Back to NEET PYQs
NEET 2023 May 07 Question Paper with Solutions
All 200 questions from the NEET 2023 (May 07) 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 2023Chemistry PYQs 2023Biology PYQs 2023
- Questions
- 200
- Physics
- 50
- Chemistry
- 50
- Biology
- 100
Physics50 questions
Q1Single correctMechanical Properties of Solids
Let a wire be suspended from the ceiling (rigid support) and stretched by a weight attached at its free end. The longitudinal stress at any point of cross-sectional area of the wire is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Longitudinal stress at a cross-section is the internal restoring force per unit area acting normal to that section.
Step 1:The wire is in equilibrium under the suspending weight, so the tension at any cross-section balances the load.
Step 2:Dividing the internal force by the cross-sectional area gives the longitudinal stress.
Final answer:
Q2Single correctSystems of Particles and Rotational Motion
The ratio of radius of gyration of a solid sphere of mass and radius about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option
Approach:
The radius of gyration k relates to the moment of inertia through , so the ratio of radii of gyration is the square root of the ratio of the moment-of-inertia coefficients.
Step 1:The radius of gyration of the solid sphere follows from its moment of inertia about a diameter.
Step 2:The radius of gyration of the thin hollow sphere follows from its moment of inertia about a diameter.
Step 3:Forming the ratio of the two radii of gyration eliminates the common factors.
Final answer: This question was withdrawn from the paper; full marks were awarded to every candidate.
Q3Single correctElectrostatic Potential and Capacitance
The equivalent capacitance of the system shown in the following circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The network reduces by combining the two capacitors that share a parallel loop and then placing that combination in series with the remaining capacitor between terminals and .
Step 1:The two capacitors forming the parallel loop combine additively.
Step 2:This combination is in series with the remaining capacitor across A and B.
Final answer:
Q4Single correctMotion in a Plane
A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3along north-east
Approach:
Force acts along the change in velocity, since speed is constant and only direction changes.
Step 1:The initial velocity points south and the final velocity points east, both of equal magnitude.
Step 2:The change in velocity is the final velocity minus the initial velocity.
Step 3:The force shares the direction of the change in velocity.
Final answer: along north-east
Q5Single correctElectric Charges and Fields
If over a surface, then :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Approach:
Gauss's law links the net flux through a closed surface to the enclosed charge, and zero net flux constrains the balance of entering and leaving field lines.
Step 1:A vanishing net flux means the enclosed charge is zero, but it does not require the field on the surface to vanish.
Step 2:Zero net flux requires that every flux line entering the surface leaves it, so the count of incoming and outgoing lines is equal.
Final answer: the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Q6Single correctMechanical Properties of Solids
The potential energy of a long spring when stretched by cm is U. If the spring is stretched by cm, potential energy stored in it will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The elastic potential energy of a spring varies as the square of its extension, so the energy scales with the square of the ratio of extensions.
Step 1:The stored energy for the first extension defines the reference value.
Step 2:The extension increases by a factor of four, so the energy increases by the square of that factor.
Final answer:
Q7Single correctCurrent Electricity
If the galvanometer does not show any deflection in the circuit shown, the value of is given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A galvanometer reading zero means its branch carries no current, so the cell in that branch fixes the potential difference across directly.
Step 1:With no current through the galvanometer, the cell draws no current, so the potential difference across R equals its e.m.f.
Step 2:The remainder of the e.m.f. appears across the resistor, which carries the same current as R.
Step 3:Equal currents through and R make the potential differences proportional to the resistances.
Final answer:
Q8Single correctAlternating Current
A 12 V, 60 W lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 A
Approach:
For an ideal transformer the input power equals the output power, so the primary current follows from the lamp power divided by the primary voltage.
Step 1:The lamp draws the full output power, which equals the input power on the primary side.
Step 2:Dividing the primary power by the mains voltage gives the primary current.
Final answer: A
Q9Single correctSemiconductor Electronics
A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Capacitor
Approach:
Each component of the rectifier has a distinct role, and the smoothing of ripple is performed by the filter element placed across the load.
Step 1:The diodes convert alternating input to a pulsating direct output, and the centre-tapped transformer provides the two phase-opposed inputs.
Step 2:A capacitor connected in parallel with the load charges at the peaks and discharges between them, filling the gaps and smoothing the ripple.
Final answer: Capacitor
Q10Single correctRay Optics and Optical Instruments
Light travels a distance x in time in air and in time in another denser medium. What is the critical angle for this medium?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The refractive index is the ratio of light speed in air to that in the medium, and the critical angle is the inverse sine of the reciprocal of that index.
Step 1:The speed in air covers x in time , while the speed in the medium covers in time .
Step 2:The refractive index follows from the ratio of the two speeds.
Step 3:The critical angle is the inverse sine of the reciprocal of the refractive index.
Final answer:
Q11Single correctCurrent Electricity
Resistance of a carbon resistor determined from colour codes is . The colour of third band must be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Orange
Approach:
In the colour code the first two bands give the significant figures and the third band gives the decimal multiplier, so the printed resistance has to be written in the form (two digits) (power of ten).
Step 1:Express the stated resistance with two significant figures and a power of ten.
Step 2:The first two bands are therefore Red and Red, and the third band carries the multiplier .
Step 3:In the standard colour code the digit 3, and hence the multiplier , is Orange.
Final answer: Orange
Q12Single correctSemiconductor Electronics
Given below are two statements :
Statement I : Photovoltaic devices can convert optical radiation into electricity.
Statement II : Zener diode is designed to operate under reverse bias in breakdown region.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Photovoltaic devices can convert optical radiation into electricity.
Statement II : Zener diode is designed to operate under reverse bias in breakdown region.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct.
Approach:
Each statement is judged against the operating principle of the named device.
Step 1:A photovoltaic device generates an electromotive force when illuminated, converting light energy into electrical energy, so Statement I is correct.
Step 2:A Zener diode is fabricated to operate in the reverse breakdown region, where it maintains a nearly constant voltage, so Statement II is correct.
Final answer: Both Statement I and Statement II are correct.
Q13Single correctElectromagnetic Induction
The magnetic energy stored in an inductor of inductance carrying a current of A is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The energy stored in an inductor is half the inductance times the square of the current.
Step 1:Substituting the inductance and current into the energy expression evaluates the stored energy.
Step 2:Carrying out the arithmetic gives the energy in joules.
Final answer:
Q14Single correctSystems of Particles and Rotational Motion
The angular acceleration of a body, moving along the circumference of a circle, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4along the axis of rotation
Approach:
Angular acceleration is the rate of change of angular velocity, which is an axial vector directed along the rotation axis.
Step 1:The angular velocity of a body in circular motion is an axial vector pointing along the axis of rotation.
Step 2:The time derivative of an axial vector remains directed along the same axis, so the angular acceleration is axial as well.
Final answer: along the axis of rotation
Q15Single correctThermodynamics
A Carnot engine has an efficiency of 50% when its source is at a temperature C. The temperature of the sink is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 C
Approach:
The Carnot efficiency depends on the absolute temperatures of source and sink, so the sink temperature follows from the efficiency and source temperature.
Step 1:The source temperature is converted to the absolute scale.
Step 2:Rearranging the efficiency relation gives the sink temperature in kelvin.
Step 3:Converting back to the Celsius scale gives the sink temperature.
Final answer: C
Q16Single correctGravitation
Two bodies of mass and are placed at a distance . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( = gravitational constant) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The null-field point is located by equating the field magnitudes of the two masses, then the potential there is the sum of the contributions from both masses.
Step 1:Setting the field of mass at distance equal to the field of mass at distance locates the null point.
Step 2:The potential at the null point sums the potentials due to each mass at its respective distance.
Step 3:Adding the two contributions gives the total potential.
Final answer:
Q17Single correctMotion in a Straight Line
A vehicle travels half the distance with speed and the remaining distance with speed . Its average speed is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Average speed is the total distance divided by the total time, where each half-distance contributes its own travel time.
Step 1:For a total distance , the first half takes time and the second half takes time .
Step 2:Dividing the total distance by the total time gives the average speed.
Final answer:
Q18Single correctMechanical Properties of Fluids
The amount of energy required to form a soap bubble of radius cm from a soap solution is nearly : (surface tension of soap solution N )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 J
Approach:
A soap bubble has two surfaces, so the energy needed equals the surface tension times twice the spherical surface area.
Step 1:The bubble presents an inner and an outer surface, doubling the area on which work is done.
Step 2:Substituting the radius and surface tension evaluates the required energy.
Step 3:Carrying out the arithmetic gives the energy.
Final answer: J
Q19Single correctDual Nature of Radiation and Matter
The minimum wavelength of -rays produced by an electron accelerated through a potential difference of volts is proportional to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The shortest X-ray wavelength occurs when the entire kinetic energy of the electron converts into one photon, fixing the wavelength through the Duane–Hunt relation.
Step 1:At the minimum wavelength the electron transfers all its kinetic energy to a single photon.
Step 2:Rearranging for the wavelength shows its dependence on the accelerating potential.
Final answer:
Q20Single correctNuclei
The half life of a radioactive substance is minutes. In how much time, the activity of substance drops to of its initial value?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 minutes
Approach:
Activity falls by one half each half-life, so the number of half-lives needed follows from the ratio of final to initial activity.
Step 1:The target ratio is expressed as a power of one half to count the half-lives.
Step 2:The total time is the number of half-lives multiplied by the half-life duration.
Final answer: minutes
Q21Single correctUnits and Measurements
A metal wire has mass g, radius mm and length cm. The maximum possible percentage error in the measurement of density will nearly be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Density depends on mass, radius and length, so the maximum fractional error is the sum of the individual fractional errors weighted by their powers in the density formula.
Step 1:The fractional errors in mass, radius and length are computed from the given uncertainties.
Step 2:The radius contributes twice its fractional error because the area depends on the square of the radius.
Step 3:Expressing the fractional error as a percentage gives the maximum error in density.
Final answer:
Q22Single correctElectromagnetic Waves
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of Hz and amplitude V . Then the amplitude of oscillating magnetic field is : (Speed of light in free space m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 T
Approach:
In an electromagnetic wave the amplitudes of the electric and magnetic fields are related by the speed of light.
Step 1:The magnetic field amplitude is the electric field amplitude divided by the speed of light.
Step 2:Carrying out the division gives the magnetic field amplitude.
Final answer: T
Q23Single correctKinetic Theory
The temperature of a gas is C. To what temperature the gas should be heated so that the rms speed is increased by times?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 C
Approach:
The root-mean-square speed of a gas molecule varies as the square root of the absolute temperature, so a known factor increase in speed fixes the factor increase in absolute temperature.
Step 1:Convert the initial temperature to the absolute scale.
Step 2:Increasing the speed by three times makes the final speed four times the initial speed.
Step 3:Squaring the speed ratio gives the ratio of absolute temperatures.
Step 4:Convert the final absolute temperature back to the Celsius scale.
Final answer: C
Q24Single correctAlternating Current
An ac source is connected to a capacitor C. Due to decrease in its operating frequency :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3displacement current decreases.
Approach:
Capacitive reactance varies inversely with frequency, so a frequency decrease raises the reactance and lowers the current, which equals the displacement current in the capacitor.
Step 1:Lowering the frequency raises the capacitive reactance because reactance is inversely proportional to frequency.
Step 2:A larger reactance reduces the conduction current, and the displacement current in the capacitor equals this conduction current.
Final answer: displacement current decreases.
Q25Single correctWave Optics
For Young's double slit experiment, two statements are given below:
Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false.
Approach:
The angular fringe separation depends on the wavelength and the slit spacing but not on the screen distance, so each statement is tested against this relation.
Step 1:The angular separation contains no dependence on the screen distance, so moving the screen leaves it unchanged, making Statement I true.
Step 2:A larger wavelength increases the angular separation because is proportional to wavelength, so the claim of a decrease in Statement II is false.
Final answer: Statement I is true but Statement II is false.
Q26Single correctAtoms
In hydrogen spectrum, the shortest wavelength in the Balmer series is . The shortest wavelength in the Bracket series is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The shortest wavelength (series limit) of a hydrogen spectral series corresponds to the transition from to the lower level . Comparing the Balmer limit () with the Brackett limit () gives the required ratio.
Step 1:For the Balmer series limit the lower level is , giving the shortest Balmer wavelength.
Step 2:For the Brackett series limit the lower level is , giving the shortest Brackett wavelength .
Step 3:Dividing the two results expresses in terms of .
Final answer:
Q27Single correctDual Nature of Radiation and Matter
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cs only
Approach:
Photoemission occurs only when the incident photon energy exceeds the work function of the surface. The incident energy is compared against each work function.
Step 1:The incident photon energy is 2.20 eV.
Step 2:Comparing with each work function: Cs (2.14 eV) is below 2.20 eV, while K (2.30 eV) and Na (2.75 eV) are above 2.20 eV.
Step 3:Only the Caesium surface has a work function smaller than the incident energy, so only Cs emits photoelectrons.
Final answer: Cs only
Q28Single correctUnits and Measurements
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Random errors
Approach:
Errors are classified by their origin. Fluctuations that are irregular and unpredictable in sign and size define random errors.
Step 1:Instrumental, personal and least-count errors are systematic or fixed in character, tied to the device, observer or resolution.
Step 2:Unpredictable variations in temperature and supply voltage cause readings to scatter irregularly about the true value, which is the defining feature of random errors.
Final answer: Random errors
Q29Single correctAlternating Current
In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 F and resistance R is 100 . The frequency at which resonance occurs is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 41.59 kHz
Approach:
At resonance in a series LCR circuit the inductive and capacitive reactances cancel, fixing the angular frequency from L and C; the resonance frequency follows by dividing by .
Step 1:Computing the product LC from and .
Step 2:Evaluating the resonant angular frequency.
Step 3:Converting to ordinary frequency.
Final answer: 1.59 kHz
Q30Single correctMechanical Properties of Fluids
The venturi-meter works on :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Bernoulli's principle
Approach:
A venturi-meter measures flow rate from the pressure difference between a wide and a constricted section, which is governed by the relation between speed and pressure in a flowing fluid.
Step 1:In the narrow throat the fluid speed rises, so its pressure falls; the resulting pressure drop is read to find the flow rate.
Step 2:This speed-pressure relation for a streamline flow is Bernoulli's principle; Huygen's principle and the axis theorems are unrelated to fluid flow.
Final answer: Bernoulli's principle
Q31Single correctWaves
The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 22 : 1
Approach:
The fundamental frequency of an open pipe is while that of a closed pipe of the same length is ; their ratio follows directly.
Step 1:An open pipe supports a fundamental with an antinode at each end, giving .
Step 2:A closed pipe has a node at the closed end and an antinode at the open end, giving .
Step 3:Forming the ratio for equal lengths cancels and .
Final answer: 2 : 1
Q32Single correctElectric Charges and Fields
An electric dipole is placed at an angle of with an electric field of intensity . It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 42 mC
Approach:
The torque on a dipole is with dipole moment . Solving for the charge gives the answer.
Step 1:Rearranging the torque relation for the charge.
Step 2:Substituting , , and .
Step 3:Evaluating the quotient.
Final answer: 2 mC
Q33Single correctCurrent Electricity
The magnitude and direction of the current in the following circuit is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.5 A from to through
Approach:
The two cells sit in the same branch with their positive terminals facing the same junction, so they oppose; the net e.m.f. then drives a single current through the total loop resistance.
Step 1:Both cells have their positive plates facing the junction , so their e.m.f.s subtract.
Step 2:Add the resistances around the single loop.
Step 3:Divide the net e.m.f. by the total resistance; the stronger cell sets the sense.
Final answer: 0.5 A from to through
Q34Single correctMagnetism and Matter
The net magnetic flux through any closed surface is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Zero
Approach:
Gauss's law for magnetism states that magnetic field lines form closed loops, so the flux entering a closed surface equals the flux leaving it.
Step 1:Magnetic monopoles do not exist, so magnetic field lines have no sources or sinks and always close on themselves.
Step 2:The net flux through any closed surface therefore vanishes.
Final answer: Zero
Q35Single correctMotion in a Plane
A bullet is fired from a gun at the speed of in the direction above the horizontal. The maximum height attained by the bullet is (, ) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 31000 m
Approach:
The maximum height of a projectile depends on the vertical component of the launch speed through .
Step 1:Substituting , and .
Step 2:Evaluating the numerator and dividing.
Final answer: 1000 m
Q36Single correctRay Optics and Optical Instruments
Two thin lenses are of same focal lengths (), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Infinite
Approach:
For thin lenses in contact the equivalent power is the sum of the individual powers; equal and opposite focal lengths give zero net power.
Step 1:The convex lens has focal length and the concave lens .
Step 2:Adding the reciprocals gives zero net power.
Final answer: Infinite
Q37Single correctAlternating Current
The net impedance of circuit (as shown in figure) will be :
[Figure: A series a.c. circuit driven by a source containing a inductor, a resistor, and a capacitor in series.]
[Figure: A series a.c. circuit driven by a source containing a inductor, a resistor, and a capacitor in series.]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For a series -- branch across an a.c. source, compute the inductive and capacitive reactances at the supply frequency and combine them with the resistance in the impedance triangle.
Step 1:Evaluate the inductive reactance for at .
Step 2:Evaluate the capacitive reactance for at the same frequency.
Step 3:Combine the net reactance with the resistance.
Final answer:
Q38Single correctOscillations
The x-t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Read the amplitude and period off the graph to get the angular frequency, note the displacement at the instant asked for, then use the defining relation between acceleration and displacement in simple harmonic motion.
Step 1:The graph completes one cycle in and reaches a maximum displacement of .
Step 2:At the curve is at its crest, so the displacement equals the amplitude.
Step 3:Substitute into the acceleration relation.
Final answer:
Q39Single correctElectrostatic Potential and Capacitance
An electric dipole is placed as shown in the figure. The electric potential (in ) at point P due to the dipole is ( permittivity of free space and ) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Electric potential is a scalar, so add the potentials of the two point charges using their actual distances from measured along the common line.
Step 1:The charges sit on either side of the centre O, and P lies from O on the same line beyond .
Step 2:Add the two contributions, the nearer charge being positive and the farther one negative.
Step 3:Evaluate the bracket and express the result in the stated unit of .
Final answer:
Q40Single correctWork, Energy and Power
A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes . Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 127 cm
Approach:
Assuming a constant retarding force, the kinematic relation links the distance penetrated to the change in the square of the speed. The total stopping distance follows from .
Step 1:Over the first 24 cm the speed falls from to .
Step 2:The total stopping distance corresponds to the speed reaching zero.
Step 3:Dividing the two relations eliminates and .
Final answer: 27 cm
Q41Single correctRay Optics and Optical Instruments
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
[Figure: a biconvex lens of refractive index is embedded in a surrounding medium of refractive index ; its left and right surfaces have radii .]
[Figure: a biconvex lens of refractive index is embedded in a surrounding medium of refractive index ; its left and right surfaces have radii .]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A converging piece of glass placed in a denser surrounding medium behaves as a diverging lens, and its focal length follows from the lens-maker relation written for the two spherical surfaces.
Step 1:Take the surface radii with sign: the first surface is convex towards the incoming light and the second is concave towards it.
Step 2:The lens index is lower than that of the medium, so the leading bracket is negative.
Step 3:Multiply the two brackets and invert.
Final answer:
Q42Single correctSemiconductor Electronics
For the following logic circuit, the truth table is :
[Figure: inputs and each pass through a NOT gate; the two inverted signals are the inputs of a NAND gate whose output is .]
[Figure: inputs and each pass through a NOT gate; the two inverted signals are the inputs of a NAND gate whose output is .]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Write the Boolean output of the network gate by gate and simplify with De Morgan's theorem, then tabulate the simplified expression.
Step 1:The NOT gates deliver and to the NAND gate.
Step 2:Apply the NAND operation and simplify with De Morgan's theorem.
Step 3:Tabulate for the four input combinations.
Final answer:
Q43Single correctMotion in a Straight Line
A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity . The ball strikes the water surface after . The height of bridge above water surface is (Take ) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 364 m
Approach:
Taking the upward direction as positive and the bridge as origin, the displacement of the ball at the moment it reaches the water is . Applying yields the height.
Step 1:With upward positive, , and .
Step 2:The negative displacement of 64 m below the start means the water surface is 64 m below the bridge.
Final answer: 64 m
Q44Single correctCurrent Electricity
10 resistors, each of resistance are connected in series to a battery of emf and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased times. The value of is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2100
Approach:
The current from the battery is divided by the equivalent resistance. Comparing the parallel current to the series current gives the factor .
Step 1:In series the current is ; in parallel the current is .
Step 2:The ratio of parallel to series current gives .
Final answer: 100
Q45Single correctMoving Charges and Magnetism
A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field . The magnitude of the magnetic force acting on the wire is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The magnetic force on a straight current element is with . The cross product is evaluated and its magnitude taken.
Step 1:With , the cross product retains only the and components of .
Step 2:Taking the magnitude of the resulting vector.
Final answer:
Q46Single correctGravitation
A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity represents :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a satellite skimming the surface, gravity provides the centripetal force. Expressing the earth's mass through its density gives the period in terms of and .
Step 1:Substituting into the orbital period for cancels the radius.
Step 2:The combination therefore equals the square of the period.
Final answer:
Q47Single correctLaws of Motion
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 () :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The body stays at rest relative to the car only while static friction can supply the needed force; the maximum acceleration is set by the limiting friction .
Step 1:The maximum friction force per unit mass equals , which is the largest acceleration that friction can impart.
Final answer:
Q48Single correctCurrent Electricity
The resistance of platinum wire at is and at . The temperature coefficient of resistance of the wire is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The temperature coefficient relates resistances at two temperatures by . Solving for uses the two given resistance values.
Step 1:Rearranging for in terms of the two resistances and the temperature change.
Step 2:Substituting , and .
Step 3:Expressing in power-of-ten form.
Final answer:
Q49Single correctAtoms
The radius of inner most orbit of hydrogen atom is . What is the radius of third allowed orbit of hydrogen atom?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
In the Bohr model the orbit radius scales as , so the third orbit radius is nine times the first orbit radius.
Step 1:The innermost orbit corresponds to with .
Step 2:For the third orbit , the radius is times the first.
Final answer:
Q50Single correctMoving Charges and Magnetism
A very long conducting wire is bent in a semi-circular shape from to as shown in figure. The magnetic field at point for steady current configuration is given by :
[Figure: a current arrives from the left along the upper straight wire and enters the semicircle at ; the semicircle of radius bulges to the left with its centre at , and the current leaves at along the lower straight wire, flowing back to the left.]
[Figure: a current arrives from the left along the upper straight wire and enters the semicircle at ; the semicircle of radius bulges to the left with its centre at , and the current leaves at along the lower straight wire, flowing back to the left.]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 pointed away from page
Approach:
Split the wire into the semicircular arc and the two semi-infinite straight sections, find each contribution at the centre , and add them as vectors.
Step 1:The semicircle subtends at P; traversed from A round to B its field at the centre points out of the page.
Step 2:Each straight section ends level with at a perpendicular distance , and both give a field into the page.
Step 3:Subtract the opposing contributions and factor out the arc term.
Final answer: pointed away from page
Chemistry46 questions
Q51Single correctAmines
Which of the following reactions will NOT give primary amine as the product?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Each route is checked for the class of amine it furnishes. Reduction of an isocyanide gives a secondary amine, whereas the other routes lead to primary amines.
Step 1:An amide treated with bromine and alkali undergoes the Hofmann bromamide degradation, losing one carbon and producing a primary amine.
Step 2:A nitrile reduced by lithium aluminium hydride adds two hydrogen atoms across the triple bond to give a primary amine.
Step 3:An isocyanide carries the nitrogen attached to carbon already; reduction converts it to a secondary amine since the alkyl group remains bonded to nitrogen.
Step 4:An amide reduced by lithium aluminium hydride retains the carbon framework and yields a primary amine.
Final answer:
Q52Single correctThe p-Block Elements
Choose the correct answer from the options given below :
| List - I | List - II |
|---|---|
| A. Coke | I. Carbon atoms are hybridised. |
| B. Diamond | II. Used as a dry lubricant |
| C. Fullerene | III. Used as a reducing agent |
| D. Graphite | IV. Cage like molecules |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Each allotrope or carbon form is paired with the property that defines its structure or use.
Step 1:Coke is impure carbon obtained from coal and serves to reduce metal oxides in metallurgy.
Step 2:Diamond has every carbon atom hybridised in a three dimensional tetrahedral network.
Step 3:Fullerene consists of closed cage like molecules such as .
Step 4:Graphite has layered sheets that slide over one another, making it a dry lubricant.
Final answer: A-III, B-I, C-IV, D-II
Q53Single correctThe s-Block Elements
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.
Reason R : The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.
Reason R : The deep blue solution is due to the formation of amide.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false.
Approach:
Judge the assertion and the reason separately against the chemistry of alkali metals in liquid ammonia, then read off the required combination.
Step 1:Sodium dissolving in liquid ammonia releases its valence electron into the solvent, and the resulting ammoniated electrons absorb in the red to give the familiar deep blue colour; the unpaired electrons make the solution paramagnetic.
Step 2:The colour comes from those ammoniated electrons, not from sodium amide. Sodamide forms only on standing or on catalysis, and its solutions are colourless.
Step 3:A true assertion together with a false reason fixes the required combination.
Final answer: A is true but R is false.
Q54Single correctOrganic Chemistry: Some Basic Principles and Techniques
In Lassaigne's extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with due to the formation of -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
When nitrogen and sulphur occur together, the fused sodium converts them into thiocyanate, which forms a coloured complex with ferric ions.
Step 1:Sodium fusion with a compound containing both nitrogen and sulphur produces sodium thiocyanate rather than separate cyanide and sulphide.
Step 2:Thiocyanate reacts with ferric ion to form a blood red complex.
Step 3:The other listed species correspond to Prussian blue or nitroprusside tests, which do not give this colour for combined nitrogen and sulphur.
Final answer:
Q55Single correctElectrochemistry
The conductivity of centimolar solution of KCl at is oh c and the resistance of the cell containing the solution at is ohm. The value of cell constant is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 c
Approach:
The cell constant equals the product of conductivity and resistance, since conductivity is the conductance multiplied by the cell constant.
Step 1:Conductivity relates to conductance and cell constant through , which rearranges to give the cell constant as conductivity times resistance.
Step 2:Substituting the conductivity and resistance gives the cell constant.
Final answer: c
Q56Single correctChemical Kinetics
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : A reaction can have zero activation energy.
Reason R : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : A reaction can have zero activation energy.
Reason R : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A is false but R is true.
Approach:
Each statement is judged against the definition and physical meaning of activation energy.
Step 1:Every reaction requires its reactant molecules to attain a threshold energy, so the activation energy cannot be zero; the assertion therefore fails.
Step 2:Activation energy is correctly defined as the minimum extra energy that reactant molecules must absorb to reach the threshold value, so the reason holds.
Step 3:A false assertion paired with a true reason fixes the required combination.
Final answer: A is false but R is true.
Q57Single correctSurface Chemistry
Which one is an example of heterogenous catalysis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.
Approach:
A heterogeneously catalysed reaction has its catalyst in a phase different from the reactants, so the phase of the catalyst in each reaction settles the matter.
Step 1:Oxidation of sulphur dioxide by nitrogen oxides, acid catalysed hydrolysis of sugar, and ozone decomposition by nitrogen monoxide all proceed with the catalyst in the same phase as the reactants.
Step 2:Ammonia synthesis runs over solid finely divided iron while the reactants are gases, placing catalyst and reactants in different phases.
Final answer: Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron.
Q59Single correctAmines
Identify the product in the following reaction:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Benzene the unsubstituted ring (drawn structure, option 2)
Approach:
The diazonium salt is carried through bromination by the Sandmeyer reaction, formation of a Grignard reagent, and aqueous workup; the final hydrolysis of the carbon to magnesium bond regenerates the carbon to hydrogen bond.
Step 1:Cuprous bromide with hydrobromic acid replaces the diazonium group with bromine through the Sandmeyer reaction, giving bromobenzene.
Step 2:Magnesium in dry ether inserts into the carbon to bromine bond, producing phenylmagnesium bromide.
Step 3:Water protonates the Grignard reagent, cleaving the carbon to magnesium bond and placing a hydrogen on the ring carbon to give benzene.
Final answer: Benzene the unsubstituted ring (drawn structure, option 2)
Q60Single correctThe p-Block Elements
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Helium is used to dilute oxygen in diving apparatus.
Reason R : Helium has high solubility in .
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : Helium is used to dilute oxygen in diving apparatus.
Reason R : Helium has high solubility in .
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false.
Approach:
Each statement is checked against the use of helium in breathing mixtures and its solubility behaviour.
Step 1:Helium dilutes oxygen in diving gas mixtures because it is far less soluble in blood than nitrogen and prevents bends, so the assertion is correct.
Step 2:Helium has low, not high, solubility, and the relevant solubility is in blood rather than in oxygen, so the reason is incorrect.
Step 3:A true assertion with a false reason fixes the required combination.
Final answer: A is true but R is false.
Q61Single correctThe Solid State
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy of tetrahedral voids. If the formula of the compound is , then the value of is in option
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Counting atoms of B in a close packed array and the tetrahedral voids occupied by A fixes the formula, from which follows.
Step 1:In a cubic close packed lattice the number of tetrahedral voids is twice the number of close packed atoms, so for atoms of B there are tetrahedral voids.
Step 2:Atoms of A fill one third of these voids, giving the number of A atoms.
Step 3:The simplest whole number ratio gives the formula, and the subscripts are added.
Final answer:
Q62Single correctBiomolecules
Given below are two statements :
Statement I : A unit formed by the attachment of a base to position of sugar is known as nucleoside.
Statement II : When nucleoside is linked to phosphorous acid at -position of sugar moiety, we get nucleotide.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : A unit formed by the attachment of a base to position of sugar is known as nucleoside.
Statement II : When nucleoside is linked to phosphorous acid at -position of sugar moiety, we get nucleotide.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false.
Approach:
Each statement is tested against the definitions of nucleoside and nucleotide.
Step 1:A nucleoside is the unit formed when a base joins the position of the sugar, so the first statement is correct.
Step 2:A nucleotide is formed when a nucleoside is linked to phosphoric acid, not phosphorous acid, at the position, so the second statement is incorrect as printed.
Step 3:A true first statement with a false second statement fixes the required combination.
Final answer: Statement I is true but Statement II is false.
Q63Single correctStructure of Atom
The relation between , ( = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number (l), is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The number of magnetic quantum number values for a given azimuthal quantum number is known, and the relation is rearranged for the azimuthal quantum number.
Step 1:For a given azimuthal quantum number the magnetic quantum number ranges from to , giving permissible values.
Step 2:Rearranging for the azimuthal quantum number isolates .
Final answer:
Q64Single correctChemical Bonding and Molecular Structure
Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is
:
:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The number of electrons in the outermost shell of the central atom is counted for each species, and those departing from eight are tallied.
Step 1:In ammonia the nitrogen shares three pairs and holds one lone pair, completing eight electrons; in carbon tetrachloride carbon shares four pairs, giving eight electrons.
Step 2:Aluminium in aluminium chloride has six electrons, beryllium in beryllium chloride has four electrons, and phosphorus in phosphorus pentachloride has ten electrons, so each lacks an exact octet.
Step 3:Three species fall outside the octet around the central atom.
Final answer:
Q65Single correctChemical Bonding and Molecular Structure
The correct order of energies of molecular orbitals of molecule, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
For diatomic molecules up to nitrogen, s-p mixing raises the orbital above the degenerate orbitals, fixing the energy order.
Step 1:Nitrogen lies among the lighter diatomics where mixing of and orbitals is significant, so the bonding orbital lies above the bonding and orbitals.
Step 2:The full sequence begins with the inner and orbitals, then the degenerate bonding orbitals, the , the antibonding pair, and finally the antibonding .
Final answer:
Q66Single correctChemical Bonding and Molecular Structure
The number of bonds, bonds and lone pair of electrons in pyridine, respectively are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Pyridine is a six membered aromatic ring with five carbon atoms, one nitrogen atom and five hydrogen atoms; the bonds and lone pairs are counted from its structure.
Step 1:The ring contains six framework bonds joining the five carbons and one nitrogen, and five carbon to hydrogen bonds, all of which are sigma bonds.
Step 2:The aromatic ring carries three alternating double bonds, each contributing one pi bond.
Step 3:The nitrogen atom holds one lone pair in the plane of the ring, available for donation.
Final answer:
Q67Single correctStates of Matter
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include :
A. dipole - dipole forces.
B. dipole - induced dipole forces.
C. hydrogen bonding.
D. covalent bonding.
E. dispersion forces.
Choose the most appropriate answer from the options given below :
A. dipole - dipole forces.
B. dipole - induced dipole forces.
C. hydrogen bonding.
D. covalent bonding.
E. dispersion forces.
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, B, C, E are correct.
Approach:
Each listed interaction is classified as intermolecular or intramolecular, retaining only the intermolecular ones.
Step 1:Dipole to dipole, dipole to induced dipole, hydrogen bonding and dispersion forces all act between molecules and count as intermolecular forces.
Step 2:Covalent bonding holds atoms together within a molecule and is an intramolecular force, so it is excluded.
Step 3:The remaining intermolecular set fixes the answer.
Final answer: A, B, C, E are correct.
Q68Single correctHydrogen
Which of the following statements are NOT correct?
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H-H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below :
A. Hydrogen is used to reduce heavy metal oxides to metals.
B. Heavy water is used to study reaction mechanism.
C. Hydrogen is used to make saturated fats from oils.
D. The H-H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any element.
E. Hydrogen reduces oxides of metals that are more active than iron.
Choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3D, E only
Approach:
Each statement about hydrogen and heavy water is checked, and the incorrect ones are selected.
Step 1:Hydrogen reduces heavy metal oxides to metals, heavy water serves to study reaction mechanisms, and hydrogenation of oils gives saturated fats, so these statements are correct.
Step 2:The H-H bond dissociation enthalpy is among the highest of single bonds rather than the lowest, so this statement is incorrect.
Step 3:Hydrogen reduces oxides of metals less active than itself, not those more active than iron, so this statement is incorrect.
Final answer: D, E only
Q69Single correctPolymers
Which amongst the following molecules on polymerization produces neoprene?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Chloroprene C=C(Cl)CH=C (drawn structure, option 2)
Approach:
Neoprene is the polymer of chloroprene, so the monomer carrying a chlorine on the second carbon of 1,3-butadiene is identified.
Step 1:Neoprene is the polymer obtained from chloroprene, which is 2-chloro-1,3-butadiene.
Step 2:Only the diene carrying chlorine on the second carbon of butadiene is chloroprene; 1,3-butadiene polymerises to polybutadiene, the enyne gives a different polymer, and isoprene gives a natural-rubber type polymer.
Final answer: Chloroprene C=C(Cl)CH=C (drawn structure, option 2)
Q70Single correctChemistry in Everyday Life
Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Veronal
Approach:
Each listed tranquilizer is matched to its chemical class, retaining the barbituric acid derivative.
Step 1:Veronal is a derivative of barbituric acid and belongs to the barbiturate class of tranquilizers.
Step 2:Chlordiazepoxide and valium are benzodiazepines, and meprobamate is a separate non barbiturate tranquilizer, so none of these belong to barbiturates.
Final answer: Veronal
Q71Single correctClassification of Elements and Periodicity
The element expected to form largest ion to achieve the nearest noble gas configuration is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3N
Approach:
Each element forms an ion isoelectronic with neon; the ion with the highest negative charge has the largest radius because of the greatest electron to proton ratio.
Step 1:Oxygen, fluorine, nitrogen and sodium each reach the neon configuration by gaining or losing electrons, forming an isoelectronic series.
Step 2:Within an isoelectronic set the radius increases as the nuclear charge decreases, so the ion from the element with the smallest nuclear charge and the highest negative charge is the largest.
Final answer: N
Q72Single correctSome Basic Concepts of Chemistry
Select the correct statements from the following:
A. Atoms of all elements are composed of two fundamental particles.
B. The mass of the electron is kg.
C. All the isotopes of a given element show same chemical properties.
D. Protons and electrons are collectively known as nucleons.
E. Dalton's atomic theory, regarded the atom as an ultimate particle of matter.
Choose the correct answer from the options given below :
A. Atoms of all elements are composed of two fundamental particles.
B. The mass of the electron is kg.
C. All the isotopes of a given element show same chemical properties.
D. Protons and electrons are collectively known as nucleons.
E. Dalton's atomic theory, regarded the atom as an ultimate particle of matter.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4B, C and E only
Approach:
Each statement on atomic constitution is checked against established facts, and the correct ones are selected.
Step 1:Atoms are made of three fundamental particles, protons, neutrons and electrons, so the claim of only two is wrong; and nucleons are protons and neutrons, not protons and electrons, so that claim is also wrong.
Step 2:The electron mass is kg, isotopes of an element share the same chemical properties, and Dalton treated the atom as the ultimate indivisible particle of matter.
Final answer: B, C and E only
Q73Single correctHaloalkanes and Haloarenes
Consider the following reaction and identify the product (P).
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12-Bromo-2-methylbutane (drawn structure, option 1)
Approach:
Protonation of the alcohol and loss of water generate a secondary carbocation that rearranges by a hydride shift to a more stable tertiary carbocation before bromide attacks.
Step 1:Hydrobromic acid protonates the hydroxyl group, and water leaves to give a secondary carbocation at the second carbon.
Step 2:A hydride shifts from the adjacent tertiary carbon to the cation centre, producing a more stable tertiary carbocation.
Step 3:Bromide adds to the tertiary carbocation, giving 2-bromo-2-methylbutane.
Final answer: 2-Bromo-2-methylbutane (drawn structure, option 1)
Q74Single correctThe d- and f-Block Elements
The stability of is more than salts in aqueous solution due to -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3hydration energy.
Approach:
The greater stability of the dipositive copper ion in water is traced to the energy released on hydration, which outweighs the higher second ionisation enthalpy.
Step 1:The dipositive copper ion is smaller and more highly charged than the unipositive ion, so it releases far more energy on hydration.
Step 2:This large hydration energy more than compensates for the second ionisation enthalpy, making the dipositive ion the stable species in aqueous solution.
Final answer: hydration energy.
Q75Single correctThe s-Block Elements
Which one of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The daily requirement of Mg and Ca in the human body is estimated to be g.
Approach:
Each statement on the biological roles of magnesium and calcium is checked against established physiology.
Step 1:The daily requirement of magnesium and calcium together is in the range of about to g, so the first statement matches the accepted figure.
Step 2:Enzymes using ATP in phosphate transfer require magnesium, not calcium, as the cofactor; bone is a dynamic tissue continually remodelled; and the role in neuromuscular function and interneuronal transmission belongs to calcium rather than magnesium.
Final answer: The daily requirement of Mg and Ca in the human body is estimated to be g.
Q76Single correctAldehydes, Ketones and Carboxylic Acids
Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Heating a sodium salt of a carboxylic acid with soda lime (NaOH and CaO) removes the carboxyl group as carbonate, leaving the hydrocarbon with one less carbon. The product mass for two moles is then computed.
Step 1:Sodium ethanoate undergoes decarboxylation with soda lime to give methane.
Step 2:The molar mass of methane is found by adding atomic masses.
Step 3:The mass of two moles is the molar mass multiplied by two.
Final answer:
Q77Single correctChemical Bonding and Molecular Structure
Amongst the given options which of the following molecules / ion acts as a Lewis acid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A Lewis acid is an electron-pair acceptor. The species with an incomplete octet and a vacant orbital available to accept a lone pair is identified.
Step 1:Boron in boron trifluoride has only six electrons in its valence shell and a vacant p-orbital, making it electron deficient.
Step 2:Ammonia, water and the hydroxide ion all carry lone pairs and act as electron-pair donors, so they are Lewis bases.
Step 3:The electron-deficient boron trifluoride accepts an electron pair and therefore behaves as a Lewis acid.
Final answer:
Q78Single correctAldehydes, Ketones and Carboxylic Acids
Identify product (A) in the following reaction :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both carbonyls reduced to ethyl groups (drawn structure, option 1)
Approach:
Zinc amalgam with concentrated hydrochloric acid is the Clemmensen reagent, which reduces a carbonyl group of an aldehyde or ketone to a methylene group. The transformation of each acetyl group is traced.
Step 1:Clemmensen reduction converts a ketonic carbonyl into a methylene group, releasing water.
Step 2:Each acetyl group is reduced to an ethyl group .
Step 3:Loss of two water molecules accompanies reduction of the two carbonyls, matching the printed .
Final answer: Both carbonyls reduced to ethyl groups (drawn structure, option 1)
Q79Single correctThe p-Block Elements
Taking stability as the factor, which one of the following represents relationship?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Down group 13 the inert pair effect makes the lower (+1) oxidation state increasingly stable relative to the (+3) state. Stability of each halide pair is compared on this basis.
Step 1:Thallium is the heaviest stable member of group 13, where the inert pair effect is strongest, so the +1 oxidation state is more stable than the +3 state.
(stability)
Step 2:Consequently the monohalide thallium(I) iodide is more stable than thallium(III) iodide.
Step 3:For aluminium and, to a large extent, indium the +3 state remains the stable one, so the other listed orderings are reversed.
Final answer:
Q80Single correctCoordination Compounds
Homoleptic complex from the following complexes is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Potassium trioxalatoaluminate (III)
Approach:
A homoleptic complex is one in which the central metal ion is bound to only one kind of donor ligand. Each named complex is examined for the variety of its coordinated ligands.
Step 1:Potassium trioxalatoaluminate(III) contains aluminium coordinated to three oxalate ligands and nothing else, so a single ligand type surrounds the metal.
Step 2:Each of the remaining complexes mixes different donor ligands such as ammine with chlorido, nitrito, carbonato or aqua, making them heteroleptic.
etc.
Step 3:Only the trioxalatoaluminate(III) ion satisfies the homoleptic criterion.
Final answer: Potassium trioxalatoaluminate (III)
Q82Single correctStates of Matter
Which amongst the following options is graphical representation of Boyle's Law?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2P against 1/V giving straight lines through the origin (drawn graph, option 2)
Approach:
Boyle's law states that at constant temperature pressure is inversely proportional to volume. Each graph is tested against this relationship to find the one consistent with it.
Step 1:Boyle's law gives a constant product of pressure and volume at fixed temperature, so pressure varies as the reciprocal of volume.
Step 2:A plot of pressure against the reciprocal of volume is therefore a straight line passing through the origin, with slope equal to the constant .
Step 3:Since the slope nRT increases with temperature, the higher-temperature lines are steeper, which is the ordering .
Final answer: P against 1/V giving straight lines through the origin (drawn graph, option 2)
Q83Single correctSome Basic Concepts of Chemistry
The option for the mass of produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 g
Approach:
The pure calcium carbonate present in the sample is found from the stated purity, converted to moles, and one mole of carbon dioxide is released per mole of carbonate. The mass of carbon dioxide follows.
Step 1:The pure limestone in a 20 g, 20% pure sample is 20% of 20 g.
Step 2:The moles of calcium carbonate follow from its molar mass of 100 g/mol.
Step 3:Each mole of carbonate yields one mole of carbon dioxide, whose molar mass is 44 g/mol.
Final answer: g
Q84Single correctChemical Kinetics
For a certain reaction, the rate = , when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3increase by a factor of nine.
Approach:
The rate depends on the square of the concentration of A. Multiplying that concentration by three multiplies the rate by three squared, with the concentration of B held fixed.
Step 1:The reaction is second order in A, so the rate scales with the square of the concentration of A.
Step 2:Replacing the concentration of A by three times its value, with B unchanged, gives a new rate proportional to the square of three.
Step 3:The initial rate therefore increases by a factor of nine.
Final answer: increase by a factor of nine.
Q85Single correctElectrochemistry
Given below are two statements : one is labelled as and the other is labelled as :
In equation , value of depends on n.
is an intensive property and is an extensive property.
In the light of the above statements, choose the answer from the options given below :
In equation , value of depends on n.
is an intensive property and is an extensive property.
In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both and are true and is the correct explanation of .
Approach:
The relation between reaction Gibbs energy and cell potential is examined, together with the intensive or extensive nature of each quantity, to judge both statements and their logical link.
Step 1:The reaction Gibbs energy is the product of the number of electrons transferred, the Faraday constant and the cell potential, so it carries the factor n and depends on it. Assertion A is true.
Step 2:Cell potential does not change when the reaction is scaled, marking it as intensive, whereas reaction Gibbs energy scales with the amount of reaction and is extensive. Reason R is true.
intensive, extensive
Step 3:Because n appears in the equation precisely to convert the intensive potential into the extensive Gibbs energy, the extensive nature of explains its dependence on n, so R correctly explains A.
Final answer: Both and are true and is the correct explanation of .
Q86Single correctThe p-Block Elements
Match List - I with List - II : Choose the correct answer from the options given below :
| List - I (Oxoacids of Sulphur) | List - II (Bonds) |
|---|---|
| A. Peroxodisulphuric acid | I. Two S-OH, Four S=O, One S-O-S |
| B. Sulphuric acid | II. Two S-OH, One S=O |
| C. Pyrosulphuric acid | III. Two S-OH, Four S=O, One S-O-O-S |
| D. Sulphurous acid | IV. Two S-OH, Two S=O |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
The bonding around sulphur in each oxoacid is recalled, distinguishing the peroxo linkage, the S-O-S bridge and the number of terminal S=O and S-OH bonds, and each acid is paired with its description.
Step 1:Peroxodisulphuric acid contains the peroxide -O-O- bridge between two sulphur centres along with two S-OH and four S=O bonds, matching description III.
Step 2:Sulphuric acid has two S-OH and two S=O bonds, matching description IV.
Step 3:Pyrosulphuric acid carries a single S-O-S bridge with two S-OH and four S=O bonds, matching description I, while sulphurous acid has two S-OH and one S=O, matching description II.
Final answer: A-III, B-IV, C-I, D-II
Q87Single correctThe d- and f-Block Elements
Which of the following statements are ?
A. All the transition metals except scandium form oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in to .
C. Basic character increases from to to .
D. dissolves in acids to give salts.
E. CrO is basic but is amphoteric.
Choose the answer from the options given below :
A. All the transition metals except scandium form oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in to .
C. Basic character increases from to to .
D. dissolves in acids to give salts.
E. CrO is basic but is amphoteric.
Choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C and D only
Approach:
Each statement on transition metal oxides is checked against known facts about ionic character, oxidation states, acid-base nature and dissolution behaviour to identify the incorrect ones.
Step 1:Basic character of metal oxides rises as the oxidation state falls, so it increases from the higher oxide toward the lower oxide; stating an increase in the listed direction as a general fact within statement C is incorrect as framed in the question.
Step 2:Vanadium in the +4 oxide dissolves in acids to give vanadyl species rather than the +5 orthovanadate ion, so statement D is incorrect.
, not
Step 3:Statements A, B and E correctly describe ionic MO oxides, the span of highest oxidation states and the basic-to-amphoteric trend of chromium oxides, leaving C and D as the incorrect pair.
Final answer: C and D only
Q88Single correctCoordination Compounds
Which complex compound is most stable?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Stability of a coordination complex is enhanced by chelation, since a bidentate ligand forming a ring locks the metal more firmly than separate monodentate donors. The complex containing the chelating ligand is selected.
Step 1:Ethylenediamine is a bidentate ligand that binds the metal through two nitrogen donors, forming a stable five-membered chelate ring.
Step 2:The complex bearing two ethylenediamine ligands gains extra stability from the chelate effect relative to complexes carrying only monodentate ammine or aqua donors.
Step 3:The remaining cations hold only monodentate ligands and lack this chelate stabilisation.
etc.
Final answer:
Q90Single correctThe Solid State
What fraction of one edge centred octahedral void lies in one unit cell of fcc?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
An octahedral void located at the centre of an edge of the face-centred cubic cell is shared among the cubes that meet along that edge. The fraction belonging to one cell is the reciprocal of the number of sharing cells.
Step 1:An edge of a cubic cell is common to four neighbouring unit cells arranged around it.
Step 2:A void sitting at the midpoint of that edge is therefore divided equally among those four cells.
Step 3:Hence one edge-centred octahedral void contributes a quarter to a single fcc unit cell.
Final answer:
Q91Single correctThermodynamics
Which amongst the following options is the relation between change in enthalpy and change in internal energy?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Enthalpy is internal energy plus the pressure-volume product. For gaseous reactions at constant temperature and pressure the volume work is expressed through the change in moles of gas, giving the standard relation.
Step 1:Enthalpy change equals internal energy change plus the change in the pressure-volume product.
Step 2:For ideal gases at constant temperature the pressure-volume term becomes the change in moles of gas times RT.
Step 3:Substituting gives the relation linking enthalpy and internal energy changes.
Final answer:
Q92Single correctRedox Reactions
On balancing the given redox reaction,
the coefficients a, b and c are found to be, respectively -
the coefficients a, b and c are found to be, respectively -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The dichromate-sulphite reaction is balanced by matching electrons gained by chromium with electrons lost by sulphur, then balancing charge with hydrogen ions.
Step 1:One dichromate ion gains six electrons as two chromium centres go from +6 to +3, fixing a as one.
Step 2:Each sulphite to sulphate conversion releases two electrons, so three sulphite ions supply the six electrons, fixing b as three.
Step 3:Balancing the oxygen and hydrogen requires eight hydrogen ions, giving four water molecules and fixing c as eight.
Final answer:
Q93Single correctEquilibrium
The equilibrium concentrations of the species in the reaction are 2, 3, 10 and 6 mol , respectively at 300 K. for the reaction is (R = 2 cal / mol K)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 cal
Approach:
The equilibrium constant is the ratio of product to reactant concentrations. The standard Gibbs energy follows from the relation linking it to the natural logarithm of that constant.
Step 1:The equilibrium constant is computed from the given concentrations.
Step 2:The standard Gibbs energy is the negative of RT times the natural logarithm of the constant.
Step 3:Evaluating with ln 10 equal to 2.303 gives the standard Gibbs energy in calories.
Final answer: cal
Q94Single correctSurface Chemistry
Pumice stone is an example of -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3solid sol
Approach:
Colloidal systems are classified by the physical states of the dispersed phase and the dispersion medium. Pumice stone is matched to its colloidal class on this basis.
Step 1:Pumice stone is a porous solid formed by gas trapped within a solidified mineral matrix.
Step 2:A colloid whose dispersion medium is a solid and whose dispersed phase is a gas is termed a solid sol.
Step 3:Pumice stone therefore belongs to the solid sol class.
Final answer: solid sol
Q95Single correctAldehydes, Ketones and Carboxylic Acids
Identify the major product obtained in the following reaction :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The ring bearing COC with the aldehyde oxidised to COO (drawn structure, option 3)
Approach:
Tollens reagent oxidises an aldehyde to a carboxylate but leaves a ketone untouched. The fate of each carbonyl group on the ring is traced under the given conditions.
Step 1:The diamminesilver(I) complex is a mild oxidant that converts an aldehyde group into a carboxylate ion.
Step 2:This reagent does not oxidise a ketone, so the acetyl group on the ring is retained unchanged.
Step 3:The product carries the unchanged acetyl group at one position and the new carboxylate at the adjacent position.
Final answer: The ring bearing COC with the aldehyde oxidised to COO (drawn structure, option 3)
Q96Single correctHaloalkanes and Haloarenes
Identify the final product [D] obtained in the following sequence of reactions.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ethylbenzene a benzene ring bearing an ethyl group (drawn structure, option 1)
Approach:
Each step is carried through in turn: reduction, dehydration, hydrobromination and a Wurtz-Fittig coupling with bromobenzene to build the final aromatic product.
Step 1:Lithium aluminium hydride reduces acetaldehyde to ethanol, which is dehydrated by hot sulphuric acid to ethene.
Step 2:Addition of hydrogen bromide to ethene gives bromoethane as [C].
Step 3:Sodium in dry ether couples bromoethane with bromobenzene through the Wurtz-Fittig reaction to give ethylbenzene.
Final answer: Ethylbenzene a benzene ring bearing an ethyl group (drawn structure, option 1)
Q97Single correctAlcohols, Phenols and Ethers
Which amongst the following will be most readily dehydrated under acidic conditions ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pentane-2,4-diol (drawn structure, option 2)
Approach:
Acid-catalysed dehydration proceeds through a carbocation, so it is fastest for the alcohol whose intermediate cation is best stabilised. The electron-withdrawing nitro group destabilises an adjacent cation, while a structure free of such a group dehydrates most readily.
Step 1:Dehydration under acid passes through a carbocation, and the rate increases with the stability of that cation.
Step 2:A nitro group is strongly electron withdrawing and destabilises a nearby carbocation, slowing dehydration of the nitro-bearing alcohols.
withdraws electron density
Step 3:The diol free of a nitro group forms its carbocation without this destabilisation, so it dehydrates most readily.
easier loss of water
Final answer: Pentane-2,4-diol (drawn structure, option 2)
Q98Single correctEnvironmental Chemistry
Given below are two statements :
The nutrient deficient water bodies lead to eutrophication.
Eutrophication leads to decrease in the level of oxygen in the water bodies.
In the light of the above statements, choose the answer from the options given below :
The nutrient deficient water bodies lead to eutrophication.
Eutrophication leads to decrease in the level of oxygen in the water bodies.
In the light of the above statements, choose the answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 is incorrect but is true.
Approach:
Each statement on eutrophication is judged against its accepted definition, which concerns nutrient enrichment rather than nutrient deficiency and its effect on dissolved oxygen.
Step 1:Eutrophication arises from nutrient enrichment of a water body, not from nutrient deficiency, so the first statement is incorrect.
Step 2:Excess nutrients drive algal blooms whose decay consumes dissolved oxygen, lowering the oxygen level, so the second statement is correct.
Step 3:Hence the first statement is incorrect while the second is true.
Final answer: is incorrect but is true.
Q100Single correctGeneral Principles and Processes of Isolation of Elements
The reaction that does take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The reactions of the blast furnace are placed in their characteristic temperature zones. The reduction step that belongs to the cooler upper zone rather than the 900-1500 K range is identified.
Step 1:Reduction of haematite by carbon monoxide to ferrous oxide proceeds in the cooler upper region of the furnace, below the stated range.
Step 2:Within the 900-1500 K range carbon monoxide reduces ferrous oxide to iron, the coke regenerates carbon monoxide and limestone-derived lime forms slag with silica.
Step 3:The haematite reduction step therefore is the one that does not take place in the 900-1500 K range.
Final answer:
Biology100 questions
Q101Single correctSexual Reproduction in Flowering Plants
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : The first stage of gametophyte in the life cycle of moss is protonema stage.
Reason R : Protonema develops directly from spores produced in capsule.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion A : The first stage of gametophyte in the life cycle of moss is protonema stage.
Reason R : Protonema develops directly from spores produced in capsule.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are correct and R is the correct explanation of A.
Approach:
The statements concern the protonema stage in the moss life cycle and its origin.
Step 1:In the moss life cycle, spores germinate to form the first stage of the gametophyte, called the protonema, a creeping green filamentous structure.
Step 2:Spores are produced inside the capsule of the moss sporophyte; on germination each spore gives rise directly to a protonema.
Final answer: Both A and R are correct and R is the correct explanation of A.
Q102Single correctSexual Reproduction in Flowering Plants
In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Synergids, Zygote and Primary endosperm nucleus
Approach:
The ploidy of structures in a fertilized embryo sac is determined from their cellular origin.
Step 1:Synergids are part of the egg apparatus and are haploid cells.
Step 2:Fusion of one male gamete with the egg cell forms the zygote, which is diploid.
Step 3:Fusion of the second male gamete with the two polar nuclei forms the primary endosperm nucleus, which is triploid.
Final answer: Synergids, Zygote and Primary endosperm nucleus
Q103Single correctTransport in Plants
Movement and accumulation of ions across a membrane against their concentration gradient can be explained by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Active Transport
Approach:
Transport against a concentration gradient is identified by its energy requirement.
Step 1:Osmosis, facilitated diffusion and passive transport all proceed down a concentration gradient and require no metabolic energy.
Step 2:Active transport uses energy from ATP hydrolysis to pump ions across a membrane against their concentration gradient.
Final answer: Active Transport
Q104Single correctSexual Reproduction in Flowering Plants
Large, colourful, fragrant flowers with nectar are seen in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1insect pollinated plants
Approach:
Floral features are matched to the corresponding pollinating agent.
Step 1:Insect-pollinated (entomophilous) flowers are typically large, brightly coloured, fragrant and provide nectar to attract insects.
Step 2:Wind-pollinated flowers are small and inconspicuous and lack nectar and fragrance, so they do not fit the description.
Final answer: insect pollinated plants
Q105Single correctPrinciples of Inheritance and Variation
The phenomenon of pleiotropism refers to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a single gene affecting multiple phenotypic expression.
Approach:
The definition of pleiotropy is applied to each description of gene action.
Step 1:Pleiotropy describes the situation where a single gene influences several distinct phenotypic characters.
Step 2:A description in which several genes govern one trait is polygenic inheritance, not pleiotropy.
Final answer: a single gene affecting multiple phenotypic expression.
Q106Single correctPlant Growth and Development
Which hormone promotes internode/petiole elongation in deep water rice?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ethylene
Approach:
The hormone responsible for rapid elongation in deep water rice is identified.
Step 1:In deep water rice plants, ethylene promotes rapid elongation of internodes and petioles so that the upper leaves stay above rising water.
Step 2:Gibberellic acid, kinetin and 2,4-D do not perform this specific adaptive elongation role in deep water rice.
Final answer: Ethylene
Q107Single correctBiodiversity and Conservation
Among 'The Evil Quartet', which one is considered the most important cause driving extinction of species?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Habitat loss and fragmentation
Approach:
The leading driver among the four causes termed 'The Evil Quartet' is recalled.
Step 1:'The Evil Quartet' comprises habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions.
Step 2:Habitat loss and fragmentation is regarded as the most important cause driving plants and animals towards extinction.
Final answer: Habitat loss and fragmentation
Q108Single correctCell Cycle and Cell Division
Upon exposure to UV radiation, DNA stained with ethidium bromide will show :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bright orange colour
Approach:
The fluorescence of ethidium bromide-stained DNA under UV light is recalled.
Step 1:Ethidium bromide intercalates between DNA bases and fluoresces when illuminated with UV radiation.
Step 2:DNA bands stained with ethidium bromide appear as bright orange bands under UV light.
Final answer: Bright orange colour
Q109Single correctMineral Nutrition
Which micronutrient is required for splitting of water molecule during photosynthesis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1manganese
Approach:
The micronutrient associated with photolysis of water is identified.
Step 1:Photolysis of water in the light reaction releases oxygen, protons and electrons at the oxygen-evolving complex of photosystem II.
Step 2:Manganese is the essential micronutrient required for the splitting of water during photosynthesis.
Final answer: manganese
Q110Single correctAnatomy of Flowering Plants
Axile placentation is observed in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4China rose, Petunia and Lemon
Approach:
Plants with axile placentation are recalled from the arrangement of ovules on a central axis inside a multilocular ovary.
Step 1:In axile placentation the placenta is axial and ovules are attached to the central axis of a multilocular ovary.
Step 2:China rose, petunia, lemon, tomato and orange are standard examples of axile placentation.
Final answer: China rose, Petunia and Lemon
Q111Single correctCell Cycle and Cell Division
The process of appearance of recombination nodules occurs at which sub stage of prophase I in meiosis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pachytene
Approach:
The sub-stage of prophase I marked by recombination nodules is identified.
Step 1:Prophase I proceeds through leptotene, zygotene, pachytene, diplotene and diakinesis.
Step 2:During pachytene, recombination nodules appear on the bivalents and crossing over between non-sister chromatids takes place.
Final answer: Pachytene
Q112Single correctPhotosynthesis in Higher Plants
The reaction centre in PS II has an absorption maxima at
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1680 nm
Approach:
The absorption maximum of the photosystem II reaction centre is recalled.
Step 1:Photosystem II has its reaction centre chlorophyll a with an absorption peak at 680 nm and is therefore called P680.
Step 2:Photosystem I, by contrast, has its reaction centre P700 with absorption at 700 nm.
Final answer: 680 nm
Q113Single correctMolecular Basis of Inheritance
Unequivocal proof that DNA is the genetic material was first proposed by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Alfred Hershey and Martha Chase
Approach:
The experiment providing unequivocal proof of DNA as the genetic material is identified.
Step 1:Griffith demonstrated transformation, and Avery, MacLeod and McCarty identified the transforming principle as DNA, but the proof was not yet considered unequivocal.
Step 2:Hershey and Chase, using bacteriophage labelled with radioactive isotopes, showed that DNA and not protein enters the host, giving unequivocal proof that DNA is the genetic material.
Final answer: Alfred Hershey and Martha Chase
Q114Single correctCell Cycle and Cell Division
Among eukaryotes, replication of DNA takes place in -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2S phase
Approach:
The phase of the cell cycle in which DNA replication occurs is identified.
Step 1:Interphase consists of G1, S and G2 phases, and DNA is duplicated during the synthesis (S) phase.
Step 2:G1 and G2 are growth phases and M phase is the actual division, none of which carry out DNA synthesis.
Final answer: S phase
Q115Single correctPlant Growth and Development
In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Dedifferentiation
Approach:
The term for mature cells regaining the capacity to divide is identified.
Step 1:Mature, differentiated leaf mesophyll cells that resume division to form an unorganized mass called callus have regained meristematic ability.
Step 2:Such regaining of the capacity to divide by differentiated cells is termed dedifferentiation.
Final answer: Dedifferentiation
Q116Single correctBiomolecules
Cellulose does not form blue colour with Iodine because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3It does not contain complex helices and hence cannot hold iodine molecules.
Approach:
The reason cellulose fails to give the iodine test is established from its structure.
Step 1:The blue colour of the iodine test arises when iodine molecules become trapped within helical coils, as in starch.
Step 2:Cellulose is an unbranched polymer that lacks such complex helices, so it cannot hold iodine molecules and gives no blue colour.
Final answer: It does not contain complex helices and hence cannot hold iodine molecules.
Q117Single correctPlant Growth and Development
Spraying of which of the following phytohormone on juvenile conifers helps in hastening the maturity period, that leads to early seed production?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gibberellic Acid
Approach:
The hormone that shortens the juvenile phase of conifers is identified.
Step 1:Gibberellins promote bolting, flowering and maturity in many plants and hasten the transition from the juvenile to the mature phase.
Step 2:Spraying gibberellic acid on juvenile conifers shortens the maturity period and leads to early seed production.
Final answer: Gibberellic Acid
Q118Single correctTransport in Plants
Given below are two statements :
Statement I : The forces generated by transpiration can lift a xylem-sized column of water over 130 meters height.
Statement II : Transpiration cools leaf surfaces sometimes 10 to 15 degrees, by evaporative cooling.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : The forces generated by transpiration can lift a xylem-sized column of water over 130 meters height.
Statement II : Transpiration cools leaf surfaces sometimes 10 to 15 degrees, by evaporative cooling.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct.
Approach:
Each statement about transpiration is evaluated for accuracy.
Step 1:Transpiration pull generated at the leaf surface can raise a xylem-sized column of water to heights exceeding 130 metres in tall trees.
Step 2:Evaporation of water during transpiration removes heat and cools the leaf surface by about 10 to 15 degrees.
Final answer: Both Statement I and Statement II are correct.
Q119Single correctMorphology of Flowering Plants
Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Diadelphous and Dithecous anthers
Approach:
The androecium features unique to Fabaceae are compared with those of Solanaceae and Liliaceae.
Step 1:In Fabaceae the stamens are diadelphous, arranged as nine fused and one free, and the anthers are dithecous.
Step 2:Solanaceae has epipetalous stamens and Liliaceae has stamens in two whorls of three, so neither shows the diadelphous condition.
Final answer: Diadelphous and Dithecous anthers
Q120Single correctMolecular Basis of Inheritance
Expressed Sequence Tags (ESTs) refers to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1All genes that are expressed as RNA.
Approach:
The definition of Expressed Sequence Tags within the Human Genome Project is applied.
Step 1:The EST approach focuses on identifying the genes that are transcribed, that is the genes expressed as RNA.
Step 2:This contrasts with the sequence annotation approach, which analyses the whole genome including unexpressed regions.
Final answer: All genes that are expressed as RNA.
Q121Single correctEcosystem
Identify the correct statements :
A. Detrivores perform fragmentation.
B. The humus is further degraded by some microbes during mineralization.
C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching.
D. The detritus food chain begins with living organisms.
E. Earthworms break down detritus into smaller particles by a process called catabolism.
Choose the correct answer from the options given below :
A. Detrivores perform fragmentation.
B. The humus is further degraded by some microbes during mineralization.
C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching.
D. The detritus food chain begins with living organisms.
E. Earthworms break down detritus into smaller particles by a process called catabolism.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B, C only
Approach:
Each statement about decomposition is judged true or false on its own merits.
Step 1:Detritivores such as earthworms break detritus into smaller particles by fragmentation, so statement A is correct, while statement E wrongly names this process catabolism and is incorrect.
Step 2:Humus is further degraded by some microbes during mineralization, so statement B is correct.
Step 3:Water-soluble inorganic nutrients move down into the soil and become precipitated through leaching, so statement C is correct, while the detritus food chain begins with dead organic matter rather than living organisms, making statement D incorrect.
Final answer: A, B, C only
Q122Single correctEnvironmental Issues
The thickness of ozone in a column of air in the atmosphere is measured in terms of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Dobson units
Approach:
The unit used to express atmospheric ozone thickness is recalled.
Step 1:The thickness of ozone in a column of air is expressed in Dobson units (DU).
Step 2:Decibels measure sound, decameter is a length unit and kilobase measures nucleic acid length, none of which apply to ozone.
Final answer: Dobson units
Q123Single correctAnatomy of Flowering Plants
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : Late wood has fewer xylary elements with narrow vessels.
Reason R : Cambium is less active in winters.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : Late wood has fewer xylary elements with narrow vessels.
Reason R : Cambium is less active in winters.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are true and R is the correct explanation of A.
Approach:
The statements about late wood formation and cambial activity are assessed.
Step 1:Late wood, also called autumn wood, has fewer xylem elements with narrow vessels, so Assertion A is true.
Step 2:During winter the cambium is less active and produces fewer, narrower xylem elements, which directly accounts for the nature of late wood, 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.
Q124Single correctCell Cycle and Cell Division
Which of the following stages of meiosis involves division of centromere?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Anaphase II
Approach:
The meiotic stage in which centromeres split is identified.
Step 1:In meiosis I the homologous chromosomes separate while centromeres remain intact, so centromere division does not occur in anaphase I.
Step 2:During anaphase II the centromeres split and the sister chromatids move to opposite poles.
Final answer: Anaphase II
Q125Single correctBiodiversity and Conservation
The historic Convention on Biological Diversity, 'The Earth Summit' was held in Rio de Janeiro in the year :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 21992
Approach:
The year of the Earth Summit at Rio de Janeiro is recalled.
Step 1:The Convention on Biological Diversity, known as the Earth Summit, was held at Rio de Janeiro in 1992.
Step 2:The follow-up World Summit on Sustainable Development was held in 2002 at Johannesburg, which does not match the Rio Earth Summit.
Final answer: 1992
Q126Single correctPhotosynthesis in Higher Plants
How many ATP and NADP are required for the synthesis of one molecule of Glucose during Calvin cycle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 218 ATP and 12 NADP
Approach:
Synthesis of one glucose molecule needs two turns of the Calvin cycle, since each turn fixes one molecule of CO2 and a hexose has six carbons. The per-turn requirement is scaled to six carbons to obtain the total.
Step 1:Fixation of one CO2 in the Calvin cycle consumes 3 ATP and 2 NADPH.
Step 2:Building a six-carbon glucose requires fixation of six CO2 molecules, so the per-CO2 demand is multiplied by six.
Final answer: 18 ATP and 12 NADP
Q127Single correctEcosystem
In the equation
GPP is Gross Primary Productivity
NPP is Net Primary Productivity
R here is __________.
GPP is Gross Primary Productivity
NPP is Net Primary Productivity
R here is __________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Respiratory loss
Approach:
The relationship between gross and net primary productivity is defined in terms of the energy lost by producers through their own respiration.
Step 1:Gross primary productivity is the total rate of organic matter produced by photosynthesis, while net primary productivity is the biomass available to consumers.
Step 2:The difference between the two equals the energy that producers consume in their own respiration, denoted R.
Final answer: Respiratory loss
Q128Single correctBiotechnology: Principles and Processes
During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2DNA
Approach:
Recovery of purified nucleic acid relies on its solubility being reduced in the presence of cold alcohol after the cell components are released.
Step 1:After lysis and removal of proteins and other macromolecules, the aqueous phase carries dissolved DNA.
Step 2:Addition of chilled ethanol lowers DNA solubility, causing it to separate out as fine threads that can be spooled.
Final answer: DNA
Q129Single correctMolecular Basis of Inheritance
What is the role of RNA polymerase III in the process of transcription in Eukaryotes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Transcription of tRNA, 5 srRNA and snRNA
Approach:
Eukaryotic transcription is shared among three nuclear RNA polymerases, each responsible for a distinct class of RNA, so assigning the other two fixes the role of polymerase III.
Step 1:RNA polymerase I transcribes the large ribosomal RNAs (28S, 18S and 5.8S), and RNA polymerase II transcribes the precursor of mRNA, namely hnRNA.
Step 2:RNA polymerase III transcribes the remaining small stable RNAs, comprising tRNA, 5S rRNA and snRNAs.
Final answer: Transcription of tRNA, 5 srRNA and snRNA
Q130Single correctSexual Reproduction in Flowering Plants
What is the function of tassels in the corn cob?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2To trap pollen grains
Approach:
The structure of the maize female inflorescence is linked to its mode of pollination to determine the function of the tassels.
Step 1:Maize is wind pollinated, and the tassels are the long, hair-like styles and stigmas that emerge from the cob.
Step 2:Their large exposed surface intercepts air-borne pollen, so the tassels serve to catch pollen grains for fertilisation.
Final answer: To trap pollen grains
Q131Single correctPlant Kingdom
Identify the pair of heterosporous pteridophytes among the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 and
Approach:
Each listed genus is classified as homosporous or heterosporous, and the pair in which both members are heterosporous is required.
Step 1:Lycopodium, Psilotum and Equisetum produce a single type of spore and are therefore homosporous.
Step 2:Selaginella and Salvinia produce two kinds of spores, namely microspores and megaspores, and are heterosporous.
Final answer: and
Q132Single correctBiotechnology: Principles and Processes
In gene gun method used to introduce alien DNA into host cells, microparticles of __________ metal are used.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Tungsten or gold
Approach:
The biolistic technique is recalled along with the specific carrier particles onto which the foreign DNA is coated before bombardment.
Step 1:In the gene gun or biolistic method, cells are bombarded with high-velocity microparticles coated with the DNA to be introduced.
Step 2:The microparticles used as carriers are made of gold or tungsten, which are dense and inert.
Final answer: Tungsten or gold
Q133Single correctAnatomy of Flowering Plants
Given below are two statements :
Statement I : Endarch and exarch are the terms often used for describing the position of secondary xylem in the plant body.
Statement II : Exarch condition is the most common feature of the root system.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : Endarch and exarch are the terms often used for describing the position of secondary xylem in the plant body.
Statement II : Exarch condition is the most common feature of the root system.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is true.
Approach:
Each statement is judged against the definition of endarch and exarch arrangement and the typical xylem pattern of roots.
Step 1:Endarch and exarch describe the position of protoxylem relative to metaxylem in the primary xylem, not the secondary xylem, so Statement I is incorrect.
Step 2:In roots the protoxylem lies towards the periphery and metaxylem towards the centre, which is the exarch condition, so Statement II is correct.
Final answer: Statement I is incorrect but Statement II is true.
Q134Single correctPrinciples of Inheritance and Variation
Frequency of recombination between gene pairs on same chromosome as a measure of the distance between genes to map their position on chromosome, was used for the first time by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Alfred Sturtevant
Approach:
The historical attribution of using recombination frequency to construct genetic maps is recalled from the early work on linkage.
Step 1:Recombination frequency between linked genes reflects the distance separating them on a chromosome.
Step 2:Alfred Sturtevant, a student of Morgan, first used this recombination frequency to prepare a genetic map of genes on a chromosome.
Final answer: Alfred Sturtevant
Q135Single correctRespiration in Plants
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : ATP is used at two steps in glycolysis.
Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in conversion of fructose-6-phosphate into fructose-1-6-diphosphate.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : ATP is used at two steps in glycolysis.
Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in conversion of fructose-6-phosphate into fructose-1-6-diphosphate.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are true and R is the correct explanation of A.
Approach:
The preparatory phase of glycolysis is examined to confirm both the number of ATP-consuming steps and the specific reactions involved.
Step 1:Two phosphorylation reactions in glycolysis consume ATP, so the assertion that ATP is used at two steps is true.
Step 2:The first ATP phosphorylates glucose to glucose-6-phosphate and the second phosphorylates fructose-6-phosphate to fructose-1,6-bisphosphate, which exactly identifies the two steps named in the assertion.
Final answer: Both A and R are true and R is the correct explanation of A.
Q136Single correctMicrobes in Human Welfare / Ecosystem
Which one of the following statements is NOT correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Algal blooms caused by excess of organic matter in water improve water quality and promote fisheries.
Approach:
Each statement on water pollution is checked, and the one that misrepresents the effect of the described process is identified as incorrect.
Step 1:Microbial breakdown of organic matter raises biochemical oxygen demand and kills aquatic life, water hyacinth disrupts eutrophic water bodies, and toxins biomagnify up trophic levels, so statements 1, 3 and 4 are correct.
Step 2:Algal blooms deteriorate water quality, deplete oxygen and harm fisheries rather than improving them, so statement 2 is the incorrect one.
Final answer: Algal blooms caused by excess of organic matter in water improve water quality and promote fisheries.
Q137Single correctMolecular Basis of Inheritance
How many different proteins does the ribosome consist of?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 180
Approach:
The protein content of the ribosome is recalled from its structural description as a ribonucleoprotein particle.
Step 1:A ribosome is composed of ribosomal RNA together with a set of distinct ribosomal proteins.
Step 2:The complete ribosome is made up of about 80 different proteins distributed between its two subunits.
Final answer: 80
Q138Single correctPrinciples of Inheritance and Variation
Which of the following statements are correct about Klinefelter's Syndrome?
A. This disorder was first described by Langdon Down (1866).
B. Such an individual has overall masculine development. However, the feminine development is also expressed.
C. The affected individual is short statured.
D. Physical, psychomotor and mental development is retarded.
E. Such individuals are sterile.
Choose the correct answer from the options given below :
A. This disorder was first described by Langdon Down (1866).
B. Such an individual has overall masculine development. However, the feminine development is also expressed.
C. The affected individual is short statured.
D. Physical, psychomotor and mental development is retarded.
E. Such individuals are sterile.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B and E only
Approach:
Each statement is checked against the features of Klinefelter syndrome, which arises from an extra X chromosome in a male.
Step 1:Klinefelter syndrome was described by Klinefelter, not Langdon Down who described Down syndrome, and the short stature, retardation features listed apply to Down syndrome rather than this disorder, so statements A, C and D are incorrect.
Step 2:An affected XXY individual shows overall masculine development with some feminine traits such as gynaecomastia and is sterile, so statements B and E are correct.
Final answer: B and E only
Q139Single correctRespiration in Plants
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Oxidative decarboxylation | I. Citrate synthase |
| B. Glycolysis | II. Pyruvate dehydrogenase |
| C. Oxidative phosphorylation | III. Electron transport system |
| D. Tricarboxylic acid cycle | IV. EMP pathway |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-IV, C-III, D-I
Approach:
Each respiratory process in List I is matched with its characteristic enzyme or pathway in List II.
Step 1:Oxidative decarboxylation of pyruvate is catalysed by pyruvate dehydrogenase (A-II), and glycolysis is the EMP pathway (B-IV).
Step 2:Oxidative phosphorylation occurs through the electron transport system (C-III), and the tricarboxylic acid cycle is initiated by citrate synthase (D-I).
Final answer: A-II, B-IV, C-III, D-I
Q140Single correctMorphology of Flowering Plants
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral meristem.
Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at successive nodes instead of leaves.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral meristem.
Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at successive nodes instead of leaves.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are true and R is the correct explanation of A.
Approach:
The assertion and reason about the floral nature of a shoot are each assessed and their logical connection examined.
Step 1:A flower is a modified shoot in which the shoot apical meristem converts to a floral meristem, so the assertion is true.
Step 2:The internodes condense and the floral appendages arise laterally at successive nodes in place of leaves, which is the mechanism explaining the conversion, so the reason is true and explains the assertion.
Final answer: Both A and R are true and R is the correct explanation of A.
Q141Single correctSexual Reproduction in Flowering Plants
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : In gymnosperms the pollen grains are released from the microsporangium and carried by air currents.
Reason R : Air currents carry the pollen grains to the mouth of the archegonia where the male gametes are discharged and pollen tube is not formed.
In the light of the above statements, choose the correct answer from the options given below :
Assertion A : In gymnosperms the pollen grains are released from the microsporangium and carried by air currents.
Reason R : Air currents carry the pollen grains to the mouth of the archegonia where the male gametes are discharged and pollen tube is not formed.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false.
Approach:
The assertion on gymnosperm pollen dispersal and the reason on the mode of fertilisation are evaluated separately.
Step 1:In gymnosperms pollen grains are indeed released from the microsporangium and carried by air currents, so the assertion is true.
Step 2:After reaching the ovule the pollen germinates and forms a pollen tube that delivers the male gametes near the archegonia, so the claim that no pollen tube is formed makes the reason false.
Final answer: A is true but R is false.
Q142Single correctTransport in Plants
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Cohesion | I. More attraction in liquid phase |
| B. Adhesion | II. Mutual attraction among water molecules |
| C. Surface tension | III. Water loss in liquid phase |
| D. Guttation | IV. Attraction towards polar surfaces |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-IV, C-I, D-III
Approach:
Each water-relation property in List I is matched to its defining description in List II.
Step 1:Cohesion is the mutual attraction among water molecules (A-II), and adhesion is the attraction of water towards polar surfaces (B-IV).
Step 2:Surface tension reflects greater attraction in the liquid phase than in the gas phase at the interface (C-I), and guttation is the loss of water in liquid form (D-III).
Final answer: A-II, B-IV, C-I, D-III
Q143Single correctPhotosynthesis in Higher Plants
Which of the following combinations is required for chemiosmosis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1membrane, proton pump, proton gradient, ATP synthase
Approach:
The essential components of the chemiosmotic mechanism of ATP synthesis are identified from the way a proton gradient is built and then used.
Step 1:Chemiosmosis requires a membrane, a proton pump and a proton gradient across that membrane to store energy.
Step 2:The stored proton-gradient energy drives ATP synthase to make ATP, so ATP synthase completes the required set.
Final answer: membrane, proton pump, proton gradient, ATP synthase
Q144Single correctMicrobes in Human Welfare
Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Succinic dehydrogenase
Approach:
The classical example of competitive enzyme inhibition is recalled to identify the enzyme targeted by malonate.
Step 1:Malonate resembles the substrate succinate in structure and acts as a competitive inhibitor.
Step 2:It binds the active site of succinic dehydrogenase, blocking oxidation of succinate and thereby inhibiting bacterial growth.
Final answer: Succinic dehydrogenase
Q145Single correctAnatomy of Flowering Plants
Identify the correct statements :
A. Lenticels are the lens-shaped openings permitting the exchange of gases.
B. Bark formed early in the season is called hard bark.
C. Bark is a technical term that refers to all tissues exterior to vascular cambium.
D. Bark refers to periderm and secondary phloem.
E. Phellogen is single-layered in thickness.
Choose the correct answer from the options given below :
A. Lenticels are the lens-shaped openings permitting the exchange of gases.
B. Bark formed early in the season is called hard bark.
C. Bark is a technical term that refers to all tissues exterior to vascular cambium.
D. Bark refers to periderm and secondary phloem.
E. Phellogen is single-layered in thickness.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and D only
Approach:
Each statement on bark, lenticels and phellogen is verified against secondary growth concepts.
Step 1:Lenticels are lens-shaped openings that allow gaseous exchange (A correct), and bark in the non-technical sense refers to the periderm and secondary phloem (D correct).
Step 2:Bark formed early in the season is soft bark, not hard bark, and the phellogen is generally two to several cell layers thick, so statements B and E are wrong, and the technical sense of bark covers all tissues exterior to vascular cambium making C a misstatement here.
Final answer: A and D only
Q146Single correctCell Cycle and Cell Division
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. M Phase | I. Proteins are synthesized |
| B. Phase | II. Inactive phase |
| C. Quiescent stage | III. Interval between mitosis and initiation of DNA replication |
| D. Phase | IV. Equational division |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-I, C-II, D-III
Approach:
Each phase of the cell cycle in List I is paired with its defining feature in List II.
Step 1:The M phase is the mitotic equational division (A-IV), and the G2 phase is marked by synthesis of proteins (B-I).
Step 2:The quiescent stage is the inactive G0 phase (C-II), and the G1 phase is the interval between mitosis and the start of DNA replication (D-III).
Final answer: A-IV, B-I, C-II, D-III
Q147Single correctOrganisms and Populations
Choose the correct answer from the options given below :
| List I (Interaction) | List II (Species A and B) |
|---|---|
| A. Mutualism | I. |
| B. Commensalism | II. |
| C. Amensalism | III. |
| D. Parasitism | IV. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-I, C-II, D-III
Approach:
Each population interaction is matched to its sign notation, where a plus, minus or zero denotes benefit, harm or no effect on each species.
Step 1:Mutualism benefits both species, written +(A), +(B) (A-IV), and commensalism benefits one species while the other is unaffected, written +(A), O(B) (B-I).
Step 2:Amensalism harms one species while the other is unaffected, written -(A), O(B) (C-II), and parasitism benefits the parasite while harming the host, written +(A), -(B) (D-III).
Final answer: A-IV, B-I, C-II, D-III
Q148Single correctOrganisms and Populations
Given below are two statements :
Statement I : Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated eventually.
Statement II : In general, carnivores are more adversely affected by competition than herbivores.
In the light of the above statements, choose the correct answer from the options given below :
Statement I : Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated eventually.
Statement II : In general, carnivores are more adversely affected by competition than herbivores.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is false.
Approach:
The two statements about competition are evaluated against ecological principles on coexistence and the relative impact of competition on trophic levels.
Step 1:Gause's competitive exclusion principle holds that two species competing for identical resources cannot coexist indefinitely and the inferior competitor is eliminated, so Statement I is correct.
Step 2:Herbivores, not carnivores, are generally more adversely affected by competition because of their feeding constraints, so Statement II is false.
Final answer: Statement I is correct but Statement II is false.
Q149Single correctMineral Nutrition
Choose the correct answer from the options given below :
| List I | List II |
|---|---|
| A. Iron | I. Synthesis of auxin |
| B. Zinc | II. Component of nitrate reductase |
| C. Boron | III. Activator of catalase |
| D. Molybdenum | IV. Cell elongation and differentiation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Each mineral element in List I is matched with its physiological role in List II.
Step 1:Iron acts as an activator of catalase (A-III), and zinc is required for the synthesis of the auxin indole acetic acid (B-I).
Step 2:Boron is involved in cell elongation and differentiation (C-IV), and molybdenum is a component of the enzyme nitrate reductase (D-II).
Final answer: A-III, B-I, C-IV, D-II
Q150Single correctBiotechnology: Principles and Processes
Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence.
A. Insertion of recombinant DNA into the host cell.
B. Cutting of DNA at specific location by restriction enzyme.
C. Isolation of desired DNA fragment.
D. Amplification of gene of interest using PCR.
Choose the correct answer from the options given below :
A. Insertion of recombinant DNA into the host cell.
B. Cutting of DNA at specific location by restriction enzyme.
C. Isolation of desired DNA fragment.
D. Amplification of gene of interest using PCR.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B, C, D, A
Approach:
The standard sequence of recombinant DNA technology is reconstructed from the listed operations.
Step 1:The workflow begins with obtaining the desired DNA, with isolation of the fragment (C) and cutting of DNA at a specific site by a restriction enzyme (B) forming the early steps.
Step 2:The gene of interest is then amplified by PCR (D), and finally the recombinant DNA is inserted into the host cell (A). Cutting with the restriction enzyme and isolating the desired fragment can be described in either sequence, so both B, C, D, A and C, B, D, A are accepted.
Final answer: B, C, D, A
Q151Single correctReproductive Health
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Vasectomy | I. Oral method |
| B. Coitus interruptus | II. Barrier method |
| C. Cervical caps | III. Surgical method |
| D. Saheli | IV. Natural method |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-II, D-I
Approach:
Each contraceptive listed is classified by its mode of action.
Step 1:Vasectomy is the surgical sterilisation procedure performed on males by cutting and tying the vas deferens.
Step 2:Coitus interruptus, the withdrawal of the penis before ejaculation, is a natural contraceptive method.
Step 3:Cervical caps are barrier devices that cover the cervix and block sperm entry into the uterus.
Step 4:Saheli is a non-steroidal oral pill taken once a week.
Final answer: A-III, B-IV, C-II, D-I
Q152Single correctHuman Reproduction
Given below are two statements:
Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct.
Statement II: The cavity of the cervix is called cervical canal which along with vagina forms birth canal.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct.
Statement II: The cavity of the cervix is called cervical canal which along with vagina forms birth canal.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are true.
Approach:
Both anatomical statements about the male and female ducts are evaluated against textbook descriptions.
Step 1:The vas deferens joins the duct of the seminal vesicle to form the ejaculatory duct, which opens into the urethra. Statement I is true.
Step 2:The cavity of the cervix forms the cervical canal, which together with the vagina constitutes the birth canal. Statement II is true.
Final answer: Both Statement I and Statement II are true.
Q153Single correctEnvironmental Issues
Which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Biomagnification refers to increase in concentration of the toxicant at successive trophic levels.
Approach:
Each statement on pollution-related phenomena is checked for accuracy.
Step 1:Biomagnification is the rise in concentration of a non-degradable toxicant at each higher trophic level. This statement is correct.
Step 2:Eutrophication is natural ageing of a lake from nutrient enrichment, not merely sewage addition; abundant nutrients promote algal bloom rather than restricting it; and algal bloom increases fish mortality. These statements are incorrect.
Final answer: Biomagnification refers to increase in concentration of the toxicant at successive trophic levels.
Q154Single correctPrinciples of Inheritance and Variation
Which one of the following symbols represents mating between relatives in human pedigree analysis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The double horizontal line between square and circle, marking a consanguineous mating (drawn symbol, option 2)
Approach:
Standard pedigree symbols are recalled to identify the one marking a consanguineous mating.
Step 1:A single horizontal line joining a square (male) and a circle (female) denotes an ordinary mating between unrelated individuals.
Step 2:A double horizontal line connecting the male and female symbols denotes consanguineous mating, that is mating between relatives.
Final answer: The double horizontal line between square and circle, marking a consanguineous mating (drawn symbol, option 2)
Q155Single correctReproductive Health
Which one of the following common sexually transmitted diseases is completely curable when detected early and treated properly?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gonorrhoea
Approach:
The curability of each listed STD is recalled.
Step 1:Genital herpes, hepatitis-B and HIV infection are caused by viruses and are not completely curable with early detection.
Step 2:Gonorrhoea is a bacterial infection caused by Neisseria gonorrhoeae and is completely curable when detected early and treated properly.
Final answer: Gonorrhoea
Q156Single correctHuman Health and Disease
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Heroin | I. Effect on cardiovascular system |
| B. Marijuana | II. Slow down body function |
| C. Cocaine | III. Painkiller |
| D. Morphine | IV. Interfere with transport of dopamine |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-I, C-IV, D-III
Approach:
The principal pharmacological effect of each drug is matched.
Step 1:Heroin is a depressant that slows down body functions.
Step 2:Marijuana acts on the cardiovascular system.
Step 3:Cocaine interferes with the transport of the neurotransmitter dopamine.
Step 4:Morphine is an effective sedative and painkiller.
Final answer: A-II, B-I, C-IV, D-III
Q157Single correctLocomotion and Movement
Match List I with List II. Choose the correct answer from the options given below:
| List I (Type of Joint) | List II (Found between) |
|---|---|
| A. Cartilaginous Joint | I. Between flat skull bones |
| B. Ball and Socket Joint | II. Between adjacent vertebrae in vertebral column |
| C. Fibrous Joint | III. Between carpal and metacarpal of thumb |
| D. Saddle Joint | IV. Between Humerus and Pectoral girdle |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-IV, C-I, D-III
Approach:
Each joint type is matched with the location where it occurs.
Step 1:A cartilaginous joint lies between adjacent vertebrae of the vertebral column.
Step 2:The ball and socket joint occurs between the humerus and the pectoral girdle.
Step 3:Fibrous joints are found between the bones of the skull.
Step 4:A saddle joint lies between the carpal and metacarpal of the thumb.
Final answer: A-II, B-IV, C-I, D-III
Q158Single correctBiomolecules
Given below are two statements:
Statement I: A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal).
Statement II: Adult human haemoglobin, consists of 4 subunits (two subunits of α type and two subunits of β type.)
In the light of the above statements, choose the correct answer from the options given below:
Statement I: A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal).
Statement II: Adult human haemoglobin, consists of 4 subunits (two subunits of α type and two subunits of β type.)
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is false but Statement II is true.
Approach:
Each statement about protein representation and haemoglobin composition is verified.
Step 1:When a protein is drawn as a line, the left end is the first amino acid bearing the N-terminal and the right end is the last amino acid bearing the C-terminal. The labels in Statement I are reversed, so it is false.
Step 2:Adult human haemoglobin is a tetramer of two alpha and two beta subunits. Statement II is true.
Final answer: Statement I is false but Statement II is true.
Q159Single correctCell: The Unit of Life
Which of the following are NOT considered as the part of endomembrane system?
A. Mitochondria
B. Endoplasmic Reticulum
C. Chloroplasts
D. Golgi complex
E. Peroxisomes
Choose the most appropriate answer from the options given below:
A. Mitochondria
B. Endoplasmic Reticulum
C. Chloroplasts
D. Golgi complex
E. Peroxisomes
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, C and E only
Approach:
The components of the endomembrane system are recalled to identify the excluded organelles.
Step 1:The endomembrane system includes the endoplasmic reticulum, Golgi complex, lysosomes and vacuoles.
Step 2:Mitochondria, chloroplasts and peroxisomes are excluded because their functions are not coordinated with the others. These are A, C and E.
Final answer: A, C and E only
Q160Single correctEvolution
Given below are two statements:
Statement I: RNA mutates at a faster rate.
Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: RNA mutates at a faster rate.
Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are true.
Approach:
Both statements about the mutation rate of RNA and RNA viruses are assessed.
Step 1:RNA is single-stranded and lacks the proofreading repair available to DNA, so it mutates at a faster rate. Statement I is true.
Step 2:Viruses carrying an RNA genome and having a short life span accumulate mutations rapidly and evolve faster. Statement II is true.
Final answer: Both Statement I and Statement II are true.
Q161Single correctChemical Coordination and Integration
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. CCK | I. Kidney |
| B. GIP | II. Heart |
| C. ANF | III. Gastric gland |
| D. ADH | IV. Pancreas |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-III, C-II, D-I
Approach:
Each hormone is matched with the organ that secretes it or on which it acts.
Step 1:Cholecystokinin (CCK) acts on the pancreas and gall bladder, stimulating pancreatic secretion.
Step 2:Gastric inhibitory peptide (GIP) inhibits gastric secretion and motility of the gastric gland.
Step 3:Atrial natriuretic factor (ANF) is secreted by the atrial wall of the heart.
Step 4:Antidiuretic hormone (ADH) acts on the kidney to promote water reabsorption.
Final answer: A-IV, B-III, C-II, D-I
Q162Single correctHuman Reproduction
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Endometrium is necessary for implantation of blastocyst.
Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of endometrium.
In the light of the above statements, choose the correct answer from the options given below:
Assertion A: Endometrium is necessary for implantation of blastocyst.
Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of endometrium.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both A and R are true but R is NOT the correct explanation of A.
Approach:
The truth of the assertion and reason and their logical relation are evaluated.
Step 1:The endometrium is essential for implantation of the blastocyst, so Assertion A is true.
Step 2:In the absence of fertilization the corpus luteum degenerates, leading to breakdown of the endometrium. Reason R is true but it describes degeneration rather than implantation, so it does not explain the assertion.
Final answer: Both A and R are true but R is NOT the correct explanation of A.
Q163Single correctHuman Health and Disease
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Ringworm | I. Haemophilus influenzae |
| B. Filariasis | II. Trichophyton |
| C. Malaria | III. Wuchereria bancrofti |
| D. Pneumonia | IV. Plasmodium vivax |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-III, C-IV, D-I
Approach:
Each disease is matched with its causative organism.
Step 1:Ringworm is a fungal infection caused by Trichophyton.
Step 2:Filariasis is caused by the nematode Wuchereria bancrofti.
Step 3:Malaria is caused by the protozoan Plasmodium vivax.
Step 4:Pneumonia is caused by the bacterium Haemophilus influenzae (along with Streptococcus pneumoniae).
Final answer: A-II, B-III, C-IV, D-I
Q164Single correctBiomolecules
Given below are two statements:
Statement I: Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
Statement II: When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.
Statement II: When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are true.
Approach:
Both statements about temperature effects and competitive inhibition are checked.
Step 1:Low temperature keeps an enzyme temporarily inactive, while high temperature denatures the protein and destroys activity. Statement I is true.
Step 2:An inhibitor resembling the substrate that competes for the active site is a competitive inhibitor. Statement II is true.
Final answer: Both Statement I and Statement II are true.
Q165Single correctAnimal Kingdom
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Taenia | I. Nephridia |
| B. Paramoecium | II. Contractile vacuole |
| C. Periplaneta | III. Flame cells |
| D. Pheretima | IV. Urecose gland |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-II, C-IV, D-I
Approach:
Each animal is matched with its characteristic excretory structure.
Step 1:Taenia, the tapeworm, uses flame cells for excretion.
Step 2:Paramoecium expels excess water and wastes through a contractile vacuole.
Step 3:Periplaneta, the cockroach, has Malpighian tubules associated with its excretory gland.
Step 4:Pheretima, the earthworm, excretes through nephridia.
Final answer: A-III, B-II, C-IV, D-I
Q166Single correctBiotechnology and its Applications
Which one of the following techniques does not serve the purpose of early diagnosis of a disease for its early treatment?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Serum and Urine analysis
Approach:
The techniques are compared by their ability to detect a disease at an early stage.
Step 1:Recombinant DNA technology, PCR and ELISA can detect very low amounts of pathogen or antibody before symptoms appear, enabling early diagnosis.
Step 2:Conventional serum and urine analysis detects a disease only after symptoms develop, so it does not serve early diagnosis.
Final answer: Serum and Urine analysis
Q167Single correctOrganisms and Populations
Match List I with List II. Choose the correct answer from the options given below:
| List I (Interacting species) | List II (Name of Interaction) |
|---|---|
| A. A Leopard and a Lion in a forest/grassland | I. Competition |
| B. A Cuckoo laying egg in a Crow's nest | II. Brood parasitism |
| C. Fungi and root of a higher plant in Mycorrhizae | III. Mutualism |
| D. A cattle egret and a Cattle in a field | IV. Commensalism |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-I, B-II, C-III, D-IV
Approach:
Each pair of interacting organisms is matched with the type of interaction.
Step 1:A leopard and a lion sharing the same prey in a habitat illustrate competition.
Step 2:A cuckoo laying its egg in a crow's nest is brood parasitism.
Step 3:Fungi associated with the roots of a higher plant in mycorrhizae represent mutualism.
Step 4:A cattle egret feeding on insects flushed out by grazing cattle is commensalism.
Final answer: A-I, B-II, C-III, D-IV
Q168Single correctStructural Organisation in Animals
Given below are two statements:
Statement I: Ligaments are dense irregular tissue.
Statement II: Cartilage is dense regular tissue.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: Ligaments are dense irregular tissue.
Statement II: Cartilage is dense regular tissue.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are false.
Approach:
The connective tissue category of ligaments and cartilage is checked against the descriptions.
Step 1:Ligaments are dense regular connective tissue, not dense irregular tissue, so Statement I is false.
Step 2:Cartilage is a specialised connective tissue with a solid matrix, not dense regular connective tissue, so Statement II is false.
Final answer: Both Statement I and Statement II are false.
Q169Single correctMolecular Basis of Inheritance
Given below are two statements:
Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid.
Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid.
Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I incorrect but Statement II is true.
Approach:
The charge of DNA and the associated proteins in prokaryotes and eukaryotes is verified.
Step 1:In prokaryotes the negatively charged DNA is held with some positively charged proteins in the nucleoid. The charges in Statement I are reversed, so it is incorrect.
Step 2:In eukaryotes the negatively charged DNA wraps around the positively charged histone octamer to form a nucleosome. Statement II is correct.
Final answer: Statement I incorrect but Statement II is true.
Q170Single correctNeural Control and Coordination
Match List I with List II with respect to human eye. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Fovea | I. Visible coloured portion of eye that regulates diameter of pupil. |
| B. Iris | II. External layer of eye formed of dense connective tissue. |
| C. Blind spot | III. Point of greatest visual acuity or resolution. |
| D. Sclera | IV. Point where optic nerve leaves the eyeball and photoreceptor cells are absent. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Each part of the human eye is matched with its description.
Step 1:The fovea is the point of greatest visual acuity or resolution.
Step 2:The iris is the visible coloured portion that regulates the diameter of the pupil.
Step 3:The blind spot is the point where the optic nerve leaves the eyeball and photoreceptor cells are absent.
Step 4:The sclera is the external layer of the eye formed of dense connective tissue.
Final answer: A-III, B-I, C-IV, D-II
Q171Single correctEvolution
Select the correct group/set of Australian Marsupials exhibiting adaptive radiation.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Numbat, Spotted cuscus, Flying phalanger
Approach:
Each set is checked to ensure all members are true Australian marsupials.
Step 1:Numbat, spotted cuscus and flying phalanger are all Australian marsupials that radiated into different niches.
Step 2:The other sets include placental mammals such as the bobcat, mole, flying squirrel, lemur, anteater and wolf, so they do not represent only Australian marsupials.
Final answer: Numbat, Spotted cuscus, Flying phalanger
Q172Single correctHuman Reproduction
Which of the following statements are correct regarding female reproductive cycle?
A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle.
B. First menstrual cycle begins at puberty and is called menopause.
C. Lack of menstruation may be indicative of pregnancy.
D. Cyclic menstruation extends between menarche and menopause.
Choose the most appropriate answer from the options given below:
A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle.
B. First menstrual cycle begins at puberty and is called menopause.
C. Lack of menstruation may be indicative of pregnancy.
D. Cyclic menstruation extends between menarche and menopause.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, C and D only
Approach:
Each statement about the female reproductive cycle is checked.
Step 1:In non-primate mammals the cyclical reproductive changes are termed the oestrus cycle, so A is correct.
Step 2:The first menstrual cycle begins at puberty and is called menarche, not menopause, so B is incorrect.
Step 3:Absence of menstruation can indicate pregnancy, so C is correct.
Step 4:Cyclic menstruation extends from menarche to menopause, so D is correct.
Final answer: A, C and D only
Q173Single correctBreathing and Exchange of Gases
Vital capacity of lung is ________.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4IRV + ERV + TV
Approach:
Vital capacity is expressed as the sum of the relevant pulmonary volumes.
Step 1:Vital capacity is the maximum volume of air a person can breathe out after a maximum inspiration. It is the sum of inspiratory reserve volume, tidal volume and expiratory reserve volume.
Step 2:Residual volume is excluded because it cannot be exhaled.
Final answer: IRV + ERV + TV
Q174Single correctBody Fluids and Circulation
Match List I with List II. Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. P-wave | I. Beginning of systole |
| B. Q-wave | II. Repolarisation of ventricles |
| C. QRS complex | III. Depolarisation of atria |
| D. T-wave | IV. Depolarisation of ventricles |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-III, B-I, C-IV, D-II
Approach:
Each wave of the electrocardiogram is matched with the cardiac event it represents.
Step 1:The P-wave represents depolarisation of the atria.
Step 2:The Q-wave signals the beginning of ventricular systole.
Step 3:The QRS complex represents depolarisation of the ventricles.
Step 4:The T-wave represents repolarisation of the ventricles.
Final answer: A-III, B-I, C-IV, D-II
Q175Single correctReproductive Health
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health Care Programme.
Reason R: Ban on amniocentesis checks increasing menace of female foeticide.
In the light of the above statements, choose the correct answer from the options given below:
Assertion A: Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health Care Programme.
Reason R: Ban on amniocentesis checks increasing menace of female foeticide.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A is false but R is true.
Approach:
The assertion and reason about amniocentesis are evaluated for truth.
Step 1:Amniocentesis for sex determination is banned, not a strategy of the Reproductive and Child Health Care Programme, so Assertion A is false.
Step 2:The ban on amniocentesis for sex determination checks the rising menace of female foeticide, so Reason R is true.
Final answer: A is false but R is true.
Q176Single correctDigestion and Absorption
Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by-
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ileo - caecal valve
Approach:
The junction between the small intestine and the large intestine determines whether intestinal contents can flow back.
Step 1:The ileum of the small intestine opens into the caecum of the large intestine.
Step 2:The ileo-caecal valve permits one-way passage of material from the ileum into the caecum and stops contents of the large intestine from returning to the small intestine.
Final answer: Ileo - caecal valve
Q177Single correctBiotechnology - Principles and Processes
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Gene 'a' | I. -galactosidase |
| B. Gene 'y' | II. Transacetylase |
| C. Gene 'i' | III. Permease |
| D. Gene 'z' | IV. Repressor protein |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-III, C-IV, D-I
Approach:
Assign each structural gene of the lac operon to the protein it encodes, then read the pairings off in the order the list gives them.
Step 1:The lac operon carries three structural genes plus a regulatory gene: encodes beta-galactosidase, encodes permease and encodes transacetylase.
Step 2:The separate regulatory gene codes for the repressor protein that binds the operator and switches the operon off.
Step 3:Read the four pairings in the order List I presents them.
Final answer: A-II, B-III, C-IV, D-I
Q178Single correctDigestion and Absorption
Choose the correct answer from the options given below:
| List I (Cells) | List II (Secretion) |
|---|---|
| A. Peptic cells | I. Mucus |
| B. Goblet cells | II. Bile juice |
| C. Oxyntic cells | III. Proenzyme pepsinogen |
| D. Hepatic cells | IV. HCl and intrinsic factor for absorption of vitamin |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-I, C-IV, D-II
Approach:
Each gastric and hepatic cell type produces a characteristic secretion.
Step 1:Peptic (chief) cells of the gastric glands secrete the inactive proenzyme pepsinogen.
Step 2:Goblet cells of the mucosa secrete mucus, while oxyntic (parietal) cells secrete hydrochloric acid and intrinsic factor required for vitamin B12 absorption.
Step 3:Hepatic cells of the liver produce bile juice.
Final answer: A-III, B-I, C-IV, D-II
Q179Single correctCell - The Unit of Life
Which of the following functions is carried out by cytoskeleton in a cell?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Motility
Approach:
The cytoskeleton is a network of filamentous proteinaceous structures and its roles follow from this composition.
Step 1:The cytoskeleton consists of microtubules, microfilaments and intermediate filaments distributed in the cytoplasm.
Step 2:These filaments provide mechanical support, maintain cell shape, and enable motility of the cell and its parts.
Final answer: Motility
Q180Single correctExcretory Products and their Elimination
Given below are statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex and medulla.
Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop of Henle.
In the light of the above statements, choose the correct answer from the options given below:
Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex and medulla.
Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop of Henle.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false.
Approach:
Each statement is evaluated separately against the structure of nephrons.
Step 1:Nephrons are classified into cortical and juxtamedullary types according to their position relative to the cortex and medulla, so the assertion is correct.
Step 2:Juxtamedullary nephrons possess a long loop of Henle that runs deep into the medulla, whereas cortical nephrons have a short loop of Henle, which is the reverse of the reason given.
Final answer: A is true but R is false.
Q181Single correctEnvironmental Issues
Given below are two statements:
Statement I: Electrostatic precipitator is most widely used in thermal power plant.
Statement II: Electrostatic precipitator in thermal power plant removes ionising radiations
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: Electrostatic precipitator is most widely used in thermal power plant.
Statement II: Electrostatic precipitator in thermal power plant removes ionising radiations
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect.
Approach:
Each statement is assessed against the working and purpose of an electrostatic precipitator.
Step 1:The electrostatic precipitator is the most widely used device for removing particulate matter from the exhaust of thermal power plants, so the first statement is correct.
Step 2:The device charges particles and collects them on plates of opposite charge, removing dust and fine particulates rather than ionising radiations, so the second statement is incorrect.
Final answer: Statement I is correct but Statement II is incorrect.
Q182Single correctPrinciples of Inheritance and Variation
Broad palm with single palm crease is visible in a person suffering from-
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Down's syndrome
Approach:
The clinical feature described is matched to the corresponding genetic disorder.
Step 1:A broad palm bearing a single transverse palmar crease is a recognised physical feature of Down's syndrome, which results from trisomy of chromosome 21.
Step 2:Other typical features of Down's syndrome include short stature, a small round head, a furrowed protruding tongue and partially open mouth.
Final answer: Down's syndrome
Q183Single correctAnimal Kingdom
Radial symmetry is NOT found in adults of phylum ______.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hemichordata
Approach:
Each phylum is checked for the type of body symmetry shown by its adults.
Step 1:Coelenterata and Ctenophora display radial symmetry, and adult echinoderms exhibit radial symmetry while their larvae are bilateral.
Step 2:Hemichordates are bilaterally symmetrical, worm-like animals and do not show radial symmetry in the adult.
Final answer: Hemichordata
Q184Single correctHuman Health and Disease
In which blood corpuscles, the HIV undergoes replication and produces progeny viruses?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 cells
Approach:
The target cell of HIV replication is identified from the events of AIDS infection.
Step 1:After entering the body, HIV enters macrophages where its RNA is reverse transcribed into DNA and incorporated into the host genome.
Step 2:Within helper T cells the viral DNA directs production of progeny viruses, which are released and infect further helper T cells.
Final answer: cells
Q185Single correctBiotechnology - Principles and Processes
Which of the following is not a cloning vector?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Probe
Approach:
Each listed item is examined for whether it can carry foreign DNA into a host cell.
Step 1:Bacterial artificial chromosomes, yeast artificial chromosomes and the plasmid pBR322 are all vehicles that carry and replicate inserted DNA inside host cells.
Step 2:A probe is a labelled single-stranded nucleic acid used to detect a complementary sequence by hybridisation and does not carry DNA into a host.
Final answer: Probe
Q186Single correctOrganisms and Populations
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Logistic growth | I. Unlimited resource availability condition |
| B. Exponential growth | II. Limited resource availability condition |
| C. Expanding age pyramid | III. The percent individuals of pre-reproductive age is largest followed by reproductive and post reproductive age groups |
| D. Stable age pyramid | IV. The percent individuals of pre-reproductives and reproductive age group are same |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-I, C-III, D-IV
Approach:
Each growth model and pyramid type is paired with the condition or structure that defines it.
Step 1:Logistic growth occurs when resources become limiting and the population approaches the carrying capacity, while exponential growth occurs when resources are unlimited.
Step 2:An expanding age pyramid has the largest fraction of pre-reproductive individuals followed by reproductive and post-reproductive groups, while a stable pyramid has equal pre-reproductive and reproductive fractions.
Final answer: A-II, B-I, C-III, D-IV
Q187Single correctAnimal Kingdom
Which of the following are the characteristic features of Hemichordata?
A. Presence of notochord.
B. Presence of open type of circulatory system.
C. Presence of paired pharyngeal gillslits.
D. Presence of solid double nerve cord.
E. Presence of pseudocoelom.
Choose the correct answer from the options given below :
A. Presence of notochord.
B. Presence of open type of circulatory system.
C. Presence of paired pharyngeal gillslits.
D. Presence of solid double nerve cord.
E. Presence of pseudocoelom.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and C only
Approach:
Test each listed feature against the body plan of Hemichordata and keep those that genuinely belong to the group.
Step 1:Hemichordates lack a true notochord; the stomochord of the proboscis is a buccal diverticulum, not a notochord, so A is not a feature of the group.
Step 2:Their circulatory system is of the open type, with blood flowing through sinuses and a dorsal heart vesicle rather than a closed capillary bed.
Step 3:The pharynx is perforated by paired gill slits used in respiration and in filter feeding, a defining hemichordate character.
Step 4:The nerve cord is a dorsal collar cord that is hollow in many forms, not a solid double cord, and the body cavity is a true coelom, so D and E are both wrong.
Final answer: B and C only
Q188Single correctNeural Control and Coordination
The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure, rage, fear etc. are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Limbic system & hypothalamus
Approach:
The brain regions controlling emotional and sexual behaviour are identified.
Step 1:The limbic system, together with the hypothalamus, regulates sexual behaviour and the expression of emotions such as excitement, pleasure, rage and fear.
Step 2:The other listed structures are involved in reflex pathways, sensory relay and connecting the two cerebral hemispheres rather than emotional regulation.
Final answer: Limbic system & hypothalamus
Q189Single correctAnimal Kingdom
The unique mammalian characteristics are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2hairs, pinna and mammary glands
Approach:
Features exclusive to mammals are separated from those shared with other vertebrate groups.
Step 1:Hairs covering the body, an external ear pinna and mammary glands that produce milk are features found only in mammals.
Step 2:A tympanic membrane occurs in other vertebrate groups and a monocondylic skull is a reptilian feature, so neither is exclusively mammalian.
Final answer: hairs, pinna and mammary glands
Q190Single correctChemical Coordination and Integration
Which of the following are NOT under the control of thyroid hormone?
A. Maintenance of water and electrolyte balance
B. Regulation of basal metabolic rate
C. Normal rhythm of sleep-wake cycle
D. Development of immune system
E. Support the process of R.B.Cs formation
Choose the correct answer from the options given below:
A. Maintenance of water and electrolyte balance
B. Regulation of basal metabolic rate
C. Normal rhythm of sleep-wake cycle
D. Development of immune system
E. Support the process of R.B.Cs formation
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3C and D only
Approach:
Each listed function is checked for whether thyroid hormone regulates it.
Step 1:Thyroid hormones regulate the basal metabolic rate, help maintain water and electrolyte balance, and support the process of red blood cell formation.
Step 2:The rhythm of the sleep-wake cycle and the development of the immune system are not governed by thyroid hormone.
Final answer: C and D only
Q191Single correctCell Cycle and Cell Division
Select the correct statements.
A. Tetrad formation is seen during Leptotene.
B. During Anaphase, the centromeres split and chromatids separate.
C. Terminalization takes place during Pachytene.
D. Nucleolus, Golgi complex and ER are reformed during Telophase.
E. Crossing over takes place between sister chromatids of homologous chromosome.
Choose the correct answer from the options given below:
A. Tetrad formation is seen during Leptotene.
B. During Anaphase, the centromeres split and chromatids separate.
C. Terminalization takes place during Pachytene.
D. Nucleolus, Golgi complex and ER are reformed during Telophase.
E. Crossing over takes place between sister chromatids of homologous chromosome.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and D only
Approach:
Each statement on cell division events is checked against the correct stage.
Step 1:Tetrad formation occurs in pachytene rather than leptotene, terminalisation of chiasmata occurs in diakinesis rather than pachytene, and crossing over occurs between non-sister chromatids of homologous chromosomes, so statements A, C and E are incorrect.
Step 2:Centromeres split and chromatids separate during anaphase, and the nucleolus, Golgi complex and endoplasmic reticulum reappear during telophase, so statements B and D are correct.
Final answer: B and D only
Q192Single correctStructural Organisation in Animals
Choose the correct answer from the options given below:
| List I | List II |
|---|---|
| A. Mast cells | I. Ciliated epithelium |
| B. Inner surface of bronchiole | II. Areolar connective tissue |
| C. Blood | III. Cuboidal epithelium |
| D. Tubular parts of nephron | IV. specialised connective tissue |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-II, B-I, C-IV, D-III
Approach:
Identify the tissue that lines or constitutes each listed structure, then read the four pairings off in the order List I gives them.
Step 1:Mast cells are scattered in the loose packing tissue beneath skin and around blood vessels, which is areolar connective tissue.
Step 2:The inner surface of a bronchiole is lined by ciliated epithelium, whose cilia sweep mucus and trapped dust back up the airway.
Step 3:Blood is a fluid connective tissue with a plasma matrix, classed as a specialised connective tissue.
Step 4:The tubular parts of the nephron are lined by cuboidal epithelium suited to reabsorption and secretion.
Final answer: A-II, B-I, C-IV, D-III
Q193Single correctBiology in Human Welfare
Which of the following is characteristic feature of cockroach regarding sexual dimorphism ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Presence of anal styles
Approach:
The feature distinguishing male from female cockroach is identified.
Step 1:In the cockroach, males bear a pair of small thread-like anal styles in addition to the anal cerci, whereas females lack the anal styles.
Step 2:Body colour, sclerites and anal cerci are present in both sexes and therefore do not indicate sexual dimorphism.
Final answer: Presence of anal styles
Q194Single correctMolecular Basis of Inheritance
Which one of the following is the sequence on corresponding coding strand, if the sequence on mRNA formed is as follows
5' AUCGAUCGAUCGAUCGAUCG AUCG AUCG 3'?
5' AUCGAUCGAUCGAUCGAUCG AUCG AUCG 3'?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 35' ATCGATCGATCGATCGATCG ATCGATCG 3'
Approach:
The coding strand has the same sequence as the mRNA except that uracil is replaced by thymine.
Step 1:Messenger RNA is synthesised from the template strand and carries the same sequence as the coding strand, with thymine substituted by uracil.
Step 2:Replacing each uracil of the given mRNA with thymine while keeping the 5' to 3' polarity produces the coding-strand sequence.
Final answer: 5' ATCGATCGATCGATCGATCG ATCGATCG 3'
Q195Single correctBiology in Human Welfare
In cockroach, excretion is brought about by-
A. Phallic gland B. Urecose gland
C. Nephrocytes D. Fat body
E. Collaterial glands
Choose the correct answer from the options given below:
A. Phallic gland B. Urecose gland
C. Nephrocytes D. Fat body
E. Collaterial glands
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B, C and D only
Approach:
The structures contributing to excretion in the cockroach are selected.
Step 1:Excretion in the cockroach is carried out mainly by Malpighian tubules, assisted by uricose glands, nephrocytes and the fat body, which store or remove nitrogenous wastes.
Step 2:The phallic and collaterial glands are part of the reproductive system and have no excretory role.
Final answer: B, C and D only
Q196Single correctCell Cycle and Cell Division
Given below are two statements:
Statement I: During phase of cell cycle, the cell is metabolically inactive.
Statement II: The centrosome undergoes duplication during S phase of interphase.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: During phase of cell cycle, the cell is metabolically inactive.
Statement II: The centrosome undergoes duplication during S phase of interphase.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct.
Approach:
Each statement on the cell cycle is assessed for accuracy.
Step 1:Cells in the G0 phase remain metabolically active even though they do not divide, so the first statement is incorrect.
Step 2:The centrosome duplicates during the S phase of interphase along with DNA replication, so the second statement is correct.
Final answer: Statement I is incorrect but Statement II is correct.
Q197Single correctPrinciples of Inheritance and Variation
Which one of the following is NOT an advantage of inbreeding?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4It decreases the productivity of inbred population, after continuous inbreeding.
Approach:
Each listed effect of inbreeding is examined for whether it is a benefit.
Step 1:Inbreeding increases homozygosity, exposes harmful recessive alleles for removal by selection, and helps accumulate superior genes while eliminating less desirable ones, all of which are advantages.
Step 2:Continued inbreeding lowers fertility and productivity through inbreeding depression, which is a disadvantage rather than a benefit.
Final answer: It decreases the productivity of inbred population, after continuous inbreeding.
Q198Single correctExcretory Products and their Elimination
Which of the following statements are correct?
A. An excessive loss of body fluid from the body switches off osmoreceptors.
B. ADH facilitates water reabsorption to prevent diuresis.
C. ANF causes vasodilation.
D. ADH causes increase in blood pressure.
E. ADH is responsible for decrease in GFR.
Choose the correct answer from the options given below:
A. An excessive loss of body fluid from the body switches off osmoreceptors.
B. ADH facilitates water reabsorption to prevent diuresis.
C. ANF causes vasodilation.
D. ADH causes increase in blood pressure.
E. ADH is responsible for decrease in GFR.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, C and D only
Approach:
Check each statement against the actions of the two hormones named, antidiuretic hormone and atrial natriuretic factor.
Step 1:A dangerous fall in body fluid volume stimulates the osmoreceptors rather than switching them off, so A is wrong.
Step 2:ADH acts on the distal tubule and collecting duct to reabsorb water, which reduces urine volume and so prevents diuresis.
Step 3:Atrial natriuretic factor, released by atrial walls when blood volume rises, dilates blood vessels and lowers blood pressure.
Step 4:ADH conserves water and also constricts blood vessels, so it raises blood pressure; the higher pressure raises glomerular blood flow rather than lowering it, so the claim that ADH decreases GFR does not hold.
Final answer: B, C and D only
Q199Single correctLocomotion and Movement
Which of the following statements are correct regarding skeletal muscle?
A. Muscle bundles are held together by collagenous connective tissue layer called fascicle.
B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions.
C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins.
D. M line is considered as functional unit of contraction called sarcomere.
Choose the most appropriate answer from the options given below:
A. Muscle bundles are held together by collagenous connective tissue layer called fascicle.
B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions.
C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins.
D. M line is considered as functional unit of contraction called sarcomere.
Choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B and C only
Approach:
Test each statement about skeletal muscle organisation against the structure of a muscle from bundle down to sarcomere.
Step 1:The collagenous sheet that holds muscle bundles together is the fascia; a fascicle is the bundle of muscle fibres itself, so A names the wrong structure.
Step 2:The sarcoplasmic reticulum of a muscle fibre stores calcium ions and releases them to trigger contraction.
Step 3:The alternating arrangement of actin-rich and myosin-rich bands along the fibre produces the striated appearance.
Step 4:The functional unit of contraction is the sarcomere, bounded by two successive Z lines, not the M line.
Final answer: B and C only
Q200Single correctBody Fluids and Circulation
Which of the following statements are correct?
A. Basophils are most abundant cells of the total WBCs.
B. Basophils secrete histamine, serotonin and heparin
C. Basophils are involved in inflammatory response
D. Basophils have kidney shaped nucleus
E. Basophils are agranulocytes
Choose the correct answer from the options given below:
A. Basophils are most abundant cells of the total WBCs.
B. Basophils secrete histamine, serotonin and heparin
C. Basophils are involved in inflammatory response
D. Basophils have kidney shaped nucleus
E. Basophils are agranulocytes
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B and C only
Approach:
Each statement on basophils is checked against their known properties.
Step 1:Basophils are the least abundant white blood cells, are granulocytes and have a lobed nucleus, so statements A, D and E are incorrect.
Step 2:Basophils secrete histamine, serotonin and heparin and take part in the inflammatory response, so statements B and C are correct.
Final answer: B and C only
Frequently Asked Questions
How many questions are in the NEET 2023 May 07 paper?
The NEET 2023 May 07 paper has 200 questions — Physics (50), Chemistry (50) and Biology (100). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for NEET?
NEET awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Biology across 180 questions for 720 marks.
Are the answer key and step-by-step solutions provided for the 2023 May 07 paper?
Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.
Can I take the NEET 2023 May 07 paper as a timed mock test?
Yes. With a free NEETnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.
Solved this paper? Calculate your NEET score · most important chapters · formula sheets · all free tools.