approach:
Each type of inheritance is matched with its standard textbook example.
Step 1:Flower colour in Antirrhinum is the classic example of incomplete dominance, where the heterozygote shows an intermediate phenotype.
Step 2:Human ABO blood groups illustrate co-dominance, where alleles I-A and I-B are expressed together.
Step 3:Phenylketonuria is a pleiotropic condition where a single gene affects several traits.
Step 4:Human skin colour is governed by several genes, an example of polygenic inheritance.
Final answer: A-II, B-I, C-IV, D-III
Q108Single correctSexual Reproduction in Flowering Plants
Which part of the ovule stores reserve food materials?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Nucellus
Approach:
The nutritive tissue of the ovule that holds reserve food is identified.
Step 1:The nucellus is a mass of cells enclosed within the integument and has abundant food reserves.
Step 2:Integuments are protective coverings, the placenta bears the ovule, and the funicle is the stalk, none of which serve as the reserve food store.
Final answer: Nucellus
Q109Single correctAnatomy of Flowering Plants
Which one of the following is found in Gymnosperms?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Vessels
Approach:
The conducting element absent from typical gymnosperm xylem is determined.
Step 1:Gymnosperm xylem lacks vessels and conducts water through tracheids.
Step 2:Gymnosperm phloem contains sieve cells and albuminous cells in place of sieve tube elements and companion cells.
Final answer: Vessels
Q110Single correctBiodiversity and Conservation
Which of the following is included under - conservation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Botanical garden
Approach:
The listed conservation methods are sorted into in-situ and ex-situ categories.
Step 1:Wildlife sanctuaries, biosphere reserves and national parks protect species in their natural habitat, an in-situ strategy.
Step 2:A botanical garden conserves plants outside their natural habitat, an ex-situ strategy.
Final answer: Botanical garden
Q111Single correctBiotechnology and its Applications
Given below are two statements:
Statement I : The Indian Government has set up GEAC, which will make decisions regarding the validity of GM research.
Statement II : Biopiracy is the term used to refer to the use of bio-resources by native people.
In the light of the above statements, choose the answer from the options given below :
Statement I : The Indian Government has set up GEAC, which will make decisions regarding the validity of GM research.
Statement II : Biopiracy is the term used to refer to the use of bio-resources by native people.
In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Approach:
Each statement on GM regulation and biopiracy is verified against its definition.
Step 1:The Indian Government has set up GEAC (Genetic Engineering Approval Committee) to decide the validity of GM research and the safety of introducing GM organisms, so Statement I is true.
Step 2:Biopiracy refers to exploitation of bioresources by multinational companies and others without compensatory payment to the native people, not use by native people, so Statement II is false.
Final answer: Statement I is true but Statement II is false
Q112Single correctSexual Reproduction in Flowering Plants
Pollen grains remain preserved as fossils due to the presence of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Exine layer
Approach:
The pollen wall layer responsible for its resistance to decay and fossilisation is identified.
Step 1:The exine is the outer pollen wall layer made of sporopollenin, one of the most resistant organic materials known.
Step 2:Sporopollenin resists physical, chemical and biological degradation, so pollen grains are well preserved as fossils.
Final answer: Exine layer
Q113Single correctPlant Kingdom
Identify the pair :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sphenopsida
Approach:
Each genus is checked against the pteridophyte class it is assigned to.
Step 1:Equisetum belongs to Sphenopsida, while Adiantum belongs to Pteropsida, so the pairing of Sphenopsida with Adiantum is incorrect.
Step 2:Dryopteris is correctly placed in Pteropsida, Psilotum in Psilopsida and Selaginella in Lycopsida.
Final answer: Sphenopsida
Q114Single correctPlant Kingdom
Which of the following is the match?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gymnosperms : , ,
Approach:
Each plant group is checked against the genera listed with it.
Step 1:Cedrus, Pinus and Sequoia are all gymnosperms, making option 1 correct.
Step 2:Sequoia is a gymnosperm, so it is wrongly listed under angiosperms; Polysiphonia is a red alga, not a bryophyte; and Ginkgo is a gymnosperm, not a pteridophyte.
Final answer: Gymnosperms : , ,
Q115Single correctMolecular Basis of Inheritance
Given below are two statements regarding RNA polymerase in prokaryotes.
Statement I : In prokaryotes, RNA polymerase is capable of catalysing the process of elongation during transcription.
Statement II : RNA polymerase associate transiently with Rho factor to initiate transcription.
In the light of the above statements, choose the answer from the options given below :
Statement I : In prokaryotes, RNA polymerase is capable of catalysing the process of elongation during transcription.
Statement II : RNA polymerase associate transiently with Rho factor to initiate transcription.
In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Approach:
The roles of RNA polymerase and its associated factors in prokaryotic transcription are examined.
Step 1:Prokaryotic RNA polymerase catalyses the elongation phase of transcription on its own, so Statement I is correct.
Step 2:RNA polymerase associates with the sigma factor to initiate transcription and with the rho factor to terminate it, so Statement II is incorrect.
Final answer: Statement I is true but Statement II is false
Q116Single correctMolecular Basis of Inheritance
Which of the following is a nucleotide?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Adenylic acid
Approach:
Each option is classified as a base, nucleoside or nucleotide based on its components.
Step 1:A nucleotide contains a nitrogenous base, a sugar and a phosphate group; adenylic acid (adenosine monophosphate) has all three.
Step 2:Uridine and guanosine are nucleosides lacking phosphate, while guanine is a free nitrogenous base.
Final answer: Adenylic acid
Q117Single correctMorphology of Flowering Plants
Match List-I with List-II :
| List-I | List-II |
|---|---|
| A. Vexillary aestivation | I. Brinjal |
| B. Epipetalous stamens | II. Peach |
| C. Epiphyllous stamens | III. Pea |
| D. Perigynous flower | IV. Lily |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Each floral feature is matched with the plant that exemplifies it.
Step 1:Vexillary aestivation is seen in the pea flower.
Step 2:Epipetalous stamens, attached to the petals, are found in brinjal.
Step 3:Epiphyllous stamens, attached to the perianth, are found in lily.
Step 4:Perigynous flowers are seen in peach.
Final answer: A-III, B-I, C-IV, D-II
Q118Single correctMorphology of Flowering Plants
Match List-I with List-II :
| List-I | List-II |
|---|---|
| A. China rose | I. Free central |
| B. Mustard | II. Basal |
| C. Primrose | III. Axile |
| D. Marigold | IV. Parietal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-I, D-II
Approach:
Each plant is matched with its type of placentation.
Step 1:China rose exhibits axile placentation.
Step 2:Mustard exhibits parietal placentation.
Step 3:Primrose exhibits free central placentation.
Step 4:Marigold exhibits basal placentation.
Final answer: A-III, B-IV, C-I, D-II
Q119Single correctAnatomy of Flowering Plants
Which of the following helps in maintenance of the pressure gradient in sieve tubes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Companion cells
Approach:
The phloem cell type that supports the pressure gradient driving translocation is identified.
Step 1:Companion cells are closely associated with sieve tube elements and help maintain the pressure gradient required for translocation.
Step 2:Albuminous cells, sieve cells and phloem parenchyma do not perform this specific role for sieve tubes.
Final answer: Companion cells
Q120Single correctCell: The Unit of Life
Mesosome in a cell is a :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Special structure formed by extension of plasma membrane
Approach:
The structural origin of the mesosome in a prokaryotic cell is recalled.
Step 1:Mesosomes are formed by extensions of the plasma membrane into the cell as a special infolded structure.
Step 2:A membrane-bound vesicle, a polyribosome chain on a single mRNA, and a chromosome are distinct structures and do not describe the mesosome.
Final answer: Special structure formed by extension of plasma membrane
Q121Single correctPlant Growth and Development
Match List-I with List-II :
| List-I | List-II |
|---|---|
| A. Abscisic acid | I. Promotes female flowers in cucumber |
| B. Ethylene | II. Helps seeds to withstand desiccation |
| C. Gibberellin | III. Helps in nutrient mobilisation |
| D. Cytokinin | IV. Promotes bolting in beet, cabbage etc |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-I, C-IV, D-III
Approach:
Each plant growth regulator is matched with its characteristic physiological effect.
Step 1:Abscisic acid helps seeds withstand desiccation.
Step 2:Ethylene promotes female flowers in cucumber.
Step 3:Gibberellin promotes bolting in beet, cabbage and similar plants.
Step 4:Cytokinin helps in nutrient mobilisation.
Final answer: A-II, B-I, C-IV, D-III
Q122Single correctBiotechnology and its Applications
Match List-I with List-II :
| List-I | List-II |
|---|---|
| A. Genetically engineered Human Insulin | I. Gene therapy |
| B. GM Cotton | II. |
| C. ADA Deficiency | III. Antigen-antibody interaction |
| D. ELISA | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-I, D-III
Approach:
Each biotechnology application is matched with the organism or principle behind it.
Step 1:Genetically engineered human insulin is produced using E. coli as the host cell.
Step 2:GM cotton is created using genes from the bacterium Bacillus thuringiensis.
Step 3:ADA deficiency is treated by gene therapy.
Step 4:ELISA is based on antigen-antibody interaction.
Final answer: A-II, B-IV, C-I, D-III
Q123Single correctRespiration in Plants
Match List-I with List-II :
| List-I | List-II |
|---|---|
| A. ETS Complex I | I. NADH Dehydrogenase |
| B. ETS Complex II | II. Cytochrome b |
| C. ETS Complex III | III. Cytochrome C oxidase |
| D. ETS Complex IV | IV. Succinate Dehydrogenase |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-I, B-IV, C-II, D-III
Approach:
Each complex of the electron transport system is matched with its enzyme or complex name.
Step 1:ETS complex I is NADH dehydrogenase.
Step 2:ETS complex II is succinate dehydrogenase.
Step 3:ETS complex III is the cytochrome bc1 complex.
Step 4:ETS complex IV is the cytochrome c oxidase complex.
Final answer: A-I, B-IV, C-II, D-III
Q124Single correctBiodiversity and Conservation
Cryopreservation technique is used for :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Preservation of gametes in viable and fertile condition for a long period
Approach:
The purpose of cryopreservation as an ex-situ conservation method is identified.
Step 1:Cryopreservation is an ex-situ method in which gametes of threatened species are preserved in viable and fertile condition for a long period.
Step 2:Protection of environment, biodiversity hotspots and in-situ conservation do not describe the cryopreservation technique.
Final answer: Preservation of gametes in viable and fertile condition for a long period
Q125Single correctRespiration in Plants
Which of the following are correct about cellular respiration?
A. Cellular respiration is the breaking of C-C bonds of complex organic molecules by oxidation.
B. The entire cellular respiration takes place in Mitochondria.
C. Fermentation takes place under anaerobic condition in germinating seeds.
D. The fate of pyruvate formed during glycolysis depends on the type of organism also.
E. Water is formed during respiration as a result of accepting electrons and getting reduced.
Choose the correct answer from the options given below:
A. Cellular respiration is the breaking of C-C bonds of complex organic molecules by oxidation.
B. The entire cellular respiration takes place in Mitochondria.
C. Fermentation takes place under anaerobic condition in germinating seeds.
D. The fate of pyruvate formed during glycolysis depends on the type of organism also.
E. Water is formed during respiration as a result of accepting electrons and getting reduced.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, C, D, E only
Approach:
Each statement about cellular respiration is judged correct or incorrect.
Step 1:Cellular respiration breaks C-C bonds of complex organic molecules by oxidation, fermentation occurs anaerobically in germinating seeds, the fate of pyruvate depends on the organism, and water forms when oxygen accepts electrons and is reduced; statements A, C, D and E are correct.
Step 2:The entire cellular respiration does not occur only in mitochondria, since glycolysis takes place in the cytoplasm, so statement B is incorrect.
Final answer: A, C, D, E only
Q126Single correctMolecular Basis of Inheritance
Given below are two statements:
Statement I: In eukaryotes there are three RNA polymerases in the nucleus in addition to the RNA polymerase found in the organelles.
Statement II: All the three RNA polymerases in eukaryotic nucleus have different roles.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: In eukaryotes there are three RNA polymerases in the nucleus in addition to the RNA polymerase found in the organelles.
Statement II: All the three RNA polymerases in eukaryotic nucleus have different roles.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both Statement I and Statement II are correct
Approach:
Each statement about eukaryotic RNA polymerases is evaluated against the established model of nuclear transcription.
Step 1:In eukaryotes the nucleus contains at least three RNA polymerases, which is in addition to the RNA polymerase present in the organelles.
Step 2:RNA polymerase I transcribes rRNAs (28S, 18S and 5.8S), RNA polymerase III is responsible for transcription of tRNA, 5S rRNA and snRNAs, and RNA polymerase II transcribes the precursor of mRNA, so the three enzymes have distinct roles.
Final answer: Both Statement I and Statement II are correct
Q127Single correctMolecular Basis of Inheritance
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Histones | I.. Loosely packed chromatin |
| B.. Nucleosome | II.. Densely packed Chromatin |
| C.. Euchromatin | III.. Positively charged basic proteins |
| D.. Heterochromatin | IV.. DNA wrapped around histone octamer |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-I, D-II
Approach:
Each item in List-I is paired with its defining feature in List-II.
Step 1:Histones are positively charged basic proteins rich in lysine and arginine.
Step 2:A nucleosome consists of DNA wrapped around a histone octamer.
Step 3:Euchromatin is the loosely packed and transcriptionally active form of chromatin.
Step 4:Heterochromatin is the densely packed and transcriptionally inactive form of chromatin.
Final answer: A-III, B-IV, C-I, D-II
Q128Single correctCell Cycle and Cell Division
Given below are two statements:
Statement I: Failure of segregation of chromatids during cell cycle resulting in the gain or loss of whole set of chromosome in an organism is known as aneuploidy.
Statement II: Failure of cytokinesis after anaphase stage of cell division results in the gain or loss of a chromosome to called polyploidy.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Failure of segregation of chromatids during cell cycle resulting in the gain or loss of whole set of chromosome in an organism is known as aneuploidy.
Statement II: Failure of cytokinesis after anaphase stage of cell division results in the gain or loss of a chromosome to called polyploidy.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both Statement I and Statement II are false
Approach:
Each statement is checked against the correct definitions of aneuploidy and polyploidy.
Step 1:Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome, a condition called aneuploidy, not the gain or loss of a whole set of chromosomes.
Step 2:Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism, a phenomenon known as polyploidy, not the gain or loss of a single chromosome.
Final answer: Both Statement I and Statement II are false
Q129Single correctCell Cycle and Cell Division
Recombination between homologous chromosomes is completed by the end of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pachytene
Approach:
The stage at which crossing over and recombination are completed during meiotic prophase I is identified.
Step 1:During pachytene of prophase I, crossing over occurs between non-sister chromatids of homologous chromosomes through the action of recombination nodules and the enzyme recombinase.
Final answer: Pachytene
Q130Single correctCell Cycle and Cell Division
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Metacentric chromosome | I.. Chromosome has a terminal centromere |
| B.. Sub-metacentric chromosome | II.. Middle centromere forming two equal arms of chromosome |
| C.. Acrocentric chromosome | III.. Centromere is slightly away from the middle of chromosome resulting into two unequal arms |
| D.. Telocentric chromosome | IV.. Centromere is situated close to its end forming one extremely short and one very long arm |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Each chromosome type is matched with the position of its centromere and the resulting arm lengths.
Step 1:A metacentric chromosome has a centromere in the middle forming two equal arms.
Step 2:A sub-metacentric chromosome has its centromere slightly away from the middle, giving two unequal arms.
Step 3:An acrocentric chromosome has its centromere situated close to its end, forming one extremely short and one very long arm.
Step 4:A telocentric chromosome has a terminal centromere.
Final answer: A-II, B-III, C-IV, D-I
Q131Single correctBiotechnology: Principles and Processes
Ligases is a class of enzymes responsible for catalysing the linking together of two compounds.
Which of the following bonds is not catalysed by it?
Which of the following bonds is not catalysed by it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C – C
Approach:
The types of bonds formed by ligase enzymes are identified to find the one not catalysed.
Step 1:Ligases are enzymes that catalyse the linking together of two compounds by joining bonds such as C – O, C – S, C – N and P – O.
Step 2:The formation of a C – C bond is not among the bonds joined by ligases.
Final answer: C – C
Q132Single correctPlant Growth and Development
F. Skoog observed that callus proliferated from the internodal segments of tobacco stem when auxin was supplied with one of the following except :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Abscisic acid
Approach:
The factors that promoted callus proliferation in Skoog's experiment are recalled to identify the exception.
Step 1:F. Skoog observed that callus proliferated from the internodal segments of tobacco stem only when, in addition to auxin, the nutrient medium was supplemented with extract of vascular tissues, yeast extract, coconut milk or DNA.
Step 2:Miller and co-workers later identified and crystallised the active cytokinin substance that they termed kinetin. Abscisic acid was not one of these growth-promoting factors.
Final answer: Abscisic acid
Q133Single correctPlant Growth and Development
Given below are some statements about plant growth regulators.
A. All GAs are acidic in nature.
B. Auxins are antagonists to GAs.
C. Zeatin was isolated from coconut milk.
D. Ethylene induces flowering in Mango.
E. Abscisic acid induces parthenocarpy.
Choose the correct set of statements from the options given below:
A. All GAs are acidic in nature.
B. Auxins are antagonists to GAs.
C. Zeatin was isolated from coconut milk.
D. Ethylene induces flowering in Mango.
E. Abscisic acid induces parthenocarpy.
Choose the correct set of statements from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, C, D
Approach:
Each statement about plant growth regulators is evaluated for correctness.
Step 1:All gibberellins are acidic in nature.
Step 2:Abscisic acid, not auxin, is antagonistic to gibberellic acid.
Step 3:Zeatin, a natural cytokinin, was isolated from coconut milk and corn kernels.
Step 4:Ethylene promotes flowering in mango.
Step 5:Auxins, not abscisic acid, induce parthenocarpy in fruits such as tomatoes.
Final answer: A, C, D
Q134Single correctBiotechnology: Principles and Processes
Identify the incorrect statement related to gel electrophoresis.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Separated DNA fragments can be directly seen under UV radiation
Approach:
Each statement about agarose gel electrophoresis is checked to find the incorrect one.
Step 1:Separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide, followed by exposure to UV radiation. They cannot be seen in visible light or without staining.
Step 2:The separated bands of DNA can be cut out from the agarose gel and the DNA extracted from the gel piece, a step called elution.
Step 3:DNA is negatively charged, so it moves towards the positive electrode, the anode.
Step 4:The sieving effect of the agarose matrix separates fragments according to their size.
Final answer: Separated DNA fragments can be directly seen under UV radiation
Q135Single correctSexual Reproduction in Flowering Plants
Which of the following examples show monocarpellary, unilocular ovary with many ovules?
A. Sesbania
B. Brinjal
C. Indigofera
D. Tobacco
E. Asparagus
Choose the correct answer from the options given below :
A. Sesbania
B. Brinjal
C. Indigofera
D. Tobacco
E. Asparagus
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A and C only
Approach:
The given plants are grouped by family and the gynoecium features of a monocarpellary, unilocular ovary with many ovules are matched.
Step 1:A monocarpellary, unilocular ovary with many ovules is the characteristic feature of members of the family Fabaceae.
Step 2:Sesbania and Indigofera belong to the family Fabaceae.
Step 3:Brinjal and tobacco belong to the family Solanaceae, while Asparagus belongs to the family Liliaceae.
Final answer: A and C only
Q136Single correctMolecular Basis of Inheritance
Given below are two statements :
Statement I: In the lac operon, the z gene codes for beta-galactosidase which is primarily responsible for the hydrolysis of lactose into galactose and glucose.
Statement II: In addition to lactose, glucose or galactose can also induce lac operon.
In the light of the above statements, choose the correct answer from the options given below :
Statement I: In the lac operon, the z gene codes for beta-galactosidase which is primarily responsible for the hydrolysis of lactose into galactose and glucose.
Statement II: In addition to lactose, glucose or galactose can also induce lac operon.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Statement I is true but Statement II is false
Approach:
Each statement about the lac operon is evaluated for accuracy.
Step 1:In the lac operon the z gene codes for beta-galactosidase, which is primarily responsible for the hydrolysis of the disaccharide lactose into its monomeric units, galactose and glucose.
Step 2:Glucose and galactose cannot act as the inducers of the lac operon. Rather, lactose or allolactose acts as the inducer of the lac operon.
Final answer: Statement I is true but Statement II is false
Q137Single correctSexual Reproduction in Flowering Plants
The part marked as 'x' in the given figure is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Thalamus
Approach:
The labelled structure in the false fruit figure is identified from its position.
Step 1:The given figure is of a false fruit, the strawberry, in which the fleshy edible part develops from the thalamus rather than from the ovary.
Step 2:The part marked 'x', forming the bulk of the fleshy structure on which the true fruits are embedded, represents the thalamus.
Final answer: Thalamus
Q138Single correctAnatomy of Flowering Plants
Given below are two statements :
Statement I: In a dicotyledonous leaf, the adaxial epidermis generally bears more stomata than the abaxial epidermis.
Statement II: In a dicotyledonous leaf, the adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: In a dicotyledonous leaf, the adaxial epidermis generally bears more stomata than the abaxial epidermis.
Statement II: In a dicotyledonous leaf, the adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Statement I is false but Statement II is true
Approach:
Each statement on dicot leaf anatomy is checked against the structure of the dorsiventral leaf.
Step 1:In a dicotyledonous leaf, the adaxial (upper surface) epidermis generally bears very few or even no stomata, having fewer stomata in comparison to the abaxial (lower) surface.
Step 2:The adaxially placed palisade parenchyma is made up of elongated cells that are arranged vertically and parallel to each other.
Final answer: Statement I is false but Statement II is true
Q139Single correctBiomolecules
Which of the following are not fatty acids?
A. Glutamic acid
B. Arachidonic acid
C. Palmitic acid
D. Lecithin
E. Aspartic acid
Choose the correct answer from the options given below :
A. Glutamic acid
B. Arachidonic acid
C. Palmitic acid
D. Lecithin
E. Aspartic acid
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, D and E only
Approach:
Each listed molecule is classified to identify those that are not fatty acids.
Step 1:Glutamic acid and aspartic acid are amino acids.
Step 2:Lecithin is a phospholipid.
Step 3:Palmitic acid and arachidonic acid are fatty acids.
Final answer: A, D and E only
Q140Single correctEcosystem
Consider the pyramid of energy of an ecosystem given below:
If is equivalent to 1000 J, what is the value at ?
If is equivalent to 1000 J, what is the value at ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 J
Approach:
The 10 percent law of energy transfer is applied successively from the highest trophic level down to the base of the pyramid.
Step 1:According to the 10 percent law, only 10 percent of the energy is transferred to each trophic level from the lower trophic level. With equivalent to 1000 J, the value at is obtained by multiplying by ten.
Step 2:Proceeding to the next lower level gives the value at .
Step 3:Applying the same factor once more gives the energy at the base level .
Final answer: J
Q141Single correctPhotosynthesis in Higher Plants
Which one of the following products diffuses out of the chloroplast during photosynthesis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The fate of each light reaction product is examined to find which one leaves the chloroplast.
Step 1:The products of the light reaction are ATP, NADPH and oxygen.
Step 2:Among these products, oxygen diffuses out of the chloroplast, while ATP and NADPH are used to drive the processes leading to the synthesis of food, more accurately, sugars.
Final answer:
Q142Single correctBiotechnology: Principles and Processes
Recombinant DNA molecule can be created normally by cutting the vector DNA and source DNA respectively with:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hind II, Hind II
Approach:
The requirement for compatible ends in recombinant DNA formation is used to select the correct enzyme pair.
Step 1:On cutting the vector and the source DNA with the same restriction enzyme, the same kind of complementary sticky ends are generated, which can then be joined.
Step 2:In the other options the restriction enzymes used for cutting the vector DNA and source DNA are different, so the recombinant vector molecule cannot be created from them.
Final answer: Hind II, Hind II
Q143Single correctEcosystem
Which one of the following is not a limitation of ecological pyramids?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3It accommodates a food web
Approach:
Each option is compared with the known limitations of ecological pyramids to find the one that is not a limitation.
Step 1:The recognised limitations of ecological pyramids are that they do not take into account the same species belonging to two or more trophic levels, they assume a simple food chain that almost never exists in nature, they do not accommodate a food web, and saprophytes are not given any place even though they play a vital role in the ecosystem.
Step 2:Ecological pyramids do not accommodate a food web, so the statement that they accommodate a food web is not a limitation but an incorrect claim.
Final answer: It accommodates a food web
Q144Single correctBiotechnology and its Applications
The Bt toxin in genetically engineered Bt cotton kills the pest by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Creating pores in the midgut
Approach:
The mechanism of action of the activated Bt toxin in the insect gut is recalled.
Step 1:The activated Bt toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis, eventually causing death of the insect.
Step 2:Bt toxin does not kill the pest by affecting its respiratory system, nervous system or the pH of body fluids.
Final answer: Creating pores in the midgut
Q145Single correctBiological Classification
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Euglenoid | I.. Parasitic |
| B.. Dinoflagellate | II.. Saprophytic |
| C.. Slime mould | III.. Photosynthetic |
| D.. | IV.. Switching between photosynthetic and heterotrophic mode |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-III, C-II, D-I
Approach:
Each protist is matched with its characteristic mode of nutrition.
Step 1:Euglenoids are photosynthetic in the presence of sunlight, but when deprived of sunlight they behave like heterotrophs, switching between photosynthetic and heterotrophic modes.
Step 2:Dinoflagellates are mostly marine and photosynthetic.
Step 3:Slime moulds are saprophytic protists.
Step 4:Plasmodium is a malarial parasite that causes malaria, hence parasitic.
Final answer: A-IV, B-III, C-II, D-I
Q146Single correctBiomolecules
Which of the following graphs depicts the effect of substrate concentration on velocity of enzyme catalysed reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Velocity rising with substrate concentration and saturating at Vmax (drawn graph, option 1)
Approach:
The expected dependence of reaction velocity on substrate concentration is compared with each graph.
Step 1:With an increase in substrate concentration, the velocity of the enzymatic reaction rises at first. The reaction ultimately reaches maximum velocity (Vmax), which is not exceeded by any further rise in the concentration of the substrate.
Step 2:This plateau occurs because the enzyme molecules are fewer than substrate molecules, and after saturation there are no free enzyme molecules left to bind the additional substrate molecules.
Step 3:Option (2) is incorrect as velocity rises continuously without saturation. In option (3) after reaching Vmax the velocity declines, and in option (4) velocity declines from high values on increasing substrate concentration, so options (3) and (4) are incorrect.
Final answer: Velocity rising with substrate concentration and saturating at Vmax (drawn graph, option 1)
Q147Single correctOrganisms and Populations
When will the population density increase, under special conditions?
When the number of :
When the number of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Births plus number of immigrants is more than the sum of number of deaths and number of emigrants.
Approach:
The population growth equation is used to determine the condition under which density increases over a time interval.
Step 1:The population density at time relates to the density at time through the births , immigrants , deaths and emigrants .
Step 2:Population density increases if the number of births plus the number of immigrants is more than the number of deaths plus the number of emigrants .
Final answer: Births plus number of immigrants is more than the sum of number of deaths and number of emigrants.
Q148Single correctPrinciples of Inheritance and Variation
When a tall pea plant with round seeds was selfed, it produced the progeny of :
(a) Tall plants with round seeds and
(b) Tall plants with wrinkled seeds.
Identify the genotype of the parent plant.
(a) Tall plants with round seeds and
(b) Tall plants with wrinkled seeds.
Identify the genotype of the parent plant.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4TTRr
Approach:
The progeny phenotypes are analysed trait by trait to deduce the parental genotype on selfing.
Step 1:On selfing, all progeny are tall, so the parent is homozygous tall, TT, because no dwarf plants appear.
Step 2:The progeny include both round and wrinkled seeds, so the round-seeded parent must be heterozygous, Rr, for the seed shape trait.
Step 3:Combining the two traits, the genotype of the parent plant is TTRr.
Final answer: TTRr
Q149Single correctBiodiversity and Conservation
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A.. Biodiversity hotspot | I.. Khasi and Jantia hills in Meghalaya |
| B.. Sacred groves | II.. World Summit on Sustainable Development 2002 |
| C.. Johannesburg, South Africa | III.. |
| D.. Alien species invasion | IV.. Western Ghats |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-I, C-II, D-III
Approach:
Each conservation-related term is matched with its corresponding example or event.
Step 1:Biodiversity hotspots are regions with very high levels of species richness and a high degree of endemism, such as the Western Ghats and Sri Lanka, the Indo-Burma region and the Himalaya.
Step 2:Sacred groves are tracts of forest set aside and protected on cultural and religious grounds, found in the Khasi and Jaintia hills in Meghalaya.
Step 3:The World Summit on Sustainable Development was held in 2002 in Johannesburg, South Africa.
Step 4:Parthenium is an alien species that has invaded and spread in India.
Final answer: A-IV, B-I, C-II, D-III
Q150Single correctPhotosynthesis in Higher Plants
Observe the given figure. Identify the different stages labelled with alphabets by selecting the correct option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-Carboxylation, B-Reduction, C-Regeneration
Approach:
Each labelled stage of the Calvin cycle is identified from the substrate and product positions in the diagram.
Step 1:The Calvin cycle proceeds in three stages. Stage A is carboxylation, during which CO2 combines with ribulose-1,5-bisphosphate to form 3-phosphoglycerate.
Step 2:Stage B is reduction, during which carbohydrate is formed at the expense of the photochemically made ATP and NADPH, converting 3-phosphoglycerate to triose phosphate.
Step 3:Stage C is regeneration, during which the CO2 acceptor ribulose-1,5-bisphosphate is formed again so that the cycle continues.
Final answer: A-Carboxylation, B-Reduction, C-Regeneration
Q151Single correctOrganisms and Populations
Match List-I with List-II: Chose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Predator | I. |
| B. Mutualism | II. |
| C. Parasitism | III. Female wasp and fig |
| D. Sexual deceit | IV. Plasmodium |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Each interaction is paired with its representative example based on the nature of the relationship between the species involved.
Step 1:In predation, one species is benefited (+) while the other is detrimental (-). Pisaster is a predator that plays a key role in the rocky intertidal communities of the American Pacific coast.
Step 2:In the fig tree, a tight one-to-one relationship exists with its partner wasp species, which serves as an example of mutualism.
Step 3:Plasmodium is an endoparasite in humans that causes malaria, representing parasitism.
Step 4:The Mediterranean orchid Ophrys employs sexual deceit to achieve pollination, done by a species of bee.
Final answer: A-II, B-III, C-IV, D-I
Q152Single correctLocomotion and Movement
Match List-I with List-II: Chose the correct answer from the options given below:
| List I (Location of Joint) | List II (Type of Joint) |
|---|---|
| A. Joint between humerus and pectoral girdle | I. Gliding joint |
| B. Knee joint | II. Ball and Socket joint |
| C. Joint between atlas and axis | III. Hinge joint |
| D. Joint between carpals | IV. Pivot joint |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each joint location is matched with its functional joint type based on the movement it permits.
Step 1:The joint between the humerus and the pectoral girdle is a ball and socket joint allowing movement in all planes.
Step 2:The knee joint is a hinge joint permitting movement in one plane only.
Step 3:The joint between the atlas and axis is a pivot joint allowing rotation.
Step 4:The joint between carpals is a gliding joint permitting limited sliding movement.
Final answer: A-II, B-III, C-IV, D-I
Q153Single correctBiotechnology and its Applications
Following are the steps involved in action of toxin in Bt. Cotton
A. The inactive toxin converted into active form due to alkaline pH of gut of insect.
B. Bacillus\ thuringiensis produce crystals with toxic insecticidal proteins.
C. The alkaline pH solubilises the crystals.
D. The activated toxin binds to the surface of midgut cells, creates pores and causes death of the insect.
E. The toxin proteins exist as inactive protoxins in bacteria.
Choose the correct sequence of steps from the option given below:
A. The inactive toxin converted into active form due to alkaline pH of gut of insect.
B. Bacillus\ thuringiensis produce crystals with toxic insecticidal proteins.
C. The alkaline pH solubilises the crystals.
D. The activated toxin binds to the surface of midgut cells, creates pores and causes death of the insect.
E. The toxin proteins exist as inactive protoxins in bacteria.
Choose the correct sequence of steps from the option given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The steps are arranged following the natural sequence of Bt toxin action from crystal production to insect death.
Step 1:Bacillus thuringiensis produces crystals with toxic insecticidal proteins.
Step 2:The toxin proteins exist as inactive protoxins in the bacteria.
Step 3:The alkaline pH of the insect gut solubilises the crystals.
Step 4:The inactive toxin is converted into active form due to the alkaline pH of the gut of the insect.
Step 5:The activated toxin binds to the surface of midgut cells, creates pores and causes death of the insect.
Final answer:
Q154Single correctEvolution
Match List-I with List-II: Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Gene pool | I. Stable within a generation |
| B. Genetic drift | II. Change in gene frequency by chance |
| C. Gene flow | III. Transfer of genes into or out of population |
| D. Gene frequency | IV. Total number of genes and their alleles |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-II, C-III, D-I
Approach:
Each population-genetics term is matched with its correct definition.
Step 1:A gene pool is the total number of genes and their alleles in a population.
Step 2:Genetic drift is the change in gene frequency that occurs by chance.
Step 3:Gene flow is the transfer of genes into or out of a population due to migration.
Step 4:According to the Hardy-Weinberg principle, gene frequencies remain stable within a generation.
Final answer: A-IV, B-II, C-III, D-I
Q155Single correctEvolution
Which evolutionary phenomenon is depicted by the sketch given in figure?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Adaptive radiation
Approach:
The figure shows four finch heads with differently shaped beaks, indicating divergence from a common ancestral form.
Step 1:The sketch depicts adaptive radiation, as many forms of finches with altered beaks arise from the original seed-eating features, enabling them to become insectivorous and vegetarian finches.
Step 2:Artificial selection involves traits favoured by humans, so option (1) is incorrect; genetic drift is a change in gene frequency by chance, so option (2) is incorrect.
Step 3:Convergent evolution occurs when different structures with different origins serve the same function, which is divergent evolution here, so option (3) is incorrect.
Final answer: Adaptive radiation
Q156Single correctBody Fluids and Circulation
A person with blood group ARh can receive the blood transfusion from which of the following types?
A. BRh
B. ABRh
C. ORh
D. ARh
E. ARh
Choose the correct answer from the options given below :
A. BRh
B. ABRh
C. ORh
D. ARh
E. ARh
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C and D only
Approach:
The recipient is blood group A and Rh-negative, so any donor blood must lack the B antigen and the Rh antigen to avoid agglutination.
Step 1:A recipient who is A blood group with Rh(-ve) antigen cannot receive blood from any blood group with Rh(+ve) antigen, and as an A blood group individual has anti-B antibodies in plasma, the donor blood must be O (Rh -ve) or A (Rh -ve) only.
Step 2:Options BRh-negative, ABRh-negative and ARh-positive are not correct answers because B, AB and A blood groups will lead to serious clumping reactions.
Final answer: C and D only
Q157Single correctBiomolecules
Enzymes that catalyse the removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds, are known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Lyases
Approach:
The defining feature given is the non-hydrolytic removal of groups leaving double bonds, which identifies a specific enzyme class.
Step 1:Lyases are the group of enzymes that catalyse the removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.
Step 2:Transferases catalyse the transfer of a group G (other than hydrogen) between a pair of substrates S and S', so option (1) is incorrect.
Step 3:Oxidoreductases catalyse oxidoreduction between two substrates S and S', and dehydrogenases are a type of oxidoreductase, so options (2) and (3) are incorrect.
Final answer: Lyases
Q158Single correctCell Cycle and Cell Division
Match List-I with List-II. Choose the correct answer from the options given below :
| List-I (Event) | List-II (Stage of Prophase-I (Meiosis-I)) |
|---|---|
| A. Chiasmata formation | I. Pachytene |
| B. Crossing over | II. Diakinesis |
| C. Synaptonemal complex formation | III. Diplotene |
| D. Terminalisation of chiasmata | IV. Zygotene |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Each event of prophase-I is mapped to the substage of meiotic prophase-I in which it occurs.
Step 1:Zygotene is the stage during which synaptonemal complex formation occurs.
Step 2:In pachytene the four chromatids of each bivalent chromosome become distinct and crossing over occurs.
Step 3:Diplotene is recognised by the dissolution of the synaptonemal complex, and the X-shaped sites of crossing over are called chiasmata, so chiasmata formation occurs in diplotene.
Step 4:Diakinesis is marked by terminalisation of chiasmata.
Final answer: A-III, B-I, C-IV, D-II
Q159Single correctBiomolecules
Match List-I with List-II. Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A. Primary structure of protein | I. Human haemoglobin |
| B. Secondary structure of protein | II. Disulphide bonds |
| C. Tertiary structure of protein | III. Polypeptide chain |
| D. Quaternary structure of protein | IV. Alpha helix and β sheet |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-II, D-I
Approach:
Each level of protein organisation is matched with its defining feature or example.
Step 1:The primary structure is the polypeptide chain that carries positional information of amino acids.
Step 2:The secondary structure consists of the alpha helix and β sheet.
Step 3:The tertiary structure forms a hollow woollen ball like shape with hydrogen and disulphide bonds.
Step 4:The quaternary structure is the assembly of more than one polypeptide, seen in adult human haemoglobin.
Final answer: A-III, B-IV, C-II, D-I
Q160Single correctChemical Coordination and Integration
Match List-I with List-II. Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A. Epinephrine | I. Hyperglycemia |
| B. Thyroxine | II. Smooth muscle contraction |
| C. Oxytocin | III. Basal metabolic rate |
| D. Glucagon | IV. Emergency hormone |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-III, C-II, D-I
Approach:
Each hormone is matched with its characteristic physiological action.
Step 1:Epinephrine is an emergency hormone, also known as adrenaline, secreted from the adrenal medulla.
Step 2:Thyroxine is secreted from the thyroid gland and plays an important role in regulation of basal metabolic rate.
Step 3:Oxytocin is a peptide hormone synthesised by the hypothalamus and released by the posterior pituitary; it causes smooth muscle contraction.
Step 4:Glucagon acts on α-cells of pancreas, acts mainly on hepatocytes and stimulates glycogenolysis resulting in an increased blood sugar level or hyperglycemia.
Final answer: A-IV, B-III, C-II, D-I
Q161Single correctStructural Organisation in Animals
Which of the following statements is about the type of junction and their role in our body?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Tight junctions help to stop substances from leaking across a tissue.
Approach:
Each statement is evaluated against the established function of the named cell junction.
Step 1:Tight junctions help to stop substances from leaking across a tissue, so option (2) is correct.
Step 2:Adhering junctions perform cementing to keep neighbouring cells together, so option (1) is incorrect and the function in option (3) belongs to adhering junctions rather than tight junctions.
Step 3:Gap junctions facilitate the cells to communicate with each other by connecting the cytoplasm of adjoining cells for rapid transfer of ions, small molecules and sometimes big molecules, so option (4) is incorrect.
Final answer: Tight junctions help to stop substances from leaking across a tissue.
Q162Single correctBiotechnology - Principles and Processes
Select the restriction endonuclease enzymes whose restriction sites are present for the tetracycline resistance (te) gene in the pBR322 cloning vector.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Bam HI and Sal I
Approach:
The pBR322 vector map identifies which restriction sites fall within the tetracycline resistance gene.
Step 1:The restriction sites for restriction endonucleases Bam HI and Sal I are present for the gene in the pBR322 cloning vector.
Step 2:The restriction sites of Pst I and Pvu I are present within the gene in the pBR322 cloning vector, so options (2), (3) and (4) are incorrect.
Final answer: Bam HI and Sal I
Q163Single correctAnimal Kingdom
Match List-I with List-II. Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Chondrichthyes | I. |
| B. Cyclostomata | II. |
| C. Osteichthyes | III. |
| D. Amphibia | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-III, C-I, D-IV
Approach:
Each vertebrate class is paired with a representative genus based on its diagnostic features.
Step 1:Carcharodon is a cartilaginous fish belonging to the class Chondrichthyes.
Step 2:Myxine is a jawless vertebrate belonging to the class Cyclostomata.
Step 3:Clarias is a bony fish belonging to the class Osteichthyes.
Step 4:Ichthyophis is a limbless animal belonging to the class Amphibia.
Final answer: A-II, B-III, C-I, D-IV
Q164Single correctHuman Reproduction
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: During menstrual cycle, the ovulation takes place approximately on 1 day.
Reason R: Rapid secretion of LH in the middle of menstrual cycle induces rupture of Graafian follicle and thereby the release of ovum.
In the light of the above statements, choose the most appropriate answer from the options given below.
Assertion A: During menstrual cycle, the ovulation takes place approximately on 1 day.
Reason R: Rapid secretion of LH in the middle of menstrual cycle induces rupture of Graafian follicle and thereby the release of ovum.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both A and R are correct and R is the correct explanation of A.
Approach:
Both the assertion and the reason are assessed for truth and for whether the reason explains the assertion.
Step 1:In a 28 days menstrual cycle of a human female, rapid secretion of LH leading to its maximum level during the mid-cycle (14th day) called LH surge induces the rupture of the Graafian follicle and thereby the release of ovum, that is ovulation.
Step 2:Both A and R are correct and R is the correct explanation of A.
Final answer: Both A and R are correct and R is the correct explanation of A.
Q165Single correctEvolution
Match List-I with List-II with respect to convergent evolution: Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Lemur | I. Flying phalanger |
| B. Bobcat | II. Numbat |
| C. Anteater | III. Spotted cuscus |
| D. Flying squirrels | IV. Tasmanian tiger cat |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-II, D-I
Approach:
Each placental mammal is matched with the Australian marsupial that shows convergent evolution with it.
Step 1:Lemur is a placental mammal that shows convergent evolution with the spotted cuscus, which is an Australian marsupial.
Step 2:Bobcat is a placental mammal that shows convergent evolution with the Tasmanian tiger cat, which is an Australian marsupial.
Step 3:Anteater is a placental mammal that shows convergent evolution with Numbat, which is an Australian marsupial.
Step 4:Flying squirrel is a placental mammal that shows convergent evolution with the flying phalanger, which is an Australian marsupial.
Final answer: A-III, B-IV, C-II, D-I
Q166Single correctCell Cycle and Cell Division
Match List-I with List-II. Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Cells are metabolically active and proliferate | I. phase |
| B. DNA replication takes place | II. phase |
| C. Proteins are synthesised | III. phase |
| D. Quiescent stage with metabolically active cells | IV. S phase |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-I, D-III
Approach:
Each described cellular activity is matched with the corresponding phase of the cell cycle.
Step 1:During G1 phase the cell is metabolically active and continuously grows but does not replicate its DNA.
Step 2:S or synthesis phase marks the period during which DNA synthesis or replication takes place.
Step 3:During the G2 phase, proteins are synthesised in preparation for mitosis while cell growth continues.
Step 4:Cells that do not divide further exit G1 phase to enter an inactive stage called quiescent stage (G0) of the cell cycle; cells in this stage remain metabolically active but no longer proliferate unless called on to do so.
Final answer: A-II, B-IV, C-I, D-III
Q167Single correctOrganisms and Populations
Match List-I with List-II. Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Migratory flamingoes and resident fish in South American lakes | I. Interference competition |
| B. Abingdon tortoise became extinct after introduction of goats in their habitat | II. Competitive release |
| C. expands its distributional range in the absence of | III. Resource Partitioning |
| D. Five closely related species of Warblers feeding in different locations on same tree | IV. Interspecific competition |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-I, C-II, D-III
Approach:
Each ecological example is matched with the type of competitive interaction it illustrates.
Step 1:In some shallow South American lakes, visiting flamingoes and resident fishes compete for their food, which is interspecific competition.
Step 2:The Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats, which is an example of interference competition.
Step 3:The larger and competitively superior barnacle Balanus dominates the intertidal areas and excludes the smaller barnacle Chathamalus from that zone, so its expansion in the absence of Balanus shows competitive release.
Step 4:If two species compete for the same resource, they could avoid competition by choosing different feeding times or feeding patterns, and the Warblers feeding in different locations on the same tree shows resource partitioning.
Final answer: A-IV, B-I, C-II, D-III
Q168Single correctMicrobes in Human Welfare
Match List-I with List-II relating to microbes and their products: Choose the answer from the options given below:
| List-I (Microbes) | List-II (Products) |
|---|---|
| A. | I. Citric acid |
| B. Trichoderma\ polysporum | II. Clot buster |
| C. Monascus\ purpureus | III. Cyclosporin A |
| D. Aspergillus\ niger | IV. Statins |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each microbe is matched with the commercially important product it yields.
Step 1:Streptokinase is produced by Streptococcus and is used as a clot buster for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.
Step 2:Cyclosporin A, used as an immuno-suppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum.
Step 3:Statins produced by Monascus purpureus have been commercialized as blood-cholesterol lowering agents.
Step 4:Citric acid is produced by Aspergillus niger.
Final answer: A-II, B-III, C-IV, D-I
Q169Single correctCell - The Unit of Life
Match List-I with List-II. Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Particles | I. Chromosomes |
| B. Histones | II. Cilia |
| C. Axoneme | III. Golgi apparatus |
| D. Cisternae | IV. Mitochondria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-I, C-II, D-III
Approach:
Each cellular structure is matched with the organelle or component to which it belongs.
Step 1:F1 particles or oxysomes are found on the inner face of the inner membrane of mitochondria.
Step 2:Histones are basic proteins around which chromatin fibres condense to form chromosomes.
Step 3:The core of cilia is known as axoneme.
Step 4:Golgi complex cisternae are sac like structures.
Final answer: A-IV, B-I, C-II, D-III
Q170Single correctReproductive Health
Match List-I with List-II relating to examples of various kind of IUDs and barrier: Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Copper releasing IUD | I. Vaults |
| B. Non-medicated IUD | II. Multiload 375 |
| C. Contraceptive barrier | III. LNG-20 |
| D. Hormone releasing IUD | IV. Lippes loop |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-I, D-III
Approach:
Each category of intrauterine device or barrier is matched with its named example.
Step 1:Multiload 375 is a copper releasing IUD which suppresses sperm motility and fertilizing capacity of sperms.
Step 2:Lippes loop is a non-medicated intra uterine device.
Step 3:Vaults are a barrier type of contraceptives which prevent physical meeting of ovum and sperms.
Step 4:LNG-20 is a hormone releasing IUD that makes the uterus unsuitable for implantation and the cervix hostile to the sperms.
Final answer: A-II, B-IV, C-I, D-III
Q171Single correctHuman Health and Disease
Given below are two statements:
Statement I: Antibiotics are chemicals produced by microbes that kill other microbes.
Statement II: Antibodies are chemicals formed in body that eliminate microbes.
In the light of the above statements, choose the most appropriate answer from the options given below.
Statement I: Antibiotics are chemicals produced by microbes that kill other microbes.
Statement II: Antibodies are chemicals formed in body that eliminate microbes.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both Statement I and Statement II are correct
Approach:
Each statement is evaluated against the definitions of antibiotics and antibodies.
Step 1:Antibiotics are chemicals produced by microorganisms with the capacity to inhibit the growth and eventually destroy bacterial and other microorganisms in low concentration, so statement I is correct.
Step 2:Antibodies are immunoglobulins produced in the body in response to any attack from pathogens; they facilitate killing of microbes by various mechanisms and provide immunity to the body, so statement II is also correct.
Step 3:Both statement I and statement II are correct.
Final answer: Both Statement I and Statement II are correct
Q172Single correctReproduction in Organisms
Arrange the following parts in human Mammary gland, traversing the route of milk ejection.
A. Mammary duct
B. Lactiferous duct
C. Mammary alveolus
D. Ampulla
E. Mammary tubule
Choose the answer from the options given below:
A. Mammary duct
B. Lactiferous duct
C. Mammary alveolus
D. Ampulla
E. Mammary tubule
Choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The parts are ordered following the path milk travels from its site of secretion to the nipple.
Step 1:The correct route of milk ejection via mammary glands in humans is mammary alveolus, then mammary tubule, then mammary duct, then mammary ampulla, then lactiferous duct.
Step 2:Options (1), (2) and (4) represent the wrong pathway.
Final answer:
Q173Single correctBiotechnology - Principles and Processes
Which of the following are correct about ?
A. Cut the DNA with blunt end
B. Cut the DNA with sticky end
C. Recognise a specific palindromic sequence
D. Cut the DNA between the base G and A when it encounters the DNA sequence 'GAATTC'
E. Exonuclease
Choose the answer from the options given below:
A. Cut the DNA with blunt end
B. Cut the DNA with sticky end
C. Recognise a specific palindromic sequence
D. Cut the DNA between the base G and A when it encounters the DNA sequence 'GAATTC'
E. Exonuclease
Choose the answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B, C, D only
Approach:
Each statement about EcoRI is checked against the known cutting behaviour and classification of the enzyme.
Step 1:EcoRI does not cut the DNA with blunt end; instead, it cuts the DNA with sticky or cohesive or staggered ends on each strand.
Step 2:EcoRI is a restriction endonuclease that recognises a specific palindromic sequence and cuts at a specific site within the DNA, known as the restriction site, so it is not an exonuclease and statements A and E are incorrect.
Step 3:The recognition sequence for EcoRI is 5'-ATTC-3' and 3'--5', and it cuts the DNA between bases G and A only when the sequence GAATTC is present in the DNA.
Final answer: B, C, D only
Q174Single correctStructural Organisation in Animals
Which of the following is/are present in female cockroach?
A. Collateral gland
B. Mushroom gland
C. Spermatheca
D. Anal style
E. Phallic gland
Choose the most appropriate answer from the options given below:
A. Collateral gland
B. Mushroom gland
C. Spermatheca
D. Anal style
E. Phallic gland
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A and C only
Approach:
Each listed structure is checked for whether it belongs to the female or the male reproductive system of the cockroach.
Step 1:The collateral gland is present in female cockroaches as a pair of glands that secrete the hard egg case or ootheca.
Step 2:The mushroom gland is absent in female cockroaches and is present in males in the 6th to 7th abdominal segments.
Step 3:Spermatheca is present in female cockroaches in the 6th abdominal segment.
Step 4:Anal styles are absent in female cockroaches and are present in males, projecting backwards from the 9th sternum; the phallic gland is also absent in female cockroaches and present in males as a large club shaped gland below the ejaculatory duct.
Final answer: A and C only
Q175Single correctHuman Health and Disease
Match List-I with List-II: Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Malignant tumors | I. Destroy tumors |
| B. MALT | II. AIDS |
| C. NACO | III. Metastasis |
| D. α-Interferons | IV. Lymphoid tissue |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-II, D-I
Approach:
Each term is paired with the property, disease, tissue or function with which it is associated.
Step 1:Malignant tumors exhibit the property of metastasis.
Step 2:MALT stands for Mucosa Associated Lymphoid Tissue.
Step 3:NACO stands for National AIDS Control Organisation.
Step 4:α-Interferons are biological response modifiers given to cancer patients that activate their immune system and help in destroying the tumor.
Final answer: A-III, B-IV, C-II, D-I
Q176Single correctAnimal Kingdom
Open Circulatory system is present in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3, ,
Approach:
Open circulatory systems lack continuous closed vessels, so blood bathes the tissues directly in body sinuses; identifying which animals show this pattern requires recalling their phyla.
Step 1:Anopheles is an arthropod, Limax is a mollusc and Limulus is an arthropod; arthropods and most molluscs have an open circulatory system.
Step 2:In option (1) Nereis is an annelid with a closed circulatory system, so the set is mixed.
Step 3:In option (2) Hirudinaria is an annelid with a closed circulatory system, so the set is mixed.
Step 4:In option (4) Pheretima is an annelid with a closed circulatory system, so the set is mixed.
Final answer: , ,
Q177Single correctStructural Organisation in Animals
In which of the following connective tissues, the cells secrete fibres of collagen or elastin?
A. Cartilage
B. Bone
C. Adipose tissue
D. Blood
E. Areolar tissue
Choose the most appropriate answer from the options given below :
A. Cartilage
B. Bone
C. Adipose tissue
D. Blood
E. Areolar tissue
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B, C and E only
Approach:
Fibroblasts and related cells of connective tissues proper and supportive tissues secrete collagen or elastin fibres; blood lacks such fibre-secreting cells.
Step 1:Cartilage, bone, adipose tissue and areolar tissue contain cells that secrete fibres of collagen or elastin into a structured matrix.
Step 2:Blood is a fluid connective tissue whose matrix is plasma and which lacks fibroblast-like fibre-secreting cells.
Final answer: A, B, C and E only
Q178Single correctAnimal Kingdom
Which of the following pairs is an match?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Platyhelminthes-Diploblastic organisation
Approach:
Each pair links a taxon to a structural feature; the incorrect match is found by comparing the stated feature with the established characteristic of the group.
Step 1:Platyhelminthes are triploblastic organisms, not diploblastic, so this pairing is wrong.
Step 2:Annelids and arthropods show bilateral symmetry, sponges are acoelomate, and coelenterates and ctenophores show radial symmetry, all of which are correct.
Final answer: Platyhelminthes-Diploblastic organisation
Q179Single correctBreathing and Exchange of Gases
Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Residual Volume | I. Maximum volume of air that can be breathed in after forced expiration |
| B. Vital Capacity | II. Volume of air inspired or expired during normal respiration |
| C. Expiratory Capacity | III. Volume of air remaining in lungs after forcible expiration |
| D. Tidal Volume | IV. Total volume of air expired after normal expiration |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Each respiratory term is matched to its definition based on standard pulmonary volume and capacity concepts.
Step 1:Residual volume is the volume of air remaining in the lungs after forcible expiration, matching III.
Step 2:Vital capacity is the maximum volume of air that can be breathed in after forced expiration, matching I.
Step 3:Expiratory capacity is the total volume of air expired after a normal expiration, matching IV.
Step 4:Tidal volume is the volume of air inspired or expired during normal respiration, matching II.
Final answer: A-III, B-I, C-IV, D-II
Q180Single correctEvolution
Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Living Fossil | I. Elongated canine teeth |
| B. Connecting Link | II. Vermiform appendix |
| C. Vestigial Organ | III. |
| D. Atavism | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-III, C-II, D-I
Approach:
Each evolutionary term is matched to its representative example using standard textbook associations.
Step 1:Latimeria is a living fossil, the bony fish coelacanth, matching IV.
Step 2:Echidna is a connecting link between reptiles and mammals, matching III.
Step 3:The vermiform appendix is a vestigial organ, a remnant of a once functional structure, matching II.
Step 4:Atavism is the reappearance of an ancestral trait such as elongated canine teeth, matching I.
Final answer: A-IV, B-III, C-II, D-I
Q181Single correctNeural Control and Coordination
Choose the answer from the options given below:
| List-I | List-II |
|---|---|
| A. Schwann cells | I. Neurotransmitter |
| B. Synaptic knob | II. Cerebral cortex |
| C. Bipolar neurons | III. Myelin sheath |
| D. Multipolar neurons | IV. Retina |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Each neural component is matched to its associated structure or product based on neuron anatomy and physiology.
Step 1:The myelinated nerve fibres are enveloped with Schwann cells which form a myelin sheath around the axon, matching III.
Step 2:The synaptic knob possesses synaptic vesicles containing chemicals called neurotransmitters, matching I.
Step 3:Bipolar neurons have one axon and one dendrite and are found in the retina of the eye, matching IV.
Step 4:Multipolar neurons have one axon and many dendrites, found in the cerebral cortex, matching II.
Final answer: A-III, B-I, C-IV, D-II
Q182Single correctExcretory Products and their Elimination
Diuresis is prevented by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Vasopressin from Neurohypophysis
Approach:
Diuresis means increased urine output; the hormone that promotes water reabsorption and thus reduces urine volume prevents it.
Step 1:Vasopressin or anti-diuretic hormone from the neurohypophysis facilitates water reabsorption from the latter part of renal tubules, reducing urine output.
Step 2:Renin converts angiotensinogen to angiotensin I and further to angiotensin II, a vasoconstrictor that raises GFR rather than preventing diuresis.
Step 3:ANF causes vasodilation and thereby decreases GFR; aldosterone is released from the adrenal cortex, not the medulla.
Final answer: Vasopressin from Neurohypophysis
Q183Single correctHuman Health and Disease
Following is the list of STDs. Select the diseases which are not completely curable.
A. Genital warts
B. Genital herpes
C. Syphilis
D. Hepatitis-B
E. Trichomoniasis
Choose the correct answer from given below:
A. Genital warts
B. Genital herpes
C. Syphilis
D. Hepatitis-B
E. Trichomoniasis
Choose the correct answer from given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and D only
Approach:
Most STDs are completely curable if detected early; viral STDs such as genital herpes and hepatitis-B are not completely curable.
Step 1:Genital herpes and hepatitis-B are caused by viruses and are not completely curable.
Step 2:Genital warts, syphilis (a bacterial STD) and trichomoniasis (a protozoan STD) are completely curable upon proper detection and treatment.
Final answer: B and D only
Q184Single correctEvolution
What is the correct order (old to recent) of periods in Paleozoic era?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Silurian, Devonian, Carboniferous, Permian
Approach:
The periods of the Paleozoic era are arranged from oldest to most recent according to the geological time scale.
Step 1:Within the Paleozoic era the sequence from old to recent runs Silurian, Devonian, Carboniferous, Permian.
Step 2:Option (1) places Permian before Carboniferous, option (3) places Permian first, and option (4) places Devonian last, all of which are out of order.
Final answer: Silurian, Devonian, Carboniferous, Permian
Q185Single correctBody Fluids and Circulation
'Lub' sound of Heart is caused by the____________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4closure of the tricuspid and bicuspid valves
Approach:
The first heart sound 'Lub' arises from the closure of the atrioventricular valves at the onset of ventricular systole.
Step 1:The 'Lub' is the first heart sound produced by the closing of the tricuspid (right atrium and right ventricle) and bicuspid or mitral (left atrium and left ventricle) valves during ventricular systole.
Step 2:Closure of the semilunar valves produces the second heart sound 'Dub', and opening of valves does not produce heart sounds.
Final answer: closure of the tricuspid and bicuspid valves
Q186Single correctHuman Reproduction
Choose the correct answer from the option given below :
| List I (Structures) | List II (Features) |
|---|---|
| A. Mons pubis | I. A fleshy fold of tissue surrounding the vaginal opening |
| B. Clitoris | II. Fatty cushion of cells covered by skin and hair |
| C. Hymen | III. Tiny finger-like structure above labia minora |
| D. Labina majora | IV. A thin membrane-like structure covering vaginal opening |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each structure of the female external genitalia is matched to its descriptive feature.
Step 1:Mons pubis is a cushion of fatty tissue covered by skin and pubic hair, matching II.
Step 2:Clitoris is a tiny finger-like structure which lies at the upper junction of the two labia minora, matching III.
Step 3:Hymen is a membrane that partially covers the opening of the vagina, matching IV.
Step 4:Labia majora are fleshy folds of tissue which extend down from the mons pubis and surround the vaginal opening, matching I.
Final answer: A-II, B-III, C-IV, D-I
Q187Single correctPrinciples of Inheritance and Variation
Aneuploidy is a chromosomal disorder where chromosome number is not the exact copy of its haploid set of chromosomes, due to :
A. Substitution
B. Addition
C. Deletion
D. Translocation
E. Inversion
Choose the most appropriate answer from the options given below :
A. Substitution
B. Addition
C. Deletion
D. Translocation
E. Inversion
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and C only
Approach:
Aneuploidy results from failure of segregation of chromatids during cell division, leading to a gain or loss of chromosomes; this corresponds to addition or deletion of a chromosome.
Step 1:Failure of segregation of chromatids during the cell division cycle results in the gain or loss of a chromosome, called aneuploidy.
Step 2:Gain corresponds to addition and loss corresponds to deletion of a chromosome, so B and C are correct.
Final answer: B and C only
Q188Single correctBiotechnology - Principles and Processes
Given below are two statements :
Statement I : RNA interference takes place in all Eukaryotic organisms as method of cellular defense.
Statement II : RNAi involves the silencing of a specific mRNA due to a complementary single-stranded RNA molecule that binds and prevents translation of mRNA
In the light of the above statements, choose the answer from the options given below.
Statement I : RNA interference takes place in all Eukaryotic organisms as method of cellular defense.
Statement II : RNAi involves the silencing of a specific mRNA due to a complementary single-stranded RNA molecule that binds and prevents translation of mRNA
In the light of the above statements, choose the answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Statement I is true but Statement II is false.
Approach:
Each statement on RNA interference is assessed against the established mechanism of RNAi-mediated gene silencing.
Step 1:Statement I correctly states that RNA interference takes place in all eukaryotic organisms as a method of cellular defense.
Step 2:Statement II is incorrect because the silencing of a specific mRNA in RNAi involves a complementary double-stranded RNA molecule rather than a single-stranded one.
Final answer: Statement I is true but Statement II is false.
Q189Single correctChemical Coordination and Integration
Identify the wrong statements :
A. Erythropoietin is produced by juxtaglomerular cells of the kidney
B. Leydig cells produce Androgens
C. Atrial Natriuretic factor, a peptide hormone is secreted by the seminiferous tubules of the testes
D. Cholecystokinin is produced by gastrointestinal tract
E. Gastrin acts on intestinal wall and helps in the production of pepsinogen
Choose the most appropriate answer from the options given below :
A. Erythropoietin is produced by juxtaglomerular cells of the kidney
B. Leydig cells produce Androgens
C. Atrial Natriuretic factor, a peptide hormone is secreted by the seminiferous tubules of the testes
D. Cholecystokinin is produced by gastrointestinal tract
E. Gastrin acts on intestinal wall and helps in the production of pepsinogen
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C and E only
Approach:
Each statement on hormone source and action is checked against established endocrine physiology to identify the wrong ones.
Step 1:Atrial natriuretic factor is secreted from the atria of the heart, not the seminiferous tubules of the testes, so statement C is wrong.
Step 2:Gastrin acts on the gastric glands and stimulates the secretion of hydrochloric acid and pepsinogen, not on the intestinal wall, so statement E is wrong.
Step 3:Statements A, B and D correctly describe erythropoietin from juxtaglomerular cells, androgens from Leydig cells and cholecystokinin from the gastrointestinal tract.
Final answer: C and E only
Q190Single correctBiotechnology - Principles and Processes
Following are the steps involved in the process of PCR.
A. Annealing
B. Amplification (1 billion times)
C. Denaturation
D. Treatment with Taq polymerase and deoxynucleotides
E. Extension
Choose the correct sequence of steps of PCR from the options given below :
A. Annealing
B. Amplification (1 billion times)
C. Denaturation
D. Treatment with Taq polymerase and deoxynucleotides
E. Extension
Choose the correct sequence of steps of PCR from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The polymerase chain reaction proceeds through denaturation, annealing of primers, extension by Taq polymerase, and repeated cycling to amplify the DNA.
Step 1:Denaturation separates the DNA strands and is the first step.
Step 2:Annealing of primers follows, then treatment with Taq polymerase and deoxynucleotides, then extension.
Step 3:Repeated cycling brings amplification by about one billion times as the final outcome.
Final answer:
Q191Single correctExcretory Products and their Elimination
Given below are two statements :
Statements I: Concentrated urine is formed due to counter current mechanism in nephron.
Statement II: Counter current mechanism helps to maintain osmotic gradient in the medullary interstitium.
In the light of the above statements, choose the most appropriate answer from the options given below.
Statements I: Concentrated urine is formed due to counter current mechanism in nephron.
Statement II: Counter current mechanism helps to maintain osmotic gradient in the medullary interstitium.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both Statement I and Statement II are correct.
Approach:
Both statements describe the role of the counter current mechanism in concentrating urine and maintaining the medullary osmotic gradient.
Step 1:Mammals have the ability to produce concentrated urine; the Henle's loop and vasa recta play a significant role in this counter current mechanism.
Step 2:The proximity between the Henle's loop and vasa recta maintains an increasing osmolarity from about 300 mOsmol in the cortex to about 1200 mOsmol in the inner medulla, sustaining the medullary gradient.
Final answer: Both Statement I and Statement II are correct.
Q192Single correctCell - The Unit of Life
Given below are two statements:
Statement I: Concentrically arranged cisternae of Golgi complex are arranged near the nucleus with distinct convex cis or maturing and concave trans or forming face.
Statement II: A number of proteins are modified in the cisternae of Golgi complex before they are released from cis face.
In the light of the above statements, choose the answer from the option given below.
Statement I: Concentrically arranged cisternae of Golgi complex are arranged near the nucleus with distinct convex cis or maturing and concave trans or forming face.
Statement II: A number of proteins are modified in the cisternae of Golgi complex before they are released from cis face.
In the light of the above statements, choose the answer from the option given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both Statement I and Statement II are false
Approach:
Each statement on Golgi complex orientation and protein release is checked against the correct faces of the organelle.
Step 1:The Golgi cisternae are concentrically arranged near the nucleus with the convex cis or the forming face and the concave trans or maturing face, so Statement I reverses the labels and is false.
Step 2:Proteins synthesised by ribosomes on the endoplasmic reticulum are modified in the cisternae of the Golgi apparatus before they are released from its trans face, not the cis face, so Statement II is false.
Final answer: Both Statement I and Statement II are false
Q193Single correctHuman Reproduction
Choose the correct answer from the option given below:
| List I | List II |
|---|---|
| A. Parturition | I. Several antibodies for new-born babies |
| B. Placenta | II. Collection of ovum after ovulation |
| C. Colostrum | III. Foetal ejection reflex |
| D. Fimbriae | IV. Secretion of the hormone hCG |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-IV, C-I, D-II
Approach:
Each reproductive term is matched to its associated feature based on the events of pregnancy, parturition and lactation.
Step 1:Parturition is induced by a neuroendocrine mechanism whose signals trigger the foetal ejection reflex, matching III.
Step 2:Placenta acts as an endocrine tissue and produces hormones such as hCG, matching IV.
Step 3:Colostrum is the milk produced during the initial days of lactation and contains several antibodies necessary for new-born babies, matching I.
Step 4:Fimbriae help in the collection of the ovum after ovulation, matching II.
Final answer: A-III, B-IV, C-I, D-II
Q194Single correctAnimal Kingdom
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Members of subphylum vertebrata possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.
Reason R: Thus all chordates are vertebrates but not all vertebrates are chordates.
In the light of the above statements choose the correct answer from the option given below.
Assertion A: Members of subphylum vertebrata possess notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.
Reason R: Thus all chordates are vertebrates but not all vertebrates are chordates.
In the light of the above statements choose the correct answer from the option given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A is true but R is false.
Approach:
Each part of the assertion-reason pair is checked against the classification of phylum Chordata and its subphyla.
Step 1:Members of subphylum Vertebrata possess a notochord during the embryonic period, and it is later replaced by a cartilaginous or bony vertebral column in the adult, so Assertion A is true.
Step 2:Phylum Chordata is divided into the three subphyla Urochordata, Cephalochordata and Vertebrata; thus all vertebrates are chordates but not all chordates are vertebrates, which is the reverse of what Reason R states, making R false.
Final answer: A is true but R is false.
Q195Single correctPrinciples of Inheritance and Variation
The mother has A+ blood group the father has B+ and the child is A+. What can be the possibility of genotypes of all three, respectively?
A.
B.
C.
D.
E.
Choose the correct answer from the option given below:
A.
B.
C.
D.
E.
Choose the correct answer from the option given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B and E
Approach:
The child of A blood group has genotype , so each parent must be able to contribute the appropriate allele; valid parental genotypes are deduced from the child's makeup.
Step 1:The child blood group A means an i allele came from the father, so the father with B blood group has genotype .
Step 2:The mother with A blood group can have genotype or , and the child is .
or
Step 3:The possible genotypes for mother, father and child respectively are (B) or (E).
Final answer: B and E
Q196Single correctOrganisms and Populations
What do 'a' and 'b' represent in the following population growth curve?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1'a' represents exponential growth when responses are not limiting the growth; and 'b' represents logistic growth when responses are limiting the growth.
Approach:
The two curves are identified from their shapes and governing equations, distinguishing unlimited exponential growth from resource-limited logistic growth.
Step 1:Curve 'a' follows the exponential equation and forms a J-shaped curve, representing growth when responses are not limiting the growth.
Step 2:Curve 'b' follows the logistic equation and forms an S-shaped curve, representing growth when responses are limiting the growth.
Final answer: 'a' represents exponential growth when responses are not limiting the growth; and 'b' represents logistic growth when responses are limiting the growth.
Q197Single correctLocomotion and Movement
Select the correct statements regarding mechanism of muscle contraction.
A. It is initiated by a signal sent by CNS via sensory neuron.
B. Neurotransmitter generates action potential in the sarcolemma.
C. Increased C level leads to the binding of calcium with troponin on action filaments.
D. Masking of active site for actin is activated.
E. Utilising the energy from ATP hydrolysis to form cross bridge.
Choose the most appropriate answer from the options given below:
A. It is initiated by a signal sent by CNS via sensory neuron.
B. Neurotransmitter generates action potential in the sarcolemma.
C. Increased C level leads to the binding of calcium with troponin on action filaments.
D. Masking of active site for actin is activated.
E. Utilising the energy from ATP hydrolysis to form cross bridge.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B, C and E only
Approach:
Each statement on the steps of muscle contraction is checked against the sliding filament mechanism to select the correct ones.
Step 1:Muscle contraction is initiated by a signal sent by the CNS via a motor neuron, not a sensory neuron, so statement A is wrong.
Step 2:The neurotransmitter acetylcholine released at the neuromuscular junction generates an action potential in the sarcolemma, so statement B is correct.
Step 3:Increased calcium binds to troponin on actin filaments and unmasks the active sites for myosin rather than masking them, so statement C is correct and statement D is wrong.
Step 4:Energy from ATP hydrolysis is utilised by the myosin head to bind the exposed active sites on actin and form a cross bridge, so statement E is correct.
Final answer: B, C and E only
Q198Single correctStructural Organisation in Animals
Choose the correct answer from the options given below :
| List - I | List - II |
|---|---|
| A. Squamous Epithelium | I. Goblet cells of alimentary canal |
| B. Ciliated Epithelium | II. Inner lining of pancreatic ducts |
| C. Glandular Epithelium | III. Walls of blood vessels |
| D. Compound Epithelium | IV. Inner surface of Fallopian tubes |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-IV, C-I, D-II
Approach:
Each epithelial type is matched to the body location where it is characteristically found.
Step 1:Simple squamous epithelium forms a thin diffusion boundary and is found in the walls of blood vessels, matching III.
Step 2:Ciliated epithelium moves particles in a specific direction and lines the inner surface of the Fallopian tubes, matching IV.
Step 3:Glandular epithelium includes the isolated goblet cells of the alimentary canal, matching I.
Step 4:Compound epithelium of more than one layer of cells is found in the inner lining of pancreatic ducts, matching II.
Final answer: A-III, B-IV, C-I, D-II
Q199Single correctHuman Health and Disease
Choose the correct answer from the options given below :
| List - I | List - II |
|---|---|
| A. B-Lymphocytes | I. Passive immunity |
| B. Interferons | II. Cell mediated immunity |
| C. T-Lymphocytes | III. Produce an army of proteins in response to pathogens |
| D. Colostrum | IV. Innate immunity |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-IV, C-II, D-I
Approach:
Each immune component is matched to the type of immunity or function it provides.
Step 1:B-lymphocytes produce an army of proteins called antibodies in response to pathogens, matching III.
Step 2:Interferons form the cytokine barrier of innate immunity and protect non-infected cells from further viral infection, matching IV.
Step 3:T-lymphocytes mediate cell mediated immunity, matching II.
Step 4:Colostrum provides natural passive immunity to the infant, matching I.
Final answer: A-III, B-IV, C-II, D-I
Q200Single correctBreathing and Exchange of Gases
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: During the transportation of gases, about 20-25 percent of C is carried by Haemoglobin as carbamino-haemoglobin.
Reason R: This binding is related to high pC and low p in tissues.
In the light of the above statements, choose the correct answer from the options given below.
Assertion A: During the transportation of gases, about 20-25 percent of C is carried by Haemoglobin as carbamino-haemoglobin.
Reason R: This binding is related to high pC and low p in tissues.
In the light of the above statements, choose the correct answer from the options given below.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both A and R are true and R is the correct explanation of A.
Approach:
Both the assertion on carbamino-haemoglobin transport and the reason on partial pressures are checked against the mechanism of carbon dioxide transport.
Step 1:Nearly 20-25 percent of C is transported by RBCs bound to haemoglobin as carbamino-haemoglobin, while about 70 percent is carried as bicarbonate and about 7 percent in dissolved state in plasma, so Assertion A is true.
Step 2:The binding of C with haemoglobin is related to the partial pressures of the gases; high pC and low p in tissues favour more binding of C, so Reason R is true and explains Assertion A.
Final answer: Both A and R are true and R is the correct explanation of A.
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