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![X, of formula C2H6O, is heated with copper at 573 K to give A. A treated with [Ag(NH3)2]+ and -OH under heat gives a silver mirror. A treated with -OH and heat gives Y. A treated with NH2-NH-CO-NH2 gives Z.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2Ffaff200f-937c-4543-a81c-a547772cc033%2Ffaff200f-937c-4543-a81c-a547772cc033%2Fimages%2FQ61_reaction_scheme.webp)



NEET 2017 May 07 Question Paper with Solutions
All 179 questions from the NEET 2017 (May 07) paper — Physics (44), Chemistry (45) and Biology (90) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2017Chemistry PYQs 2017Biology PYQs 2017
- Questions
- 179
- Physics
- 44
- Chemistry
- 45
- Biology
- 90
Physics44 questions
Q1Single correctCurrent Electricity
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a condition of no current flow through the galvanometer
Approach:
Identify the principle that makes a potentiometer superior to a voltmeter for measuring EMF.
Step 1:A voltmeter draws current from the source, so it reads terminal voltage, not true EMF.
Step 2:A potentiometer is balanced when the galvanometer shows no deflection, meaning zero current is drawn from the cell under test.
Step 3:With no current drawn, the internal resistance causes no drop and the true EMF is measured.
Final answer: a condition of no current flow through the galvanometer
Q2Single correctKinetic Theory of Gases
A gas mixture consists of 2 moles of and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Sum the internal energies of a diatomic gas and a monatomic gas using the equipartition theorem.
Step 1:Oxygen is diatomic with f = 5 (3 translational + 2 rotational).
Step 2:Argon is monatomic with f = 3.
Step 3:Add the two contributions.
Final answer:
Q3Single correctNuclei
Radioactive material 'A' has decay constant '' and material 'B' has decay constant ''. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'A' to that of 'B' will be ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Apply the radioactive decay law to both materials and form the ratio.
Step 1:Write the number of nuclei remaining for each material with equal initial nuclei.
Step 2:Form the ratio A to B. Because A decays with the larger constant it depletes faster, so this ratio falls below one as time passes.
Step 3:Set the ratio equal to 1/e and match the exponents.
Final answer:
Q4Single correctMechanical Properties of Fluids
A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on that side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Equate pressures at the bottom of the U-tube from the oil column and the water column.
Step 1:Water rises 65 mm on the water side, so the water level difference between the two arms is 2 × 65 = 130 mm.
Step 2:Oil stands 10 mm above the water level on its side, and the depressed water on that side is 65 mm below original, giving an oil column of 130 + 10 = 140 mm.
Step 3:Equate pressures of the two liquid columns measured to the common interface level.
Step 4:Solve for the oil density with water density 1000.
Final answer:
Q6Single correctDual Nature of Radiation and Matter
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Relate thermal kinetic energy to momentum, then apply the de Broglie relation.
Step 1:Equate average thermal kinetic energy to translational kinetic energy.
Step 2:Take square root to find momentum.
Step 3:Substitute into the de Broglie relation.
Final answer:
Q7Single correctSystems of Particles and Rotational Motion
One end of string of length is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v' the net force on the particle (directed towards centre) will be (T represents the tension in the string) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Identify the physical force that provides the centripetal force for uniform circular motion.
Step 1:On a smooth horizontal table, the only horizontal force directed toward the centre is the string tension.
Step 2:The net centripetal force equals this single physical force; it numerically equals mv²/l, but the net force IS the tension.
Final answer:
Q8Single correctCurrent Electricity
Figure shows a circuit that contains three identical resistors with resistance each, two identical inductors with inductance mH each, and an ideal battery with emf V. The current 'I' through the battery just after the switch closed is......
(1)
(2)
(3)
(4)
SolutionAnswer: Option
Approach:
Apply inductor behaviour at the instant a switch is closed, then reduce the resistive network.
Step 1:Just after closing, inductors oppose sudden current change and behave as open branches, blocking the two inductor arms.
Step 2:With inductor branches open, the battery drives current through the single available resistor of 9 ohm.
Step 3:NTA declared this question a bonus; full marks were awarded to all candidates.
Q9Single correctMotion in a Plane
The x and y coordinates of the particle at any time are and respectively, where x and y are in metres and t in seconds. The acceleration of the particle at s is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Differentiate each coordinate twice to obtain acceleration components.
Step 1:Differentiate x twice.
Step 2:Differentiate y twice.
Step 3:The net acceleration is constant and along x.
(along -x)
Final answer:
Q10Single correctGravitation
Suppose the charge of a proton and an electron differ slightly. One of them is , the other is . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then is of the order of [Given mass of hydrogen kg]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Set the net charge per atom as Δe, equate the electrostatic repulsion to the gravitational attraction.
Step 1:Each atom carries net charge Δe; balance the two inverse-square forces so d cancels.
Step 2:Solve for Δe.
Step 3:Substitute numerical values.
Final answer:
Q11Single correctThermal Properties of Matter
Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are and . The thermal conductivity of the composite rod will be :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Identify the geometry as parallel slabs of equal area sharing the same temperature gradient.
Step 1:The two rods are welded along their length so they conduct in parallel between the same end temperatures.
Step 2:With equal cross-sectional areas, the effective conductivity is the average.
Final answer:
Q12Single correctElectrostatic Potential and Capacitance
The diagrams below show regions of equipotentials.
A positive charge is moved from A to B in each diagram.
A positive charge is moved from A to B in each diagram.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2In all the four cases the work done is the same.
Approach:
Use the fact that work to move a charge depends only on the potential difference between endpoints.
Step 1:Points A and B lie on the same labelled equipotentials in every diagram, so the potential difference is identical.
Step 2:Work depends only on this potential difference, independent of path or field pattern.
Final answer: In all the four cases the work done is the same.
Q13Single correctAtoms
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The last line of a series corresponds to the series limit (transition from infinity).
Step 1:Last line of Balmer has n1=2, n2=infinity.
Step 2:Last line of Lyman has n1=1, n2=infinity.
Step 3:Form the ratio.
Final answer:
Q14Single correctWave Optics
Young's double slit experiment is first performed in air and then in a medium other than air. It is found that bright fringe in the medium lies where dark fringe lies in air. The refractive index of the medium is nearly :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Equate the position of the 8th bright fringe in the medium with the 5th dark fringe in air.
Step 1:8th bright in medium uses wavelength λ/μ.
Step 2:5th dark in air corresponds to (2×5−1)/2 = 9/2.
Step 3:Equate the two positions and solve for μ.
Final answer:
Q15Single correctOscillations
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Equate the magnitudes of SHM velocity and acceleration at the given displacement to find angular frequency.
Step 1:Set velocity magnitude equal to acceleration magnitude at x = 2 cm.
Step 2:Substitute A = 3, x = 2.
Step 3:Compute the time period.
Final answer:
Q16Single correctThermodynamics
Thermodynamic processes are indicated in the following diagram.
Match the following
Match the following
| Column-1 | Column-2 |
|---|---|
| P.. Process I | a.. Adiabatic |
| Q.. Process II | b.. Isobaric |
| R.. Process III | c.. Isochoric |
| S.. Process IV | d.. Isothermal |

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2P c, Q a, R d, S b
Approach:
Identify each curve on the P-V diagram by its characteristic shape and label.
Step 1:A vertical line at constant volume is isochoric, matching Process I.
Step 2:The steepest curve is adiabatic, matching Process II.
Step 3:The less steep hyperbola is isothermal, matching Process III; the horizontal line at constant pressure is isobaric, matching Process IV.
Final answer: P c, Q a, R d, S b
Q17Single correctElectrostatic Potential and Capacitance
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2decreases by a factor of 2
Approach:
Use charge conservation since the battery is removed, then compare energies before and after.
Step 1:Initial energy with charge Q on capacitance C.
Step 2:After connecting an identical capacitor, charge Q is conserved but capacitance doubles.
Step 3:Compare final to initial energy.
Final answer: decreases by a factor of 2
Q18Single correctDual Nature of Radiation and Matter
The photoelectric threshold wavelength of silver is m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength m is :
(Given eVs and m)
(Given eVs and m)
(1)
(2)
(3)
(4)
SolutionAnswer: Option
Approach:
Apply Einstein's photoelectric equation to find the kinetic energy and hence the electron speed.
Step 1:Compute the maximum kinetic energy from the difference of photon energies.
Step 2:Convert to joules and solve for velocity.
Step 3:Options 1 and 2 are numerically equal (); NTA declared this item a bonus.
Q19Single correctPhysical World and Measurement
A physical quantity of the dimensions of length that can be formed out of c, G and is [c is velocity of light, G is universal constant of gravitation and e is charge] :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Assign dimensions to each constant and combine them to obtain length.
Step 1:Dimensions: c is L , G is , and /(4πε0) is M .
Step 2:The product G × /(4πε0) has dimensions ; its square root is .
Step 3:Dividing by ( ) yields pure length.
Final answer:
Q20Single correctWaves
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s] :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Apply the Doppler effect with both source and observer moving toward each other.
Step 1:Observer (second car) moves toward source at 16.5 m/s; source (first car) moves toward observer at 22 m/s.
Step 2:Substitute into the Doppler relation.
Step 3:Evaluate.
Final answer:
Q21Single correctSemiconductor Electronics
In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is 3 k. If current gain is 100 and the base resistance is 2 k, the voltage and power gain of the amplifier is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and
Approach:
Compute voltage gain from current gain and resistance ratio, then power gain as the product.
Step 1:Voltage gain equals current gain times resistance ratio.
Step 2:Power gain equals current gain times voltage gain.
Final answer: and
Q22Single correctSemiconductor Electronics
Which one of the following represents forward bias diode ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The branch with 0 V on the diode's anode side and 2 V beyond the resistor (drawn circuit, option 1)
Approach:
Forward bias requires the p-side (anode) at a higher potential than the n-side (cathode).
Step 1:Each option shows the anode (left) connected to one potential and the cathode (right, through R) to another.
Step 2:For 0 V on the anode side and -2 V on the cathode side, the anode is higher, satisfying forward bias.
Final answer: The branch with 0 V on the diode's anode side and 2 V beyond the resistor (drawn circuit, option 1)
Q23Single correctMechanical Properties of Solids
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k". Then k' : k" is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the spring constants of each cut piece, then combine in series and in parallel.
Step 1:Lengths in ratio 1:2:3 give spring constants 6k, 3k, 2k (inverse of length fractions 1/6, 2/6, 3/6).
Step 2:Series combination of these pieces returns the original spring constant k.
Step 3:Parallel combination sums the constants.
Step 4:Form the ratio.
Final answer:
Q24Single correctSemiconductor Electronics
The given electrical network is equivalent to:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3NOR gate
Approach:
Trace the Boolean expression of the network gate by gate, then reduce it to a single equivalent gate.
Step 1:Inputs A and B feed a NOR gate.
Step 2:The second NOR gate has both of its inputs tied to , so it acts as an inverter.
Step 3:The final NOT gate inverts once more, returning the NOR of A and B.
Final answer: NOR gate
Q25Single correctGravitation
The acceleration due to gravity at a height above the earth is the same as at a depth d below the surface of earth. Then:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Equate the height and depth variations of g to first order in and .
Step 1:For small height, gravity decreases as .
Step 2:At depth, gravity decreases as .
Step 3:Setting gives , hence . Substituting .
Final answer:
Q26Single correctSystem of Particles and Rotational Motion
Which of the following statements are correct?
(a) Centre of mass of a body always coincides with the centre of gravity of the body
(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero
(c) A couple on a body produce both translational and rotational motion in a body
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
(a) Centre of mass of a body always coincides with the centre of gravity of the body
(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero
(c) A couple on a body produce both translational and rotational motion in a body
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(b) and (d)
Approach:
Assess each statement against the definitions of centre of mass, centre of gravity, couple, and mechanical advantage.
Step 1:Statement (a) is false because centre of mass coincides with centre of gravity only in a uniform gravitational field, not always.
Step 2:Statement (b) is true: the centre of mass is the point about which the net gravitational torque vanishes in a uniform field.
Step 3:Statement (c) is false: a couple produces only pure rotation with no net translational force.
Step 4:Statement (d) is true: mechanical advantage above one means a small effort lifts a large load.
Final answer: (b) and (d)
Q27Single correctThermodynamics
A carnot engine having an efficiency of as heat engine, is used as a refrigerator. If the work done on the system is , the amount of energy absorbed from the reservoir at lower temperature is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Relate engine efficiency to the heat ratios, then apply energy conservation for the refrigerator cycle.
Step 1:Efficiency gives .
Step 2:Run as a refrigerator with work input ; energy balance gives .
Step 3:From , substituting yields , so and .
Final answer:
Q28Single correctMagnetism and Matter
If and are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip is given by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Resolve the horizontal component of earth's field in two perpendicular vertical planes and relate apparent dips to the true dip.
Step 1:In a plane making angle with the magnetic meridian, the apparent dip satisfies .
Step 2:In the perpendicular plane the angle is , giving .
Step 3:Squaring and adding eliminates using .
Final answer:
Q29Single correctMoving Charges and Magnetism
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current 'I' along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire 'B' is given by:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Compute the force per unit length on wire B from each of the two other wires placed at a right angle, then vector-add the two perpendicular contributions.
Step 1:Wires A and C are each at distance d from B along directions making a angle at B (as shown). Force per length from each wire is .
Step 2:The two forces are mutually perpendicular, so the resultant has magnitude times one of them.
Step 3:Simplifying gives the net force per length.
Final answer:
Q30Single correctLaws of Motion / Gravitation
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2move towards each other
Approach:
Identify the only force acting between two isolated masses in free space.
Step 1:In gravitational free space, the only interaction between the two astronauts is their mutual gravitational attraction.
Step 2:This attractive force pulls each toward the other, however small it may be.
Step 3:Therefore the two astronauts gradually move towards each other.
Final answer: move towards each other
Q31Single correctElectromagnetic Waves
In an electromagnetic wave in free space the root mean square value of the electric field is . The peak value of the magnetic field is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Convert rms electric field to rms magnetic field using , then convert to peak value.
Step 1:Find the rms magnetic field from .
Step 2:The peak magnetic field is times the rms value.
Step 3:Evaluating gives the peak field.
Final answer:
Q32Single correctMechanical Properties of Solids
The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', the fractional decrease in radius is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Relate volumetric strain to fractional radius change for a sphere, then apply the definition of bulk modulus.
Step 1:For a sphere, since volume scales as .
Step 2:Bulk modulus gives under uniform pressure.
Step 3:Equating the two expressions gives , so the fractional decrease in radius is .
Final answer:
Q33Single correctWave Optics
The ratio of resolving powers of an optical microscope for two wavelengths and is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Use the inverse proportionality of resolving power to wavelength for a microscope.
Step 1:Resolving power of a microscope is inversely proportional to wavelength.
Step 2:The ratio becomes .
Step 3:Simplifying gives the ratio .
Final answer:
Q34Single correctWork, Energy and Power
Consider a drop of rain water having mass falling from a height of . It hits the ground with a speed of . Take 'g' constant with a value . The work done by the (i) gravitational force and the (ii) resistive force of air is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(i) (ii)
Approach:
Compute gravitational work from mgh and resistive work from the work-energy theorem.
Step 1:Gravitational work is .
Step 2:Kinetic energy on impact is .
Step 3:By work-energy theorem .
Final answer: (i) (ii)
Q35Single correctThermal Properties of Matter
A spherical black body with a radius of radiates power at . If the radius were halved and the temperature doubled, the power radiated in watt would be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 41800
Approach:
Apply the Stefan-Boltzmann law with power proportional to area times the fourth power of temperature.
Step 1:Power scales as .
Step 2:Halving radius gives factor ; doubling temperature gives factor .
Step 3:Thus .
Final answer: 1800
Q36Single correctLaws of Motion
Two blocks A and B of masses and respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Find the spring tension in equilibrium, then apply Newton's second law to each block immediately after the string is cut.
Step 1:Before cutting, the spring supports the total weight, so , which does not change instantly when the string is cut.
Step 2:For block A (mass ): net upward force , giving upward.
Step 3:For block B (mass m): only gravity acts after the string is cut, so downward.
Final answer:
Q37Single correctWave Optics
Two Polaroids and are placed with their axis perpendicular to each other. Unpolarised light is incident on . A third polaroid is kept in between and such that its axis makes an angle with that of . The intensity of transmitted light through is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply Malus's law successively through the three polaroids, starting with unpolarised light losing half its intensity at the first polaroid.
Step 1:After , unpolarised light becomes .
Step 2: at to transmits .
Step 3: is at to (since ), so .
Final answer:
Q38Single correctElectromagnetic Induction
A long solenoid of diameter has turns per meter. At the centre of the solenoid, a coil of turns and radius is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to from in . If the resistance of the coil is , the total charge flowing through the coil during this time is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the flux change through the coil from the solenoid field, then use induced charge equals flux change over resistance.
Step 1:The solenoid field is , giving a change as current falls from 4 A to 0.
Step 2:The flux change through one coil turn over area multiplied by 100 turns gives the total linked flux change.
Step 3:Induced charge with . Evaluating, , , hence .
Final answer:
Q39Single correctSystem of Particles and Rotational Motion
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities and . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Use conservation of angular momentum to find the common angular velocity, then compute the difference between initial and final rotational kinetic energies.
Step 1:Common angular velocity from conservation of angular momentum is .
Step 2:Initial energy is and final energy is .
Step 3:Substituting and simplifying gives the loss .
Final answer:
Q40Single correctMotion in a Straight Line
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time . On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time . The time taken by her to walk up on the moving escalator will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Express walking speed and escalator speed as length divided by respective times, then add velocities for the combined motion.
Step 1:Let escalator length be L. Walking speed is and escalator speed is .
Step 2:When both act together, speeds add: .
Step 3:Solving for t gives .
Final answer:
Q41Single correctSystem of Particles and Rotational Motion
A rope is wound around a hollow cylinder of mass and radius . What is the angular acceleration of the cylinder if the rope is pulled with a force of ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Compute the torque from the applied force and the moment of inertia of a hollow cylinder, then apply the rotational form of Newton's second law.
Step 1:Torque about the axis is .
Step 2:Moment of inertia of a hollow cylinder is .
Step 3:Angular acceleration .
Final answer:
Q42Single correctRay Optics
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle , the spot of the light is found to move through a distance y on the scale. The angle is given by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Use the property that rotating a mirror by an angle rotates the reflected ray by twice that angle, then relate the arc displacement to the small angle.
Step 1:Rotating the mirror by deviates the reflected ray by .
Step 2:On the scale at distance x, the spot shifts by for small angles.
Step 3:Solving for gives .
Final answer:
Q43Single correctWaves
The two nearest harmonics of a tube closed at one end and open at other end are and . What is the fundamental frequency of the system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
A closed pipe supports only odd harmonics, so the spacing between successive allowed harmonics equals twice the fundamental frequency.
Step 1:A pipe closed at one end produces only odd harmonics, spaced by .
Step 2:The difference of the two nearest harmonics is .
Step 3:Therefore , giving .
Final answer:
Q44Single correctRay Optics
A thin prism having refracting angle is made of glass of refractive index . This prism is combined with another thin prism of glass of refractive index . This combination produces dispersion without deviation. The refracting angle of second prism should be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For dispersion without deviation the net deviation is zero; equate the deviations produced by the two thin prisms.
Step 1:Deviation of the first prism is .
Step 2:For zero net deviation, with .
Step 3:Solving gives .
Final answer:
Q45Single correctCurrent Electricity
The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Use that the volume of the wire stays constant on stretching, so the cross-section decreases as length increases, and apply the resistance formula.
Step 1:Stretching to times the length keeps volume constant, so the new area is .
Step 2:Resistance scales as , so .
Step 3:Therefore the new resistance is .
Final answer:
Chemistry45 questions
Q46Single correctHydrocarbons
With respect to the conformers of ethane, which of the following statements is true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both bond angles and bond length remains same
Approach:
Conformers arise only from rotation about the carbon-carbon single bond and differ in the relative spatial arrangement of the hydrogen atoms.
Step 1:Conformations of ethane interconvert by rotation about the C-C sigma bond.
Step 2:Since no bond is broken or formed, the bond lengths of all C-H and C-C bonds and all bond angles stay fixed; only the dihedral (torsion) angle changes.
Final answer: Both bond angles and bond length remains same
Q47Single correctThe p-Block Elements
Which of the following pairs of compounds is isoelectronic and isostructural?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Isoelectronic species have the same number of valence electrons; isostructural species have the same shape and hybridisation about the central atom.
Step 1:Both species must have the same number of bond pairs and lone pairs on the central atom.
Step 2:In xenon has 2 bond pairs and 3 lone pairs (5 electron domains, ), giving a linear shape.
Step 3:In iodine has 2 bond pairs and 3 lone pairs (5 electron domains, ), giving a linear shape with the same valence electron count as .
Final answer:
Q48Single correctThe d- and f-Block Elements
and both when dissolved in water containing ions the pair of species formed is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Both mercury(II) and iodine form complex iodide species in the presence of excess iodide ions.
Step 1:Mercury(II) chloride reacts with excess iodide to form the soluble tetraiodomercurate(II) complex.
Step 2:Molecular iodine combines with iodide to form the triiodide ion.
Final answer:
Q49Single correctChemistry in Everyday Life
Mixture of chloroxylenol and terpineol acts as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2antiseptic
Approach:
Identify the pharmacological class of a known commercial mixture.
Step 1:A mixture of chloroxylenol and terpineol is the active basis of Dettol.
Step 2:This mixture kills or prevents the growth of microorganisms on living tissue, which is the defining action of an antiseptic.
Final answer: antiseptic
Q50Single correctThe Solid State
Which is the incorrect statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal.
Approach:
Evaluate each statement about solid-state defects and electrical properties and identify the false one.
Step 1: shows metal deficiency due to non-stoichiometry, so statement (1) is correct.
Step 2:Schottky defect removes equal numbers of cations and anions, lowering density, so statement (2) is correct.
Step 3:NaCl is an insulator, silicon a semiconductor, silver a conductor and quartz piezoelectric, so statement (3) is correct.
Step 4:Frenkel defect occurs when there is a large difference between cation and anion sizes, not when the sizes are almost equal, so statement (4) is the incorrect statement.
Final answer: Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal.
Q51Single correctEquilibrium
Concentration of the ions in a saturated solution of is . Solubility product of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Relate the silver-ion concentration to the oxalate concentration through the dissolution stoichiometry, then compute the solubility product.
Step 1:Dissolution of silver oxalate gives two silver ions per oxalate ion.
Step 2:The oxalate concentration is half the silver-ion concentration.
Step 3:Substituting into the solubility-product expression.
Final answer:
Q52Single correctAldehydes, Ketones and Carboxylic Acids
Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The alpha,beta-unsaturated ketone two cyclohexane rings joined by a C=C with one C=O (drawn structure, option 2)
Approach:
Cyclohexanone undergoes self aldol addition through its alpha-carbon, and subsequent heating eliminates water to give an alpha,beta-unsaturated ketone.
Step 1:The alpha-carbon of one cyclohexanone attacks the carbonyl carbon of another, forming a beta-hydroxy ketone (aldol product) in which the two rings are joined.
Step 2:Heating causes dehydration across the alpha,beta-positions to give a conjugated alpha,beta-unsaturated ketone in which one ring bears the carbonyl and the other ring is linked by a double bond.
Final answer: The alpha,beta-unsaturated ketone two cyclohexane rings joined by a C=C with one C=O (drawn structure, option 2)
Q53Single correctChemical Bonding and Molecular Structure
The species, having bond angles of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Use VSEPR theory to find which species has trigonal planar geometry with bond angles of .
Step 1:Boron in has three bond pairs and no lone pair, giving hybridisation.
Step 2:Three electron domains arrange in a trigonal planar geometry with bond angles of , whereas and are pyramidal and is T-shaped.
Final answer:
Q54Single correctSolutions
If molality of the dilute solution is doubled, the value of molal depression constant will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4unchanged
Approach:
Examine whether the molal depression constant depends on solute concentration.
Step 1:The molal depression constant is a cryoscopic property determined solely by the solvent.
Step 2:Changing the concentration of the solute alters but does not change .
Final answer: unchanged
Q55Single correctAlcohols, Phenols and Ethers
Which one is the most acidic compound?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The phenol carrying three N groups 2,4,6-trinitrophenol, picric acid (drawn structure, option 4)
Approach:
Acidity of a phenol increases as electron-withdrawing groups on the ring stabilise the phenoxide ion.
Step 1:Electron-withdrawing nitro groups at ortho and para positions delocalise the negative charge of the phenoxide ion, increasing acidity, while a methyl substituent donates electrons and decreases acidity.
Step 2:The compound bearing three nitro groups (2,4,6-trinitrophenol, picric acid) provides the greatest stabilisation of the phenoxide and is therefore the most acidic.
Final answer: The phenol carrying three N groups 2,4,6-trinitrophenol, picric acid (drawn structure, option 4)
Q56Single correctThe p-Block Elements
It is because of inability of electrons of the valence shell to participate in bonding that:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 is reducing while is oxidising
Approach:
Apply the inert pair effect down group 14 to predict the stability of the +2 and +4 oxidation states for tin and lead.
Step 1:The inert pair effect makes the +2 state increasingly stable down the group, so for lead the +2 state is the stable one and readily gains electrons, acting as an oxidising agent.
Step 2:For tin the +4 state is more stable, so tends to lose electrons and is reducing.
Final answer: is reducing while is oxidising
Q57Single correctHydrocarbons
Predict the correct intermediate and product in the following reaction:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A is the enol C(OH)=C and B is acetone (drawn structures, option 4)
Approach:
Acid-catalysed hydration of an alkyne with mercuric sulfate follows Markovnikov addition to give an enol intermediate that tautomerises to a ketone.
Step 1:Water adds across the triple bond of propyne following Markovnikov's rule, placing the hydroxyl on the more substituted carbon to give the enol as intermediate A.
Step 2:The enol tautomerises to the more stable keto form, giving propan-2-one (acetone) as product B.
Final answer: A is the enol C(OH)=C and B is acetone (drawn structures, option 4)
Q58Single correctChemical Kinetics
Which one of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.
Approach:
Recall the role of a catalyst in equilibrium and kinetics and identify the false statement.
Step 1:A catalyst lowers the activation energy of both forward and backward reactions equally and so does not alter the position of equilibrium or the equilibrium constant.
Step 2:Therefore the statement that the equilibrium constant is changed by a catalyst is not correct.
Final answer: The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.
Q59Single correctStructure of Atom
Which one is the wrong statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The energy of orbital is less than the energy of orbital in case of Hydrogen like atoms.
Approach:
Evaluate each atomic-structure statement and find the false one, focusing on orbital energies in hydrogen-like species.
Step 1:In hydrogen-like atoms the orbital energy depends only on the principal quantum number , so the and orbitals are degenerate.
Step 2:The claim that the energy is less than the energy in hydrogen-like atoms is therefore the wrong statement.
Final answer: The energy of orbital is less than the energy of orbital in case of Hydrogen like atoms.
Q60Single correctThermodynamics
A gas is allowed to expand in a well insulated container against a constant external pressure of from an initial volume of to a final volume of . The change in internal energy of the gas in joules will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For an insulated (adiabatic) container the heat exchange is zero, so the change in internal energy equals the work done on the gas.
Step 1:The container is insulated, so the process is adiabatic and , giving .
Step 2:Work of expansion against constant external pressure: .
Step 3:Converting using : .
Final answer:
Q61Single correctAldehydes, Ketones and Carboxylic Acids
Consider the reactions given in the diagram and identify A, X, Y and Z:
![X, of formula C2H6O, is heated with copper at 573 K to give A. A treated with [Ag(NH3)2]+ and -OH under heat gives a silver mirror. A treated with -OH and heat gives Y. A treated with NH2-NH-CO-NH2 gives Z.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2Ffaff200f-937c-4543-a81c-a547772cc033%2Ffaff200f-937c-4543-a81c-a547772cc033%2Fimages%2FQ61_reaction_scheme.webp)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.
Approach:
Trace the sequence: oxidation of ethanol over copper, Tollens test for an aldehyde, and reaction of the carbonyl with semicarbazide.
Step 1:Ethanol passed over heated copper at 573 K is oxidised (dehydrogenated) to ethanal (acetaldehyde).
Step 2:Ethanal gives a positive Tollens (silver mirror) test, confirming the aldehyde, and on aldol condensation followed by dehydration yields but-2-enal (Y).
Step 3:Reaction of the carbonyl with semicarbazide forms a semicarbazone (Z).
Final answer: A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone.
Q62Single correctHydrocarbons
Which one is the correct order of acidity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Acidity of a C-H bond increases with the s-character of the hybrid orbital holding the electron pair on carbon.
Step 1:An sp carbon (alkyne) has the highest s-character and holds its lone pair closest to the nucleus, giving the most stable carbanion and the greatest acidity, followed by (alkene) and then (alkane).
Step 2:Among the alkynes, terminal ethyne with two terminal C-H bonds is more acidic than propyne, so the order is ethyne > propyne > ethene > ethane.
Final answer:
Q63Single correctElectrochemistry
In the electrochemical cell:
, the emf of this Daniell cell is . When the concentration of is changed to and that of changed to , the emf changes to . From the following, which one is the relationship between and ? (Given, )
, the emf of this Daniell cell is . When the concentration of is changed to and that of changed to , the emf changes to . From the following, which one is the relationship between and ? (Given, )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply the Nernst equation to the Daniell cell for both sets of ion concentrations and compare the resulting emfs.
Step 1:For the first case the ratio , so the log term is negative and raises the emf above .
Step 2:For the second case the ratio is , so the log term is positive and lowers the emf below .
Step 3:Comparing the two results gives .
Final answer:
Q64Single correctAmines
The correct increasing order of basic strength for the following compounds is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4II < I < III
Approach:
Basic strength of an aniline depends on the availability of the nitrogen lone pair, which electron-donating groups increase and electron-withdrawing groups decrease.
Step 1:The nitro group in compound II is strongly electron-withdrawing and reduces the electron density on nitrogen, making it the least basic.
Step 2:The methyl group in compound III is electron-donating and increases the electron density on nitrogen, making it more basic than aniline (I).
Step 3:Therefore the increasing order of basicity is II < I < III.
Final answer: II < I < III
Q65Single correctThe p-Block Elements
In which pair of ions both the species contain bond?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Examine the structures of the sulfur oxoanions to identify which contain a direct sulfur-sulfur bond.
Step 1:In thiosulfate, , one sulfur replaces a terminal oxygen of sulfate, giving a central S bonded to another S, so it contains an S-S bond.
Step 2:In tetrathionate, , the two central sulfur atoms are linked directly, giving an S-S-S-S chain that includes S-S bonds.
Step 3:Pyrosulfate has an S-O-S bridge and peroxodisulfate has an O-O bridge, so neither contains a direct S-S bond.
Final answer:
Q66Single correctCoordination Compounds
The correct order of the stoichiometries of formed when in excess is treated with the complexes: , , respectively is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 33 AgCl 2 AgCl 1 AgCl
Approach:
Only chloride ions that are ionisable (outside the coordination sphere) precipitate as AgCl, so count the free chloride ions in each complex.
Step 1:In all six ammonia molecules occupy the coordination sphere, leaving all three chlorides ionisable and giving 3 AgCl.
Step 2:In one chloride enters the coordination sphere, leaving two ionisable chlorides and giving 2 AgCl.
Step 3:In two chlorides are coordinated, leaving one ionisable chloride and giving 1 AgCl.
Final answer: 3 AgCl 2 AgCl 1 AgCl
Q67Single correctThe p-Block Elements
Match the interhalogen compounds of column I with the geometry in column II and assign the correct code.
| Column I | Column II |
|---|---|
| a. | i. T-shape |
| b. | ii. Pentagonal bipyramidal |
| c. | iii. Linear |
| d. | iv. Square-pyramidal |
| v. Tetrahedral |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Use VSEPR theory to assign the shape of each interhalogen type from the number of bond pairs and lone pairs on the central halogen.
Step 1: has one bond pair and a linear arrangement.
Step 2: has three bond pairs and two lone pairs, giving a T-shape.
Step 3: has five bond pairs and one lone pair, giving a square-pyramidal shape; has seven bond pairs and is pentagonal bipyramidal.
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q68Single correctThe d- and f-Block Elements
The reason for greater range of oxidation states in actinoids is attributed to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3, and levels having comparable energies
Approach:
Relate the variability of oxidation states in actinoids to the relative energies of the participating electronic levels.
Step 1:In actinoids the , and orbitals lie close together in energy.
Step 2:Because these levels have comparable energies, a larger number of electrons can participate in bonding, producing a greater range of oxidation states than in lanthanoids.
Final answer: , and levels having comparable energies
Q69Single correctEquilibrium
A 20 litre container at 400 K contains at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of attains its maximum value, will be:
(Given that : , )
(Given that : , )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The pressure of CO2 rises as the volume shrinks until it reaches the maximum value fixed by the equilibrium Kp; beyond this point further CO2 converts to SrCO3. Initial moles of CO2 are conserved.
Step 1:The maximum partial pressure CO2 can reach equals Kp of the decomposition equilibrium.
Step 2:Apply the isothermal relation between the initial state and the state of maximum pressure for the fixed amount of CO2 gas.
Step 3:Solving for the volume at which CO2 attains its maximum pressure.
Final answer:
Q70Single correctOrganic Chemistry - Some Basic Principles and Techniques
The statement regarding electrophile is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Approach:
Classify electrophiles by charge and bonding behaviour, then select the option consistent with their definition.
Step 1:An electrophile is an electron-deficient species; it may be a cation or a neutral molecule with an electron-deficient centre.
Step 2:Being electron-deficient, an electrophile accepts a lone pair from an electron-rich nucleophile to form a covalent bond.
Step 3:An electrophile may be a neutral molecule or a positively charged species, and it bonds to a centre that supplies the electron pair.
Final answer: Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Q71Single correctp-Block Elements
Which of the following is a sink for CO?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Micro organisms present in the soil
Approach:
Identify the natural agent that removes carbon monoxide from the atmosphere.
Step 1:A sink for a pollutant is the system that removes it from the environment.
Step 2:Certain soil micro-organisms metabolise atmospheric CO, lowering its concentration.
Step 3:Haemoglobin binds CO as a poison rather than removing it from the environment, so it is not the sink.
Final answer: Micro organisms present in the soil
Q72Single correctClassification of Elements and Periodicity in Properties
The element Z = 114 has been discovered recently. It will belong to which of the following family/ group and electronic configuration?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Carbon family,
Approach:
Build the electronic configuration for Z = 114 starting from the [Rn] core and assign the group from the outermost electrons.
Step 1:Filling beyond the radon core: 5f and 6d are completed, then 7s and 7p are filled.
Step 2:The outermost shell holds four electrons in the ns2 np2 arrangement, characteristic of group 14.
Step 3:Placement in group 14 together with the configuration ending in identifies the element as a member of the carbon family.
Final answer: Carbon family,
Q73Single correctCoordination Compounds
Correct increasing order for the wavelengths of absorption in the visible region for the complexes of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Rank the ligands by crystal field strength; stronger fields give larger splitting energy, hence shorter absorbed wavelength. Wavelength order is the reverse of field strength.
Step 1:From the spectrochemical series the field strength of these ligands decreases as en > NH3 > H2O.
Step 2:Larger field splitting corresponds to higher absorbed energy and shorter wavelength, so wavelength increases in the reverse order of field strength.
Step 3:Increasing wavelength therefore runs en complex, then ammine, then aqua complex.
Final answer:
Q74Single correctBiomolecules
Which of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Denaturation makes the proteins more active.
Approach:
Evaluate each biochemical statement and identify the incorrect one.
Step 1:Insulin regulates blood glucose, ovalbumin is the egg-white storage protein, and thrombin with fibrinogen drive clotting; these statements are correct.
Step 2:Denaturation disrupts secondary and tertiary structure, causing loss of biological activity rather than enhancement.
Step 3:The statement claiming denaturation increases activity is therefore the incorrect one.
Final answer: Denaturation makes the proteins more active.
Q75Single correctCoordination Compounds
An example of a sigma bonded organometallic compound is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Grignard's reagent
Approach:
Distinguish sigma-bonded organometallics from pi-bonded sandwich compounds.
Step 1:Grignard reagent RMgX contains a direct metal-carbon sigma bond between magnesium and the alkyl group.
Step 2:Ferrocene, ruthenocene and cobaltocene are metallocenes with pi bonding between the metal and cyclopentadienyl rings.
Step 3:Only Grignard reagent fits the sigma-bonded category.
Final answer: Grignard's reagent
Q76Single correctSolutions
Which of the following is dependent on temperature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Molarity
Approach:
Determine which concentration term involves volume, since volume changes with temperature.
Step 1:Molality, mole fraction and weight percentage are defined using masses, which do not vary with temperature.
Step 2:Molarity depends on the solution volume, which expands or contracts with temperature.
Step 3:Hence molarity is the temperature-dependent quantity.
Final answer: Molarity
Q77Single correctThermodynamics
For a given reaction, and . The reaction is spontaneous at: (Assume that and do not vary with temperature)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Spontaneity requires the Gibbs free energy change to be negative; for endothermic, entropy-increasing reactions this sets a temperature threshold.
Step 1:The reaction is spontaneous when the free energy change is negative.
Step 2:Convert enthalpy to joules and substitute the values.
Step 3:Rounding gives the threshold temperature, above which the reaction is spontaneous.
Final answer:
Q78Single correctOrganic Chemistry - Some Basic Principles and Techniques
The most suitable method of separation of mixture of ortho and para - nitrophenols is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Steam distillation
Approach:
Use the difference in volatility arising from intramolecular versus intermolecular hydrogen bonding in the two isomers.
Step 1:ortho-Nitrophenol forms intramolecular hydrogen bonds and is volatile in steam.
Step 2:para-Nitrophenol forms intermolecular hydrogen bonds, is less volatile and stays behind.
Step 3:This volatility difference makes steam distillation the suitable separation method.
Final answer: Steam distillation
Q79Single correctChemical Bonding and Molecular Structure
Which one of the following pairs of species have the same bond order?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Count valence electrons for each species and compute bond order using molecular orbital theory; isoelectronic species share the same bond order.
Step 1:CO and CN- each have 14 electrons, making them isoelectronic.
Step 2:A 14-electron diatomic of this type carries a bond order of three.
Step 3:The other pairs differ in electron count and bond order, so the matching pair is CN- and CO.
Final answer:
Q80Single correctHaloarenes and Amines
Identify A and predict the type of reaction.

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The ring with OC and N meta to each other, and substitution reaction (drawn structure, option 1)
Approach:
Sodamide on an aryl halide proceeds through a benzyne (elimination-addition) pathway; track the position of the OCH3 directing group to place the incoming nucleophile.
Step 1:NaNH2 removes a proton adjacent to the C-Br bond and eliminates bromide to generate a benzyne intermediate.
Step 2:The amide ion adds across the benzyne; addition at the carbon meta to the methoxy group leaves the carbanion on the carbon next to OCH3, where the methoxy group's inductive withdrawal stabilises it best, so the amino group ends up meta to OCH3.
Step 3:The amino group occupies the same ring position the bromine had, so the net change is an ordinary substitution and not a cine substitution.
Final answer: The ring with OC and N meta to each other, and substitution reaction (drawn structure, option 1)
Q81Single correctChemical Kinetics
A first order reaction has a specific reaction rate of . How much time will it take for 20 g of the reactant to reduce to 5 g?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Apply the integrated first order rate law with the initial and final amounts to solve for time.
Step 1:The amount falls from 20 g to 5 g, a factor of four, equivalent to two half-lives.
Step 2:Substitute the ratio and rate constant into the integrated law.
Step 3:Evaluate to obtain the elapsed time.
Final answer:
Q82Single correctRedox Reactions and p-Block Elements
Name the gas that can readily decolourise acidified solution:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Identify the reducing gas that reduces the purple permanganate ion, decolourising the solution.
Step 1:Acidified KMnO4 is decolourised by a reducing agent that converts Mn(VII) to colourless Mn(II).
Step 2:Sulphur dioxide is oxidised to sulphate while reducing permanganate.
Step 3:CO2, NO2 and P2O5 lack this reducing action, so SO2 is the answer.
Final answer:
Q83Single correctAlcohols, Phenols and Ethers
The heating of phenyl-methyl ethers with HI produces.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3phenol
Approach:
Apply the rule for cleavage of aryl alkyl ethers by hydrogen iodide, where the alkyl-oxygen bond breaks.
Step 1:In anisole (phenyl methyl ether) HI attacks the weaker alkyl-oxygen bond rather than the aryl-oxygen bond.
Step 2:Cleavage gives methyl iodide and the aryl oxygen retains the ring as phenol.
Step 3:The aromatic product is phenol.
Final answer: phenol
Q84Single correctCoordination Compounds
Pick out the correct statement with respect to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3It is hybridised and octahedral
Approach:
Determine the oxidation state and d-electron count of manganese, then use the strong field cyanide ligand to assign hybridisation and geometry.
Step 1:Six cyanide ligands with overall charge 3- place manganese in the +3 state, giving a d4 configuration.
Step 2:Cyanide is a strong field ligand, so inner d orbitals are used, giving d2sp3 hybridisation.
Step 3:Six-coordinate d2sp3 hybridisation corresponds to an octahedral geometry.
Final answer: It is hybridised and octahedral
Q85Single correctThe s-Block Elements
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Li
Approach:
Relate ionic mobility in solution to the size of the moving species, which is the hydrated ion rather than the bare ion.
Step 1:In aqueous solution an alkali metal ion drags a shell of water molecules with it, so the species that migrates is the hydrated ion. Mobility falls as that hydrated radius grows.
Step 2:Charge density decides how tightly water is held. Going down the group the bare ion grows, so its charge density falls and it holds fewer water molecules: Li+ is the smallest bare ion and therefore the most heavily hydrated, while Rb+ is the least.
(hydrated radius)
Step 3:The largest hydrated ion moves most sluggishly through the solution, so lithium has the lowest ionic mobility of the four.
Final answer: Li
Q86Single correctEquilibrium
The equilibrium constants of the following are:
The equilibrium constant (K) of the reaction:
, will be:
The equilibrium constant (K) of the reaction:
, will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Express the target reaction as a combination of the three given equilibria; reversing a reaction inverts its constant and multiplying a reaction raises its constant to the corresponding power.
Step 1:Reverse the ammonia synthesis to consume NH3, which inverts K1.
Step 2:Add the NO formation step and three times the water formation step to supply NO and H2O.
Step 3:Multiplying the contributions gives the overall constant.
Final answer:
Q87Single correctAmines
Which of the following reactions is appropriate for converting acetamide to methanamine?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hoffmann hypobromamide reaction
Approach:
Select the named reaction that converts a primary amide into an amine with one fewer carbon.
Step 1:Converting acetamide (two carbons) to methanamine (one carbon) requires loss of one carbon as the amide is degraded.
Step 2:The Hoffmann bromamide degradation treats an amide with bromine and alkali to give a primary amine with one fewer carbon.
Step 3:This matches the required transformation.
Final answer: Hoffmann hypobromamide reaction
Q88Single correctChemical Kinetics
Mechanism of a hypothetical reaction is given below
(i) (fast)
(ii) (slow)
(iii) (fast)
The overall order of the reaction will be
(i) (fast)
(ii) (slow)
(iii) (fast)
The overall order of the reaction will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Write the rate law from the slow step, then substitute the concentration of the intermediate X obtained from the fast pre-equilibrium.
Step 1:The slow step fixes the rate as first order in X and first order in Y2.
Step 2:The fast dissociation of X2 gives X in terms of the square root of the X2 concentration.
Step 3:Substituting yields an overall order of one-half plus one.
Final answer:
Q89Single correctAldehydes, Ketones and Carboxylic Acids
The IUPAC name of the compound shown is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 13-keto-2-methylhex-4-enal
Approach:
Identify the principal characteristic group, number the chain to give the aldehyde the lowest locant, and place the keto, methyl and double bond positions.
Step 1:The aldehyde group is the senior characteristic group and takes carbon one, giving the suffix -al.
Step 2:Numbering along the six-carbon chain places the methyl at carbon two, the keto group at carbon three, and the double bond starting at carbon four.
Step 3:Assembling the locants gives the IUPAC name.
Final answer: 3-keto-2-methylhex-4-enal
Q90Single correctGeneral Principles and Processes of Isolation of Elements
Extraction of gold and silver involves leaching with ion. Silver is later recovered by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4displacement with Zn
Approach:
Recall the cyanide leaching process and the reducing metal used to recover silver from its cyanide complex.
Step 1:Cyanide leaching dissolves silver as a soluble dicyanoargentate complex.
Step 2:Zinc, being more reactive, displaces silver from this complex.
Step 3:The recovery method is displacement with zinc.
Final answer: displacement with Zn
Biology90 questions
Q91Single correctSexual Reproduction in Flowering Plants
Double fertilization is exhibited by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Angiosperms
Approach:
Double fertilization is the hallmark of angiosperms, where one male gamete fuses with the egg (syngamy) and the second fuses with the two polar nuclei (triple fusion).
Step 1:Two male gametes are released from each pollen tube into the embryo sac.
Step 2:One gamete fuses with the egg to form the zygote; the other fuses with the central cell to form the triploid primary endosperm nucleus.
Step 3:This dual fusion, termed double fertilization, occurs only in flowering plants.
Final answer: Angiosperms
Q92Single correctBiological Classification
Which of the following are found in extreme saline conditions ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Archaebacteria
Approach:
Archaebacteria are extremophiles that thrive in harsh habitats, including the halophiles found in highly saline environments.
Step 1:Archaebacteria are grouped into halophiles, thermoacidophiles and methanogens based on habitat.
Step 2:Their cell wall lacks peptidoglycan and contains branched-chain lipids, allowing survival in extreme salt concentrations.
Step 3:Eubacteria, cyanobacteria and mycobacteria are not characteristic of extreme saline habitats.
Final answer: Archaebacteria
Q93Single correctMineral Nutrition / Biological Classification
Select the mismatch :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Rhodospirillum — Mycorrhiza
Approach:
The task is to find the incorrectly paired entry. Rhodospirillum is a free-living nitrogen-fixing bacterium and has no association with mycorrhiza.
Step 1:Frankia forms nitrogen-fixing root nodules in non-leguminous Alnus.
Step 2:Anabaena is a nitrogen fixer, and Rhizobium nodulates the legume Alfalfa.
Step 3:Rhodospirillum is a free-living anaerobic nitrogen fixer; mycorrhiza is a fungus-root association, so the pair Rhodospirillum—Mycorrhiza is mismatched.
Final answer: Rhodospirillum — Mycorrhiza
Q94Single correctBiotechnology - Principles and Processes
What is the criterion for DNA fragments movement on agarose gel during gel electrophoresis ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The smaller the fragment size, the farther it moves
Approach:
In agarose gel electrophoresis DNA fragments separate by size as they migrate toward the anode through the gel sieve.
Step 1:DNA is negatively charged and therefore migrates toward the positive electrode under the electric field.
Step 2:The agarose matrix retards larger fragments more than smaller ones.
Step 3:Smaller fragments encounter less resistance and travel a greater distance from the well.
Final answer: The smaller the fragment size, the farther it moves
Q95Single correctSexual Reproduction in Flowering Plants
Attractants and rewards are required for :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Entomophily
Approach:
Attractants such as colour, fragrance and rewards like nectar and pollen are floral adaptations to lure animal pollinators, characteristic of entomophily.
Step 1:Insect-pollinated flowers provide nectar and pollen as rewards and display bright colours and scent as attractants.
Step 2:Anemophily (wind) and hydrophily (water) rely on abiotic agents and need no floral rewards.
Step 3:Cleistogamy is self-pollination within closed flowers, requiring no pollinator.
Final answer: Entomophily
Q96Single correctAnatomy of Flowering Plants
Which of the following is made up of dead cells ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Phellem
Approach:
Phellem (cork) cells are dead at maturity and suberised, whereas the other tissues retain living protoplasts.
Step 1:Phellem is produced outward by the cork cambium and its cells lose protoplasm and become suberised.
Step 2:Xylem parenchyma and collenchyma are living tissues.
Step 3:Phloem as a tissue contains living sieve elements and companion cells.
Final answer: Phellem
Q97Single correctDigestion and Absorption
Which cells of 'Crypts of Lieberkuhn' secrete antibacterial lysozyme ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Paneth cells
Approach:
Paneth cells, located at the base of the crypts of Lieberkuhn, secrete antibacterial agents including lysozyme.
Step 1:Crypts of Lieberkuhn are intestinal glands lined by several specialised cell types.
Step 2:Paneth cells at the crypt base release lysozyme and defensins, contributing to innate gut immunity.
Step 3:Argentaffin cells secrete hormones, zymogen cells secrete enzymes, and Kupffer cells are hepatic macrophages.
Final answer: Paneth cells
Q98Single correctBody Fluids and Circulation
Adult human RBCs are enucleate. Which of the following statement(s) is/are most appropriate explanation for this feature ?
(a) They do not need to reproduce
(b) They are somatic cells
(c) They do not metabolize
(d) All their internal space is available for oxygen transport
(a) They do not need to reproduce
(b) They are somatic cells
(c) They do not metabolize
(d) All their internal space is available for oxygen transport
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a), (c) and (d)
Approach:
Loss of the nucleus in mature RBCs is explained by statements that follow from being terminally differentiated oxygen carriers.
Step 1:Mature RBCs do not divide, so a nucleus for reproduction is unnecessary (a).
Step 2:Without a nucleus and mitochondria a mature RBC cannot respire aerobically and so consumes none of the oxygen it carries (c), and the space freed by losing those organelles is given over to haemoglobin, maximising oxygen carriage (d).
Step 3:Being somatic (b) is true of many cells and does not specifically explain enucleation, so it is excluded.
Final answer: (a), (c) and (d)
Q99Single correctBody Fluids and Circulation
The hepatic portal vein drains blood to liver from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Intestine
Approach:
The hepatic portal vein is a special venous channel carrying nutrient-rich blood from the digestive tract to the liver.
Step 1:Blood absorbed from the intestinal capillaries collects into the hepatic portal vein.
Step 2:This vein delivers the absorbed nutrients to the liver before the blood returns to general circulation.
Step 3:The intestinal capillary bed is the source of that blood, so the hepatic portal vein carries blood from the intestine to the liver.
Final answer: Intestine
Q100Single correctMolecular Basis of Inheritance
The final proof for DNA as the genetic material came from the experiments of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hershey and Chase
Approach:
The unambiguous proof that DNA is the genetic material came from the bacteriophage experiments of Hershey and Chase using radioactive labelling.
Step 1:Hershey and Chase labelled phage DNA with radioactive phosphorus and protein with radioactive sulphur.
Step 2:Only the radioactive DNA entered the bacterial cells and directed production of new phages.
Step 3:Griffith showed transformation and Avery's group identified the transforming principle, but the conclusive proof is credited to Hershey and Chase.
Final answer: Hershey and Chase
Q101Single correctBiological Classification
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Mycoplasma
Approach:
The described organism lacking a cell wall and being the smallest living cell that survives anaerobically matches Mycoplasma.
Step 1:Mycoplasmas are the smallest known living cells and completely lack a cell wall.
Step 2:They are pathogenic to both plants and animals and can survive without oxygen.
Step 3:Bacillus, Pseudomonas and Nostoc all possess cell walls, so they are excluded.
Final answer: Mycoplasma
Q102Single correctCell Cycle and Cell Division
Which of the following options gives the correct sequence of events during mitosis ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2condensation nuclear membrane disassembly arrangement at equator centromere division segregation telophase
Approach:
The proper order of mitotic events runs through prophase, prometaphase, metaphase, anaphase and telophase, without any crossing over.
Step 1:Chromatin condenses in prophase, then the nuclear membrane disassembles in prometaphase.
Step 2:Chromosomes arrange at the equator in metaphase; centromeres split and chromatids segregate in anaphase.
Step 3:Telophase reforms the nuclei. Crossing over does not occur in mitosis, and the nuclear membrane must disassemble before the chromosomes can align at the equator, so a sequence that omits that step is incomplete.
Final answer: condensation nuclear membrane disassembly arrangement at equator centromere division segregation telophase
Q103Single correctBiomolecules
Which one of the following statements is correct with reference to enzymes ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Holoenzyme = Apoenzyme + Coenzyme
Approach:
The complete catalytically active enzyme is the holoenzyme, formed by the protein apoenzyme together with its coenzyme.
Step 1:The apoenzyme is the protein portion that is inactive on its own.
Step 2:A coenzyme is a non-protein organic cofactor that binds the apoenzyme.
Step 3:The combination gives the active holoenzyme, so Holoenzyme = Apoenzyme + Coenzyme.
Final answer: Holoenzyme = Apoenzyme + Coenzyme
Q104Single correctMolecular Basis of Inheritance
During DNA replication, Okazaki fragments are used to elongate :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The lagging strand away from the replication fork.
Approach:
Okazaki fragments are short stretches synthesised discontinuously on the lagging strand in the direction away from the replication fork.
Step 1:DNA polymerase extends only in the 5' to 3' direction, so the lagging strand is built in pieces.
Step 2:Each Okazaki fragment is synthesised moving away from the advancing replication fork.
Step 3:The leading strand is made continuously toward the fork, so it does not use Okazaki fragments.
Final answer: The lagging strand away from the replication fork.
Q105Single correctBiomolecules
Which of the following are not polymeric ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Lipids
Approach:
Nucleic acids, proteins and polysaccharides are true polymers of repeating monomers, whereas lipids are not polymeric.
Step 1:Nucleic acids are polymers of nucleotides and proteins are polymers of amino acids.
Step 2:Polysaccharides are polymers of monosaccharide units.
Step 3:Lipids are assemblies of fatty acids and glycerol that are not formed by repeated polymerisation, so they are not polymeric.
Final answer: Lipids
Q106Single correctEnvironmental Issues / Biodiversity and Conservation
The region of Biosphere Reserve which is legally protected and where no human activity is allowed is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Core zone
Approach:
The innermost legally protected part of a biosphere reserve where no human activity is permitted is the core zone.
Step 1:A biosphere reserve is organised into core, buffer and transition zones.
Step 2:The core zone is undisturbed and legally protected, prohibiting human activity.
Step 3:Buffer and transition zones permit regulated and cooperative human use, so they are excluded.
Final answer: Core zone
Q107Single correctSexual Reproduction in Flowering Plants
A dioecious flowering plant prevents both :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Autogamy and geitonogamy
Approach:
In dioecious plants the sexes are on separate individuals, so transfer of pollen within a flower or between flowers of one plant is impossible.
Step 1:Autogamy is self-pollination within the same flower, prevented because each plant bears only one sex.
Step 2:Geitonogamy is pollination between flowers of the same plant, also impossible since a plant is unisexual.
Step 3:Xenogamy (cross-pollination between different plants) is still required and therefore allowed.
Final answer: Autogamy and geitonogamy
Q108Single correctChemical Coordination and Integration
A temporary endocrine gland in the human body is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Corpus luteum
Approach:
The corpus luteum forms transiently from the ruptured ovarian follicle and secretes progesterone, making it a temporary endocrine gland.
Step 1:After ovulation the remaining follicular cells develop into the corpus luteum.
Step 2:It secretes progesterone to maintain the uterine lining and degenerates if pregnancy does not occur.
Step 3:The pineal gland is permanent, while corpus cardiacum and corpus allatum are insect endocrine structures.
Final answer: Corpus luteum
Q109Single correctHuman Health and Disease
Match the following sexually transmitted diseases (Column-I) with their causative agents (Column-II) and select the correct option.
| Column-I | Column-II |
|---|---|
| a. Gonorrhea | i. HIV |
| b. Syphilis | ii. Neisseria |
| c. Genital Warts | iii. Treponema |
| d. AIDS | iv. Human Papilloma - Virus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each sexually transmitted disease is matched to its specific pathogen.
Step 1:Gonorrhea is caused by Neisseria (ii) and syphilis by Treponema (iii).
Step 2:Genital warts are caused by Human Papilloma Virus (iv) and AIDS by HIV (i).
Step 3:The completed matching is (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q110Single correctHuman Health and Disease
Transplantation of tissues/organs fails often due to non-acceptance by the patient's body. Which type of immune-response is responsible for such rejections ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cell-mediated immune response
Approach:
Graft rejection is driven primarily by T-lymphocytes, that is, the cell-mediated arm of the immune system.
Step 1:Transplanted tissue carries foreign histocompatibility antigens recognised as non-self.
Step 2:T-cells become activated against the graft and destroy it.
Step 3:This T-cell driven reaction is the cell-mediated immune response.
Final answer: Cell-mediated immune response
Q111Single correctMolecular Basis of Inheritance
Spliceosomes are not found in cells of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bacteria
Approach:
Spliceosomes carry out removal of introns from precursor mRNA, a process restricted to eukaryotes; bacteria lack them.
Step 1:Spliceosomes are ribonucleoprotein complexes that splice introns out of eukaryotic pre-mRNA.
Step 2:Plants, fungi and animals are eukaryotes and possess spliceosomes.
Step 3:Bacteria are prokaryotes whose genes typically lack introns, so they have no spliceosomes.
Final answer: Bacteria
Q112Single correctBiological Classification
An example of colonial alga is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Volvox
Approach:
Volvox is a classic colonial green alga in which many flagellated cells form a hollow spherical colony.
Step 1:Volvox cells are arranged in a coordinated spherical colony embedded in a gelatinous matrix.
Step 2:Chlorella is unicellular, while Ulothrix and Spirogyra are filamentous.
Step 3:Volvox alone forms a colony, so it is the colonial alga.
Final answer: Volvox
Q113Single correctAnimal Kingdom / Evolution
Which of the following represents order of 'Horse' ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Perissodactyla
Approach:
The horse belongs to the order Perissodactyla (odd-toed ungulates) within the taxonomic hierarchy.
Step 1:Equidae is the family of the horse, not its order.
Step 2:Caballus and ferus are specific/subspecific epithets, not an order.
Step 3:The order of the horse is Perissodactyla.
Final answer: Perissodactyla
Q114Single correctCell Structure and Function
Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Mitochondrion
Approach:
Identify the organelle that performs aerobic respiration to release energy stored in carbohydrates as ATP.
Step 1:Glucose oxidation through glycolysis, the Krebs cycle and oxidative phosphorylation releases energy. The Krebs cycle and electron transport occur within the mitochondrion.
Step 2:Lysosomes are digestive, ribosomes synthesise protein, and chloroplasts capture light energy rather than extract energy from carbohydrates.
Final answer: Mitochondrion
Q115Single correctBiotechnology and its Applications
The process of separation and purification of expressed protein before marketing is called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Downstream processing
Approach:
Match the post-production separation and purification stage of a biotechnological product to its standard name.
Step 1:After the biosynthetic stage in a bioreactor, the product is separated and purified through a series of steps collectively termed downstream processing.
Step 2:Upstream processing refers to media and inoculum preparation prior to fermentation, which precedes purification.
Final answer: Downstream processing
Q116Single correctEcosystem
Mycorrhizae are the example of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Mutualism
Approach:
Classify the mycorrhizal association by the benefit each partner derives.
Step 1:In mycorrhizae a fungus colonises plant roots; the fungus absorbs mineral nutrients and water for the plant while the plant supplies the fungus with sugars.
Step 2:An interaction where both species gain is mutualism, distinct from amensalism or antibiosis where one partner is harmed.
Final answer: Mutualism
Q117Single correctBiological Classification
Viroids differ from viruses in having :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4RNA molecules without protein coat
Approach:
Recall the composition of a viroid and contrast it with that of a virus.
Step 1:A viroid is a free, circular RNA molecule that lacks the protein coat present in viruses.
Step 2:Viruses possess a nucleic acid core enclosed in a protein coat, which the viroid lacks.
Final answer: RNA molecules without protein coat
Q118Single correctAnatomy of Flowering Plants
Root hairs develop from the region of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Maturation
Approach:
Locate the root zone in which epidermal cells differentiate to form root hairs.
Step 1:The root tip has the region of meristematic activity, the region of elongation and the region of maturation in succession from the apex.
Step 2:In the region of maturation the epidermal cells mature and some form fine root hairs for absorption.
Final answer: Maturation
Q119Single correctSexual Reproduction in Flowering Plants
Coconut fruit is a :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Drupe
Approach:
Classify coconut by its pericarp structure.
Step 1:Coconut has a fibrous mesocarp and a hard stony endocarp surrounding the seed, with a thin outer epicarp.
Step 2:Fruits with a hard, stony endocarp are drupes.
Final answer: Drupe
Q120Single correctPlant Kingdom
Plants which produce characteristic pneumatophores and show vivipary belong to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Halophytes
Approach:
Match the adaptations of pneumatophores and viviparous germination to the appropriate ecological group.
Step 1:Mangroves grow in saline, waterlogged coastal soils and develop negatively geotropic pneumatophores for aeration along with viviparous seed germination.
Step 2:Salt-tolerant plants of saline habitats are halophytes.
Final answer: Halophytes
Q121Single correctBiodiversity and Conservation
Which one of the following is related to Ex-situ conservation of threatened animals and plants ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Wildlife Safari parks
Approach:
Distinguish ex-situ conservation, which protects organisms away from their natural habitat, from in-situ approaches.
Step 1:Ex-situ conservation maintains threatened species outside their natural homes in zoos, botanical gardens, wildlife safari parks and gene banks.
Step 2:Biodiversity hot spots, the Amazon rainforest and the Himalayan region conserve species within their natural habitats, making them in-situ.
Final answer: Wildlife Safari parks
Q122Single correctPlant Kingdom
Select the mismatch :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 — Dioecious
Approach:
Test each genus against the stated character to find the incorrect pairing.
Step 1:Pinus is monoecious, bearing both male and female cones on the same plant, so pairing it with dioecious is incorrect.
Step 2:Cycas is dioecious, Salvinia is heterosporous and Equisetum is homosporous, so these pairings are correct.
Final answer: — Dioecious
Q123Single correctPlant Physiology
Which of the following facilitates opening of stomatal aperture ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Radial orientation of cellulose microfibrils in the cell wall of guard cells
Approach:
Determine the structural feature of guard cells that drives stomatal opening upon turgor increase.
Step 1:When guard cells gain turgor through water uptake they swell, and the orientation of their wall microfibrils controls the resulting shape change.
Step 2:The radial arrangement of cellulose microfibrils makes the cells bow outward so the pore opens, while loss of turgor closes it.
Final answer: Radial orientation of cellulose microfibrils in the cell wall of guard cells
Q124Single correctMolecular Basis of Inheritance
The association of histone H1 with a nucleosome indicates :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The DNA is condensed into a Chromatin Fibre.
Approach:
Recall the role of histone H1 in higher-order chromatin packaging.
Step 1:The nucleosome core is wrapped by DNA around an octamer of histones, and histone H1 binds the linker DNA at the point where DNA enters and exits the core.
Step 2:Binding of H1 packs adjacent nucleosomes together, condensing the beads-on-a-string form into a compact chromatin fibre.
Final answer: The DNA is condensed into a Chromatin Fibre.
Q125Single correctMolecular Basis of Inheritance
DNA fragments are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Negatively charged
Approach:
Deduce the net charge of DNA from its phosphate backbone.
Step 1:Each nucleotide of DNA carries a phosphate group, and these phosphate groups are negatively charged at cellular pH.
Step 2:Because the phosphate charges dominate, DNA fragments migrate toward the anode in gel electrophoresis, confirming a net negative charge.
Final answer: Negatively charged
Q126Single correctHuman Reproduction
Capacitation occurs in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Female Reproductive tract
Approach:
Recall where sperm acquire fertilising ability after ejaculation.
Step 1:Sperm released into the female tract undergo capacitation, a maturation in which membrane changes prepare them for the acrosomal reaction.
Step 2:This process is completed in the female reproductive tract rather than within the male ducts.
Final answer: Female Reproductive tract
Q127Single correctEcosystem
Which ecosystem has the maximum biomass ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Forest ecosystem
Approach:
Compare standing biomass across the listed ecosystems.
Step 1:Forests contain large, long-lived trees that accumulate a great deal of organic matter as woody tissue.
Step 2:Grasslands and aquatic ponds or lakes hold far smaller standing biomass than forests.
Final answer: Forest ecosystem
Q128Single correctPrinciples of Inheritance and Variation
A disease caused by an autosomal primary non-disjunction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Down's Syndrome
Approach:
Identify the disorder arising from non-disjunction of an autosome rather than a sex chromosome or a gene mutation.
Step 1:Down's syndrome results from an extra copy of autosome 21 caused by non-disjunction during gamete formation.
Step 2:Klinefelter's and Turner's syndromes arise from sex-chromosome non-disjunction, and sickle cell anaemia is a point mutation, so they do not fit.
Final answer: Down's Syndrome
Q129Single correctPlant Kingdom
Life cycle of and respectively are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Haplodiplontic, Diplontic
Approach:
Assign the life-cycle pattern of each brown alga based on its alternation of generations.
Step 1:Ectocarpus shows an alternation of an independent haploid gametophyte and a diploid sporophyte, which is the haplodiplontic pattern.
Step 2:Fucus has a dominant diploid plant body with only the gametes being haploid, giving a diplontic life cycle.
Final answer: Haplodiplontic, Diplontic
Q130Single correctMolecular Basis of Inheritance
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 333
Approach:
Count how many triplet codons lie downstream of the deletion, since a single-base deletion shifts the reading frame from that point onward.
Step 1:The 999 bases form 333 codons, with base 901 starting the 301st codon since bases 901 to 903 make codon 301.
Step 2:Deleting base 901 shifts the reading frame for every base from position 901 onward, so all codons from the 301st to the 333rd are read differently.
Final answer: 33
Q131Single correctLocomotion and Movement
The pivot joint between atlas and axis is a type of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3synovial joint
Approach:
Classify the atlas-axis pivot joint by the broader joint category to which pivot joints belong.
Step 1:The atlas and axis form a pivot joint that allows rotation of the head, and pivot joints are freely movable with a fluid-filled cavity.
Step 2:Freely movable joints with a synovial cavity are synovial joints, of which the pivot joint is one subtype.
Final answer: synovial joint
Q132Single correctBiotechnology - Principles and Processes
A gene whose expression helps to identify the transformed cell is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Selectable marker
Approach:
Identify the gene whose expression distinguishes transformed cells from non-transformed ones.
Step 1:A selectable marker, often an antibiotic resistance gene, allows transformed cells to survive a selective agent while non-transformed cells perish.
Step 2:A vector or plasmid carries DNA and a structural gene encodes a protein product, none of which serve to identify transformants.
Final answer: Selectable marker
Q133Single correctEcosystem
Presence of plants arranged into well defined vertical layers depending on their height can be seen best in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Tropical Rain Forest
Approach:
Choose the community showing the most distinct vertical stratification of vegetation.
Step 1:Tropical rain forests display a clear layering of emergent trees, canopy, understorey, shrub and ground vegetation according to height.
Step 2:Savannahs and grasslands have far fewer strata, so stratification is best seen in the tropical rain forest.
Final answer: Tropical Rain Forest
Q134Single correctPrinciples of Inheritance and Variation
The genotypes of a Husband and Wife are and .
Among the blood types of their children, how many different genotypes and phenotypes are possible ?
Among the blood types of their children, how many different genotypes and phenotypes are possible ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 34 genotypes ; 3 phenotypes
Approach:
Construct the cross between the parental genotypes and tabulate the offspring genotypes and corresponding blood-group phenotypes.
Step 1:Crossing the parents produces four genotype combinations among the children.
Step 2:Genotypes and both give blood group A, while gives AB and gives B.
Final answer: 4 genotypes ; 3 phenotypes
Q135Single correctPlant Kingdom
Zygotic meiosis is characteristic of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Match zygotic meiosis, in which the diploid zygote divides meiotically, to the appropriate organism.
Step 1:In a haplontic life cycle the only diploid stage is the zygote, which immediately undergoes meiosis to restore the haploid condition.
Step 2:Chlamydomonas is haplontic, whereas Marchantia and Funaria are haplodiplontic and Fucus is diplontic.
Final answer:
Q136Single correctMicrobes in Human Welfare
Which of the following is matched for the product produced by them ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Saccharomyces\ cerevisiae : Ethanol
Approach:
Match each microorganism with the industrial product it yields.
Step 1:Saccharomyces cerevisiae (brewer's/baker's yeast) ferments sugars to produce ethanol and carbon dioxide.
Step 2:Acetobacter aceti produces acetic acid, Penicillium notatum produces the antibiotic penicillin, and Lactobacillus (not Methanobacterium) produces lactic acid; Methanobacterium produces methane.
Final answer: Saccharomyces\ cerevisiae : Ethanol
Q137Single correctBody Fluids and Circulation
Frog's heart when taken out of the body continues to beat for sometime.
Select the best option from the following statements.
(a) Frog is a poikilotherm.
(b) Frog does not have any coronary circulation.
(c) Heart is "myogenic" in nature.
(d) Heart is autoexcitable.
Options :
Select the best option from the following statements.
(a) Frog is a poikilotherm.
(b) Frog does not have any coronary circulation.
(c) Heart is "myogenic" in nature.
(d) Heart is autoexcitable.
Options :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(c) and (d)
Approach:
Identify the property responsible for a vertebrate heart continuing to beat after isolation.
Step 1:The vertebrate heartbeat originates from cardiac muscle itself, making the heart myogenic, so impulse generation does not depend on external nerve supply.
Step 2:Because the pacemaker tissue generates impulses on its own, the isolated heart remains autoexcitable and keeps beating for some time outside the body.
Step 3:Statements (c) and (d) together explain the continued beating; poikilothermy and coronary circulation are unrelated to the intrinsic rhythm.
Final answer: (c) and (d)
Q138Single correctRespiration and Cellular Respiration
Which statement is for Krebs' cycle ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The cycle starts with condensation of acetyl group (acetyl CoA) with pyruvic acid to yield citric acid
Approach:
Test each statement against the established biochemistry of the citric acid cycle.
Step 1:The Krebs' cycle begins when acetyl CoA condenses with oxaloacetic acid (a four-carbon compound), not with pyruvic acid, to form citric acid.
Step 2:Three NADH are produced (isocitrate, alpha-ketoglutarate, malate steps), one FADH2 is produced (succinate step), and GTP is formed during the succinyl CoA to succinate conversion, so options 1, 2 and 3 are correct.
Final answer: The cycle starts with condensation of acetyl group (acetyl CoA) with pyruvic acid to yield citric acid
Q139Single correctAnimal Kingdom
In case of poriferans, the spongocoel is lined with flagellated cells called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3choanocytes
Approach:
Recall the cell type that lines the internal cavity of sponges.
Step 1:In sponges, the central cavity (spongocoel) and canal system are lined by flagellated collar cells known as choanocytes, which drive water currents and trap food.
Step 2:Ostia are incurrent pores, osculum is the excurrent opening, and mesenchymal cells lie in the mesohyl, so these do not line the spongocoel.
Final answer: choanocytes
Q140Single correctMolecular Basis of Inheritance
Which of the following RNAs should be most abundant in animal cell ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1r-RNA
Approach:
Compare the relative cellular proportions of the major RNA classes.
Step 1:Ribosomal RNA (r-RNA) constitutes about 80 percent of total cellular RNA, being a structural component of the numerous ribosomes.
Step 2:Transfer RNA accounts for about 15 percent, while messenger RNA and micro RNA together make up only a small fraction.
Final answer: r-RNA
Q141Single correctAnimal Kingdom
Which among these is the combination of aquatic mammals ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Whales, Dolphins, Seals
Approach:
Identify which listed combination contains only mammals.
Step 1:Whales, dolphins and seals are all aquatic mammals belonging to class Mammalia.
Step 2:Sharks and Trygon (a stingray) are cartilaginous fishes (Chondrichthyes), so any combination including them is incorrect.
Final answer: Whales, Dolphins, Seals
Q142Single correctPhotosynthesis in Higher Plants
With reference to factors affecting the rate of photosynthesis, which of the following statement is not correct ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 plants respond to higher temperatures with enhanced photosynthesis while plants have much lower temperature optimum
Approach:
Evaluate each statement against the known physiology of C3 and C4 photosynthesis.
Step 1:C4 plants, not C3 plants, show higher temperature optima and respond to higher temperatures with enhanced photosynthesis; the statement reverses this relationship.
Step 2:Light saturation near 10 percent of full sunlight, enhancement of fixation by raised CO2 up to about 0.05 percent, and tomato grown in CO2-enriched greenhouses are all correct statements.
Final answer: plants respond to higher temperatures with enhanced photosynthesis while plants have much lower temperature optimum
Q143Single correctOrganisms and Populations
Asymptote in a logistic growth curve is obtained when :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Determine the condition under which logistic growth levels off at carrying capacity.
Step 1:In logistic growth the population size N rises until it equals the carrying capacity K, at which point the growth rate becomes zero and the curve flattens into an asymptote.
when
Step 2:The term (K - N)/K reaches zero only when N equals K, giving the horizontal asymptote of the sigmoid curve.
Final answer:
Q144Single correctLocomotion and Movement
Out of 'X' pairs of ribs in humans only 'Y' pairs are true ribs. Select the option that correctly represents values of X and Y and provides their explanation.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 — True ribs are attached dorsally to vertebral column and ventrally to the sternum
Approach:
Recall the number and attachment of human ribs.
Step 1:Humans possess 12 pairs of ribs, of which the first seven pairs are true (vertebrosternal) ribs.
Step 2:True ribs articulate dorsally with the thoracic vertebrae and ventrally with the sternum, while the remaining pairs are false or floating ribs.
Final answer: — True ribs are attached dorsally to vertebral column and ventrally to the sternum
Q145Single correctBiotechnology - Principles and Processes
The DNA fragments separated on an agarose gel can be visualised after staining with :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ethidium bromide
Approach:
Recall the standard dye used to detect DNA bands in agarose gel electrophoresis.
Step 1:DNA fragments separated on an agarose gel are stained with ethidium bromide, which intercalates between bases and fluoresces orange under ultraviolet light.
Step 2:Bromophenol blue is a tracking dye, while acetocarmine and aniline blue are cytological stains, none of which makes DNA bands visible under UV.
Final answer: Ethidium bromide
Q146Single correctSexual Reproduction in Flowering Plants
Functional megaspore in an angiosperm develops into :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Embryo sac
Approach:
Trace the fate of the functional megaspore during megagametogenesis.
Step 1:After meiosis in the megaspore mother cell, three megaspores degenerate and the single functional megaspore enlarges and undergoes mitotic divisions.
Step 2:These divisions form the seven-celled, eight-nucleate female gametophyte, the embryo sac.
Final answer: Embryo sac
Q147Single correctPrinciples of Inheritance and Variation
Among the following characters, which one was not considered by Mendel in his experiments on pea ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Trichomes - Glandular or non-glandular
Approach:
Recall the seven contrasting characters of garden pea studied by Mendel.
Step 1:Mendel studied stem height, seed colour, seed shape, pod shape, pod colour, flower colour and flower position in Pisum sativum.
Step 2:Trichomes (glandular or non-glandular) were not among the traits Mendel investigated.
Final answer: Trichomes - Glandular or non-glandular
Q148Single correctBreathing and Exchange of Gases
Lungs are made up of air-filled sacs, the alveoli. They do not collapse even after forceful expiration, because of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Residual Volume
Approach:
Identify the lung volume that remains after maximal expiration.
Step 1:Residual volume is the air that stays in the lungs even after the most forceful expiration, keeping the alveoli partly inflated.
Step 2:Tidal, inspiratory reserve and expiratory reserve volumes are all exchangeable air and do not keep the alveoli open after expiration.
Final answer: Residual Volume
Q149Single correctChemical Coordination and Integration
GnRH, a hypothalamic hormone, needed in reproduction, acts on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2anterior pituitary gland and stimulates secretion of LH and FSH.
Approach:
Recall the target and action of gonadotropin-releasing hormone.
Step 1:GnRH from the hypothalamus acts on the anterior pituitary (adenohypophysis).
Step 2:It stimulates the anterior pituitary to secrete the gonadotropins luteinising hormone (LH) and follicle stimulating hormone (FSH).
Final answer: anterior pituitary gland and stimulates secretion of LH and FSH.
Q150Single correctMorphology of Flowering Plants
In Bougainvillea thorns are the modifications of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Stem
Approach:
Identify the organ from which Bougainvillea thorns arise.
Step 1:In Bougainvillea, the thorns are axillary buds modified into hard, pointed structures, making them stem modifications.
Step 2:Leaf, stipule and root modifications give rise to other structures (such as tendrils or spines in different plants) but not the Bougainvillea thorn.
Final answer: Stem
Q151Single correctPrinciples of Inheritance and Variation
Which one from those given below is the period for Mendel's hybridization experiments ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11856 - 1863
Approach:
Recall the years over which Mendel performed his pea hybridization work.
Step 1:Mendel conducted his hybridization experiments on garden pea between 1856 and 1863 at the monastery garden.
Step 2:The other date ranges do not correspond to Mendel's experimental period.
Final answer: 1856 - 1863
Q152Single correctNeural Control and Coordination
Good vision depends on adequate intake of carotene-rich food.
Select the best option from the following statements.
(a) Vitamin A derivatives are formed from carotene.
(b) The photopigments are embedded in the membrane discs of the inner segment.
(c) Retinal is a derivative of Vitamin A.
(d) Retinal is a light absorbing part of all the visual photopigments.
Options :
Select the best option from the following statements.
(a) Vitamin A derivatives are formed from carotene.
(b) The photopigments are embedded in the membrane discs of the inner segment.
(c) Retinal is a derivative of Vitamin A.
(d) Retinal is a light absorbing part of all the visual photopigments.
Options :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a), (c) and (d)
Approach:
Assess each statement on the role of carotene and retinal in vision.
Step 1:Vitamin A is derived from carotene, retinal is a derivative of vitamin A, and retinal is the light-absorbing component of all visual photopigments, so statements (a), (c) and (d) are correct.
Step 2:Statement (b) is wrong because photopigments are embedded in the membrane discs of the outer segment, not the inner segment.
Final answer: (a), (c) and (d)
Q153Single correctMicrobes in Human Welfare
Which one of the following statements is not valid for aerosols ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3They cause increased agricultural productivity
Approach:
Identify the statement that misrepresents the effects of aerosols.
Step 1:Aerosols harm human health, disturb rainfall and monsoon patterns, and reduce agricultural productivity by blocking sunlight, so these are valid.
Step 2:Aerosols decrease, rather than increase, agricultural productivity, so the statement claiming increased productivity is not valid.
Final answer: They cause increased agricultural productivity
Q154Single correctExcretory Products and their Elimination
A decrease in blood pressure/volume will not cause the release of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Atrial Natriuretic Factor
Approach:
Determine which hormone is released in response to high, not low, blood pressure/volume.
Step 1:A fall in blood pressure or volume triggers renin (via JGA), aldosterone and ADH to conserve water and salt and raise blood pressure.
Step 2:Atrial Natriuretic Factor is released when blood pressure or volume rises, causing vasodilation and sodium loss; a decrease will not cause its release.
Final answer: Atrial Natriuretic Factor
Q155Single correctStrategies for Enhancement in Food Production
Homozygous purelines in cattle can be obtained by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1mating of related individuals of same breed.
Approach:
Identify the breeding method that increases homozygosity.
Step 1:Inbreeding, the mating of closely related individuals of the same breed, increases homozygosity and is used to develop purelines.
Step 2:Outbreeding, crossbreeding and interspecific mating increase heterozygosity or produce hybrids, so they do not yield purelines.
Final answer: mating of related individuals of same breed.
Q156Single correctAnatomy of Flowering Plants
The vascular cambium normally gives rise to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Secondary xylem
Approach:
Recall the tissues produced by the vascular cambium during secondary growth.
Step 1:The vascular cambium cuts off cells towards the inside to form secondary xylem and towards the outside to form secondary phloem.
Step 2:Phelloderm and periderm are derived from the cork cambium (phellogen), and primary phloem forms from the procambium, not the vascular cambium.
Final answer: Secondary xylem
Q157Single correctExcretory Products and their Elimination
Which of the following statements is correct ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The ascending limb of loop of Henle is impermeable to water.
Approach:
Recall the permeability properties of the two limbs of the loop of Henle.
Step 1:The ascending limb of the loop of Henle is impermeable to water but actively transports electrolytes, diluting the filtrate.
Step 2:The descending limb is permeable to water and largely impermeable to electrolytes, so the remaining statements are incorrect.
Final answer: The ascending limb of loop of Henle is impermeable to water.
Q158Single correctPlant Growth and Development
Fruit and leaf drop at early stages can be prevented by the application of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Auxins
Approach:
Identify the plant growth regulator that prevents early abscission.
Step 1:Auxins inhibit the formation of the abscission layer, thereby preventing premature drop of fruits and leaves at early stages.
Step 2:Ethylene actually promotes abscission, while cytokinins and gibberellic acid have other primary roles, so they do not serve this purpose.
Final answer: Auxins
Q159Single correctAnimal Kingdom
A baby boy aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pre-molars
Approach:
The milk (deciduous) dentition of a two-year-old contains 20 teeth, while the permanent set has 32. Comparison of the two sets identifies which tooth type is missing in the milk dentition.
Step 1:The temporary or milk dentition includes incisors, canines and molars only.
Step 2:Premolars erupt only in the permanent dentition and are entirely absent from the deciduous set.
Step 3:A two-year-old child therefore lacks premolars.
Final answer: Pre-molars
Q160Single correctAnimal Kingdom
An important characteristic that Hemichordates share with Chordates is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3pharynx with gill slits
Approach:
Each option is tested against the defining features shared between Hemichordata and Chordata.
Step 1:Hemichordates possess pharyngeal gill slits, a feature also present in chordates.
Step 2:Hemichordates lack a true notochord (the stomochord is not homologous) and have a dorsal hollow nerve cord that is only partly tubular, so options on notochord and ventral cord are incorrect.
Step 3:The pharynx bearing gill slits is the common chordate-hemichordate trait.
Final answer: pharynx with gill slits
Q161Single correctEvolution
Artificial selection to obtain cows yielding higher milk output represents :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2directional as it pushes the mean of the character in one direction
Approach:
The pattern of selection is identified by how artificial selection for higher milk yield shifts the trait distribution.
Step 1:Selecting only high-yielding cows for breeding favours one extreme of the milk-output trait.
Step 2:Favouring a single extreme shifts the population mean toward that direction, which defines directional selection.
Step 3:Stabilizing favours the mean and disruptive favours both extremes, neither of which applies here.
Final answer: directional as it pushes the mean of the character in one direction
Q162Single correctHuman Reproduction
Select the correct route for the passage of sperms in male frogs :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Testes Vasa efferentia Kidney Bidder's canal Urinogenital duct Cloaca
Approach:
The anatomical sperm path in the male frog is traced from testes to cloaca, matching it against the listed sequences.
Step 1:Sperms leave the testes through the vasa efferentia, which open into the kidney.
Step 2:Within the kidney sperms enter Bidder's canal and then pass into the urinogenital duct.
Step 3:The urinogenital duct finally opens into the cloaca, completing the route.
Final answer: Testes Vasa efferentia Kidney Bidder's canal Urinogenital duct Cloaca
Q163Single correctDigestion and Absorption
Which of the following options best represents the enzyme composition of pancreatic juice?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4lipase, amylase, trypsinogen, procarboxypeptidase
Approach:
Each option is screened for enzymes that genuinely occur in pancreatic juice versus those secreted by the stomach.
Step 1:Pancreatic juice contains pancreatic lipase, pancreatic amylase, trypsinogen and procarboxypeptidase among others.
Step 2:Pepsin and rennin are gastric secretions of the stomach mucosa, not pancreatic ones, so neither belongs in a list of pancreatic enzymes.
Step 3:Only the set lipase, amylase, trypsinogen, procarboxypeptidase is fully pancreatic.
Final answer: lipase, amylase, trypsinogen, procarboxypeptidase
Q164Single correctPhotosynthesis in Higher Plants
Phosphoenol pyruvate (PEP) is the primary acceptor in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 plants
Approach:
The identity of the primary carbon-dioxide acceptor is matched to the photosynthetic pathway that uses it.
Step 1:In the C4 pathway PEP in mesophyll cells fixes carbon dioxide using PEP carboxylase to form oxaloacetate.
Step 2:In C3 plants the primary acceptor is RuBP, not PEP, so C3 options are excluded.
Step 3:PEP serves as the primary acceptor exclusively in C4 plants.
Final answer: plants
Q165Single correctMorphology of Flowering Plants
The morphological nature of the edible part of coconut is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Endosperm
Approach:
The white edible meat and coconut water are identified with the seed tissue they represent.
Step 1:Coconut water is free-nuclear liquid endosperm and the white kernel is the cellular endosperm.
Step 2:The pericarp forms the fibrous husk and hard shell, which are inedible.
Step 3:The edible part is therefore the endosperm.
Final answer: Endosperm
Q166Single correctCell Cycle and Cell Division
Anaphase Promoting Complex (APC) is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cell, which of the following is expected to occur ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Chromosomes will not segregate
Approach:
The mitotic role of the Anaphase Promoting Complex is used to predict the outcome of its failure.
Step 1:APC triggers degradation of securin, releasing separase that cleaves cohesin holding sister chromatids together.
Step 2:A defective APC leaves cohesin intact, so sister chromatids cannot separate and move to opposite poles.
Step 3:Chromosome segregation therefore fails.
Final answer: Chromosomes will not segregate
Q167Single correctBody Fluids and Circulation
MALT constitutes about ________ percent of the lymphoid tissue in human body.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 150%
Approach:
The standard textbook figure for the proportion of lymphoid tissue represented by MALT is recalled.
Step 1:Mucosa-associated lymphoid tissue is distributed along the linings of the major tracts of the body.
Step 2:MALT accounts for about 50 percent of the lymphoid tissue in the human body.
Final answer: 50%
Q168Single correctNeural Control and Coordination
Receptor sites for neurotransmitters are present on :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4post-synaptic membrane
Approach:
The site of neurotransmitter receptors is located by following signal transmission across a chemical synapse.
Step 1:Neurotransmitter is released from synaptic vesicles at the pre-synaptic membrane into the synaptic cleft.
Step 2:The released transmitter binds receptors located on the post-synaptic membrane to generate a new potential.
Step 3:Receptor sites therefore lie on the post-synaptic membrane.
Final answer: post-synaptic membrane
Q169Single correctChemical Coordination and Integration
Hypersecretion of Growth Hormone in adults does not cause further increase in height, because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Epiphyseal plates close after adolescence.
Approach:
The reason adult growth hormone excess fails to raise height is traced to the state of the long-bone growth plates.
Step 1:Linear growth in length depends on cartilage at the epiphyseal plates of long bones.
Step 2:After adolescence the epiphyseal plates ossify and fuse, ending the capacity for length increase.
Step 3:Excess growth hormone in adults therefore causes acromegaly rather than added height.
Final answer: Epiphyseal plates close after adolescence.
Q170Single correctOrganisms and Populations
Alexander Von Humboldt described for the first time :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Species area relationships
Approach:
The historical contribution attributed to Alexander Von Humboldt is matched to the listed ecological concepts.
Step 1:Alexander Von Humboldt studied the relation between species richness and the area explored in South American jungles.
Step 2:He first described the species-area relationship, in which richness rises with area up to a limit.
Final answer: Species area relationships
Q171Single correctNeural Control and Coordination
Myelin sheath is produced by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Schwann Cells and Oligodendrocytes
Approach:
The myelinating cells of the peripheral and central nervous systems are identified.
Step 1:Schwann cells form the myelin sheath around axons of the peripheral nervous system.
Step 2:Oligodendrocytes form the myelin sheath around axons of the central nervous system.
Step 3:Both cell types together produce myelin; osteoclasts and astrocytes do not.
Final answer: Schwann Cells and Oligodendrocytes
Q172Single correctReproductive Health
In case of a couple where the male is having a very low sperm count, which technique will be suitable for fertilisation ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Artificial Insemination
Approach:
Identify which assisted reproductive technique the syllabus prescribes for a male partner with a very low sperm count.
Step 1:In artificial insemination the semen is collected and introduced artificially into the vagina, or into the uterus as intra-uterine insemination. Concentrating the whole ejaculate into one delivery is what makes it workable when the count is very low.
Step 2:This is precisely the case the technique is prescribed for — a male partner unable to inseminate the female, or a very low sperm count in the ejaculate.
Step 3:Intrauterine transfer places an already-developing embryo, and GIFT transfers an ovum into the fallopian tube of a donor female, so neither addresses sperm number. Intracytoplasmic sperm injection is reserved for forming an embryo in the laboratory, a further step beyond what this couple is described as needing.
Final answer: Artificial Insemination
Q173Single correctCell - The Unit of Life
Which of the following components provides sticky character to the bacterial cell ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Glycocalyx
Approach:
The bacterial structure responsible for adhesion and sticky texture is identified.
Step 1:The glycocalyx is the outermost gel-like layer of polysaccharide surrounding the bacterial cell wall.
Step 2:As a slime layer or capsule it confers stickiness and aids attachment to surfaces.
Step 3:Cell wall, plasma membrane and nuclear membrane do not provide this property.
Final answer: Glycocalyx
Q174Single correctMolecular Basis of Inheritance
DNA replication in bacteria occurs :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Prior to fission
Approach:
The timing of bacterial DNA replication is placed within the prokaryotic division cycle.
Step 1:Bacteria divide by binary fission and lack an organised nucleus and a defined S phase like eukaryotes.
Step 2:The single circular DNA replicates fully before the cell splits, ensuring each daughter receives a copy.
Step 3:DNA replication therefore occurs prior to fission.
Final answer: Prior to fission
Q175Single correctReproductive Health
The function of copper ions in copper releasing IUD's is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1They suppress sperm motility and fertilizing capacity of sperms.
Approach:
The contraceptive action specifically attributable to released copper ions is selected.
Step 1:Copper-releasing intrauterine devices liberate copper ions into the uterine cavity.
Step 2:These ions act on sperms, reducing their motility and fertilising capacity.
Step 3:The copper ion effect therefore suppresses sperm motility and fertilising ability.
Final answer: They suppress sperm motility and fertilizing capacity of sperms.
Q176Single correctMicrobes in Human Welfare
Which of the following in sewage treatment removes suspended solids ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Primary treatment
Approach:
The stage of sewage treatment that physically removes suspended particulate matter is identified.
Step 1:Primary treatment uses physical sedimentation and filtration to settle out suspended solids.
Step 2:Secondary treatment relies on microbes to reduce organic BOD, not to remove suspended solids.
Step 3:Removal of suspended solids is therefore the function of primary treatment.
Final answer: Primary treatment
Q177Single correctTransport in Plants
The water potential of pure water is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Zero
Approach:
The reference value of water potential is recalled from the definition based on pure water.
Step 1:Water potential of pure water at standard temperature and pressure is taken as the reference zero.
Step 2:Adding solutes lowers the potential to negative values, so pure water has the highest, zero, potential.
Step 3:The water potential of pure water is therefore zero.
Final answer: Zero
Q178Single correctAnatomy of Flowering Plants
Identify the wrong statement in context of heartwood :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3It conducts water and minerals efficiently
Approach:
Each statement about heartwood is checked against its structure and function to find the false one.
Step 1:Heartwood is the central, dead, highly lignified region with deposits of tannins, resins and other organic compounds, making it durable.
Step 2:Its vessels are plugged by tyloses and deposits, so heartwood does not conduct water and minerals.
Step 3:The claim that it conducts water and minerals efficiently is therefore wrong.
Final answer: It conducts water and minerals efficiently
Q179Single correctMolecular Basis of Inheritance
Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the correct statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Thalassemia is due to less synthesis of globin molecules.
Approach:
Thalassemia and sickle cell anemia are classified as quantitative or qualitative globin defects to pick the correct statement.
Step 1:Thalassemia is a quantitative disorder where globin chains are synthesised in reduced amounts.
Step 2:Sickle cell anemia is a qualitative disorder from a single amino acid substitution in the beta-globin chain.
Step 3:Among the statements only the one describing thalassemia as less synthesis of globin is correct.
Final answer: Thalassemia is due to less synthesis of globin molecules.
Q180Single correctSexual Reproduction in Flowering Plants
Flowers which have single ovule in the ovary and are packed into inflorescence are usually pollinated by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Wind
Approach:
The pollinating agent typical of single-ovuled, closely packed inflorescence flowers is identified from their adaptations.
Step 1:Flowers with a single ovule per ovary packed into an inflorescence are a hallmark of wind-pollinated species such as grasses.
Step 2:Such flowers produce abundant light pollen and lack attractants needed for animal pollination.
Step 3:These flowers are therefore usually pollinated by wind.
Final answer: Wind
Frequently Asked Questions
How many questions are in the NEET 2017 May 07 paper?
The NEET 2017 May 07 paper has 179 questions — Physics (44), Chemistry (45) and Biology (90). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for NEET?
NEET awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Biology across 180 questions for 720 marks.
Are the answer key and step-by-step solutions provided for the 2017 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 2017 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.
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