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![Three radicals side by side. (I) H3C-C(CH3)2-CH2 with a radical dot on the CH2 that is attached to a benzene ring. (II) a central carbon carrying a radical dot bonded to three Ph groups (triphenylmethyl radical). (III) a bicyclo[2.2.1] cage with a radical dot on a ring carbon and a CH3 on the neighbouring carbon.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2F2d27d6c4-a6a1-48a7-9bac-64da39a63057%2F2d27d6c4-a6a1-48a7-9bac-64da39a63057%2Fimages%2FQ83_radicals_b1.webp)
NEET 2015 May 03 Question Paper with Solutions
All 180 questions from the NEET 2015 (May 03) paper — Physics (45), Chemistry (45) and Biology (90) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2015Chemistry PYQs 2015Biology PYQs 2015
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctUnits and Measurements
If force (E), velocity (V) and time (T) are chosen as fundamental quantities, the dimensional formula of surface tension will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The fundamental quantities are taken as E, velocity V and time T. Surface tension has dimensions of force per unit length, equivalent to energy per unit area, and is written as a product of powers of E, V and T with exponents fixed by dimensional matching.
Step 1:Surface tension equals force per unit length, which has dimensions of mass per time squared.
Step 2:Writing E as energy (M ), V as L and T as time, the relation S = gives M: a=1, L: 2a+b=0, T: -2a-b+c=-2.
Step 3:Substituting the exponents gives the dimensional formula of surface tension.
Final answer:
Q2Single correctMotion in a Plane
A Ship A is moving westwards with a speed of and a ship B south of A is moving northwards with a speed of . The time after which the distance between them becomes shortest is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The relative velocity of one ship with respect to the other determines the rate of change of separation; the shortest distance occurs when the component of relative position along the relative velocity vanishes.
Step 1:Taking east as positive x and north as positive y, ship A moves west and ship B moves north.
Step 2:B is 100 km south of A, so the initial position of B relative to A points south.
Step 3:The separation is minimum when its time derivative is zero; solving for the time gives the closest approach.
Final answer:
Q3Single correctMotion in a Straight Line
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to
where and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by
where and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Acceleration equals velocity times the derivative of velocity with respect to position, since the velocity is given as a function of position.
Step 1:Differentiating the velocity with respect to position.
Step 2:Multiplying by the velocity to obtain acceleration.
Final answer:
Q4Single correctLaws of Motion
Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg, respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The common acceleration of the three blocks follows from the applied force and total mass; the contact force between A and B equals the force needed to accelerate the blocks behind that interface.
Step 1:The whole system accelerates together under the applied force.
Step 2:The contact force between A and B drives blocks B and C, of combined mass 3 kg.
Final answer:
Q5Single correctLaws of Motion
A block A of mass rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass is suspended. The coefficient of kinetic friction between the block and the table is . When the block A is sliding on the table, the tension in the string is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Newton's second law is applied to the hanging block and to the sliding block, including kinetic friction on the table block; eliminating the common acceleration gives the string tension.
Step 1:Adding the two equations of motion eliminates the tension and gives the acceleration.
Step 2:Substituting the acceleration into the table block equation gives the tension.
Final answer:
Q6Single correctWork, Energy and Power
Two similar springs P and Q have spring constants and . They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs and are related as in case (a) and case (b), respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Spring energy is expressed first in terms of extension and then in terms of force; comparing the two springs in each case, with for similar springs, fixes the inequalities.
Step 1:For the same extension, the stored energy is proportional to the spring constant; the stiffer spring stores more energy.
Step 2:For the same force, the stored energy is inversely proportional to the spring constant; the softer spring stores more energy.
Final answer:
Q7Single correctWork, Energy and Power
A block of mass , moving in x direction with a constant speed of , is subjected to a retarding force during its travel from to . Its final KE will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The work done by the retarding force over the displacement is found by integration; subtracting this from the initial kinetic energy gives the final kinetic energy.
Step 1:The initial kinetic energy of the block.
Step 2:The work done against the retarding force is the integral of the force over the path.
Step 3:The final kinetic energy is the initial value reduced by the work done against the force.
Final answer:
Q8Single correctWork, Energy and Power
A particle of mass is driven by a machine that delivers a constant power watts. If the particle starts from rest the force on the particle at time is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Constant power equals force times velocity; the velocity as a function of time follows from the work-energy theorem under constant power, and the force is then power divided by velocity.
Step 1:Under constant power starting from rest, the kinetic energy grows linearly with time, giving the velocity.
Step 2:The force equals power divided by velocity.
Final answer:
Q9Single correctWork, Energy and Power
Two particles of masses , move with initial velocities and . On collision, one of the particles get excited to higher level, after absorbing energy . If final velocities of particles be and then we must have
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Total energy is conserved in the collision; the kinetic energy lost equals the energy absorbed for excitation, so the final kinetic energy is the initial kinetic energy reduced by that absorbed amount.
Step 1:The kinetic energy before collision exceeds that after collision by the energy absorbed for excitation.
Step 2:Rearranging places the absorbed energy on the initial side.
Final answer:
Q10Single correctSystems of Particles and Rotational Motion
The rod of weight is supported by two parallel knife edges and and is in equilibrium in a horizontal position. The knives are at a distance from each other. The centre of mass of the rod is at distance from . The normal reaction on is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Taking torques about knife edge B and using the equilibrium of the rod gives the normal reaction at A directly.
Step 1:The centre of mass is at distance from A and therefore at distance from B.
Step 2:Taking moments about B, the weight and the reaction at A balance.
Final answer:
Q11Single correctSystems of Particles and Rotational Motion
A mass m moves in a circle on a smooth horizontal plane with velocity at a radius . The mass is attached to a string which passes through a smooth hole in the plane as shown.
The tension in the string is increased gradually and finally m moves in a circle of radius . The final value of the kinetic energy is
The tension in the string is increased gradually and finally m moves in a circle of radius . The final value of the kinetic energy is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The string tension is a central force, so angular momentum about the hole is conserved; the new speed follows from this, and the kinetic energy is computed from it.
Step 1:Angular momentum about the hole is conserved as the radius is reduced to half.
Step 2:The final kinetic energy is computed from the new speed.
Final answer:
Q12Single correctSystems of Particles and Rotational Motion
Three identical spherical shells, each of mass and radius placed as shown in figure. Consider an axis XX' which is touching to two shells and passing through diameter of third shell.
Moment of inertia of the system consisting of these three spherical shells about XX' axis is
Moment of inertia of the system consisting of these three spherical shells about XX' axis is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The moment of inertia of each shell about the axis is found using the shell formula and the parallel-axis theorem where the axis is tangential, then the three contributions are summed.
Step 1:The axis passes through the diameter of the third shell, giving its moment of inertia directly.
Step 2:For each of the two shells the axis is tangential, so the parallel-axis theorem adds to the diametral value.
Step 3:Summing the three contributions gives the total moment of inertia.
Final answer:
Q13Single correctGravitation
Kepler's third law states that square of period of revolution (T) of a planet around the Sun, is proportional to third power of average distance r between Sun and planet i.e. here K is constant.
If the masses of Sun and planet are M and m respectively, then as per Newton's law of gravitation, force of attraction between them is , here G is gravitational constant. The relation between G and K is described as
If the masses of Sun and planet are M and m respectively, then as per Newton's law of gravitation, force of attraction between them is , here G is gravitational constant. The relation between G and K is described as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Equating the gravitational force to the centripetal force for circular orbital motion yields the period, which is matched to Kepler's third law to relate the constants.
Step 1:Equating gravitational and centripetal forces gives the orbital speed.
Step 2:Substituting the speed into the period expression and squaring gives the period squared.
Step 3:Comparing with identifies the constant.
Final answer:
Q14Single correctGravitation
Two spherical bodies of mass and and radii and are released in free space with initial separation between their centres equal to . If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
With no external force the centre of mass stays fixed; the two bodies move distances inversely proportional to their masses, and collision occurs when the gap between surfaces closes.
Step 1:Collision happens when the separation of centres equals the sum of radii; the centres must close the gap from to .
Step 2:The displacements are inversely proportional to the masses, so the smaller body () moves five times as far as the larger body ().
Step 3:The distance moved by the smaller body follows.
Final answer:
Q15Single correctThermal Properties of Matter
On observing light from three different starts P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If , and are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observation that
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By Wien's displacement law the wavelength of peak emission is inversely proportional to absolute temperature, so the star whose peak lies at the shortest wavelength is the hottest.
Step 1:Peak wavelength increases from violet to green to red, so the peak wavelengths satisfy ()P < ()R < ()Q.
Step 2:Since absolute temperature is inversely proportional to peak wavelength, the order of temperatures reverses.
Final answer:
Q16Single correctMechanical Properties of Fluids
The approximate depth of an ocean is . The compressibility of water is and density of water is . What fractional compression of water will be obtained at the bottom of the ocean?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The pressure at the ocean bottom is found from depth and density; the fractional compression equals the product of compressibility and pressure.
Step 1:The pressure due to the water column at the bottom is computed from density, gravity and depth.
Step 2:The fractional compression is the compressibility multiplied by this pressure.
Final answer:
Q17Single correctThermal Properties of Matter
The two ends of a metal rod are maintained at temperatures and . The rate of heat flow in the rod is found to be . If the ends are maintained at temperatures and , the rate of heat flow will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Steady-state conduction depends only on the temperature difference between the ends, not their absolute values; since the difference is unchanged, the rate of heat flow is unchanged.
Step 1:In the first case the temperature difference across the rod is the difference of its end temperatures.
Step 2:In the second case the difference is the same despite the higher temperatures.
Step 3:Since the conduction rate is proportional to the temperature difference, an unchanged difference gives an unchanged rate.
Final answer:
Q18Single correctMechanical Properties of Fluids
A wind with speed blows parallel to the roof of a house. The area of the roof is . Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be
()
()
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 upwards
Approach:
Bernoulli's principle gives a lower pressure above the roof where the wind moves; the resulting pressure difference times the roof area gives an upward force on the roof.
Step 1:The air inside the house is still, so the inside surface of the roof is at atmospheric pressure. The wind blows parallel to the outer surface, so by Bernoulli's theorem the pressure just outside is lower than atmospheric by one half rho v squared.
Step 2:This pressure difference acts over the whole area of the roof, so the net force is the pressure difference times the area.
Step 3:Because the pressure below the roof (atmospheric) now exceeds the pressure above it, the net force pushes the roof outwards, that is vertically upwards. This is why a roof is lifted off in a strong wind rather than pressed down.
Final answer: upwards
Q19Single correctThermodynamics
Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
In process AB, of heat is added to the system and in process BC, of heat is added to the system. The heat absorbed by the system in the process AC will be
In process AB, of heat is added to the system and in process BC, of heat is added to the system. The heat absorbed by the system in the process AC will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The change in internal energy is path-independent and is found from the path ABC; the heat for the direct path AC equals this internal-energy change plus the work done along AC, read from the pressure-volume diagram.
Step 1:Along AB the volume is constant, so no work is done; the work along BC at constant pressure over a volume change of is computed.
Step 2:The internal energy change from A to C follows from the first law along ABC.
Step 3:Along the direct path AC the work is the area of the trapezium under the line from A to C, equal to ; the heat absorbed is then the internal-energy change plus this work.
Final answer:
Q20Single 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 3
Approach:
The engine efficiency fixes the coefficient of performance of the reversible refrigerator; multiplying that coefficient by the work input gives the heat extracted from the cold reservoir.
Step 1:The coefficient of performance of the refrigerator is obtained from the engine efficiency.
Step 2:The heat absorbed from the cold reservoir is the coefficient of performance times the work input.
Final answer:
Q21Single correctThermodynamics
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,
The change in internal energy of the gas during the transition is
The change in internal energy of the gas during the transition is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For an ideal gas the internal energy depends only on temperature; using the diatomic expression and the ideal gas law, the change is computed from the product of pressure and volume at the two states.
Step 1:At state A the pressure is and volume , giving the product PV.
Step 2:At state B the pressure is and volume , giving its product PV.
Step 3:The change in internal energy is times the change in the PV product.
Final answer:
Q22Single correctKinetic Theory
The ratio of the specific heats in terms of degrees of freedom (n) is given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The molar heat capacities are expressed in terms of degrees of freedom; their ratio gives the adiabatic exponent.
Step 1:The heat capacity at constant volume is set by the degrees of freedom.
Step 2:The ratio of the two capacities gives the adiabatic exponent.
Final answer:
Q23Single correctOscillations
When two displacements represented by and are superimposed the motion is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Simple harmonic with amplitude
Approach:
Two simple harmonic motions of the same frequency and a 90° phase difference combine into a single simple harmonic motion whose amplitude is the resultant of the two amplitudes.
Step 1:The two displacements share the same angular frequency but differ in phase by .
Step 2:The resultant amplitude follows from the superposition formula with the cosine term vanishing for a right-angle phase difference.
Final answer: Simple harmonic with amplitude
Q24Single correctOscillations and Waves
A particle is executing SHM along a straight line. Its velocities at distances and from the mean position are and , respectively. Its time period is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The velocity of a particle in SHM relates to its displacement through the amplitude and angular frequency. Forming the relation at two positions allows elimination of the amplitude and isolation of the time period.
Step 1:The velocity at displacement x is governed by the SHM relation.
Step 2:Writing the relation at the two given positions gives two equations.
Step 3:Subtracting eliminates the amplitude.
Step 4:Substituting omega into the time period expression yields the result.
Final answer:
Q25Single correctOscillations and Waves
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both ends is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3120 cm
Approach:
Equate the fundamental frequency of the closed pipe to the second overtone (third harmonic) frequency of the open pipe and solve for the open-pipe length.
Step 1:The fundamental of a closed pipe of length 20 cm is set.
Step 2:The second overtone of an open pipe corresponds to its third harmonic.
Step 3:Equating the two frequencies and cancelling v gives the open-pipe length.
Final answer: 120 cm
Q26Single correctElectrostatics
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K which can just fill the air gap of the capacitor is now inserted in it. Which of the following is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The charge on the capacitor is not conserved.
Approach:
After disconnection the charge on an isolated capacitor stays fixed. Inserting a dielectric raises capacitance, which alters voltage and energy at constant charge. Each statement is tested against these consequences.
Step 1:The capacitor is isolated after disconnection, so the charge remains constant.
Step 2:With charge fixed and capacitance raised K times, the voltage falls.
Step 3:The energy at constant charge varies inversely with capacitance.
Step 4:The change in energy follows from the initial and final values.
Final answer: The charge on the capacitor is not conserved.
Q27Single correctElectrostatics
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply Gauss's law over a spherical surface of radius a. The radial field is uniform over this surface, so the flux relates directly to the enclosed charge.
Step 1:On the sphere of radius a the field magnitude is Aa and points radially.
Step 2:The flux through the spherical surface is the field times the area.
Step 3:Gauss's law gives the enclosed charge from the flux.
Final answer:
Q28Single correctCurrent Electricity
A potentiometer wire has length 4 m and resistance 8 . The resistance that must be connected in series with the wire and an accumulator of emf 2 V, so as to get a potential gradient 1 mV per cm on the wire is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 132
Approach:
The required potential gradient fixes the voltage across the wire, which in turn fixes the circuit current. The series resistance is then found from the emf and total resistance.
Step 1:The gradient of 1 mV per cm over 400 cm gives the potential difference across the wire.
Step 2:The current follows from this voltage across the 8 ohm wire.
Step 3:The total resistance follows from the emf and current.
Step 4:The series resistance is the total minus the wire resistance.
Final answer: 32
Q29Single correctCurrent Electricity
A, B and C are voltmeters of resistance R, 1.5R and 3R, respectively, as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are , and , respectively. Then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Voltmeter A is in series with the parallel combination of B and C. The current through A splits between B and C, and the readings are compared using their resistances and currents.
Step 1:B and C are in parallel, with combined resistance equal to A.
Step 2:The full current passes through A, so its reading is the current times R.
Step 3:The same current produces the potential difference across the parallel pair.
Step 4:All three readings are therefore equal.
Final answer:
Q30Single correctCurrent Electricity
Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2current
Approach:
Charge conservation in a steady state demands equal charge flow per unit time across every cross section. Quantities that depend on area vary, while the net flow rate is fixed.
Step 1:Conservation of charge requires the same current through every cross section in steady state.
Step 2:Current density depends inversely on area, so it varies where the area changes.
Step 3:Drift velocity and electric field both scale with current density and therefore also vary.
Final answer: current
Q31Single correctMagnetic Effects of Current
A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis, while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The field at O combines the contribution of the semicircular arc in the Y-Z plane and the two long straight segments along the X-axis. The Biot-Savart law gives each part, and vector addition yields the total field.
Step 1:The semicircular arc of radius R in the Y-Z plane produces a field along the X-axis at its centre.
Step 2:Each long straight segment behaves as a semi-infinite wire whose field at O adds along the Z-axis.
Step 3:Combining both contributions with the geometry's directions gives the total field vector.
Final answer:
Q32Single correctMagnetic Effects of Current
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A circulating electron constitutes a current loop. The equivalent current depends on the charge and rotation rate, and the field at the centre follows the loop formula.
Step 1:An electron making n rotations per second carries charge ne past a point each second.
Step 2:The field at the centre of a circular loop follows the standard expression.
Final answer:
Q33Single correctElectromagnetic Induction
A conducting square frame of side 'a' and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity 'V'. The emf induced in the frame will be proportional to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each vertical side of the frame is a moving conductor in the wire's field, producing a motional emf. The two sides sit at different distances from the wire, and the net emf is the difference of their contributions.
Step 1:The near and far vertical sides lie at distances (x - a/2) and (x + a/2) from the wire, where x is the distance of the frame centre.
Step 2:Each side gives a motional emf, and the net emf is the difference of the two.
Step 3:Combining the fractions produces a product of the two distances in the denominator.
Final answer:
Q34Single correctAlternating Current
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Power in an AC circuit equals the rms voltage times current times the power factor. Adding an inductor raises the impedance and changes both the current and the power factor for the same source voltage.
Step 1:With only the resistor the power drawn is set by the source voltage and R.
Step 2:After the inductor the impedance is Z and the power factor is R over Z.
Step 3:Eliminating the source voltage using the initial relation gives the new power in terms of P.
Final answer:
Q35Single correctDual Nature of Radiation and Matter
A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = velocity of light):
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Radiation carries momentum equal to its energy divided by the speed of light. On perfect reflection the momentum reverses, so the surface receives twice the incident momentum.
Step 1:Radiation of energy E carries momentum equal to E over C.
Step 2:Perfect reflection reverses the radiation momentum, doubling the transfer to the surface.
Final answer:
Q36Single correctRay Optics
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3-50 cm
Approach:
The system is three lenses in contact: two plano-convex glass lenses and an oil lens formed between their convex surfaces. The powers add, with the oil lens acting as a diverging biconcave element.
Step 1:Each plano-convex glass lens has one flat face and one convex face of radius 20 cm.
Step 2:The oil between the two convex surfaces forms a biconcave lens with both radii 20 cm.
Step 3:Adding the three powers gives the net focal length of the combination.
Final answer: -50 cm
Q37Single correctWave Optics
For a parallel beam of monochromatic light of wavelength '', diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The central maximum extends between the first minima on either side. The angular position of the first minimum sets the linear width on the screen.
Step 1:The first minimum occurs where the path difference across the slit equals one wavelength.
Step 2:The linear half-width on a screen at distance D is D times this angle.
Step 3:The full width spans both sides of the centre.
Final answer:
Q38Single correctWave Optics
In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10.2 mm
Approach:
The central diffraction maximum of one slit must accommodate ten interference fringes of the double slit. Equating the central diffraction width to ten fringe widths gives the slit width.
Step 1:Ten interference fringes occupy ten fringe widths on the screen.
Step 2:This span is set equal to the width of the central diffraction maximum.
Step 3:Solving for the slit width gives a fifth of the slit separation.
Final answer: 0.2 mm
Q39Single correctRay Optics
The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The prism formula relates the refractive index to the prism angle and the angle of minimum deviation. Substituting the given index and simplifying the trigonometric expression yields the minimum deviation.
Step 1:The given refractive index is the cotangent of half the prism angle.
Step 2:Substituting into the prism formula equates the two expressions for the index.
Step 3:The cosine is expressed as a sine of the complementary angle.
Step 4:Equating the arguments and solving gives the minimum deviation.
Final answer:
Q40Single correctDual Nature of Radiation and Matter
A certain metallic surface is illuminated with monochromatic light of wavelength . The stopping potential for photo-electric current for this light is . If the same surface is illuminated with light of wavelength , the stopping potential is . The threshold wavelength for this surface for photo-electric effect is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Einstein's photoelectric equation links the stopping potential to the incident wavelength and the threshold wavelength. Writing the equation for both illuminations and eliminating the work function gives the threshold wavelength.
Step 1:For wavelength lambda the stopping potential is 3V0.
Step 2:For wavelength 2 lambda the stopping potential is V0.
Step 3:Dividing the first equation by the second eliminates the constants and gives a relation between lambda and the threshold wavelength.
Step 4:Solving the relation yields the threshold wavelength.
Final answer:
Q41Single correctDual Nature of Radiation and Matter
Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Graph: P decreasing with as a rectangular hyperbola (P inversely proportional to )
Approach:
The de-Broglie relation makes momentum inversely proportional to wavelength. The correct graph is the one showing this inverse, hyperbolic dependence.
Step 1:The de-Broglie relation links momentum and wavelength inversely through Planck's constant.
Step 2:An inverse proportionality plots as a rectangular hyperbola falling from high values at small wavelength.
Final answer: Graph: P decreasing with as a rectangular hyperbola (P inversely proportional to )
Q42Single correctAtoms
Consider orbit of (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = constant, Z = 2 and h(Planck's constant) = Js]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m/s
Approach:
Bohr's model gives the orbital speed in terms of the nuclear charge and the orbit number. Substituting Z = 2 and n = 3 yields the electron speed.
Step 1:The orbital speed scales as the nuclear charge divided by the orbit number.
Step 2:Substituting the constants with Z = 2 and n = 3 gives the numerical speed.
Step 3:Evaluating the expression produces the electron speed in the third orbit.
Final answer: m/s
Q43Single correctNuclei
If radius of the nucleus is taken to be then the radius of nucleus is nearly:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The nuclear radius depends on the mass number through a cube-root law. Taking the ratio of radii for the two nuclei uses their mass numbers.
Step 1:The radius varies as the cube root of the mass number.
Step 2:The ratio of the tellurium radius to the aluminium radius uses their mass numbers 125 and 27.
Step 3:The tellurium radius follows directly from this ratio.
Final answer:
Q44Single correctSemiconductor Electronics
If in a p-n junction, a square input signal of 10 V is applied, as shown
Then the output across will be
Then the output across will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Output waveform: an upward pulse reaching 5 V (positive square pulse)
Approach:
A diode conducts only when forward biased. The input swings between +5 V and -5 V, so only the positive half passes to the load, giving a half-wave rectified output.
Step 1:The input square signal alternates between +5 V and -5 V.
Step 2:During the positive half the diode is forward biased and conducts, passing +5 V to the load.
Step 3:During the negative half the diode is reverse biased and blocks, so no output appears.
Final answer: Output waveform: an upward pulse reaching 5 V (positive square pulse)
Q45Single correctSemiconductor Electronics
Which logic gate is represented by the following combination of logic gates?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3AND
Approach:
The circuit inverts each input with a NOT gate and feeds both into a NOR gate. Applying De Morgan's theorem to the inverted inputs reveals the equivalent single gate.
Step 1:Each input is first inverted by a NOT gate.
Step 2:The NOR gate produces the inverted sum of its two inputs.
Step 3:De Morgan's theorem reduces this expression to the product of the original inputs.
Final answer: AND
Chemistry45 questions
Q46Single correctChemical Bonding and Molecular Structure
Which of the following species contains equal number of and -bonds?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The number of sigma and pi bonds in each species is counted from its Lewis structure and compared.
Step 1:In XeO4 the central xenon forms four Xe=O double bonds.
Step 2:The four sigma bonds equal the four pi bonds, so the counts match.
Step 3:HCO3- has 4 sigma and 1 pi; (CN)2 has 3 sigma and 4 pi; CH2(CN)2 has 7 sigma and 4 pi, none of which are equal.
Final answer:
Q47Single correctClassification of Elements and Periodicity in Properties
The species Ar, and contain the same number of electrons. In which order do their radii increase?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For isoelectronic species the radius increases as the nuclear charge decreases.
Step 1:All three species have 18 electrons, so they are isoelectronic.
Step 2:The nuclear charges are 20 for Ca2+, 19 for K+ and 18 for Ar.
Step 3:Higher nuclear charge contracts the electron cloud, so the radius increases in the order Ca2+ < K+ < Ar.
Final answer:
Q48Single correctThe s-Block Elements
The function of “Sodium pump” is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a constituent of this pump?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The ionic species exchanged together with sodium across the cell membrane is identified.
Step 1:The sodium pump is the Na+/K+ ATPase that moves ions across the cell membrane.
Step 2:It pumps three sodium ions out and two potassium ions into the cell per cycle.
Step 3:The companion ion of the sodium pump is therefore the potassium ion.
Final answer:
Q49Single correctThe p-Block Elements
“Metals are usually not found as nitrates in their ores”. Out of the following two
(a and b) reasons which is/are true for the above observation?
(a) Metal nitrates are highly unstable
(b) Metal nitrates are highly soluble in water
(a and b) reasons which is/are true for the above observation?
(a) Metal nitrates are highly unstable
(b) Metal nitrates are highly soluble in water
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a is false but b is true
Approach:
Each statement about metal nitrates is judged against known solubility and stability facts.
Step 1:Metal nitrates are not unstable enough to forbid their existence, so statement (a) is false.
Step 2:Nearly all metal nitrates are highly soluble in water and are leached away by rain, so statement (b) is true.
Step 3:High water solubility prevents nitrates from accumulating as ores, matching the option a false but b true.
Final answer: a is false but b is true
Q50Single correctThe s-Block Elements
Solubility of the alkaline earth’s metal sulphates in water decreases in the sequence :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The solubility trend of group 2 sulphates is linked to hydration and lattice energies down the group.
Step 1:The hydration energy of the cation falls faster than the lattice energy on descending the group.
Step 2:A larger drop in hydration energy makes the heavier sulphates less soluble.
Step 3:The solubility therefore decreases in the order Mg > Ca > Sr > Ba.
Final answer:
Q51Single correctThe d- and f-Block Elements
Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii? (Numbers in the parenthesis are atomic numbers).
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Zr (40) and Hf (72)
Approach:
The pair whose radii are equalised by the lanthanoid contraction is identified.
Step 1:The lanthanoid contraction reduces the size increase expected from the fourth to the fifth transition row.
Step 2:Zirconium and hafnium lie in the same group, with hafnium following the lanthanoids.
Step 3:The contraction makes the radii of Zr and Hf almost identical.
Zr (40) and Hf (72)
Final answer: Zr (40) and Hf (72)
Q52Single correctRedox Reactions
Which of the following processes does not involve oxidation of iron?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Formation of from Fe
Approach:
The oxidation state of iron in each process is examined to find where it is unchanged.
Step 1:In rusting, in displacement of copper and in reaction with steam, iron goes from the zero state to a positive oxidation state.
Step 2:In iron pentacarbonyl the oxidation state of iron remains zero.
Step 3:Formation of Fe(CO)5 therefore does not involve oxidation of iron.
Final answer: Formation of from Fe
Q53Single correctChemical Bonding and Molecular Structure
Which of the following pairs of ions are isoelectronic and isostructural?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each pair is tested for equal electron counts and the same shape.
Step 1:Both ClO3- and - have the same number of valence electrons.
Step 2:Both central atoms carry one lone pair and three bond pairs, giving a trigonal pyramidal shape.
Step 3:The other pairs differ in electron count or shape, so only ClO3- and - match.
Final answer:
Q54Single correctChemical Bonding and Molecular Structure
Which of the following options represents the correct bond order?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Bond order is calculated for each oxygen species from molecular orbital electron counts.
Step 1:Removing an electron from O2 gives O2+ with a bond order of 2.5.
Step 2:Neutral O2 has a bond order of 2.0 and adding an electron gives O2- with 1.5.
Step 3:Arranging these values gives the order O2- < O2 < O2+.
Final answer:
Q55Single correctThe p-Block Elements
Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4is used as a food-preservative
Approach:
Each listed property is checked for whether both oxides or only one of them displays it.
Step 1:Both NO2 and SO2 dissolve in water, act as reducing agents and contribute to acid rain.
Step 2:Only sulphur dioxide is used as a food preservative because of its antimicrobial action.
Step 3:The distinguishing property is therefore use as a food preservative.
Final answer: is used as a food-preservative
Q56Single correctChemical Bonding and Molecular Structure
Maximum bond angle at nitrogen is present in which of the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The bond angle at nitrogen is compared across the species using the electron pairs on nitrogen.
Step 1:NO2+ has no lone pair on nitrogen and is linear with a bond angle of 180 degrees.
Step 2:NO2, NO2- and NO3- all have lone or single electrons on nitrogen that reduce the angle below 180 degrees.
Step 3:The maximum bond angle therefore occurs in NO2+.
Final answer:
Q57Single correctThe d- and f-Block Elements
Magnetic moment 2.84 B.M. is given by:
(At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27)
(At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The number of unpaired electrons giving a moment of 2.84 B.M. is found and matched to an ion.
Step 1:A moment of 2.84 B.M. corresponds to two unpaired electrons.
Step 2:Ni2+ has a 3d8 configuration with exactly two unpaired electrons.
Step 3:Ti3+ has one, Cr2+ has four and Co2+ has three unpaired electrons, so only Ni2+ fits.
Final answer:
Q58Single correctCoordination Compounds
Cobalt (III) Chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at 25°C?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Only chloride present as a free counter ion is precipitated, so the complex with no ionisable chloride is selected.
Step 1:In CoCl3.3NH3 all three chloride ions occupy coordination sites as [Co(NH3)3Cl3].
Step 2:Without free chloride ions the complex cannot react with silver nitrate.
Step 3:The other complexes release one or more chloride ions and do give the test.
Final answer:
Q59Single correctCoordination Compounds
Which of these statements about is true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 has no unpaired electrons and will be in a low-spin configuration
Approach:
The spin state of the cobalt(III) cyanide complex is found from the strength of the ligand field.
Step 1:Cyanide is a strong field ligand and cobalt is in the +3 state with a 3d6 configuration.
Step 2:The strong field causes all six d electrons to pair in the t2g set.
Step 3:All six electrons occupy the lower t2g set in pairs, so the complex has zero unpaired electrons and is diamagnetic. For a d6 centre, having zero unpaired electrons IS the low-spin arrangement: the high-spin d6 case would place four electrons unpaired. So 'no unpaired electrons' and 'low-spin' necessarily describe the same state, and cannot be combined with 'high-spin'.
Final answer: has no unpaired electrons and will be in a low-spin configuration
Q60Single correctChemical Kinetics
The activation energy of a reaction can be determined from the slope of which of the following graphs?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 vs.
Approach:
The linear form of the Arrhenius equation is used to identify the graph whose slope gives activation energy.
Step 1:The logarithmic Arrhenius equation is linear in ln K against the reciprocal of temperature.
Step 2:The slope of this line equals the negative of activation energy divided by the gas constant.
Step 3:The required plot is therefore ln K against 1/T.
Final answer: vs.
Q61Single correctSolutions
Which one is not equal to zero for an ideal solution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The mixing quantities of an ideal solution are recalled to find the one that is non-zero.
Step 1:For an ideal solution the enthalpy and volume of mixing are zero and Raoult's law is obeyed exactly.
Step 2:Mixing always increases disorder, so the entropy of mixing is positive.
Step 3:The quantity that is not equal to zero is the entropy of mixing.
Final answer:
Q62Single correctSome Basic Concepts of Chemistry
A mixture of gases contains and gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 24:1
Approach:
Moles of each gas are obtained from the given mass ratio and divided by their molar masses.
Step 1:Taking 1 g of hydrogen gives 0.5 mol since its molar mass is 2 g per mole.
Step 2:Taking 4 g of oxygen gives 0.125 mol since its molar mass is 32 g per mole.
Step 3:Dividing the mole values gives a hydrogen to oxygen molar ratio of 4 to 1.
Final answer: 4:1
Q63Single correctThe Solid State
A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2127 pm
Approach:
Four atoms per unit cell indicate a face centred cubic lattice, whose radius relates to the edge length.
Step 1:Four atoms per unit cell correspond to a face centred cubic arrangement.
Step 2:Substituting the edge length of 361 pm into the FCC relation gives the radius.
Step 3:Evaluating the expression gives a radius of about 127 pm.
Final answer: 127 pm
Q64Single correctChemical Kinetics
When initial concentration of a reactant is doubled in a reaction, its half – life period is not affected. The order of the reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2First
Approach:
The dependence of half life on initial concentration is matched to the reaction order.
Step 1:The half life of a first order reaction depends only on the rate constant.
Step 2:Doubling the initial concentration leaves this half life unchanged.
Step 3:The reaction therefore follows first order kinetics.
Final answer: First
Q65Single correctEquilibrium
If the value of an equilibrium constant for a particular reaction is , then at equilibrium the system will contain
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Mostly products
Approach:
The magnitude of the equilibrium constant indicates the extent to which the reaction favours products.
Step 1:A very large equilibrium constant means the product concentrations far exceed the reactant concentrations.
Step 2:The equilibrium position lies far to the product side.
Step 3:The system therefore contains mostly products.
Final answer: Mostly products
Q66Single correctElectrochemistry
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Fuel cell
Approach:
The device that turns combustion energy of a fuel directly into electricity is identified by definition.
Step 1:A fuel cell converts the chemical energy of fuel combustion directly into electrical energy.
Step 2:Hydrogen and methane are common fuels used in such cells.
Step 3:The described device is therefore a fuel cell.
Final answer: Fuel cell
Q67Single correctSolutions
The boiling point of 0.2 mol k solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1X is undergoing dissociation in water
Approach:
Elevation in boiling point depends on the number of particles, so the higher value points to dissociation.
Step 1:Boiling point elevation is a colligative property that rises with the number of solute particles.
Step 2:At equal molality the larger elevation for X means it produces more particles than Y.
Step 3:An increase in particle number arises from dissociation, so X is dissociating in water.
Final answer: X is undergoing dissociation in water
Q68Single correctSolutions
Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of (if all are 100% ionized)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The number of ions produced on complete ionisation is counted for each electrolyte and matched to that of aluminium sulphate.
Step 1:Aluminium sulphate ionises into two aluminium and three sulphate ions, giving five particles.
Step 2:Potassium ferrocyanide ionises into four potassium ions and one complex ion, giving five particles.
Step 3:The matching van't Hoff factor of 5 belongs to potassium ferrocyanide.
Final answer:
Q69Single correctClassification of Elements and Periodicity in Properties
The number of d-electrons in () is not equal to the number of electrons in which one of the following:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2p-electrons in Cl ()
Approach:
The number of d-electrons in the given cation is counted and matched against the electron counts in the listed options.
Step 1:The configuration of iron is . Removal of two electrons from the 4s orbital gives as , so the number of d-electrons is 6.
Step 2:Magnesium () has configuration , giving six s-electrons; iron in its neutral state retains six d-electrons; neon () has six p-electrons. Each of these equals 6.
Step 3:Chlorine () has configuration , giving a total of eleven p-electrons, which differs from 6.
Final answer: p-electrons in Cl ()
Q70Single correctChemical Bonding and Molecular Structure
The correct bond order in the following species is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Bond order for each oxygen species is computed from molecular orbital electron counts and the values are arranged in increasing order.
Step 1:Neutral dioxygen has 16 electrons with bond order 2. Removing one electron from an antibonding orbital raises the bond order, while adding an electron to an antibonding orbital lowers it.
Step 2:For (14 electrons) the bond order is 3; for (15 electrons) the bond order is 2.5; for (17 electrons) the bond order is 1.5.
Step 3:Arranging the three species by increasing bond order gives .
Final answer:
Q71Single correctStructure of Atom
The angular momentum of electron in 'd' orbital is equal to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The orbital angular momentum is obtained from the azimuthal quantum number of the d orbital.
Step 1:For a d orbital the azimuthal quantum number is .
Step 2:Substituting into the angular momentum expression gives .
Step 3:Expressed in the printed form where h denotes , the magnitude equals .
Final answer:
Q72Single correctEquilibrium
The of , , and are respectively, , , , . Which one of the following salts will precipitate last if solution is added to the solution containing equal moles of , , and ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The silver ion concentration required to begin precipitation of each salt is computed, and the salt needing the highest concentration precipitates last.
Step 1:For the silver halides, the silver ion concentration needed to start precipitation is . With equal anion concentrations, a higher requires a higher silver concentration.
Step 2:For silver chromate the required silver ion concentration is , which is of the order , much larger than the values for the halides which are of the order or smaller.
Step 3:Since silver chromate demands the highest silver ion concentration to begin precipitating, it is the last salt to precipitate.
Final answer:
Q73Single correctSurface Chemistry
Which property of colloidal solution is independent of charge on the colloidal particles?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Tyndall effect
Approach:
Each listed property is assessed for its dependence on the charge carried by colloidal particles.
Step 1:Coagulation involves neutralisation of the charge on colloidal particles by added electrolytes, so it depends on the particle charge.
Step 2:Electrophoresis and electro-osmosis are migrations driven by an applied electric field acting on charged species, so both depend on the particle charge.
Step 3:The Tyndall effect is the scattering of light by colloidal particles owing to their size, an optical phenomenon independent of any charge they carry.
Final answer: Tyndall effect
Q74Single correctEquilibrium
Which of the following statements is correct for a reversible process in a state of equilibrium?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The thermodynamic relation linking standard Gibbs energy change to the equilibrium constant is identified.
Step 1:The standard Gibbs energy change relates to the equilibrium constant through .
Step 2:Converting the natural logarithm to base ten introduces the factor 2.303, giving .
Step 3:At equilibrium the actual Gibbs energy change is zero, so the correct general relation involves the standard Gibbs energy, matching the printed value of 2.30 for the conversion factor.
Final answer:
Q75Single correctChemistry in Everyday Life
Bithional is generally added to the soaps as an additive to function as a/an:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Antiseptic
Approach:
The function of bithional as a soap additive is recalled from its chemical role.
Step 1:Bithional is a chlorinated phenolic compound with germicidal activity.
Step 2:When incorporated into soaps it reduces the bacterial population responsible for body odour, acting as an antiseptic.
Step 3:The remaining options describe physical or pH-related roles unrelated to bithional's germicidal purpose.
Final answer: Antiseptic
Q76Single correctAmines
The electrolytic reduction of nitrobenzene in strongly acidic medium produces:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1P-Aminophenol
Approach:
The product of electrolytic reduction of nitrobenzene is determined from the medium-dependent reduction pathway.
Step 1:Electrolytic reduction of nitrobenzene first yields phenylhydroxylamine as an intermediate.
Step 2:In strongly acidic medium the phenylhydroxylamine undergoes an acid-catalysed rearrangement in which the hydroxyl group migrates to the para position of the ring.
Step 3:The rearrangement gives p-aminophenol as the final product.
Final answer: P-Aminophenol
Q77Single correctOrganic Chemistry - Some Basic Principles and Techniques
In Duma's method for estimation of nitrogen 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 316.76
Approach:
The collected nitrogen volume is corrected to standard conditions, converted to mass, and expressed as a percentage of the compound mass.
Step 1:The pressure of dry nitrogen is the total pressure less the aqueous tension, giving mm.
Step 2:Correcting the 40 mL of gas to standard conditions gives , which equals about 33.5 mL.
Step 3:Substituting into the percentage expression gives , which equals about 16.76 percent.
Final answer: 16.76
Q78Single correctHaloalkanes and Haloarenes
In which of the following compounds, the C-Cl bond ionization shall give most stable carbonium ion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The carbocation formed by ionization of each C-Cl bond is evaluated for stabilisation by resonance and hyperconjugation.
Step 1:Ionization of the benzylic chloride gives a benzyl cation, which is delocalised over the aromatic ring by resonance.
Step 2:The tertiary and secondary alkyl chlorides give carbocations stabilised only by hyperconjugation and induction, which is weaker than resonance.
Step 3:The chloride bearing an electron-withdrawing nitro group gives a strongly destabilised cation. Resonance delocalisation in the benzyl cation provides the greatest stabilisation.
Final answer:
Q79Single correctAlcohols, Phenols and Ethers
The reaction
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Williamson Synthesis
Approach:
The named reaction is identified from the reactants and the type of product formed.
Step 1:An alkoxide ion reacts with an alkyl halide, displacing the halide and forming an ether linkage.
Step 2:This nucleophilic substitution of an alkyl halide by an alkoxide is the defining feature of the Williamson ether synthesis.
Step 3:The Etard and Gattermann-Koch reactions concern oxidation and formylation of aromatic rings, not ether formation.
Final answer: Williamson Synthesis
Q80Single correctHydrocarbons
The reaction of with HBr produces
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The regioselectivity of HBr addition across the conjugated double bond is decided by the stability of the intermediate carbocation.
Step 1:Addition of the proton can place the positive charge on either alkene carbon; the more stable cation determines the product.
Step 2:Placing the proton on the carbon bearing the methyl group generates a benzylic carbocation on the carbon adjacent to the ring, which is stabilised by resonance with the phenyl group.
Step 3:Bromide adds to the benzylic carbon, giving .
Final answer:
Q81Single correctHydrocarbons
A single compound of the structure is obtainable from ozonolysis of which of the following cyclic compounds
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11,4-Dimethylcyclopent-1-ene
Approach:
Reverse ozonolysis is applied: the carbonyl carbons of the dicarbonyl product are rejoined into a ring and the substituent pattern is matched.
Step 1:Ozonolysis cleaves a carbon-carbon double bond and converts each alkene carbon into a carbonyl carbon. A single product from a cyclic alkene indicates the double bond lies within the ring.
Step 2:The product carries an aldehyde at one end and a methyl ketone at the other, so before cleavage one double-bond carbon bore a hydrogen and the other bore a methyl group.
Step 3:Joining the two carbonyl carbons into a ring puts the double bond between them, so the methyl that was on the ketone carbon sits on an alkene carbon (C-1) while the methyl on the central saturated carbon sits at C-4, flanked by the two CH2 groups.
Final answer: 1,4-Dimethylcyclopent-1-ene
Q82Single correctOrganic Chemistry - Some Basic Principles and Techniques
Treatment of cyclopentanone with methyl lithium gives which of the following species?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cyclopentanonyl anion
Approach:
The species generated when an organolithium base acts on a ketone bearing alpha hydrogens is identified.
Step 1:Methyl lithium is a strong base that abstracts an acidic alpha hydrogen adjacent to the carbonyl group.
Step 2:Removal of the alpha proton leaves a pair of electrons on the alpha carbon, producing a carbanion stabilised by the adjacent carbonyl.
Step 3:This negatively charged species corresponds to the cyclopentanonyl anion.
Final answer: Cyclopentanonyl anion
Q83Single correctOrganic Chemistry - Some Basic Principles and Techniques
Consider the following compounds: (I) , (II) the triphenylmethyl radical , and (III) a bicyclo[2.2.1] cage bearing a radical centre with a group on the adjacent carbon. Hyperconjugation occurs in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3III only
Approach:
Each carbocation is checked for the presence of alpha carbon-hydrogen bonds aligned with the empty orbital, which is required for hyperconjugation.
Step 1:Hyperconjugation requires carbon-hydrogen bonds on the carbon adjacent to the electron-deficient centre that can overlap with the vacant orbital.
Step 2:Structure III bears alpha hydrogens on carbons adjacent to the electron-deficient centre that are oriented for overlap, allowing hyperconjugative delocalisation.
Step 3:The triphenylmethyl radical has only phenyl groups on the radical carbon, and in structure I the neighbouring carbon is quaternary, so neither has a C-H bond adjacent to the radical centre; only structure III does.
Final answer: III only
Q84Single correctOrganic Chemistry - Some Basic Principles and Techniques
Which of the following is the most correct electron displacement of a nucleophilic reaction to take place
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Inductive arrow from , a curved arrow from the double bond toward the chlorine-bearing carbon, and a curved arrow leaving the C-Cl bond
Approach:
The curved arrows must show the pi electrons of the double bond moving toward the carbon that carries chlorine at the same time as the C-Cl bond breaks heterolytically.
Step 1:In a nucleophilic reaction on an allylic halide the pi electrons of the double bond shift in concert with departure of the leaving group.
Step 2:Only one depiction shows both movements together: the double-bond electrons shifting toward the chlorine-bearing carbon and the C-Cl bond releasing the chloride.
Step 3:The methyl group's inductive push reinforces this flow, so the displacement drawn with all three arrows is the correct one.
Final answer: Inductive arrow from , a curved arrow from the double bond toward the chlorine-bearing carbon, and a curved arrow leaving the C-Cl bond
Q85Single correctOrganic Chemistry - Some Basic Principles and Techniques
The enolic form of ethyl acetoacetate as below has:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 118 sigma bonds and 2 pi-bonds
Approach:
Every single bond in the enol structure is counted as one sigma bond and each double bond as one sigma plus one pi bond.
Step 1:The enolic form of ethyl acetoacetate contains a carbon-carbon double bond and a carbonyl double bond, contributing two pi bonds.
Step 2:Counting every carbon-hydrogen, carbon-carbon, carbon-oxygen and oxygen-hydrogen single bond, together with the sigma components of the two double bonds, gives a total of eighteen sigma bonds.
Step 3:The molecule therefore contains eighteen sigma bonds and two pi bonds.
Final answer: 18 sigma bonds and 2 pi-bonds
Q86Single correctAldehydes, Ketones and Carboxylic Acids
Given the three cyclic ketones shown, (I) 4,4-dimethylcyclohex-2-en-1-one, (II) 3,3-dimethylcyclohexan-1-one and (III) 6,6-dimethylcyclohex-2-en-1-one. Which of the given compounds can exhibit tautomerism?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4I, II and III
Approach:
Keto-enol tautomerism requires a carbonyl group bearing at least one alpha hydrogen; each structure is tested for this feature.
Step 1:Keto-enol tautomerism demands a carbonyl carbon with a hydrogen on an adjacent alpha carbon.
Step 2:In I and II a saturated ring carbon next to the carbonyl still carries hydrogen, so each enolises directly. In III both neighbours of the carbonyl are blocked - one is an alkene carbon and the other carries two methyl groups - but the CH2 in the gamma position allows the extended dienol tautomer of the conjugated enone.
Step 3:Since each compound can reach an enol form, all three show keto-enol tautomerism.
Final answer: I, II and III
Q87Single correctHydrocarbons
Given the three isomers shown, (I) 1,3,5-trimethylbenzene, (II) a non-aromatic isomer carrying two ring double bonds, two methyl groups and one exocyclic , and (III) a saturated ring carrying three separate exocyclic groups. The enthalpy of hydrogenation of these compounds will be in the order as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2III > II > I
Approach:
The enthalpy of hydrogenation is greater for less stable alkenes, so the relative stability of the double bonds in each structure is ranked.
Step 1:A more substituted and more stable double bond releases less heat on hydrogenation, while a less substituted, less stable double bond releases more heat.
Step 2:Structure I is aromatic, so its three double bonds are stabilised by delocalisation over the whole ring and it is the most stable. Structure II keeps two of its double bonds conjugated inside the ring with the exocyclic methylene, giving partial stabilisation. Structure III has three isolated, identically substituted exocyclic double bonds with no conjugation at all and is the least stable.
Step 3:Reversing the stability order gives the enthalpy of hydrogenation order III > II > I.
Final answer: III > II > I
Q88Single correctPolymers
Biodegradable polymer which can be produced from glycine and aminocaproic acid is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Nylon 2-nylon 6
Approach:
The biodegradable polymer formed from the two named monomers is identified from the carbon counts of glycine and aminocaproic acid.
Step 1:Glycine is an amino acid with two carbon atoms, and aminocaproic acid is a six-carbon amino acid.
Step 2:Their alternating condensation produces a copolyamide named after the carbon counts of the two repeating units.
Step 3:This biodegradable copolymer is nylon 2-nylon 6.
Final answer: Nylon 2-nylon 6
Q89Single correctOrganic Chemistry - Some Basic Principles and Techniques
The total number of -bond electrons in the following structure is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 28
Approach:
Each pi bond in the structure is counted and multiplied by two to give the total number of pi-bond electrons.
Step 1:Each carbon-carbon double bond contains one pi bond, and each pi bond holds two electrons.
Step 2:The structure contains four carbon-carbon double bonds.
Step 3:Multiplying four pi bonds by two electrons each gives a total of eight pi-bond electrons.
Final answer: 8
Q90Single correctAldehydes, Ketones and Carboxylic Acids
An organic compound 'X' having molecular formula yields phenyl hydrazone and gives negative response to the Iodoform test and Tollens' test. It produces n-pentane on reduction. 'X' could be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 33-pentanone
Approach:
The chemical tests are interpreted to narrow the carbonyl compound, and reduction to n-pentane fixes the carbon skeleton.
Step 1:Formation of a phenylhydrazone shows a carbonyl group, while a negative Tollens' test rules out an aldehyde, indicating a ketone.
Step 2:A negative iodoform test rules out a methyl ketone, eliminating 2-pentanone.
Step 3:Reduction to a straight chain n-pentane requires a five-carbon ketone with the carbonyl on the central carbon, which is 3-pentanone.
Final answer: 3-pentanone
Biology90 questions
Q91Single correctBiological Classification
Match the fungus with its character and class.
| List I | List II |
|---|---|
| (1). | (1). Aseptate mycelium |
| (2). | (2). Sexual reproduction absent |
| (3). | (3). Reproduction by conjugation |
| (4). | (4). Parasitic fungus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 — Sexual reproduction absent — Deuteromycetes
Approach:
Each fungal genus is paired with a defining reproductive feature and the class to which it belongs.
Step 1:Alternaria reproduces only asexually by conidia, so its sexual stage is absent.
Step 2:Fungi whose sexual stage is unknown are placed in the form-class Deuteromycetes, the fungi imperfecti.
Step 3:The other pairings are wrong: Phytophthora has aseptate mycelium but is an oomycete, Mucor reproduces by conjugation and is a zygomycete, and Agaricus is saprophytic and basidiomycetous.
Final answer: — Sexual reproduction absent — Deuteromycetes
Q92Single correctPlant Kingdom
Read the following five statements (A to E) and select the option with all correct statements:
(A) Mosses and Lichens are the first organisms to colonise a bare rock
(B) Selaginella is a homosporous pteridophyte.
(C) Coralloid roots in Cycas have VAM
(D) Main plant body in bryophytes is gametophytic, whereas in pteridophytes it is sporophytic
(E) In gymnosperms, male and female gametophytes are present within sporangia located on sporophyte.
(A) Mosses and Lichens are the first organisms to colonise a bare rock
(B) Selaginella is a homosporous pteridophyte.
(C) Coralloid roots in Cycas have VAM
(D) Main plant body in bryophytes is gametophytic, whereas in pteridophytes it is sporophytic
(E) In gymnosperms, male and female gametophytes are present within sporangia located on sporophyte.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A),(D) and (E)
Approach:
Each statement is judged true or false against textbook facts and the combination of true statements is selected.
Step 1:Mosses and lichens are pioneer colonisers of bare rock, so (A) is true.
Step 2:Selaginella is heterosporous, producing two spore sizes, so (B) is false.
Step 3:Coralloid roots of Cycas harbour nitrogen-fixing cyanobacteria, not VAM fungi, so (C) is false.
Step 4:The dominant plant body is gametophytic in bryophytes and sporophytic in pteridophytes, so (D) is true; in gymnosperms the reduced gametophytes lie within sporangia borne on the sporophyte, so (E) is true.
Final answer: (A),(D) and (E)
Q93Single correctPlant Kingdom
In which of the following gametophyte is not independent free living?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The plant whose gametophyte is retained on and nourished by the sporophyte is identified.
Step 1:Funaria and Marchantia are bryophytes whose gametophyte is the dominant free-living phase.
Step 2:Pteris is a fern whose prothallus is an independent photosynthetic gametophyte.
Step 3:Pinus is a gymnosperm in which the highly reduced gametophytes develop within the ovule and remain dependent on the sporophyte.
Final answer:
Q94Single correctPlant Kingdom
Which one of the following statements is wrong?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Mannitol is stored food in Rhodophyceae
Approach:
Each statement on algal products and storage food is evaluated and the incorrect one is chosen.
Step 1:Algin and carrageenan are hydrocolloids derived from brown and red algae, so statement 1 is correct.
Step 2:Agar is extracted from the red algae Gelidium and Gracilaria, so statement 2 is correct.
Step 3:Chlorella and Spirulina are protein-rich and used as supplementary space food, so statement 3 is correct.
Step 4:Stored food in Rhodophyceae is floridean starch, while mannitol is stored in Phaeophyceae, so statement 4 is wrong.
Final answer: Mannitol is stored food in Rhodophyceae
Q95Single correctBiological Classification
The guts of cow and buffalo possess
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Methanogens
Approach:
The organisms inhabiting the rumen of cattle are identified.
Step 1:The gut of ruminants such as cow and buffalo contains methanogenic archaebacteria.
Step 2:These methanogens act on cellulose and produce methane gas as a by-product.
Final answer: Methanogens
Q96Single correctPlant Kingdom
Male gametes are flagellated in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ectocarpus
Approach:
The alga producing motile, flagellated male gametes is selected.
Step 1:Ectocarpus is a brown alga whose gametes are biflagellate and motile.
Step 2:Polysiphonia is a red alga with non-motile gametes, Anabaena is a cyanobacterium lacking gametes, and Spirogyra reproduces by conjugation with amoeboid non-flagellate gametes.
Final answer: Ectocarpus
Q97Single correctAnatomy of Flowering Plants
Vascular bundles in monocotyledons are considered closed because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cambium is absent
Approach:
The defining feature of a closed vascular bundle is recalled.
Step 1:A vascular bundle is termed closed when it lacks intrafascicular cambium between xylem and phloem.
Step 2:Monocot bundles lack cambium and therefore show no secondary growth.
Final answer: Cambium is absent
Q98Single correctMorphology of Flowering Plants
⚥ is the floral formula of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Petunia
Approach:
The floral symbols are decoded and matched to the family and its representative genus.
Step 1:The formula shows an actinomorphic bisexual flower with gamosepalous calyx of five, gamopetalous corolla of five, five stamens and a bicarpellary syncarpous superior ovary.
Step 2:These characters define the family Solanaceae, of which Petunia is a member.
Final answer: Petunia
Q99Single correctAnatomy of Flowering Plants
A major characteristic of the monocot root is the presence of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Vasculature without cambium
Approach:
The anatomical hallmark distinguishing monocot roots is identified.
Step 1:Monocot roots are polyarch with xylem and phloem on alternate radii and lack a vascular cambium.
Step 2:Absence of cambium means monocot roots show no secondary growth.
Final answer: Vasculature without cambium
Q100Single correctMorphology of Flowering Plants
Keel is the characteristic feature of flower of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The plant bearing a keel-shaped corolla is selected.
Step 1:A keel is the boat-shaped pair of fused anterior petals in the papilionaceous corolla of legumes.
Step 2:Indigofera belongs to Fabaceae and possesses such a keel, while Tulip, Aloe and Tomato lack it.
Final answer:
Q101Single correctMorphology of Flowering Plants
Perigynous flowers are found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Rose
Approach:
The flower with a half-inferior ovary, defining the perigynous condition, is identified.
Step 1:In a perigynous flower the ovary is half inferior with the other floral parts borne on the rim of a cup-shaped thalamus.
Step 2:Rose shows this arrangement, while guava and cucumber are epigynous and China rose is hypogynous.
Final answer: Rose
Q102Single correctMorphology of Flowering Plants
Leaves become modified into spins in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The plant whose leaves are transformed into spines is selected.
Step 1:In Opuntia the leaves are reduced to spines that lower transpiration, while the flattened stem carries out photosynthesis.
Step 2:In pea the leaf tip forms a tendril, in onion the leaf base becomes fleshy scales, and in silk cotton the spines are stem outgrowths.
Final answer:
Q103Single correctCell - The Unit of Life
The structures that are formed by stacking of organised flattened membranous sacs in the chloroplasts are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Grana
Approach:
The chloroplast structure formed by stacked thylakoids is identified.
Step 1:Thylakoids are flattened membranous sacs that stack like a pile of coins to form grana.
Step 2:Cristae are mitochondrial infoldings, stroma lamellae connect grana, and stroma is the fluid matrix.
Final answer: Grana
Q104Single correctCell - The Unit of Life
The chromosomes in which centromere is situated close to one end are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Acrocentric
Approach:
Chromosomes are classified by centromere position and the type with a near-terminal centromere is chosen.
Step 1:An acrocentric chromosome has its centromere situated close to one end, giving one very long and one very short arm.
Step 2:Metacentric is median, sub-metacentric is slightly off-centre, and telocentric is exactly terminal.
Final answer: Acrocentric
Q105Single correctCell - The Unit of Life
Select the correct matching in the following pairs
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Smooth ER – Synthesis of lipids
Approach:
Each organelle function pair is checked and the correct match is selected.
Step 1:Smooth endoplasmic reticulum is the principal site of lipid and steroid synthesis, so this pairing is correct.
Step 2:Rough ER bears ribosomes and synthesises proteins, not glycogen or fatty acid oxidation, so its pairings are wrong.
Final answer: Smooth ER – Synthesis of lipids
Q106Single correctBiological Classification
True nucleus is absent in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The prokaryotic organism lacking a membrane-bound nucleus is identified.
Step 1:Anabaena is a cyanobacterium and therefore a prokaryote without a true membrane-bound nucleus.
Step 2:Mucor, Vaucheria and Volvox are eukaryotes with well-defined nuclei.
Final answer:
Q107Single correctCell - The Unit of Life
Which one of the following is not an inclusion body found in prokaryotes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Polysome
Approach:
Each structure is tested against the definition of an inclusion body and the exception is chosen.
Step 1:Phosphate, cyanophycean and glycogen granules are reserve inclusion bodies stored freely in the prokaryotic cytoplasm.
Step 2:A polysome is a chain of ribosomes translating an mRNA and is not a storage inclusion body.
Final answer: Polysome
Q108Single correctTransport in Plants
Transpiration and root pressure cause water to rise in plants by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pulling and pushing it, respectively
Approach:
The directional force exerted by each mechanism on the water column is matched in order.
Step 1:Transpiration generates a pull at the leaf surface that draws the water column upward through the xylem.
Step 2:Root pressure develops in the roots and pushes water upward from below.
Final answer: Pulling and pushing it, respectively
Q109Single correctMineral Nutrition
Minerals known to be required in large amounts for plant growth include
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Phosphorus, potassium, sulphur, calcium
Approach:
The option in which every listed element is a macronutrient is selected.
Step 1:Macronutrients required in large amounts include phosphorus, potassium, sulphur and calcium.
Step 2:The other options contain micronutrients such as manganese, copper, selenium, boron, iron and zinc.
Final answer: Phosphorus, potassium, sulphur, calcium
Q110Single correctPlant Growth and Development
What causes a green plant exposed to the light on only one side to bend toward the source of light as it grows?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Auxin accumulates on the shaded side, stimulating greater cell elongation there.
Approach:
The hormonal mechanism of bending toward unilateral light is identified.
Step 1:Unilateral light drives auxin laterally to the shaded side of the shoot.
Step 2:Higher auxin on the shaded side promotes greater cell elongation there, so the shoot curves toward the light.
Final answer: Auxin accumulates on the shaded side, stimulating greater cell elongation there.
Q111Single correctTransport in Plants
In a ring girdled plant
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The root dies first
Approach:
The effect of removing the bark ring, which carries phloem, on the plant parts is reasoned out.
Step 1:Girdling removes the phloem ring, blocking downward transport of food from leaves to roots.
Step 2:Deprived of food the root is the first part to die, while the shoot continues for a time.
Final answer: The root dies first
Q112Single correctPlant Growth and Development
Typical growth curve in plants is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sigmoid
Approach:
The characteristic shape of the plant growth curve is recalled.
Step 1:Plant growth over time passes through lag, log and stationary phases, producing an S-shaped curve.
Step 2:An S-shaped curve is described as sigmoid.
Final answer: Sigmoid
Q113Single correctTransport in Plants
Which one gives the most valid and recent explanation for stomatal movements?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Potassium influx and efflux
Approach:
The currently accepted mechanism driving guard cell turgor changes is identified.
Step 1:Active influx of potassium ions into guard cells lowers their water potential and draws water in, raising turgor so the stoma opens.
Step 2:Efflux of potassium ions reverses the turgor change and the stoma closes.
Final answer: Potassium influx and efflux
Q114Single correctSexual Reproduction in Flowering Plants
The hilum is a scar on the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Seed, where funicle was attached
Approach:
The hilum is a structural mark on the seed coat that records the seed's former attachment to the parent plant.
Step 1:The ovule is connected to the placenta of the ovary by a stalk called the funicle.
Step 2:After fertilisation the ovule matures into a seed, and the point where the funicle was joined to the body of the ovule remains as a scar on the seed.
Final answer: Seed, where funicle was attached
Q115Single correctSexual Reproduction in Flowering Plants
Which one of the following may require pollinators, but is genetically similar to autogamy?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Geitonogamy
Approach:
The mode of pollination that involves transfer of pollen between flowers of the same plant is compared with self-pollination.
Step 1:Geitonogamy is the transfer of pollen from the anther of one flower to the stigma of another flower of the same plant, and this transfer is mediated by a pollinating agent.
Step 2:Because both flowers belong to the same plant, the pollen and ovule are genetically identical, making the outcome functionally the same as autogamy.
Final answer: Geitonogamy
Q116Single correctSexual Reproduction in Flowering Plants
Which one of the following statement is not true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Honey is made by bees by digesting pollen collected from flowers
Approach:
Each statement on pollen biology is evaluated for factual accuracy to identify the incorrect one.
Step 1:Honey is produced by bees from floral nectar, not from pollen, and the nectar is processed and concentrated within the hive.
Step 2:The other three statements are correct: pollen supplements exist as tablets and syrups, pollen causes allergies such as pollinosis, and fly- and bat-pollinated flowers often emit a foul odour.
Final answer: Honey is made by bees by digesting pollen collected from flowers
Q117Single correctSexual Reproduction in Flowering Plants
Transmission tissue is characteristics feature of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Solid style
Approach:
The tissue guiding pollen tube growth through the style is identified based on style structure.
Step 1:In a solid style the central core is filled with specialised cells forming the transmitting tissue, through which the pollen tube grows toward the ovary.
Step 2:A hollow style instead has an open central canal lined with secretory cells, so a packed transmitting tissue is absent.
Final answer: Solid style
Q118Single correctReproduction in Organisms
In ginger vegetative propagation occurs through
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Rhizome
Approach:
The vegetative propagule of ginger is matched with the correct modified plant structure.
Step 1:Ginger possesses an underground horizontal stem bearing nodes and buds, which is a rhizome.
Step 2:The buds on the rhizome sprout into new shoots, allowing the plant to multiply vegetatively.
Final answer: Rhizome
Q119Single correctSexual Reproduction in Flowering Plants
Which of the following are the important floral rewards to the animal pollinators?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Nectar and pollen grains
Approach:
Floral rewards are distinguished from floral attractants to identify what pollinators actually obtain.
Step 1:Floral rewards are the nutritional substances a flower provides to a visiting pollinator, namely nectar and pollen grains.
Step 2:Colour, size and fragrance act as attractants that draw pollinators but are not consumed as a reward.
Final answer: Nectar and pollen grains
Q120Single correctPrinciples of Inheritance and Variation
How many pairs of contrasting characters in pea plants were studies by Mendel in his experiments?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Seven
Approach:
The number of contrasting character pairs Mendel studied in garden pea is recalled.
Step 1:Mendel selected garden pea, Pisum sativum, and chose seven pairs of contrasting traits for his hybridisation work.
Step 2:These include seed shape, seed colour, flower colour, pod shape, pod colour, flower position and stem height.
Final answer: Seven
Q121Single correctPrinciples of Inheritance and Variation
Which is the most common mechanism of genetic variation in the population of a sexually reproducing organism?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Recombination
Approach:
The principal source of genetic variation arising during sexual reproduction is identified.
Step 1:During meiosis, crossing over between homologous chromosomes and independent assortment produce new combinations of alleles, collectively termed recombination.
Step 2:Fusion of two genetically distinct gametes at fertilisation further increases this variation, making recombination the most common source in sexual reproduction.
Final answer: Recombination
Q122Single correctBiotechnology: Principles and Processes
A technique of micropropagation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Somatic embryogenesis
Approach:
The option representing a genuine micropropagation method is selected.
Step 1:Micropropagation is the in vitro production of large numbers of plants from small explants, and somatic embryogenesis generates embryos directly from somatic cells in culture for this purpose.
Step 2:Somatic hybridisation and protoplast fusion combine genomes of different cells, while embryo rescue recovers hybrid embryos, so these are not multiplication techniques.
Final answer: Somatic embryogenesis
Q123Single correctPrinciples of Inheritance and Variation
The movement of a gene from one linkage group to another is called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Translocation
Approach:
The chromosomal change that shifts a gene between non-homologous chromosomes is identified.
Step 1:Each chromosome forms one linkage group, so moving a gene to a different linkage group means transferring a segment to a non-homologous chromosome.
Step 2:The transfer of a chromosomal segment to a non-homologous chromosome is termed translocation.
Final answer: Translocation
Q124Single correctPrinciples of Inheritance and Variation
Multiple alleles are present
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3At the same locus of the chromosome
Approach:
The chromosomal position occupied by multiple alleles of one gene is determined.
Step 1:Multiple alleles are three or more alternative forms of the same gene, and alternative forms of a single gene occupy the same position on homologous chromosomes.
Step 2:Therefore the alleles of one gene are located at the same locus on the chromosome.
Final answer: At the same locus of the chromosome
Q125Single correctBiotechnology and its Applications
Which body of the Government of India regulates GM research and safety of introducing GM organisms for public services?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Genetic Engineering Approval committee
Approach:
The Indian regulatory authority for GMO research and release is identified.
Step 1:The Genetic Engineering Approval Committee (GEAC) is the statutory body that evaluates the validity of GM research and the safety of introducing GM organisms for public use.
Step 2:The other listed bodies handle agricultural research or internal institutional biosafety but do not grant national approval for GM release.
Final answer: Genetic Engineering Approval committee
Q126Single correctBiotechnology and its Applications
In BT cotton, the BT toxin present in plant tissue as pro-toxin is converted into active toxin due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Alkaline pH of the insect gut
Approach:
The condition activating the Bt protoxin inside the insect is identified.
Step 1:The Bt toxin exists in the plant as an inactive crystalline protoxin.
Step 2:When the insect ingests the protoxin, the alkaline pH of its midgut solubilises the crystal and converts it into the active toxin.
Final answer: Alkaline pH of the insect gut
Q127Single correctBiotechnology and its Applications
The crops engineered for glyphosate are resistant/tolerant to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Herbicides
Approach:
The category of agent that glyphosate represents is recalled to determine the engineered trait.
Step 1:Glyphosate is a widely used herbicide that inhibits an enzyme of the aromatic amino acid pathway in plants.
Step 2:Crops engineered to survive glyphosate application are therefore herbicide-tolerant crops.
Final answer: Herbicides
Q128Single correctCell: The Unit of Life
DNA is not present in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ribosomes
Approach:
The cell component lacking DNA is selected from organelles known to contain genetic material.
Step 1:The nucleus holds the main genome, while chloroplasts and mitochondria are semi-autonomous organelles carrying their own circular DNA.
Step 2:Ribosomes are made of ribosomal RNA and proteins and serve in protein synthesis, containing no DNA.
Final answer: Ribosomes
Q129Single correctBiotechnology: Principles and Processes
Which of the following enhances or induces fusion of protoplasts?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Polyethylene glycol and sodium nitrate
Approach:
The chemical fusogens used to promote protoplast fusion are identified.
Step 1:Protoplast fusion in somatic hybridisation requires agents that bring naked protoplast membranes together and promote their coalescence.
Step 2:Polyethylene glycol and sodium nitrate act as such fusogens, whereas IAA, kinetin and gibberellins are growth regulators and the chloride salts do not induce fusion.
Final answer: Polyethylene glycol and sodium nitrate
Q130Single correctEnvironmental Issues
The UN conference of Parties on climate change in the year 2011 was held in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2South Africa
Approach:
The host nation of the 2011 UN climate Conference of Parties is recalled.
Step 1:The 17th Conference of Parties under the UN Framework Convention on Climate Change was held in 2011.
Step 2:This session took place in Durban, South Africa.
Final answer: South Africa
Q131Single correctEcosystem
Vertical distribution of different species occupying different levels in a biotic community is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Stratification
Approach:
The term for vertical layering of species in a community is identified.
Step 1:Within a community, species occupy distinct vertical layers such as the canopy, understorey and ground level.
Step 2:This vertical arrangement of populations at different levels is called stratification.
Final answer: Stratification
Q132Single correctBiodiversity and Conservation
In which of the following both pairs have correct combination?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1In\ situ conservation : National Park
Ex\ situ conservation : Botanical Garden
Ex\ situ conservation : Botanical Garden
Approach:
Each conservation method is classified as in situ or ex situ to find the fully correct pair.
Step 1:In situ conservation protects species in their natural habitats, exemplified by national parks, wildlife sanctuaries and sacred groves.
Step 2:Ex situ conservation maintains species outside their habitat, exemplified by botanical gardens, seed banks, cryopreservation and tissue culture.
Final answer: In\ situ conservation : National Park
Ex\ situ conservation : Botanical Garden
Ex\ situ conservation : Botanical Garden
Q133Single correctEcosystem
Secondary succession takes place on/in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Degraded forest
Approach:
The site characteristic of secondary succession is identified by the prior presence of life and soil.
Step 1:Secondary succession begins in areas where a community previously existed but was removed, leaving soil and some biotic remnants.
Step 2:A degraded forest retains soil and propagules, so recolonisation there is secondary succession.
Final answer: Degraded forest
Q134Single correctEcosystem
The mass of living material at a trophic level at a particular time is called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Standing crop
Approach:
The ecological term for biomass present at a trophic level at a given moment is identified.
Step 1:Each trophic level holds a certain amount of living biomass at any instant.
Step 2:This mass of living material present at a trophic level at a particular time is termed the standing crop.
Final answer: Standing crop
Q135Single correctEcosystem
In an ecosystem the rate of production of organic matter during photosynthesis is termed
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Gross primary productivity
Approach:
The productivity term describing total organic matter fixed in photosynthesis is identified.
Step 1:Gross primary productivity is the rate at which producers fix organic matter through photosynthesis before any of it is consumed in respiration.
Step 2:Subtracting respiratory loss from gross primary productivity yields net primary productivity, so the total rate during photosynthesis is the gross value.
Final answer: Gross primary productivity
Q136Single correctAnimal Kingdom
Which of the following characteristics is mainly responsible for diversification of insects on land?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Exoskeleton
Approach:
Identification of the arthropod feature underlying insect success on land.
Step 1:Insects belong to phylum Arthropoda, whose body is covered by a chitinous cuticle.
Step 2:The exoskeleton restricts water loss and provides protection and rigid support against gravity on land.
Step 3:This protective, waterproof covering allowed insects to occupy varied terrestrial habitats and diversify extensively.
Final answer: Exoskeleton
Q137Single correctAnimal Kingdom
Which of the following endoparasites of humans does show viviparity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Trichinella\ spiralis
Approach:
Recall of the reproductive mode of common human nematode parasites.
Step 1:Ancylostoma, Enterobius and Ascaris release eggs and are oviparous.
Step 2:Trichinella spiralis females retain eggs internally and give birth to live larvae.
Final answer: Trichinella\ spiralis
Q138Single correctAnimal Kingdom
Which of the following represents the correct combination without any exception?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mouth ventral : grills without operculum; skin with placoid scales; persistent notochord — Chondrichthyes
Approach:
Matching diagnostic characters with the correct vertebrate class, eliminating exceptions.
Step 1:Mammalia generally bear pinnae, but exceptions such as the platypus lay eggs and aquatic mammals lack two free pairs of limbs, so the statement carries exceptions.
Step 2:Cyclostomata lack jaws and scales but are agnathans without paired appendages, so paired appendages do not apply.
Step 3:Aves have dry skin lacking glands except the oil gland, so moist and glandular skin is wrong.
Step 4:Chondrichthyes possess a ventral mouth, gill slits without operculum, placoid scales and a cartilaginous skeleton with a persistent notochord, all true without exception.
Final answer: Mouth ventral : grills without operculum; skin with placoid scales; persistent notochord — Chondrichthyes
Q139Single correctAnimal Kingdom
Which of the following animals is not viviparous?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Platypus
Approach:
Recall of egg-laying versus live-bearing mammals.
Step 1:Flying fox, elephant and whale are placental mammals that bear live young.
Step 2:The platypus is a monotreme that lays eggs.
Final answer: Platypus
Q140Single correctBody Fluids and Circulation
Erythropoiesis starts in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Liver
Approach:
Tracing the developmental site where red blood cell formation begins.
Step 1:In the human foetus, erythropoiesis first occurs in the liver and spleen during early development.
Step 2:After birth, erythropoiesis shifts to the red bone marrow.
Final answer: Liver
Q141Single correctStructural Organisation in Animals
The terga, sterna and pleura of cockroach body are joined by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Arthrodial membrane
Approach:
Recall of the connective structure between cockroach sclerites.
Step 1:The cockroach body wall is composed of hardened plates called sclerites: terga (dorsal), sterna (ventral) and pleura (lateral).
Step 2:These sclerites are connected by a thin flexible chitinous arthrodial membrane that permits movement.
Final answer: Arthrodial membrane
Q142Single correctCell: The Unit of Life
Nuclear envelope is a derivative of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Rough endoplasmic reticulum
Approach:
Determining the membrane source of the nuclear envelope.
Step 1:The outer membrane of the nuclear envelope is studded with ribosomes and is continuous with the rough endoplasmic reticulum.
Step 2:During reformation after mitosis, the nuclear envelope is reconstituted from rough endoplasmic reticulum vesicles.
Final answer: Rough endoplasmic reticulum
Q143Single correctCell: The Unit of Life
Cytochromes are found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cristae of mitochondria
Approach:
Locating cytochromes within mitochondrial sub-structure.
Step 1:Cytochromes are electron-transport chain proteins embedded in the inner mitochondrial membrane.
Step 2:The inner mitochondrial membrane folds into cristae, which house the electron transport chain.
Final answer: Cristae of mitochondria
Q144Single correctBiomolecules
Which one of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The presence of the competitive inhibitor decreases the Km of the enzyme for the substrate
Approach:
Evaluation of the kinetic effects of competitive inhibition.
Step 1:A competitive inhibitor binds reversibly at the active site forming an enzyme-inhibitor complex and is not chemically altered.
Step 2:Competitive inhibition raises the apparent Km because more substrate is required to reach half-maximal velocity.
Step 3:The claim that Km decreases is therefore incorrect.
Final answer: The presence of the competitive inhibitor decreases the Km of the enzyme for the substrate
Q145Single correctCell Cycle and Cell Division
Select the correct option:
| I | II |
|---|---|
| (a). Synapsis aligns homologous chromosomes | (i). Anaphase-II |
| (b). Synthesis of RNA and protein | (ii). Zygotene |
| (c). Action of enzyme recombinase | (iii). G₂-Phase |
| (d). Centromeres do not separate but chromatids move toward opposite poles | (iv). Anaphase-I |
| (v). Pachytene |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(iii), (c)-(v), (d)-(iv)
Approach:
Matching cell-cycle events with their correct phases.
Step 1:Synapsis, the pairing of homologous chromosomes, occurs during zygotene.
Step 2:Synthesis of RNA and protein characterises the G₂ phase of interphase.
Step 3:The enzyme recombinase mediates crossing over during pachytene.
Step 4:Separation of chromatids while centromeres remain (homologous chromosomes moving apart) occurs in anaphase-I.
Final answer: (a)-(ii), (b)-(iii), (c)-(v), (d)-(iv)
Q146Single correctCell Cycle and Cell Division
A somatic cell that has just completed the S phase of its cell cycle, as compared to gamete of the same species, has
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Twice the number of chromosomes and four times the amount of DNA
Approach:
Comparing chromosome number and DNA content of a post-S somatic cell with a haploid gamete.
Step 1:A gamete is haploid with n chromosomes and a 1C amount of DNA.
Step 2:A somatic cell is diploid with 2n chromosomes; after S phase the DNA doubles to 4C while chromosome number remains 2n.
Step 3:Relative to the gamete, the somatic cell has twice the chromosomes and four times the DNA.
Final answer: Twice the number of chromosomes and four times the amount of DNA
Q147Single correctDigestion and Absorption
Which of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Brunner's glands are present in the submucosa of stomach and secrete pepsinogen
Approach:
Verification of glandular locations and their secretions in the alimentary canal.
Step 1:Goblet cells secrete mucus in the intestinal mucosa, oxyntic cells secrete HCl in the gastric mucosa, and pancreatic acini secrete carboxypeptidase, all correct.
Step 2:Brunner's glands lie in the submucosa of the duodenum, not the stomach, and secrete mucus rather than pepsinogen.
Final answer: Brunner's glands are present in the submucosa of stomach and secrete pepsinogen
Q148Single correctDigestion and Absorption
Gastric juice of infants contains
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pepsinogen, lipase, rennin
Approach:
Recall of the enzyme composition of infant gastric juice.
Step 1:Gastric juice contains pepsinogen and gastric lipase as secretions of the stomach.
Step 2:In infants the gastric juice additionally contains rennin, which curdles milk proteins.
Step 3:Maltase, nuclease and amylase are not gastric secretions, eliminating the other options.
Final answer: Pepsinogen, lipase, rennin
Q149Single correctBreathing and Exchange of Gases
When you hold your breath, which of the following gas changes in blood would first lead to the urge to breathe?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Rising C concentration
Approach:
Identification of the primary chemical stimulus for respiratory drive.
Step 1:The respiratory centre is most sensitive to the partial pressure of carbon dioxide in the blood.
Step 2:During breath-holding, accumulating CO2 stimulates the chemoreceptors first, producing the urge to breathe before oxygen depletion becomes significant.
Final answer: Rising C concentration
Q150Single correctBody Fluids and Circulation
Blood pressure in the mammalian aorta is maximum during
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Systole of the left ventricle
Approach:
Linking aortic pressure peaks to the cardiac cycle.
Step 1:The left ventricle pumps blood into the aorta during ventricular systole.
Step 2:This forceful contraction produces the maximum aortic pressure, the systolic pressure.
Final answer: Systole of the left ventricle
Q151Single correctBody Fluids and Circulation
Which one of the following is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Blood = Plasma + RBC + WBC + Platelets
Approach:
Evaluating the compositional equations of blood and related fluids.
Step 1:Plasma is blood minus all formed elements, not merely lymphocytes, so option 1 is wrong.
Step 2:Serum is plasma without fibrinogen, so adding fibrinogen is wrong in option 2.
Step 3:Lymph lacks red blood cells, so option 3 is wrong.
Step 4:Blood consists of plasma plus the formed elements: red blood cells, white blood cells and platelets.
Final answer: Blood = Plasma + RBC + WBC + Platelets
Q152Single correctExcretory Products and their Elimination
Removal of proximal convoluted tubule from the nephron will result in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1More diluted urine
Approach:
Predicting the effect of losing PCT reabsorption on urine.
Step 1:The proximal convoluted tubule reabsorbs the bulk of water, ions and nutrients from the filtrate.
Step 2:Without the PCT, this reabsorption is lost, so more water and solutes remain in the filtrate.
Step 3:The resulting urine becomes more diluted.
Final answer: More diluted urine
Q153Single correctLocomotion and Movement
Sliding filament theory can best explained as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Actin and Myosin filaments do not shorten but slide pass each other
Approach:
Stating the molecular basis of muscle contraction per the sliding filament theory.
Step 1:During contraction the thin actin filaments slide over the thick myosin filaments toward the centre of the sarcomere.
Step 2:Neither the actin nor the myosin filaments themselves change in length; only their overlap increases.
Final answer: Actin and Myosin filaments do not shorten but slide pass each other
Q154Single correctLocomotion and Movement
Glenoid cavity articulates
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Humerus with scapula
Approach:
Identifying the bones meeting at the glenoid cavity.
Step 1:The glenoid cavity is a depression on the lateral angle of the scapula.
Step 2:The head of the humerus fits into this cavity to form the shoulder joint.
Final answer: Humerus with scapula
Q155Single correctNeural Control and Coordination
Which of the following regions of the brain is incorrectly paired with its function?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cerebellum – Language comprehension
Approach:
Checking each brain region against its true function.
Step 1:The medulla oblongata controls homeostatic functions, the corpus callosum links the two cerebral hemispheres, and the cerebrum handles calculation and thought, all correctly paired.
Step 2:The cerebellum coordinates movement and balance, not language comprehension, which is a cerebral function.
Final answer: Cerebellum – Language comprehension
Q156Single correctNeural Control and Coordination
A gymnast is able to balance his body upside down even in the total darkness because of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Vestibular apparatus
Approach:
Identifying the structure responsible for body balance independent of vision.
Step 1:The cochlea, organ of Corti and tectorial membrane are concerned with hearing.
Step 2:The vestibular apparatus of the inner ear detects position and movement of the head, maintaining balance even without sight.
Final answer: Vestibular apparatus
Q157Single correctChemical Coordination and Integration
A chemical signal that has both endocrine and neural roles is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Epinephrine
Approach:
Selecting the molecule that acts as both hormone and neurotransmitter.
Step 1:Epinephrine (adrenaline) is secreted by the adrenal medulla as a hormone into the blood.
Step 2:Epinephrine also functions as a neurotransmitter in the sympathetic nervous system.
Step 3:Melatonin, calcitonin and cortisol act only as hormones.
Final answer: Epinephrine
Q158Single correctExcretory Products and their Elimination
Which of the following does not favour the formation of large quantities of dilute urine?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Renin
Approach:
Distinguishing agents that increase urine volume from those that reduce it.
Step 1:Alcohol and caffeine inhibit ADH or act as diuretics, increasing dilute urine output.
Step 2:Atrial natriuretic factor promotes sodium and water excretion, also increasing urine volume.
Step 3:Renin activates the renin-angiotensin-aldosterone system, which conserves sodium and water and reduces urine volume.
Final answer: Renin
Q159Single correctHuman Reproduction
Capacitation refers to changes in the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Sperm before fertilisation
Approach:
Capacitation is the physiological maturation that spermatozoa undergo within the female reproductive tract before they acquire the capacity to fertilise the ovum.
Step 1:Freshly ejaculated sperms are not immediately capable of fertilising the egg. They must reside in the female reproductive tract for several hours.
Step 2:During this period the sperm plasma membrane is altered, removing cholesterol and glycoproteins, which prepares the acrosome for the acrosomal reaction.
Final answer: Sperm before fertilisation
Q160Single correctHuman Reproduction
Which of these is not an important component of initiation of parturition in humans?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Release of prolactin
Approach:
Parturition is triggered by a neuroendocrine mechanism involving the foetal-placental signals, oxytocin and prostaglandins.
Step 1:Signals from the fully developed foetus and the placenta induce mild uterine contractions, the foetal ejection reflex.
Step 2:Oxytocin from the maternal pituitary acts on the uterine muscle, while prostaglandins and a rising oestrogen-to-progesterone ratio strengthen contractions.
Step 3:Prolactin governs milk production after birth and is not a trigger of labour.
Final answer: Release of prolactin
Q161Single correctHuman Health and Disease
Which of the following viruses is not transferred through semen of an infected male?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Chikungunya virus
Approach:
Identification of the virus that is not transmitted through semen requires recalling the mode of spread of each pathogen.
Step 1:Hepatitis B virus and human immunodeficiency virus are well-known sexually transmitted agents present in semen and other body fluids.
Step 2:Ebola virus persists in semen and can be transmitted sexually for months after recovery.
Step 3:Chikungunya virus is transmitted by the bite of Aedes mosquitoes and is an arbovirus, not a sexually transmitted agent.
Final answer: Chikungunya virus
Q162Single correctHuman Reproduction
Which of the following cells during gametogenesis is normally diploid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Spermatogonia
Approach:
The ploidy of cells in gametogenesis depends on whether the cell has completed meiotic division.
Step 1:Spermatogonia are the diploid stem cells of the testis that undergo mitosis before entering meiosis.
Step 2:Spermatids form after meiosis II and are haploid, while both the primary and secondary polar bodies are haploid products of oogenesis.
Final answer: Spermatogonia
Q163Single correctReproductive Health
Hysterectomy is surgical removal of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Uterus
Approach:
The term hysterectomy is derived from the Greek root hystera, meaning uterus.
Step 1:Hysterectomy denotes the surgical removal of the uterus.
Step 2:Removal of the vas deferens is vasectomy and there are distinct procedures for the prostate and mammary glands.
Final answer: Uterus
Q164Single correctHuman Health and Disease
Which of the following is not sexually transmitted disease?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Encephalitis
Approach:
Each disease must be classified by its primary mode of transmission.
Step 1:Syphilis, AIDS and trichomoniasis are recognised sexually transmitted infections caused by a bacterium, a virus and a protozoan respectively.
Step 2:Encephalitis is an inflammation of the brain, commonly caused by mosquito-borne viruses, and is not transmitted sexually.
Final answer: Encephalitis
Q165Single correctPrinciples of Inheritance and Variation
An abnormal human baby with 'XXX' sex chromosomes was born due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Formation of abnormal ova in the mother
Approach:
An XXX karyotype arises from non-disjunction of the X chromosomes during gamete formation.
Step 1:The mother carries two X chromosomes; non-disjunction during oogenesis can produce an abnormal ovum bearing two X chromosomes (XX).
Step 2:Fertilisation of such an XX ovum by a normal X-bearing sperm yields an XXX zygote, the trisomy known as super female.
Final answer: Formation of abnormal ova in the mother
Q166Single correctPrinciples of Inheritance and Variation
Alleles are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Different molecular forms of a gene
Approach:
The definition of an allele follows from the molecular nature of genes.
Step 1:A gene occupies a fixed locus on a chromosome, and slightly different sequences of that gene give rise to alternative versions.
Step 2:These alternative forms of the same gene are termed alleles and may produce contrasting traits.
Final answer: Different molecular forms of a gene
Q167Single correctPrinciples of Inheritance and Variation
A man with blood group 'A' marries a woman with blood group 'B'. What are all the possible blood groups of their offspring?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A, B, AB and O
Approach:
The widest range of offspring blood groups occurs when both parents are heterozygous.
Step 1:The father of group A may be heterozygous with genotype i, and the mother of group B may be heterozygous with genotype i.
Step 2:The cross yields the genotypes , i, i and i i in equal proportion.
Step 3:These genotypes correspond to blood groups AB, A, B and O.
Final answer: A, B, AB and O
Q168Single correctMolecular Basis of Inheritance
Gene regulation governing lactose operon of E.\ coli that involves the lac I gene product is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Negative and inducible because repressor protein prevents transcription
Approach:
The nature of lac operon control follows from the role of the lac I repressor and the inducing substrate lactose.
Step 1:The lac I gene encodes a repressor protein that binds the operator and blocks transcription in the absence of lactose; this is negative regulation.
Step 2:When lactose is present, its isomer allolactose binds the repressor, inactivating it and switching on the operon, making the system inducible.
Final answer: Negative and inducible because repressor protein prevents transcription
Q169Single correctMolecular Basis of Inheritance
In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in the DNA are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3G 17%, A 33%, T 33%
Approach:
Base composition of double-stranded DNA is governed by Chargaff's rule, where complementary bases occur in equal amounts.
Step 1:Since guanine pairs with cytosine, the percentage of guanine equals that of cytosine.
Step 2:Guanine and cytosine together account for 34%, leaving 66% for adenine and thymine combined.
Step 3:Adenine equals thymine, so each is half of 66%.
Final answer: G 17%, A 33%, T 33%
Q170Single correctEvolution
Which of the following had the smallest brain capacity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Homo habilis
Approach:
Brain capacity increased progressively through the course of human evolution.
Step 1:Homo habilis had a cranial capacity of about 650 to 800 cubic centimetres, the smallest among the listed Homo species.
Step 2:Homo erectus reached about 900 cubic centimetres, while Neanderthals and modern Homo sapiens reached around 1400 cubic centimetres.
Final answer: Homo habilis
Q171Single correctEvolution
A population will not exist in Hardy–Weinberg equilibrium if
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Individuals mate selectively
Approach:
Hardy-Weinberg equilibrium holds only when a defined set of conditions is satisfied.
Step 1:The equilibrium requires random mating, no mutation, no migration, no selection and a large population.
Step 2:Selective or non-random mating violates the random mating assumption and disturbs allele frequencies, breaking the equilibrium.
Final answer: Individuals mate selectively
Q172Single correctHuman Health and Disease
Match each disease with its correct type of vaccine:
| List I (Disease) | List II (Type of vaccine) |
|---|---|
| (a). Tuberculosis | (i). harmless virus |
| (b). Whooping cough | (ii). Inactivated toxin |
| (c). Diphtheria | (iii). Killed bacteria |
| (d). Polio | (iv). harmless bacteria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Approach:
Each vaccine type is matched to its disease by the nature of the immunising agent used.
Step 1:The BCG vaccine for tuberculosis uses an attenuated, harmless bacterium, so tuberculosis matches harmless bacteria.
Step 2:The whooping cough vaccine uses killed Bordetella pertussis bacteria, so it matches killed bacteria.
Step 3:The diphtheria vaccine is a toxoid prepared from inactivated toxin, and the oral polio vaccine uses an attenuated harmless virus.
Final answer: (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Q173Single correctHuman Health and Disease
HIV that causes AIDS, first starts destroying
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Helper T-lymphocytes
Approach:
The target cell of HIV is identified by the receptor the virus uses to enter host cells.
Step 1:HIV binds the CD4 receptor that is abundant on helper T-lymphocytes.
Step 2:After entry the virus replicates inside helper T-lymphocytes and progressively destroys them, lowering immunity.
Final answer: Helper T-lymphocytes
Q174Single correctHuman Health and Disease
The active form of Entamoeba\ histolytica feeds upon
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Erythrocytes, mucosa and submucosa of colon
Approach:
The feeding habit of the trophozoite of Entamoeba histolytica is determined by the tissues it invades in the large intestine.
Step 1:The active trophozoite invades the wall of the colon and erodes the mucosa and submucosa, causing dysentery and ulcers.
Step 2:It also ingests red blood cells released from the damaged tissue, so it feeds on erythrocytes along with the colon lining.
Final answer: Erythrocytes, mucosa and submucosa of colon
Q175Single correctEnvironmental Issues
High value of BOD (biochemical oxygen demand) indicates that
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Water is highly polluted
Approach:
The biological oxygen demand measures the oxygen consumed by microbes while decomposing organic matter, which reflects the level of organic pollution.
Step 1:A greater amount of organic waste supports more microbial activity, which consumes more dissolved oxygen.
Step 2:A high BOD value therefore signals a high content of organic pollutants, meaning the water is highly polluted.
Final answer: Water is highly polluted
Q176Single correctOrganisms and Populations
Most animals are tree dwellers in a
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Tropical rainforest
Approach:
The biome with the greatest vertical stratification supports the largest number of arboreal animals.
Step 1:Tropical rainforests have a dense, multi-layered canopy that provides abundant food and shelter high above the ground.
Step 2:This rich canopy structure allows most of the animal life to live in the trees as arboreal dwellers.
Final answer: Tropical rainforest
Q177Single correctOrganisms and Populations
The following graph depicts changes in two populations (A and B) of herbivores in a grassy field. A possible reason for these changes is that
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Population B competed more successfully for food than population A
Approach:
The graph shows population B rising while population A declines over the same period, which points to the outcome of competition between the two herbivore populations.
Step 1:Both populations are herbivores sharing the same grassy field, so they compete for the same plant food.
Step 2:Population B increases while population A decreases, indicating that B obtained a larger share of the limited food resource.
Final answer: Population B competed more successfully for food than population A
Q178Single correctBiodiversity and Conservation
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Advanced ex\ situ conservation of biodiversity
Approach:
The conservation category is decided by whether the organism is protected within or away from its natural habitat.
Step 1:Cryopreservation stores gametes and tissues in liquid nitrogen, away from the natural habitat of the species.
Step 2:Protection of biological material away from its habitat constitutes ex situ conservation, and freezing viable gametes is an advanced form of it.
Final answer: Advanced ex\ situ conservation of biodiversity
Q179Single correctEnvironmental Issues
Rachel Carson's famous book 'Silent spring' is related to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pesticide pollution
Approach:
The theme of the book is recalled from its place in the history of environmental awareness.
Step 1:Rachel Carson's book Silent Spring exposed the harmful effects of indiscriminate use of pesticides such as DDT on birds and the environment.
Step 2:It drew public attention to pesticide pollution and helped launch the modern environmental movement.
Final answer: Pesticide pollution
Q180Single correctEnvironmental Issues
Which of the following is not one of the prime health risks associated with greater UV radiation through the atmosphere due to depletion of stratospheric ozone?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Increased liver cancer
Approach:
The health effects of increased ultraviolet exposure relate to tissues directly reached by the radiation.
Step 1:Enhanced ultraviolet B radiation damages the skin, eyes and immune system, causing skin cancer, cataract, snow blindness and weakened immunity.
Step 2:Liver cancer is not a recognised consequence of ultraviolet exposure because the radiation does not penetrate to the liver.
Final answer: Increased liver cancer
More NEET 2015 papers
Frequently Asked Questions
How many questions are in the NEET 2015 May 03 paper?
The NEET 2015 May 03 paper has 180 questions — Physics (45), Chemistry (45) and Biology (90). Every question is on this page with its correct answer and a step-by-step solution.
What is the marking scheme for NEET?
NEET awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Biology across 180 questions for 720 marks.
Are the answer key and step-by-step solutions provided for the 2015 May 03 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 2015 May 03 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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