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NEET 2019 May 05 Question Paper with Solutions
All 180 questions from the NEET 2019 (May 05) 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 2019Chemistry PYQs 2019Biology PYQs 2019
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctMechanical Properties of Solids
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes . The elastic potential energy stored in the extended wire is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The extension is produced by the suspended weight acting as a gradually increasing load. Elastic potential energy equals the average force times the extension.
Step 1:The restoring force in the wire builds from zero to the final weight Mg as the extension grows from zero to l, so the average force is half the final weight.
Step 2:Elastic potential energy stored equals the average force multiplied by the total extension l.
Final answer:
Q2Single correctLaws of Motion
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3the mass is at the lowest point
Approach:
Tension in the wire is largest where the wire must both support the weight and supply centripetal force; this occurs at the lowest point of the vertical circle.
Step 1:At the lowest point the tension acts upward while gravity acts downward, and the net upward force provides the centripetal acceleration.
Step 2:Speed is also maximum at the lowest point by energy conservation, making the centripetal term largest there. Therefore the tension peaks at the lowest point, so the wire is most likely to break there.
Final answer: the mass is at the lowest point
Q3Single correctMoving Charges and Magnetism
Ionized hydrogen atoms and -particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The radius of a charged particle moving perpendicular to a magnetic field is the momentum divided by the product of charge and field. With equal momenta and field, the radius is inversely proportional to charge.
Step 1:The ionized hydrogen carries charge e, so its radius is the momentum over e times B.
Step 2:The alpha particle carries charge 2e, so its radius is the momentum over 2e times B.
Step 3:Dividing the two radii cancels momentum and field, leaving the inverse charge ratio.
Final answer:
Q4Single correctWork, Energy and Power
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a one-dimensional elastic collision the fractional kinetic energy transferred to the target depends only on the two masses. The fraction lost by the incoming body equals four times the mass product over the sum of masses squared.
Step 1:Insert the colliding masses with body A as the projectile of mass 4m and body B as the target of mass 2m.
Step 2:Evaluate numerator and denominator and simplify.
Final answer:
Q5Single correctWave Optics
In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be . What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The angular fringe width is the wavelength over the slit separation. Immersing the apparatus in water reduces the wavelength by the refractive index, scaling the angular width by the inverse of the index.
Step 1:In air the angular fringe width is the wavelength divided by the slit separation, given as 0.2 degree.
Step 2:In water the wavelength shrinks by the refractive index, so the angular width divides by four-thirds.
Final answer:
Q6Single correctElectromagnetic Induction
In which of the following devices, the eddy current effect is not used?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Electric heater
Approach:
Eddy currents are circulating induced currents that appear in conductors exposed to changing magnetic flux. Identify which listed device operates without relying on induced eddy currents.
Step 1:An induction furnace and magnetic braking both exploit eddy currents for heating and retarding force respectively, while an electromagnet involves a steady magnetizing current that can be assisted by induced effects.
Step 2:An electric heater produces heat by Joule heating from a steady conduction current through a resistive element, with no eddy current involved.
Final answer: Electric heater
Q7Single correctMechanical Properties of Fluids
A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of N/m. The pressure inside the bubble equals at a point below the free surface of water in a container. Taking m/, density of water kg/, the value of is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 cm
Approach:
Equate the gauge pressure inside the soap bubble, which has two surfaces, to the hydrostatic gauge pressure at depth Z below the water surface, then solve for the depth.
Step 1:A soap bubble has two liquid surfaces, so its excess gauge pressure is four times the surface tension over the radius. Setting this equal to the hydrostatic pressure at depth Z gives the depth.
Step 2:Insert the surface tension, radius, density and gravity values.
Step 3:Convert the result to centimetres.
Final answer: cm
Q8Single correctElectromagnetic Waves
Which colour of the light has the longest wavelength?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Red
Approach:
Within the visible spectrum, wavelength decreases from red toward violet. Compare the listed colours to find the longest wavelength.
Step 1:In the visible band the order of increasing wavelength runs violet, blue, green, then red, so red lies at the long-wavelength end.
Final answer: Red
Q9Single correctSystem of Particles and Rotational Motion
A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 J
Approach:
The work needed to stop the disc equals its total kinetic energy, which combines translational and rotational parts. For a rolling disc the total is three-quarters of the mass times the centre of mass speed squared.
Step 1:For a disc the moment of inertia is half the mass times radius squared, and rolling enforces the speed equal to the radius times angular speed, combining the two energies into three-quarters of the mass times speed squared.
Step 2:Insert the mass 100 kg and the centre of mass speed of 0.2 m/s.
Final answer: J
Q10Single correctOscillations
The displacement of a particle executing simple harmonic motion is given by
Then the amplitude of its oscillation is given by :
Then the amplitude of its oscillation is given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The constant term only shifts the mean position and does not contribute to the amplitude. The amplitude of the combined sine and cosine terms is the resultant of two perpendicular harmonic components.
Step 1:Subtracting the constant offset leaves a pure oscillation that is the sum of a sine and a cosine of the same frequency, which differ in phase by ninety degrees.
Step 2:The resultant amplitude combines the two perpendicular components, and the cosine of ninety degrees vanishes.
Final answer:
Q11Single correctRay Optics and Optical Instruments
Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is . When the space between the two lenses is filled with glycerine (which has the same refractive index as that of glass) then the equivalent focal length is . The ratio will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Find the combined focal length of two contacting equi-convex lenses, then treat the glycerine-filled gap as a third diverging lens whose surfaces match the lens curvatures, and combine all three.
Step 1:Two identical equi-convex lenses of focal length f in contact give a combined focal length of f over two.
Step 2:Filling the gap with glycerine of the same index as glass forms a biconcave glycerine lens whose focal length has magnitude f, acting as a diverging element.
Step 3:Combine the two glass lenses and the glycerine lens in contact.
Step 4:Form the requested ratio of the two equivalent focal lengths.
Final answer:
Q12Single correctKinetic Theory
Increase in temperature of a gas filled in a container would lead to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2increase in its kinetic energy
Approach:
By kinetic theory, temperature is a direct measure of the average translational kinetic energy of gas molecules, so raising temperature raises that energy.
Step 1:The internal kinetic energy of an ideal gas is proportional to its absolute temperature, so heating raises the molecular kinetic energy.
Step 2:Mass is fixed, pressure tends to rise in a closed container, and intermolecular distance does not decrease, so the only correct outcome is increased kinetic energy.
Final answer: increase in its kinetic energy
Q13Single correctDual Nature of Radiation and Matter
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : kg)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
Use the de Broglie wavelength expression for an electron accelerated through a potential, where the wavelength in metres equals about 12.27 over the square root of the accelerating voltage divided by appropriate units.
Step 1:The de Broglie wavelength of an electron accelerated through a potential V in volts is 12.27 over the square root of V in angstrom.
Step 2:Insert the accelerating potential of ten thousand volts.
Final answer: m
Q14Single correctThermal Properties of Matter
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is : and
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 cm
Approach:
For the difference of the two rod lengths to stay constant with temperature, the absolute expansions of the two rods must be equal, giving a relation between length and coefficient of linear expansion.
Step 1:Setting the absolute expansion of copper equal to that of aluminium for the same temperature change.
Step 2:Solve for the aluminium length.
Final answer: cm
Q15Single correctRay Optics and Optical Instruments
Pick the wrong answer in the context with rainbow.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3An observer can see a rainbow when his front is towards the sun
Approach:
Recall the geometry of rainbow viewing. The observer must face away from the sun so that sunlight enters the raindrops from behind and returns toward the eye.
Step 1:A rainbow appears in the part of the sky opposite the sun, so the observer faces away from the sun rather than toward it, making the statement about facing the sun the wrong one.
Final answer: An observer can see a rainbow when his front is towards the sun
Q16Single correctGravitation
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 N
Approach:
Acceleration due to gravity at depth decreases linearly toward the centre. At half the radius the effective gravity is half its surface value, halving the weight.
Step 1:At a depth equal to half the earth radius the bracket reduces to one half, so the gravity becomes half the surface value.
Step 2:Weight scales with gravity at fixed mass, so the weight halves from 200 N.
Final answer: N
Q17Single correctCurrent Electricity
Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Each section consists of three identical bulbs that are reconfigured between full and partial glowing. Compute the equivalent resistance and hence the total power for each case, then take the ratio.
Step 1:When all six bulbs glow the network reduces to an equivalent resistance of two-thirds R, giving a total power proportional to three E squared over two R.
Step 2:With two bulbs from section A and one from section B glowing, the equivalent resistance becomes three R over two, giving a power of two E squared over three R.
Step 3:Form the ratio of the two powers.
Final answer:
Q18Single correctSemiconductor Electronics
For a p-type semiconductor, which of the following statements is true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Holes are the majority carriers and trivalent atoms are the dopants.
Approach:
Recall the doping scheme for p-type material. Trivalent impurities create electron deficiencies that act as holes, the majority carriers.
Step 1:Doping an intrinsic semiconductor with trivalent atoms produces acceptor levels that create holes, making holes the majority carriers in p-type material.
Final answer: Holes are the majority carriers and trivalent atoms are the dopants.
Q19Single correctOscillations
Average velocity of a particle executing SHM in one complete vibration is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Zero
Approach:
Average velocity is total displacement divided by total time. Over one full vibration the particle returns to its starting position, so the net displacement is zero.
Step 1:In one complete vibration the particle ends where it began, so the displacement is zero and the average velocity vanishes.
Final answer: Zero
Q20Single correctThermal Properties of Matter
The unit of thermal conductivity is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Derive the unit of the thermal conductivity coefficient from the heat conduction equation, where heat current equals conductivity times area times temperature gradient.
Step 1:Rearranging the conduction relation expresses conductivity as heat current times length over area times temperature difference.
Step 2:Substituting units of watt for heat current, metre for length, metre squared for area and kelvin for temperature gives watt per metre per kelvin.
Final answer:
Q21Single correctSystem of Particles and Rotational Motion
A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque required to stop after revolutions is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 N m
Approach:
Use the rotational work-energy theorem, equating the torque times angular displacement to the change in rotational kinetic energy of the solid cylinder.
Step 1:Express the work done by the stopping torque as the magnitude of the change in rotational kinetic energy, with final angular speed zero.
Step 2:Convert 3 rpm to radians per second and the angular displacement of two pi revolutions to radians, then divide work by angular displacement to obtain the torque.
Step 3:Evaluate the torque from work over angular displacement.
Final answer: N m
Q22Single correctWork, Energy and Power
A force acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from to m is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 J
Approach:
Work done by a position-dependent force is the integral of the force over the displacement. Integrate the given linear force from the initial to the final position.
Step 1:Integrate the force from y equal to zero to y equal to one metre.
Step 2:Evaluate the bracket at the limits.
Final answer: J
Q23Single correctCurrent Electricity
Which of the following acts as a circuit protecting device?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Fuse
Approach:
Identify the device that interrupts the circuit when current exceeds a safe limit. A fuse melts under excess current, breaking the circuit and protecting it.
Step 1:A fuse has a low melting point and melts when excess current flows, breaking the circuit due to the heat produced in it, so it serves as the circuit protecting device.
Final answer: Fuse
Q24Single correctCurrent Electricity
In the circuits shown below, the readings of voltmeters and the ammeters will be :
(Circuit 1: a resistor with voltmeter joined across it, the resistor in series with ammeter carrying current , across a cell. Circuit 2: a resistor, and in parallel with it a second resistor joined in series with voltmeter ; the pair is in series with ammeter carrying current , across a cell.)
(Circuit 1: a resistor with voltmeter joined across it, the resistor in series with ammeter carrying current , across a cell. Circuit 2: a resistor, and in parallel with it a second resistor joined in series with voltmeter ; the pair is in series with ammeter carrying current , across a cell.)

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and
Approach:
For an ideal voltmeter the resistance is infinite and for an ideal ammeter the resistance is zero, so both circuits carry the same current and develop the same voltage across the resistor.
Step 1:An ideal voltmeter draws no current, so the voltage across the resistor equals the full source potential in each circuit.
Step 2:The same holds for the second circuit since the voltmeter is ideal.
Step 3:An ideal ammeter has zero resistance, so the current in each loop is the source voltage divided by the resistance.
Final answer: and
Q25Single correctElectrostatic Potential and Capacitance
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Zero as r increases for , decreases as r increases for
Approach:
Apply Gauss's law to a hollow charged conductor to find the field inside and outside.
Step 1:For a point inside the hollow sphere the enclosed charge is zero, so the field vanishes.
Step 2:For a point outside, the whole charge Q is enclosed and the field falls off with the inverse square of distance.
Step 3:Combining both regions, the field is zero inside and decreases outside.
Final answer: Zero as r increases for , decreases as r increases for
Q26Single correctMagnetism and Matter
At a point A on the earth's surface the angle of dip, . At a point B on the earth's surface the angle of dip, . We can interpret that :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is located in the northern hemisphere and B is located in the southern hemisphere.
Approach:
Relate the sign of the angle of dip to the hemisphere using the convention that dip is positive in the northern hemisphere and negative in the southern hemisphere.
Step 1:The angle of dip is the angle between the earth's resultant magnetic field and the horizontal; the vertical component points downward in the northern hemisphere.
Step 2:In the southern hemisphere the vertical component points upward, making the dip negative.
Step 3:Point A has so it lies in the northern hemisphere, and point B has so it lies in the southern hemisphere.
Final answer: A is located in the northern hemisphere and B is located in the southern hemisphere.
Q27Single correctAtoms
The total energy of an electron in an atom in an orbit is eV. Its kinetic and potential energies are, respectively :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 eV, eV
Approach:
Use the Bohr-model relations between total energy, kinetic energy, and potential energy of the orbiting electron.
Step 1:In the Bohr model the kinetic energy equals the magnitude of the total energy.
Step 2:The potential energy is twice the total energy.
Step 3:Therefore the kinetic and potential energies are 3.4 eV and -6.8 eV respectively.
Final answer: eV, eV
Q28Single correctRay Optics and Optical Instruments
In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Apply Snell's law at the critical angle to find the refraction angle in the rarer medium.
Step 1:At the critical angle the refracted ray travels along the interface, so the angle of refraction is the limiting value.
Step 2:Snell's law confirms this since when the incidence equals the critical angle.
Step 3:Therefore the angle of refraction equals 90 degrees.
Final answer:
Q29Single correctGravitation
The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Compute the change in gravitational potential energy between the surface and a height equal to the earth's radius.
Step 1:The potential energy at the earth's surface uses distance R from the centre.
Step 2:At height the distance from the centre is .
Step 3:The work done equals the change in potential energy, and substituting gives the result.
Final answer:
Q30Single correctMotion in a Plane
When an object is shot from the bottom of a long smooth inclined plane kept at an angle with horizontal, it can travel a distance along the plane. But when the inclination is decreased to and the same object is shot with the same velocity, it can travel distance. Then will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Use the kinematic stopping distance on an incline where the retardation is the component of gravity along the plane.
Step 1:On a smooth incline the deceleration is , so the stopping distance for the same launch speed is inversely proportional to .
Step 2:Forming the ratio cancels the common factors and leaves the ratio of the sines.
Final answer:
Q31Single correctNuclei
-particle consists of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12 protons and 2 neutrons only
Approach:
Identify the alpha particle with the nucleus of a helium atom.
Step 1:An alpha particle is the nucleus of a helium atom, which carries atomic number 2 and mass number 4.
Step 2:As a bare nucleus it contains no electrons.
Final answer: 2 protons and 2 neutrons only
Q32Single correctMotion in a Plane
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path the angle at which he should make his strokes w.r.t. north is given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 west
Approach:
For the shortest path the resultant velocity must point due north, so the upstream (westward) component of the swimmer's velocity must cancel the river current.
Step 1:The river velocity is 10 m/s east and the swimmer's speed relative to water is 20 m/s.
Step 2:To make the resultant point due north, the westward component of the swimmer's velocity must equal the river speed.
Step 3:Solving for the angle measured from north gives 30 degrees toward the west.
Final answer: west
Q33Single correctLaws of Motion
A particle moving with velocity is acted by three forces shown by the vector triangle PQR. The velocity of the particle will :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Remain constant
Approach:
Recognize that three forces represented by the sides of a closed triangle taken in order sum to zero, leaving the velocity unchanged.
Step 1:Forces forming the sides of a closed triangle taken in order add vectorially to zero.
Step 2:With zero net force the acceleration is zero, so the velocity stays constant.
Final answer: Remain constant
Q34Single correctMotion in a Plane
Two particles A and B are moving in uniform circular motion in concentric circles of radii and with speed and respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Express angular speed in terms of the time period, which is identical for both particles.
Step 1:Angular speed depends only on the time period, so each particle's angular speed uses its own period.
Step 2:Since the time periods are equal, the ratio of angular speeds is unity.
Final answer:
Q35Single correctSystem of Particles and Rotational Motion
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 rad/s
Approach:
Balance the upward limiting friction against gravity, where the normal force is the centripetal force supplied by the rotating wall.
Step 1:For equilibrium the limiting friction must support the weight of the block.
Step 2:The wall supplies the normal force as the centripetal force.
Step 3:Solving for the minimum angular speed gives the threshold value.
Final answer: rad/s
Q36Single correctElectric Charges and Fields
Two parallel infinite line charges with linear charge densities C/m and C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 N/C
Approach:
Add the fields of the two oppositely charged infinite lines at the midpoint, where both fields point in the same direction.
Step 1:Each line is a distance R from the midpoint, giving the field magnitude of a single line charge.
Step 2:At the midpoint the field of the positive line points away from it and the field of the negative line points toward it, so the two add in the same direction.
Step 3:Summing the contributions gives the total field.
Final answer: N/C
Q37Single correctElectric Charges and Fields
Two point charges A and B, having charges and respectively, are placed at certain distance apart and force acting between them is F. If charge of A is transferred to B, then force between the charges becomes :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Compute the new charges after transferring a quarter of A's charge to B, then take the ratio of the new Coulomb force to the original.
Step 1:Transferring 25% of A's charge to B leaves A with three quarters of its charge.
Step 2:Adding that quarter to B changes B's charge magnitude.
Step 3:The new force is proportional to the product of the new charge magnitudes, giving nine-sixteenths of the original.
Final answer:
Q38Single correctMechanical Properties of Fluids
A small hole of area of cross-section 2 m is present near the bottom of a fully filled open tank of height 2 m. Taking , the rate of flow of water through the open hole would be nearly :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Find the efflux speed from Torricelli's law and multiply by the hole area to obtain the volume flow rate.
Step 1:The efflux speed through the hole follows from the head of 2 m.
Step 2:The volume flow rate is the product of the hole area and the efflux speed.
Step 3:Evaluating the product gives the rate of flow.
Final answer:
Q39Single correctSemiconductor Electronics
The correct Boolean operation represented by the circuit diagram drawn is :
(A supply feeds a resistor R, then the LED (Y), then a second resistor R to earth. Two two-position switches A and B, each marked 0 and 1, are joined by a left-hand wire so that closing both bridges the LED branch to earth.)
(A supply feeds a resistor R, then the LED (Y), then a second resistor R to earth. Two two-position switches A and B, each marked 0 and 1, are joined by a left-hand wire so that closing both bridges the LED branch to earth.)

(1)
(2)
(3)
(4)
SolutionAnswer: Option 3NAND
Approach:
Analyze the diode/LED arrangement to determine the output state for each combination of inputs and match it to a standard gate truth table.
Step 1:The LED glows when the voltage across it is high; this occurs unless both inputs are high.
Step 2:Tabulating the inputs gives outputs 1,1,1,0 for the input pairs (0,0),(0,1),(1,0),(1,1), which is the NAND truth table.
Final answer: NAND
Q40Single correctThermodynamics
In which of the following processes, heat is neither absorbed nor released by a system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Adiabatic
Approach:
Identify the thermodynamic process defined by zero heat exchange with the surroundings.
Step 1:An adiabatic process is defined by the absence of heat transfer between the system and surroundings.
Step 2:Therefore the process in which heat is neither absorbed nor released is adiabatic.
Final answer: Adiabatic
Q41Single correctElectromagnetic Induction
A 800 turn coil of effective area 0.05 is kept perpendicular to a magnetic field T. When the plane of the coil is rotated by around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 V
Approach:
Apply Faraday's law using the change in magnetic flux as the coil rotates from perpendicular to parallel orientation.
Step 1:Initially the plane is perpendicular to the field so the flux through the coil is maximum, and after a 90 degree rotation the flux is zero.
Step 2:Substitute the number of turns, field, and area to evaluate the flux change.
Step 3:Dividing the flux change by the time gives the induced emf.
Final answer: V
Q42Single correctOscillations
The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig.
(A circle of radius 3 m is drawn about the origin with the x-axis marked; the particle is at at the top of the circle on the axis, the sense arrow on the circle points clockwise, and the period is marked s.)
y - projection of the radius vector of rotating particle P is :
(A circle of radius 3 m is drawn about the origin with the x-axis marked; the particle is at at the top of the circle on the axis, the sense arrow on the circle points clockwise, and the period is marked s.)
y - projection of the radius vector of rotating particle P is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4, where y in m
Approach:
Project the rotating radius vector onto the y-axis using the angular frequency from the period and the given starting position.
Step 1:The angular frequency follows from the period of 4 s.
Step 2:At the y-projection is maximum, so the cosine form with amplitude 3 m applies.
Final answer: , where y in m
Q43Single correctElectrostatic Potential and Capacitance
A parallel plate capacitor of capacitance 20 F is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 260 A, 60 A
Approach:
Compute the charging current from the rate of change of voltage and recognize that the displacement current equals the conduction current.
Step 1:The conduction current equals the capacitance times the rate of change of voltage.
Step 2:The displacement current between the plates equals the conduction current in the wires.
Final answer: 60 A, 60 A
Q44Single correctUnits and Measurements
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where , will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Add the fractional errors weighted by the magnitude of each exponent in the expression for X.
Step 1:Each percentage error is multiplied by the magnitude of its exponent and the contributions are added.
Step 2:Evaluating each term gives the total maximum percentage error.
Final answer:
Q45Single correctMoving Charges and Magnetism
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field, B with the distance d from the centre of the conductor, is correctly represented by the figure :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B rises linearly from the axis to a maximum at and then falls off as
Approach:
Apply Ampere's law separately inside and outside the conductor to obtain the dependence of B on distance.
Step 1:Inside the conductor the enclosed current grows with the square of the distance, so the field increases linearly with d.
Step 2:At the surface the field reaches its maximum value.
Step 3:Outside the conductor the field falls off inversely with distance, producing a hyperbolic decrease.
Final answer: B rises linearly from the axis to a maximum at and then falls off as
Chemistry45 questions
Q46Single correctChemical Bonding and Molecular Structure
The number of sigma () and pi () bonds in pent-2-en-4-yne is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 bonds and bonds
Approach:
Draw the structure of pent-2-en-4-yne and count the sigma and pi bonds, where every single bond and the first bond of any multiple bond is sigma, and additional bonds of double/triple bonds are pi.
Step 1:The structure of pent-2-en-4-yne is CH(triple bond)C-CH=CH-CH3, with one C=C, one C(triple)C, and three C-H bonds on the saturated/terminal carbons.
Step 2:Counting sigma bonds: four C-C/C=C/C(triple)C chain sigma bonds plus six C-H sigma bonds gives ten sigma bonds.
Step 3:Counting pi bonds: the double bond contributes one pi bond and the triple bond contributes two pi bonds.
Final answer: bonds and bonds
Q47Single correctHydrocarbons / Aromatic Chemistry
The structure of intermediate A in the following reaction, is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cumene hydroperoxide: the benzene ring carries (drawn structure)
Approach:
Identify the intermediate of the cumene (isopropylbenzene) oxidation that on acidic hydrolysis yields phenol and acetone.
Step 1:In the industrial cumene process, cumene reacts with atmospheric oxygen to form cumene hydroperoxide, in which the benzylic carbon bears a -O-O-H group.
Step 2:Acid-catalysed hydrolysis of cumene hydroperoxide rearranges and cleaves the O-O bond to give phenol and acetone.
Step 3:The intermediate is cumene hydroperoxide, in which the benzylic tertiary carbon bears the group.
Final answer: Cumene hydroperoxide: the benzene ring carries (drawn structure)
Q48Single correctThe p-Block Elements
The correct structure of tribromooctaoxide is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A BrBr chain in which all eight oxygens are doubly bonded and the species is neutral: (drawn structure)
Approach:
Recall the established structure of tribromooctaoxide, B, in which bromine atoms are linked through bridging oxygen atoms and carry terminal oxygens.
Step 1:Tribromooctaoxide has the formula B with three bromine atoms connected in a chain through bridging oxygen atoms, each bromine also bearing terminal oxygen atoms.
Step 2:Tribromooctaoxide is therefore a neutral BrBrBr chain in which every one of the eight oxygens is doubly bonded to bromine.
Final answer: A BrBr chain in which all eight oxygens are doubly bonded and the species is neutral: (drawn structure)
Q49Single correctStructure of Atom
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Apply the (n+l) rule: higher (n+l) means higher energy; for equal (n+l), the orbital with larger n has higher energy.
Step 1:Compute (n+l) for each orbital: for 4d, n+l = 4 + 2 = 6; for 5p, n+l = 5 + 1 = 6; for 5f, n+l = 5 + 3 = 8; for 6p, n+l = 6 + 1 = 7.
Step 2:Order by decreasing (n+l): 5f (8) then 6p (7); for the tie at 6, 5p has larger n than 4d so 5p is higher.
Final answer:
Q50Single correctRedox Reactions
Which of the following reactions are disproportionation reaction?
(a)
(b)
(c)
(d)
Select the correct option from the following
(a)
(b)
(c)
(d)
Select the correct option from the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a) and (b) only
Approach:
A disproportionation reaction is one in which the same element in a single oxidation state is simultaneously oxidised and reduced. Examine each reaction.
Step 1:In (a), Cu in +1 state goes to +2 (oxidised) and to 0 (reduced), so it is a disproportionation.
Step 2:In (b), Mn in +6 state in MnO4(2-) goes to +7 in MnO4(-) and to +4 in MnO2, so Mn(+6) is both oxidised and reduced, a disproportionation.
Step 3:In (c), Mn stays +7 in K2MnO4? Mn changes from +7 to +6 and oxygen changes oxidation state, so the same element in one state is not simultaneously oxidised and reduced in the required sense; it is not a disproportionation of a single element.
Step 4:In (d), MnO4(-) Mn(+7) is reduced and Mn(+2) is oxidised, but two different starting oxidation states are involved, making it a comproportionation, not a disproportionation.
Step 5:Only (a) and (b) are disproportionation reactions.
Final answer: (a) and (b) only
Q51Single correctThermodynamics
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is
(Given that 1 L bar = 100 J)
(Given that 1 L bar = 100 J)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The work done BY a gas expanding against a constant external pressure equals the product of that pressure and the volume change, expressed with the sign convention appropriate to work done by the system.
Step 1:Volume change of the gas.
Step 2:Work done by the gas against the constant external pressure.
Step 3:Convert to joules using 1 L bar = 100 J.
Final answer:
Q52Single correctEnvironmental Chemistry
Among the following, the one that is not a green house gas is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Sulphur dioxide
Approach:
Recall the recognised greenhouse gases and identify the one not classified as such.
Step 1:Nitrous oxide, methane and ozone are all recognised greenhouse gases that absorb infrared radiation.
Step 2:Sulphur dioxide is an air pollutant responsible for acid rain but is not classified as a greenhouse gas.
Final answer: Sulphur dioxide
Q53Single correctElectrochemistry
For the cell reaction
V at 298 K. The standard Gibbs energy () of the cell reaction is
[Given that Faraday constant F = 96500 C mo]
V at 298 K. The standard Gibbs energy () of the cell reaction is
[Given that Faraday constant F = 96500 C mo]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Apply the relation between standard Gibbs energy and standard cell potential with the number of electrons transferred.
Step 1:The cell reaction transfers two electrons, so n = 2.
Step 2:Substituting n, F and the cell potential.
Step 3:Converting to kilojoules.
Final answer:
Q54Single correctBiomolecules / s-Block Elements
Enzymes that utilize ATP in phosphate transfer require an alkaline earth metal (M) as the cofactor. M is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mg
Approach:
Recall the alkaline earth metal that acts as the essential cofactor in ATP-dependent phosphate transfer reactions.
Step 1:Enzymes that utilize ATP in phosphate transfer require magnesium as the cofactor, with ATP functioning as the Mg-ATP complex.
Final answer: Mg
Q55Single correctHydrocarbons
The most suitable reagent for the following conversion, is

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2, Pd/C, quinoline
Approach:
Determine the reagent that converts an internal alkyne to the cis (Z) alkene by syn addition of hydrogen.
Step 1:Catalytic hydrogenation of an internal alkyne over a partially poisoned palladium catalyst proceeds by syn addition, delivering both hydrogen atoms to the same face and giving the cis alkene.
Step 2:Sodium in liquid ammonia would give the trans alkene, while the other reagents do not give cis-2-butene, so H2 over Pd/C is most suitable.
Final answer: , Pd/C, quinoline
Q56Single correctThe p-Block Elements (Group 16)
Which is the correct thermal stability order for E (E = O, S, Se, Te and Po)?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Thermal stability of group 16 hydrides depends on the E-H bond strength, which decreases down the group as atomic size increases.
Step 1:Down the group from O to Po the atomic size increases, the E-H bond becomes weaker and longer, so thermal stability decreases.
Step 2:Writing the same series in increasing order of stability gives .
Final answer:
Q57Single correctThe p-Block Elements (Group 14)
Which of the following is incorrect statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pb is covalent in nature
Approach:
Examine each statement on the bonding nature and stability of group 14 tetrahalides and identify the incorrect one.
Step 1:PbF4 and SnF4 are ionic in nature because of the small, highly electronegative fluoride combining with the metal, so the statement that PbF4 is covalent is incorrect.
Step 2:SiCl4 is readily hydrolysed, and GeX4 is generally more stable than GeX2 except where the inert pair effect dominates, so those statements are correct.
Final answer: Pb is covalent in nature
Q58Single correctThe p-Block Elements / General Principles of Metallurgy
Match the following :
| List I | List II |
|---|---|
| (a). Pure nitrogen | (i). Chlorine |
| (b). Haber process | (ii). Sulphuric acid |
| (c). Contact process | (iii). Ammonia |
| (d). Deacon's process | (iv). Sodium azide or Barium azide |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)(iv) (b)(iii) (c)(ii) (d)(i)
Approach:
Match each preparation or industrial process with its principal product.
Step 1:Pure nitrogen is obtained by thermal decomposition of sodium azide or barium azide, matching (a)-(iv).
Step 2:The Haber process produces ammonia, matching (b)-(iii).
Step 3:The contact process produces sulphuric acid, matching (c)-(ii).
Step 4:Deacon's process produces chlorine, matching (d)-(i).
Final answer: (a)(iv) (b)(iii) (c)(ii) (d)(i)
Q59Single correctChemical Bonding and Molecular Structure
Which of the following diatomic molecular species has only bonds according to Molecular Orbital Theory?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Use the molecular orbital electronic configuration to determine bond type for each species; the species whose net bonding arises solely from pi molecular orbitals is required.
Step 1:The MO configuration of C2 is , sigma*, , sigma*, = , with no net sigma bonding from the 2p set.
Step 2:The bond order of C2 is two, arising entirely from the two filled pi bonding molecular orbitals, so C2 has only pi bonds.
Final answer:
Q60Single correctClassification of Elements and Periodicity in Properties
For the second period elements the correct increasing order of first ionisation enthalpy is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
First ionisation enthalpy generally rises across a period, but two anomalies break the smooth trend: the filled 2s of Be exceeds the single 2p electron of B, and the half-filled stable 2p3 of N exceeds the paired-electron 2p4 of O.
Step 1:General left-to-right increase places Li lowest and Ne highest.
Step 2:First anomaly: removing the 2p electron of boron is easier than removing from the filled 2s of beryllium, so B falls below Be.
Step 3:Second anomaly: the half-filled 2p3 configuration of nitrogen is extra stable, so removing an electron from oxygen (2p4) is easier, placing O below N.
Step 4:Combining the baseline with both anomalies gives the full ascending order.
Final answer:
Q61Single correctPolymers
The biodegradable polymer is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Nylon-2-Nylon-6
Approach:
Identify the polymer from the list that is biodegradable.
Step 1:Nylon-2-Nylon-6 is an alternating polyamide of glycine and amino caproic acid and is a biodegradable polymer, whereas Nylon-6,6, Nylon-6 and Buna-S are not biodegradable.
Final answer: Nylon-2-Nylon-6
Q62Single correctEquilibrium
pH of a saturated solution of Ca(OH is 9. The solubility product () of Ca(OH is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Use the pH to find the hydroxide ion concentration, then express the solubility product of Ca(OH)2 in terms of the ion concentrations.
Step 1:Since pH = 9, pOH = 14 - 9 = 5 and the hydroxide concentration is M.
Step 2:Each formula unit of Ca(OH)2 gives one Ca2+ and two OH-, so the calcium concentration is half the hydroxide concentration.
Step 3:Substituting into the solubility product expression.
Final answer:
Q63Single correctChemical Kinetics
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply the integrated first order rate law to the condition of 99% completion, where the remaining reactant is 1% of the initial amount.
Step 1:For 99% completion, the concentration falls to 1% of the initial, giving the ratio [A]0/[A] = 100.
Step 2:Since log 100 = 2, rearranging gives the required time.
Final answer:
Q64Single correctBiomolecules
The non-essential amino acid among the following is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Alanine
Approach:
Distinguish between essential amino acids (which must be supplied in the diet) and non-essential amino acids (synthesised in the body).
Step 1:Valine, leucine and lysine are essential amino acids that must be obtained from the diet, whereas alanine can be synthesised by the body and is non-essential.
Final answer: Alanine
Q65Single correctHaloalkanes and Haloarenes / Aromatic Chemistry
Among the following, the reaction that proceeds through an electrophilic substitution, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Benzene with in presence of giving chlorobenzene and HCl
Approach:
Identify which reaction proceeds by generation of an electrophile that attacks the aromatic ring, replacing a ring hydrogen.
Step 1:Chlorination of benzene in the presence of the Lewis acid AlCl3 generates the chloronium electrophile Cl+ from Cl2-AlCl3, which attacks the ring and replaces a hydrogen, giving chlorobenzene and HCl.
Step 2:The diazonium reaction is a substitution proceeding through a different mechanism, the UV chlorination is a free-radical addition, and the reaction of benzyl alcohol with HCl replaces on an s carbon; only the -catalysed chlorination of benzene is electrophilic aromatic substitution.
Final answer: Benzene with in presence of giving chlorobenzene and HCl
Q66Single correctSolutions
The mixture that forms maximum boiling azeotrope is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Water + Nitric acid
Approach:
A maximum boiling azeotrope is formed by solutions showing negative deviation from Raoult's law. Identify such a mixture.
Step 1:Solutions showing negative deviation from Raoult's law form maximum boiling azeotropes.
Step 2:The water and nitric acid mixture shows negative deviation and forms a maximum boiling azeotrope, while the others either form minimum boiling azeotropes or behave ideally.
Final answer: Water + Nitric acid
Q67Single correctChemical Kinetics
For the chemical reaction
The correct option is:
The correct option is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Express the rate of reaction in terms of the change of concentration of each species divided by its stoichiometric coefficient, with negative signs for reactants and positive for products.
Step 1:From the balanced equation the rate of reaction equals the negative rate of change of nitrogen, one third the negative rate of change of hydrogen and one half the rate of formation of ammonia.
Step 2:Dividing each rate by its stoichiometric coefficient and taking reactants as negative gives .
Final answer:
Q68Single correctEquilibrium / Stoichiometry
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 330
Approach:
Use the stoichiometry of the Haber process, where three moles of hydrogen give two moles of ammonia, to find the hydrogen required for 20 moles of ammonia.
Step 1:Two moles of ammonia require three moles of hydrogen by the balanced equation.
Step 2:Scaling to 20 moles of ammonia gives the hydrogen requirement.
Final answer: 30
Q69Single correctAldehydes, Ketones and Carboxylic Acids
The compound that is most difficult to protonate is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Phenol, (drawn structure)
Approach:
Protonation occurs at the lone pair of the oxygen; the species in which that lone pair is least available is hardest to protonate. In phenol the oxygen lone pair is delocalised into the aromatic ring through resonance, decreasing electron density on oxygen.
Step 1:In simple aldehydes and ketones the carbonyl oxygen retains lone pairs that accept a proton readily.
Step 2:In phenol the oxygen lone pair enters into resonance with the benzene ring, lowering electron availability on oxygen.
Step 3:Lower electron density on the protonation site corresponds to greater difficulty of protonation.
Final answer: Phenol, (drawn structure)
Q70Single correctSolutions
For an ideal solution, the correct option is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 at constant T and P
Approach:
Apply the defining thermodynamic criteria of an ideal solution at constant temperature and pressure and test each statement.
Step 1:An ideal solution obeys Raoult's law over the whole composition range with identical solute-solute, solvent-solvent and solute-solvent interactions.
Step 2:Identical interactions also leave the total volume unchanged on mixing.
Step 3:Mixing increases disorder, so the entropy of mixing is positive and the free energy of mixing is negative; neither vanishes, so the vanishing quantity that defines ideality here is the enthalpy of mixing.
Final answer: at constant T and P
Q71Single correctEquilibrium
Conjugate base for Brønsted acids and HF are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and , respectively
Approach:
A conjugate base is formed by removal of one proton from the Brønsted acid.
Step 1:Remove one proton from water acting as an acid.
Step 2:Remove one proton from hydrogen fluoride acting as an acid.
Final answer: and , respectively
Q72Single correctSurface Chemistry
Which mixture of the solutions will lead to the formation of negatively charged colloidal sol ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 250 mL of 1 M + 50 mL of 2 M KI
Approach:
A negatively charged AgI sol forms when KI is in excess, so that I- ions are preferentially adsorbed on the AgI particles. Compare millimoles of AgNO3 and KI in each mixture.
Step 1:With 50 mL of 1 M and 50 mL of 1.5 M KI the millimoles are 50 and 75, so KI is in excess and the sol is negative.
Step 2:With 50 mL of 1 M and 50 mL of 2 M KI the millimoles are 50 and 100, so KI is in a larger excess and the sol is again negative.
Step 3:Where is in excess or exactly equimolar, iodide is not in excess and the sol is positively charged or uncharged.
Final answer: 50 mL of 1 M + 50 mL of 2 M KI
Q73Single correctChemistry in Everyday Life
Among the following, the narrow spectrum antibiotic is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Penicillin G
Approach:
Classify each antibiotic as narrow-spectrum or broad-spectrum from its known activity range.
Step 1:Penicillin G is effective mainly against Gram-positive bacteria, placing it in the narrow-spectrum class.
Step 2:Ampicillin and amoxycillin act against a wider range of bacteria, and chloramphenicol acts on many organisms, so all three are broad-spectrum.
Final answer: Penicillin G
Q74Single correctHydrocarbons
An alkene "A" on reaction with and gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3, 2-chloro-2-methylbutane (drawn structure)
Approach:
Reconstruct the alkene from the ozonolysis carbonyl fragments, then apply Markovnikov addition of HCl.
Step 1:Propanone and ethanal arise by cleaving the double bond, so joining their carbonyl carbons rebuilds alkene A as 2-methyl-2-butene.
Step 2:HCl adds by Markovnikov's rule, placing H on the carbon with more hydrogens and Cl on the more substituted carbon.
Step 3:The resulting major product is 2-chloro-2-methylbutane.
Final answer: , 2-chloro-2-methylbutane (drawn structure)
Q75Single correctCoordination Compounds
What is the correct electronic configuration of the central atom in based on crystal field theory?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Determine the oxidation state and d-electron count of iron, then fill the octahedral d-orbitals using the strong-field cyanide ligand.
Step 1:Potassium contributes +4 and each cyanide is -1, fixing iron in the +2 state.
Step 2:Fe(II) has a 3d6 configuration.
Step 3:Cyanide is a strong-field ligand causing pairing, so all six electrons occupy the lower t2g set.
Final answer:
Q76Single correctThe p-Block Elements
Identify the incorrect statement related to from the following:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 molecule is non-reactive
Approach:
Recall the trigonal bipyramidal geometry of PCl5 and its chemical behaviour, then identify which statement is false.
Step 1:PCl5 is trigonal bipyramidal with three equatorial bonds at 120 degrees and two axial bonds at 180 degrees, and the axial bonds are longer due to greater repulsion.
Step 2:PCl5 is in fact highly reactive, acting as a chlorinating agent and hydrolysing readily, so the claim of non-reactivity is false.
Final answer: molecule is non-reactive
Q77Single correctEquilibrium
Which will make basic buffer?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3100 mL of 0.1 M HCl + 200 mL of 0.1 M
Approach:
A basic buffer requires a weak base together with its salt, leaving excess weak base after partial neutralisation by a strong acid. Compute the millimoles in each mixture.
Step 1:For 100 mL of 0.1 M HCl with 200 mL of 0.1 M : HCl mmol and mmol, so the acid neutralises 10 mmol to and leaves 10 mmol of free .
Step 2:The acetic-acid/NaOH mixture with excess base leaves no weak acid, the equimolar acetic-acid/NaOH mixture gives only sodium acetate, and the HCl/NaOH mixture gives a neutral salt; none of these leaves a weak base alongside its salt.
Final answer: 100 mL of 0.1 M HCl + 200 mL of 0.1 M
Q78Single correctAmines / Carboxylic Acids
The major product of the following reaction is :

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Phthalimide: benzene ring fused to a five-membered ring carrying two C=O groups and an N-H (drawn structure)
Approach:
Trace the reaction of the ortho-dicarboxylic acid with ammonia on strong heating, which forms the diammonium salt, then the diamide, and finally cyclises with loss of water.
Step 1:Phthalic acid first reacts with ammonia to give the ammonium salt, then the diamide on heating with loss of water.
Step 2:On strong heating the diamide cyclises by intramolecular condensation, losing ammonia to form the cyclic imide.
Step 3:The cyclic imide fused to the benzene ring is phthalimide.
Final answer: Phthalimide: benzene ring fused to a five-membered ring carrying two C=O groups and an N-H (drawn structure)
Q79Single correctThe p-Block Elements
Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code:
| Column-I | Column-II |
|---|---|
| (a). | (i). Pyramidal |
| (b). | (ii). Square planar |
| (c). | (iii). Distorted octahedral |
| (d). | (iv). Square pyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) (ii) (b) (iii) (c) (iv) (d) (i)
Approach:
Assign the shape of each xenon compound from its number of bond pairs and lone pairs on xenon.
Step 1:XeF4 has four bond pairs and two lone pairs, giving a square planar shape.
Step 2:XeF6 has six bond pairs and one lone pair, giving a distorted octahedral shape.
Step 3:XeOF4 has five bond pairs and one lone pair, giving a square pyramidal shape, while XeO3 has three bond pairs and one lone pair, giving a pyramidal shape.
Final answer: (a) (ii) (b) (iii) (c) (iv) (d) (i)
Q80Single correctThe d- and f-Block Elements
The manganate and permanganate ions are tetrahedral, due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The -bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
Approach:
Identify the nature of the pi-bonding that fixes the tetrahedral geometry in the manganate and permanganate oxoanions.
Step 1:In both oxoanions manganese is surrounded by four oxygens in a tetrahedral arrangement.
Step 2:The pi-bonding arises from overlap of filled oxygen p-orbitals with vacant manganese d-orbitals, a d-p pi interaction.
Final answer: The -bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
Q81Single correctThe p-Block Elements
Which of the following species is not stable?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Assess whether each central atom can accommodate six surrounding atoms, considering atomic size and the size of the ligand.
Step 1:Germanium and tin are large enough to bind six chloride or hydroxide groups, and small fluoride fits readily around silicon, so those species are stable.
Step 2:Chloride is too large to allow six of them around the small silicon atom, so a six-coordinate chloro-silicate cannot form.
Final answer:
Q82Single correctElectrochemistry
For a cell involving one electron at 298 K, the equilibrium constant for the cell reaction is :
Given that at T = 298 K
Given that at T = 298 K
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Relate the standard cell potential to the equilibrium constant through the Nernst equation at equilibrium.
Step 1:At equilibrium the cell potential is zero, so the standard potential equals the log term.
Step 2:Substituting the values gives log Keq.
Step 3:Taking the antilog gives the equilibrium constant.
Final answer:
Q83Single correctThe s-Block Elements
Which of the following is an amphoteric hydroxide?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify which alkaline earth hydroxide reacts with both acids and bases.
Step 1:Hydroxides of strontium, calcium and magnesium are basic and react only with acids.
Step 2:Beryllium hydroxide reacts with both acid and base, forming a salt with acid and a beryllate with alkali, so it is amphoteric.
Final answer:
Q84Single correctStates of Matter
A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option about the gas and its compressibility factor (Z) is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Z < 1 and attractive forces are dominant
Approach:
Express the compressibility factor as the ratio of real to ideal molar volume and interpret its value.
Step 1:The real molar volume is 20 percent smaller, so it equals 0.8 times the ideal molar volume.
Step 2:Dividing gives the compressibility factor.
Step 3:A compressibility factor below unity indicates that attractive intermolecular forces dominate.
Final answer: Z < 1 and attractive forces are dominant
Q85Single correctThe Solid State
A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
In a close-packed lattice the number of octahedral voids equals the number of anions; multiplying that by the fractional occupancy gives the cation count, and the ratio of cations to anions fixes the formula.
Step 1:Take the number of anions A in the hcp lattice as N; the number of octahedral voids equals N.
Step 2:Cations C occupy 75% of these voids.
Step 3:Form the cation-to-anion ratio.
Step 4:Express as the simplest formula.
Final answer:
Q86Single correctThermodynamics
In which case change in entropy is negative?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify the process in which disorder of the system decreases, giving a negative entropy change.
Step 1:Evaporation, gas expansion and sublimation all increase disorder, giving positive entropy changes.
Step 2:Combining two gaseous hydrogen atoms into one molecule reduces the number of gas particles and the disorder.
Final answer:
Q87Single correctStructure of Atom
Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Balmer series
Approach:
Match each hydrogen spectral series to its spectral region.
Step 1:The Lyman series lies in the ultraviolet, and the Paschen and Brackett series lie in the infrared.
Step 2:Transitions ending at the second level form the Balmer series, which falls in the visible region.
Final answer: Balmer series
Q88Single correctThe s-Block Elements / Hydrogen
The method used to remove temporary hardness of water is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Clark's method
Approach:
Recall the specific method that removes temporary hardness caused by bicarbonates.
Step 1:Temporary hardness arises from dissolved calcium and magnesium bicarbonates.
Step 2:Clark's method adds slaked lime, which precipitates the calcium and magnesium carbonates.
Final answer: Clark's method
Q89Single correctThe d- and f-Block Elements
Which one is malachite from the following?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Recall the chemical composition of the green copper ore malachite.
Step 1:Malachite is a basic copper carbonate, the green ore of copper.
Step 2:Copper pyrites, copper hydroxide and magnetite do not correspond to malachite.
Final answer:
Q90Single correctAmines
The correct order of the basic strength of methyl substituted amines in aqueous solution is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Balance the electron-donating inductive effect, the solvation of the protonated amine by hydrogen bonding, and steric hindrance to find the aqueous basicity order.
Step 1:The inductive effect of methyl groups raises basicity, but in water solvation of the cation and steric crowding oppose this for the tertiary amine.
Step 2:The secondary amine combines good electron donation with adequate solvation, the primary amine has the best solvation but less donation, and the tertiary amine is most hindered with poorest solvation.
Final answer:
Biology90 questions
Q91Single correctEnvironmental Issues
The Earth Summit held in Rio de Janeiro in 1992 was called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2for conservation of biodiversity and sustainable utilization of its benefits
Approach:
Identify the central mandate of the 1992 Rio de Janeiro Earth Summit (United Nations Conference on Environment and Development).
Step 1:The Earth Summit (Rio Summit, 1992) brought nations together to take appropriate measures for conservation of biodiversity and the sustainable utilisation of its benefits.
Final answer: for conservation of biodiversity and sustainable utilization of its benefits
Q92Single correctReproduction in Human / Animal Husbandry
Colostrum the yellowish fluid, secreted by mother during the initial days of lactation is very essential to impart immunity to the new born infants because it contains
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Immunoglobulin A
Approach:
Recall the component of colostrum responsible for passive immunity in the newborn.
Step 1:Colostrum, the yellowish fluid secreted by the mother during initial days of lactation, is rich in antibodies (immunoglobulins), especially Immunoglobulin A (IgA).
Step 2:These maternal antibodies provide naturally acquired passive immunity to the newborn.
Final answer: Immunoglobulin A
Q93Single correctAnatomy of Flowering Plants
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Flaccidity of bulliform cells
Approach:
Relate inward curling (rolling) of grass leaves in dry weather to the water status of a specialised epidermal cell type.
Step 1:Bulliform cells are large, thin-walled epidermal cells present in grass leaves.
Step 2:On water loss these cells become flaccid, causing the leaf to roll/curl inwards and minimise transpirational water loss.
Final answer: Flaccidity of bulliform cells
Q94Single correctPrinciples of Inheritance and Variation
The shorter and longer arms of a submetacentric chromosome are referred to as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2p-arm and q-arm respectively
Approach:
Recall the standard nomenclature for the two arms of a chromosome divided by the centromere.
Step 1:In a submetacentric chromosome the centromere lies slightly off-centre, giving one short arm and one long arm.
Step 2:By convention the short arm is the 'p' arm (p = petite, i.e. short) and the long arm is the 'q' arm.
Final answer: p-arm and q-arm respectively
Q95Single correctRespiration in Plants
Respiratory Quotient (RQ) value of tripalmitin is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.7
Approach:
Compute the respiratory quotient of the fat tripalmitin from the ratio of carbon dioxide released to oxygen consumed during its oxidation.
Step 1:Oxidation of tripalmitin proceeds according to the balanced equation.
Step 2:Substituting the moles of carbon dioxide released and oxygen consumed into the RQ ratio.
Final answer: 0.7
Q96Single correctMicrobes in Human Welfare
Which of the following is a commercial blood cholesterol lowering agent?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Statin
Approach:
Identify the microbial product used commercially to lower blood cholesterol.
Step 1:Statins are obtained from the yeast (fungus) Monascus purpureus.
Step 2:Statins act by competitively inhibiting the enzyme responsible for synthesis of cholesterol, thereby lowering blood cholesterol.
Final answer: Statin
Q97Single correctDigestion and Absorption
Match the following structures with their respective location in organs
(a) Crypts of Lieberkuhn (i) Pancreas
(b) Glisson's Capsule (ii) Duodenum
(c) Islets of Langerhans (iii) Small intestine
(d) Brunner's Glands (iv) Liver
Select the correct option from the following
(a) Crypts of Lieberkuhn (i) Pancreas
(b) Glisson's Capsule (ii) Duodenum
(c) Islets of Langerhans (iii) Small intestine
(d) Brunner's Glands (iv) Liver
Select the correct option from the following
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a) (iii) (b) (iv) (c) (i) (d) (ii)
Approach:
Match each anatomical structure to the organ in which it is located.
Step 1:Crypts of Lieberkuhn are intestinal glands present in the small intestine.
Step 2:Glisson's Capsule constitutes the connective-tissue covering of the liver.
Step 3:Islets of Langerhans form the endocrine portion of the pancreas.
Step 4:Brunner's glands are found in the submucosa of the duodenum.
Final answer: (a) (iii) (b) (iv) (c) (i) (d) (ii)
Q98Single correctBiodiversity and Conservation
Which of the following is the most important cause for animals and plants being driven to extinction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Habitat loss and fragmentation
Approach:
Recall the leading cause among the major drivers of biodiversity loss ('The Evil Quartet').
Step 1:Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.
Step 2:An example is the loss of tropical rainforest, reducing cover from 14 percent to 6 percent of the land surface.
Final answer: Habitat loss and fragmentation
Q99Single correctNeural Control and Coordination
Which part of the brain is responsible for thermoregulation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hypothalamus
Approach:
Identify the brain region acting as the thermoregulatory centre.
Step 1:The hypothalamus is the thermoregulatory centre of the brain.
Step 2:It is responsible for the maintenance of a constant body temperature.
Final answer: Hypothalamus
Q100Single correctAnimal Kingdom
Consider following features
(a) Organ system level of organisation
(b) Bilateral symmetry
(c) True coelomates with segmentation of body
Select the correct option of animal groups which possess all the above characteristics
(a) Organ system level of organisation
(b) Bilateral symmetry
(c) True coelomates with segmentation of body
Select the correct option of animal groups which possess all the above characteristics
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Annelida, Arthropoda and Chordata
Approach:
Select the animal phyla that simultaneously show organ-system level of organisation, bilateral symmetry, true coelom and body segmentation.
Step 1:True segmentation is present in Annelida, Arthropoda and Chordata.
Step 2:These phyla also have organ-system level of organisation, bilateral symmetry and are true coelomates.
Final answer: Annelida, Arthropoda and Chordata
Q101Single correctAnimal Kingdom / Cockroach
Select the correct sequence of organs in the alimentary canal of cockroach starting from mouth
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pharynx Oesophagus Crop Gizzard Ileum Colon Rectum
Approach:
Trace the alimentary canal of the cockroach from the mouth posteriorly.
Step 1:The correct sequence of organs in the alimentary canal of cockroach starting from mouth follows the established order.
Step 2:Pharynx leads to Oesophagus, then Crop, then Gizzard, then Ileum, then Colon, then Rectum.
Final answer: Pharynx Oesophagus Crop Gizzard Ileum Colon Rectum
Q102Single correctEnvironmental Issues
Which of the following pairs of gases is mainly responsible for green house effect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Carbon dioxide and Methane
Approach:
Identify the pair of gases contributing most to the greenhouse effect.
Step 1:Relative contributions of the major greenhouse gases to total global warming are: carbon dioxide about 60 percent and methane about 20 percent.
Step 2:Chlorofluorocarbons contribute about 14 percent and nitrous oxide about 6 percent.
Step 3:Therefore carbon dioxide and methane are the major greenhouse gases.
Final answer: Carbon dioxide and Methane
Q103Single correctLocomotion and Movement
Which of the following muscular disorders is inherited?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Muscular dystrophy
Approach:
Determine which listed muscular disorder has a genetic (inherited) basis.
Step 1:Muscular dystrophy is a progressive degeneration of skeletal muscle that mostly occurs due to a genetic disorder.
Step 2:Tetany arises as a muscular spasm due to low calcium in body fluid; myasthenia gravis is an auto-immune disorder leading to paralysis of skeletal muscles; botulism is a case of food poisoning caused by the bacterium Clostridium Botulinum.
Final answer: Muscular dystrophy
Q104Single correctBreathing and Exchange of Gases
The ciliated epithelial cells are required to move particles or mucus in a specific direction. In humans, these cells are mainly present in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bronchioles and Fallopian tubes
Approach:
Identify the human locations lined by ciliated epithelium that propels particles or mucus.
Step 1:Bronchioles and Fallopian tubes are lined with ciliated epithelium to move particles or mucus in a specific direction.
Final answer: Bronchioles and Fallopian tubes
Q105Single correctBody Fluids and Circulation
Match the Column-I with Column-II
| Column-I | Column-II |
|---|---|
| (a). P - wave | (i). Depolarisation of ventricles |
| (b). QRS complex | (ii). Repolarisation of ventricles |
| (c). T - wave | (iii). Coronary ischemia |
| (d). Reduction in the size of T-wave | (iv). Depolarisation of atria |
| (v). Repolarisation of atria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Approach:
Assign each ECG deflection to the cardiac electrical event it records, then pair the columns.
Step 1:The P-wave of the ECG marks the spread of excitation across the atrial musculature, i.e. atrial depolarisation, pairing (a) with (iv).
Step 2:The QRS complex records the depolarisation of the ventricles, which initiates ventricular contraction, pairing (b) with (i).
Step 3:The T-wave represents repolarisation of the ventricles as the myocardium returns from the excited to the normal resting state, pairing (c) with (ii).
Step 4:A reduced amplitude of the T-wave indicates inadequate oxygen supply to the cardiac muscle, characteristic of coronary ischemia, pairing (d) with (iii).
Final answer: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
Q106Single correctBiodiversity and Conservation
Which one of the following is not a method of in\ situ conservation of biodiversity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Botanical Garden
Approach:
Classify each named conservation measure as in situ (on-site) or ex situ (off-site).
Step 1:Botanical garden is an example of ex-situ conservation (off-site conservation), where living plants (flora) are conserved in a human-managed system.
Step 2:Biosphere reserves, wildlife sanctuaries and sacred groves conserve organisms in their natural habitat and are therefore in situ methods.
Final answer: Botanical Garden
Q107Single correctOrganisms and Populations / Evolution
In a species, the weight of newborn ranges from 2 to 5 kg. 97% of the newborn with an average weight between 3 to 3.3 kg survive whereas 99% of the infants born with weight from 2 to 2.5 kg or 4.5 to 5 kg die. Which type of selection process is taking place?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Stabilizing Selection
Approach:
Determine the mode of natural selection from the survival pattern across the trait distribution.
Step 1:The given data shows stabilising selection as most of the newborn having an average weight between 3 to 3.3 kg survive and babies with less and more weight have low survival rate.
Final answer: Stabilizing Selection
Q108Single correctCell Cycle and Cell Division
The correct sequence of phases of cell cycle is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
State the ordered phases of the eukaryotic cell cycle.
Step 1:The cell cycle proceeds through interphase phases G1, S and G2 followed by the mitotic M phase.
Final answer:
Q109Single correctChemical Coordination and Integration
How does steroid hormone influence the cellular activities?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Binding to DNA and forming a gene-hormone complex
Approach:
Recall the cellular mechanism of action of steroid hormones.
Step 1:Steroid hormones directly enter into the cell and bind with intracellular receptors in the nucleus to form a hormone receptor complex.
Step 2:The hormone receptor complex interacts with the genome, regulating gene expression.
Final answer: Binding to DNA and forming a gene-hormone complex
Q110Single correctCell - The Unit of Life
Which of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Lysosomes are formed by the process of packaging in the endoplasmic reticulum
Approach:
Identify the incorrect statement about lysosomes.
Step 1:Lysosomes bud off from the trans face of Golgi bodies; the precursor lysosomal enzymes are synthesised by the rough endoplasmic reticulum and then sent to the Golgi bodies for further processing.
Step 2:Therefore the statement that lysosomes are formed by packaging in the endoplasmic reticulum is the incorrect one.
Final answer: Lysosomes are formed by the process of packaging in the endoplasmic reticulum
Q111Single correctSexual Reproduction in Flowering Plants
Which one of the following statements regarding post-fertilization development in flowering plants is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ovules develop into embryo sac
Approach:
Identify the statement that does not represent a post-fertilisation change.
Step 1:The post-fertilisation changes are: ovule develops into seed, ovary into fruit, zygote into embryo, and the central cell into endosperm.
Step 2:An ovule developing into an embryo sac is a pre-fertilisation (megasporogenesis/female gametophyte) event, so this statement is incorrect.
Final answer: Ovules develop into embryo sac
Q112Single correctBiology in Human Welfare / Plant products
Concanavalin A is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a lectin
Approach:
Classify Concanavalin A among the categories of plant secondary metabolites.
Step 1:Concanavalin A is a secondary metabolite (e.g. a lectin), which has the property to agglutinate RBCs.
Final answer: a lectin
Q113Single correctMicrobes in Human Welfare / Biotechnology
Which one of the following equipments is essentially required for growing microbes on a large scale, for industrial production of enzymes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bioreactor
Approach:
Identify the equipment used to culture microbes on a large scale for enzyme production.
Step 1:To produce enzymes in large quantity, equipment required are bioreactors.
Step 2:Large scale production involves use of bioreactors.
Final answer: Bioreactor
Q114Single correctBiomolecules
Consider the following statement :
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the correct option.
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the correct option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(A) is true but (B) is false.
Approach:
Test each statement against the definitions of prosthetic group, apoenzyme and holoenzyme.
Step 1:A prosthetic group is a non-protein cofactor that stays firmly (tightly) bound to the enzyme protein, which is what distinguishes it from a loosely and transiently associating coenzyme; statement (A) states exactly that tight-binding criterion, so it is true.
Step 2:The protein portion of a conjugate enzyme without its cofactor is the apoenzyme, whereas the complete catalytically active enzyme together with its bound cofactor is the holoenzyme; statement (B) mislabels this complete active form as the apoenzyme, so (B) is false.
Step 3:Statement (A) is therefore true and statement (B) is false.
Final answer: (A) is true but (B) is false.
Q115Single correctBiomolecules
Purines found both in DNA and RNA are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Adenine and guanine
Approach:
Identify which nitrogenous bases are purines and occur in both nucleic acids.
Step 1:Purines are the double-ring bases adenine and guanine, while pyrimidines are the single-ring bases cytosine, thymine and uracil.
Step 2:Adenine and guanine occur in both DNA and RNA, whereas thymine is restricted to DNA and uracil to RNA.
Step 3:Therefore the purines common to DNA and RNA are adenine and guanine.
Final answer: Adenine and guanine
Q116Single correctHuman Reproduction
Select the correct sequence for transport of sperm cells in male reproductive system.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Seminiferous tubules Rete testis Vasa efferentia Epididymis Vas deferens Ejaculatory duct Urethra Urethral meatus
Approach:
Trace the path of sperm from the site of production to the exterior.
Step 1:Sperms are produced in the seminiferous tubules and pass into the rete testis, then through the vasa efferentia into the epididymis.
Step 2:From the epididymis the sperm travel through the vas deferens, join the duct of the seminal vesicle to form the ejaculatory duct, and continue into the urethra.
Step 3:The urethra finally opens at the urethral meatus, completing the sequence seminiferous tubules, rete testis, vasa efferentia, epididymis, vas deferens, ejaculatory duct, urethra, urethral meatus.
Final answer: Seminiferous tubules Rete testis Vasa efferentia Epididymis Vas deferens Ejaculatory duct Urethra Urethral meatus
Q117Single correctEvolution
Match the hominids with their correct brain size :
| Column-I | Column-II |
|---|---|
| (a). | (i). 900 cc |
| (b). | (ii). 1350 cc |
| (c). | (iii). 650-800 cc |
| (d). | (iv). 1400 cc |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Recall the standard cranial capacity of each hominid in the human evolutionary sequence and pair the columns.
Step 1:Homo habilis, the earliest tool-maker of the genus, carried a cranial capacity in the range of 650-800 cc, pairing (a) with (iii).
Step 2:Homo neanderthalensis possessed a large brain of about 1400 cc, pairing (b) with (iv).
Step 3:Homo erectus had a brain size of about 900 cc, pairing (c) with (i).
Step 4:Modern Homo sapiens has an average cranial capacity near 1350 cc, pairing (d) with (ii).
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q118Single correctEvolution
Variations caused by mutation, as proposed by Hugo de Vries are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2random and directionless
Approach:
Recall the nature of mutations described in de Vries' mutation theory.
Step 1:Hugo de Vries believed evolution proceeds through mutations, which are large, sudden, discontinuous changes rather than the small continuous variations of Darwinian theory.
Step 2:These mutational variations arise without any predetermined direction, so they are random and directionless.
Step 3:Therefore the variations proposed by de Vries are random and directionless.
Final answer: random and directionless
Q119Single correctCell - The Unit of Life
Which of the following pair of organelles does not contain DNA?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Lysosomes and Vacuoles
Approach:
Identify the organelles that lack their own DNA.
Step 1:Mitochondria and chloroplasts are semi-autonomous organelles that possess their own DNA, so any pair containing either of them is excluded.
Step 2:Lysosomes are membrane-bound vesicles of hydrolytic enzymes and vacuoles are membrane-bound storage compartments; neither contains DNA.
Step 3:Therefore the pair lacking DNA is lysosomes and vacuoles.
Final answer: Lysosomes and Vacuoles
Q120Single correctBreathing and Exchange of Gases
Due to increasing air-borne allergens and pollutants, many people in urban areas are suffering from respiratory disorder causing wheezing due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2inflammation of bronchi and bronchioles
Approach:
Link the symptom of wheezing in an allergen-driven respiratory disorder to its cause.
Step 1:Air-borne allergens and pollutants trigger asthma, a respiratory disorder marked by difficulty in breathing and wheezing.
Step 2:Asthma results from an allergic inflammation of the bronchi and bronchioles, which narrows the airways and produces the wheezing sound.
Step 3:Therefore the wheezing is due to inflammation of bronchi and bronchioles.
Final answer: inflammation of bronchi and bronchioles
Q121Single correctPrinciples of Inheritance and Variation
Select the incorrect statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3In domesticated fowls, sex of progeny depends on the type of sperm rather than egg
Approach:
Examine each statement on sex determination and find the false one.
Step 1:Male Drosophila is XY and hence heterogametic, and male grasshoppers are XO so half their sperm carry no sex chromosome; both statements are correct.
Step 2:In domesticated fowls females are ZW (heterogametic) and males are ZZ, so the sex of the progeny is decided by the type of egg, not the sperm; the statement reversing this is incorrect.
Step 3:Human males are XY with the Y chromosome much shorter than the X, so the statement that the two are of comparable length is the false one.
Final answer: In domesticated fowls, sex of progeny depends on the type of sperm rather than egg
Q122Single correctBiotechnology - Principles and Processes
DNA precipitation out of a mixture of biomolecules can be achieved by treatment with
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Chilled ethanol
Approach:
Recall the reagent used to precipitate DNA during its isolation.
Step 1:During DNA isolation, the genetic material released from cells is separated from other biomolecules by the addition of a cold alcohol.
Step 2:Purified DNA precipitates out as fine threads when chilled ethanol is added to the solution.
Step 3:Therefore DNA is precipitated using chilled ethanol.
Final answer: Chilled ethanol
Q123Single correctMicrobes in Human Welfare
Select the correct group of biocontrol agents.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2, ,
Approach:
Identify the group in which every member is a biocontrol agent.
Step 1:Trichoderma is a free-living fungus used against plant pathogens, Baculoviruses (Nucleopolyhedrovirus) are used to control insects, and Bacillus thuringiensis produces an insecticidal toxin; all three are biocontrol agents.
Step 2:Other listed organisms such as Rhizobium, Nostoc, Azospirillium and Oscillatoria act as biofertilizers, and aphids are pests, so groups containing them are not purely biocontrol agents.
Step 3:Therefore the correct group of biocontrol agents is Trichoderma, Baculovirus and Bacillus thuringiensis.
Final answer: , ,
Q124Single correctStrategies for Enhancement in Food Production
Select the incorrect statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Inbreeding selects harmful recessive genes that reduce fertility and productivity
Approach:
Assess each statement on the genetic consequences of inbreeding and identify the one that is biologically incorrect.
Step 1:Mating closely related individuals raises the proportion of homozygous loci, so the statement that inbreeding increases homozygosity is correct.
Step 2:Continued inbreeding is the route used to develop pure-line homozygous stock, so the statement that it is essential to evolve purelines is correct.
Step 3:Inbreeding exposes and brings harmful recessive alleles into homozygous condition so that selection can eliminate them; it does not select for those harmful recessives, making this statement incorrect.
Step 4:By exposing variation, inbreeding permits accumulation of superior genes and removal of undesirable ones through selection, so this statement is correct.
Final answer: Inbreeding selects harmful recessive genes that reduce fertility and productivity
Q125Single correctMicrobes in Human Welfare
Match the following organisms with the products they produce. Select the correct option.
| List - I | List - II |
|---|---|
| (a). | (i). Cheese |
| (b). | (ii). Curd |
| (c). | (iii). Citric Acid |
| (d). | (iv). Bread |
| (v). Acetic Acid |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(iv), (c)-(iii), (d)-(v)
Approach:
Match each microbe with the product it characteristically produces.
Step 1:Lactobacillus converts milk to curd, so (a) matches (ii), and Saccharomyces cerevisiae (baker's yeast) leavens dough to make bread, so (b) matches (iv).
Step 2:Aspergillus niger is used for commercial production of citric acid, giving (c)-(iii), and Acetobacter aceti produces acetic acid (vinegar), giving (d)-(v).
Step 3:The complete matching is Lactobacillus with curd, Saccharomyces cerevisiae with bread, Aspergillus niger with citric acid and Acetobacter aceti with acetic acid.
Final answer: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v)
Q126Single correctTransport in Plants
What is the direction of movement of sugars in phloem?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bi-directional
Approach:
Determine the direction in which the phloem translocates sugars.
Step 1:Phloem translocation moves food from sources (regions of synthesis or storage) to sinks (regions of utilization or storage).
Step 2:Since sources and sinks can lie above or below one another and change with the season, the flow is not fixed in one direction.
Step 3:Therefore the movement of sugars in phloem is bi-directional.
Final answer: Bi-directional
Q127Single correctSexual Reproduction in Flowering Plants
In some plants, the female gamete develops into embryo without fertilization. This phenomenon is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Parthenogenesis
Approach:
Name the process in which a female gamete forms an embryo without fertilization.
Step 1:Development of an embryo from an unfertilized egg is the defining feature of the phenomenon in question.
Step 2:Autogamy and syngamy involve fusion of gametes, and parthenocarpy is the formation of seedless fruit, so these do not fit.
Step 3:Development of the female gamete into an embryo without fertilization is termed parthenogenesis.
Final answer: Parthenogenesis
Q128Single correctSexual Reproduction in Flowering Plants
Persistent nucellus in the seed is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Perisperm
Approach:
Identify the seed structure formed by residual nucellus.
Step 1:In most seeds the nucellus is consumed during development, but in some seeds a part of it persists.
Step 2:Chalaza, hilum and tegmen are other seed/ovule structures unrelated to leftover nucellus.
Step 3:The persistent nucellus in a seed is called the perisperm.
Final answer: Perisperm
Q129Single correctPrinciples of Inheritance and Variation
What map unit (Centimorgan) is adopted in the construction of genetic maps?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A unit of distance between genes on chromosomes, representing 1% cross over.
Approach:
Recall the definition of the map unit (centimorgan) used in genetic mapping.
Step 1:Genetic maps measure the relative distance between genes on a chromosome using their recombination (cross over) frequency.
Step 2:One map unit, or centimorgan, equals a recombination frequency of 1% between two genes on a chromosome.
Step 3:Therefore a centimorgan is a unit of distance between genes on chromosomes representing 1% cross over.
Final answer: A unit of distance between genes on chromosomes, representing 1% cross over.
Q130Single correctBody Fluids and Circulation
What would be the heart rate of a person if the cardiac output is 5 L, blood volume in the ventricles at the end of diastole is 100 mL and at the end of ventricular systole is 50 mL?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3100 beats per minute
Approach:
Use the relation between cardiac output, stroke volume and heart rate.
Step 1:Stroke volume is the difference between the end-diastolic volume and the end-systolic volume.
Step 2:Cardiac output of 5 L equals 5000 mL per minute, and equals stroke volume times heart rate.
Step 3:Solving for heart rate gives 100 beats per minute.
Final answer: 100 beats per minute
Q131Single correctMicrobes in Human Welfare
is a group of bacteria helpful in carrying out
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Denitrification
Approach:
Recall the role of Thiobacillus in the nitrogen cycle.
Step 1:Thiobacillus denitrificans converts nitrate into gaseous nitrogen, returning nitrogen to the atmosphere.
Step 2:This conversion of oxidised nitrogen back to free nitrogen is the process of denitrification.
Step 3:Therefore Thiobacillus carries out denitrification.
Final answer: Denitrification
Q132Single correctExcretory Products and their Elimination
Which of the following factors is responsible for the formation of concentrated urine?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
Approach:
Identify the factor that enables the kidney to produce concentrated urine.
Step 1:Concentration of urine depends on an osmotic gradient that draws water out of the filtrate as it passes through the medulla.
Step 2:The counter-current mechanism maintains a steeply increasing hyperosmolarity towards the inner medullary interstitium, allowing reabsorption of water.
Step 3:Therefore maintaining hyperosmolarity towards the inner medullary interstitium produces concentrated urine.
Final answer: Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
Q133Single correctCell - The Unit of Life
Which of the following statements regarding mitochondria is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Enzymes of electron transport are embedded in outer membrane.
Approach:
Examine each statement about mitochondrial structure and find the false one.
Step 1:The outer mitochondrial membrane is permeable to small monomers, the inner membrane is folded into cristae, and the matrix holds a single circular DNA and ribosomes; these statements are correct.
Step 2:The enzymes of the electron transport chain are located on the inner membrane, not the outer membrane, so this statement is false.
Step 3:Therefore the incorrect statement is that electron transport enzymes are embedded in the outer membrane.
Final answer: Enzymes of electron transport are embedded in outer membrane.
Q134Single correctTransport in Plants
Xylem translocates
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Water, mineral salts, some organic nitrogen and hormones
Approach:
Recall the complete list of substances transported by xylem.
Step 1:Xylem conducts water and dissolved mineral salts absorbed by the roots upward through the plant.
Step 2:Along with these, the xylem also carries some organic nitrogen and certain hormones.
Step 3:Therefore xylem translocates water, mineral salts, some organic nitrogen and hormones.
Final answer: Water, mineral salts, some organic nitrogen and hormones
Q135Single correctCell Cycle and Cell Division
Cells in phase :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3suspend the cell cycle
Approach:
Recall the nature of the quiescent G0 stage and contrast suspension with permanent exit from the cell cycle.
Step 1:Cells that do not divide further leave the cycle at G1 and pass into an inactive quiescent stage called G0.
Step 2:In G0 metabolic activity continues while DNA replication and division are held back, so the cell cycle is paused rather than ended; many such cells can re-enter the cycle on appropriate signals.
Step 3:Because activity is put on hold and can resume, cells in G0 suspend the cell cycle rather than permanently exiting or terminating it.
Final answer: suspend the cell cycle
Q136Single correctAnatomy of Flowering Plants
Which of the statements given below is not true about formation of Annual Rings in trees?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Annual rings are not prominent in trees of temperate region.
Approach:
Identify the false statement about annual ring formation in trees.
Step 1:An annual ring consists of spring wood (early wood, light and wide-vesselled) and autumn wood (late wood, dark and narrow-vesselled) formed in one growing year.
Step 2:Cambial activity is governed by seasonal climatic variation, so the differential activity produces the alternating light and dark bands.
Step 3:Distinct annual rings form where seasons are sharply marked, as in temperate regions; therefore annual rings ARE prominent in temperate trees. The statement claiming they are not prominent in temperate regions is false.
Final answer: Annual rings are not prominent in trees of temperate region.
Q137Single correctEcosystem
Which of the following ecological pyramids is generally inverted?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Pyramid of biomass in a sea
Approach:
Recall which ecological pyramid is typically inverted.
Step 1:A pyramid of energy is always upright. The pyramid of biomass in a forest and the pyramid of numbers in grassland are upright.
Step 2:In an aquatic (sea) ecosystem the standing biomass of phytoplankton at any instant is smaller than that of the zooplankton and fishes that feed on them, because the rapidly reproducing phytoplankton are continuously consumed. Thus the pyramid of biomass in a sea is inverted.
Final answer: Pyramid of biomass in a sea
Q138Single correctSexual Reproduction in Flowering Plants
Placentation in which ovules develop on the inner wall of the ovary or in peripheral part, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Parietal
Approach:
Match the placentation description to its type.
Step 1:In parietal placentation the ovules develop on the inner wall of the ovary or on peripheral placentae, as in mustard and Argemone.
Step 2:Basal has a single ovule at the base; axile has ovules on a central axis in multilocular ovaries; free central has ovules on a central axis in a unilocular ovary without septa.
Final answer: Parietal
Q139Single correctEnvironmental Issues
Which of the following protocols did aim for reducing emission of chlorofluorocarbons into the atmosphere?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Montreal Protocol
Approach:
Identify the protocol aimed at controlling ozone-depleting substances.
Step 1:The Montreal Protocol, signed in 1987, was an international treaty to control the emission of ozone-depleting substances such as chlorofluorocarbons.
Step 2:The Kyoto Protocol concerns greenhouse gas reduction; the Gothenburg Protocol targets acidification and ground-level ozone pollutants; the Geneva Protocol concerns long-range transboundary air pollution.
Final answer: Montreal Protocol
Q140Single correctReproductive Health
Which of the following contraceptive methods do involve a role of hormone?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Lactational amenorrhea, Pills Emergency contraceptives
Approach:
Select the combination in which every listed method works through hormonal action.
Step 1:Lactational amenorrhea works through suppression of gonadotropins and ovulation during intense lactation; oral pills are hormonal progestogen-estrogen combinations; emergency contraceptives also act by progestogen-estrogen combinations.
Step 2:Barrier methods and the copper-T are purely mechanical, so any combination that includes them is not entirely hormonal.
Final answer: Lactational amenorrhea, Pills Emergency contraceptives
Q141Single correctBreathing and Exchange of Gases
Tidal Volume and Expiratory Reserve Volume of an athlete is 500 mL and 1000 mL, respectively. What will be his Expiratory Capacity if the Residual Volume is 1200 mL?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11500 mL
Approach:
Apply the definition of expiratory capacity as the sum of tidal volume and expiratory reserve volume.
Step 1:Tidal volume is 500 mL and expiratory reserve volume is 1000 mL.
Step 2:Expiratory capacity is the total volume of air a person can expire after a normal inspiration, equal to tidal volume plus expiratory reserve volume.
Final answer: 1500 mL
Q142Single correctSexual Reproduction in Flowering Plants
What is the fate of the male gametes discharged in the synergid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4One fuses with the egg and other fuses with central cell nuclei.
Approach:
Recall the events of double fertilization following discharge of male gametes into the synergid.
Step 1:The pollen tube enters a synergid and releases the two male gametes there. One male gamete fuses with the egg cell (syngamy) forming the zygote.
Step 2:The second male gamete fuses with the two polar nuclei of the central cell (triple fusion) forming the primary endosperm nucleus.
Final answer: One fuses with the egg and other fuses with central cell nuclei.
Q143Single correctPlant Growth and Development
What is the site of perception of photoperiod necessary for induction of flowering in plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Leaves
Approach:
Recall where the photoperiodic stimulus for flowering is perceived.
Step 1:In photoperiodism the light/dark stimulus required for flowering is perceived by the leaves, not by the shoot apex where flowering ultimately occurs.
Step 2:A flowering hormonal signal then moves from the leaves to the shoot apical meristem to bring about floral induction.
Final answer: Leaves
Q144Single correctThe Living World
Select the correctly written scientific name of Mango which was first described by Carolus Linnaeus
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mangifera\ indica Linn.
Approach:
Apply the rules of binomial nomenclature to choose the correct scientific name.
Step 1:The genus name is capitalized and the species epithet is in lower case; both are italicized. The name of the author who first described the species is written in abbreviated form after the species name in regular (non-italic) type.
Step 2:The correct form is Mangifera indica Linn., where 'indica' is lower case and 'Linn.' is the abbreviated author citation for Linnaeus.
Final answer: Mangifera\ indica Linn.
Q145Single correctBiotechnology - Principles and Processes
Following statements describe the characteristics of the enzyme Restriction Endonuclease. Identify the incorrect statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The enzyme binds DNA at specific sites and cuts only one of the two strands.
Approach:
Identify the false statement about restriction endonuclease action.
Step 1:Restriction enzymes inspect the length of a DNA sequence, bind at a specific recognition site (a palindromic sequence) and cut the sugar-phosphate backbone at specific points on each of the two strands.
Step 2:Because the enzyme cuts BOTH strands (not only one), the statement that it cuts only one of the two strands is incorrect.
Final answer: The enzyme binds DNA at specific sites and cuts only one of the two strands.
Q146Single correctPlant Kingdom
From evolutionary point of view, retention of the female gametophyte with developing young embryo on the parent sporophyte for some time, is first observed in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pteridophytes
Approach:
Recall in which group the seed habit (retention of megaspore and embryo) first appears.
Step 1:In a few heterosporous pteridophytes the megaspore is retained on the parent sporophyte, and the developing young embryo is retained for some time. This is regarded as a precursor to the seed habit.
Step 2:This event of retention is first observed in pteridophytes; gymnosperms later exhibit fully developed naked seeds.
Final answer: Pteridophytes
Q147Single correctPrinciples of Inheritance and Variation
In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in generation pink flowers were obtained. When pink flowers were selfed, the generation showed white, red and pink flowers. Choose the incorrect statement from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Law of Segregation does not apply in this experiment
Approach:
Identify the incorrect statement about the Antirrhinum incomplete-dominance cross.
Step 1:The pink F1 arises because neither red nor white allele is fully dominant, illustrating incomplete dominance; this is an exception to the Principle of Dominance.
Step 2:Selfing the pink gives a 1 red : 2 pink : 1 white ratio, so the phenotypic and genotypic ratios coincide at 1 : 2 : 1.
Step 3:The Law of Segregation IS valid throughout because alleles still segregate; only the dominance relationship differs. Hence the statement that segregation does not apply is incorrect.
Final answer: Law of Segregation does not apply in this experiment
Q148Single correctRespiration in Plants
Conversion of glucose to glucose-6-phosphate, the first irreversible reaction of glycolysis, is catalyzed by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hexokinase
Approach:
Recall the enzyme catalyzing the first phosphorylation step of glycolysis.
Step 1:In glycolysis the conversion of glucose to glucose-6-phosphate is the first committed step and is catalyzed by hexokinase, using ATP.
Step 2:Phosphofructokinase acts later (fructose-6-phosphate to fructose-1,6-bisphosphate); aldolase cleaves the six-carbon sugar; enolase acts near the end of glycolysis.
Final answer: Hexokinase
Q149Single correctChemistry in Everyday Life
Drug called 'Heroin' is synthesized by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2acetylation of morphine
Approach:
Recall how heroin is chemically derived from morphine.
Step 1:Heroin, commonly called smack, is diacetylmorphine, obtained by acetylation of morphine.
Final answer: acetylation of morphine
Q150Single correctReproductive Health
Select the hormone-releasing Intra-Uterine Devices.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Progestasert, LNG-20
Approach:
Identify the IUDs that release hormones.
Step 1:Progestasert and LNG-20 are hormone-releasing IUDs which make the uterus unsuitable for implantation and the cervix hostile to sperms.
Step 2:Lippes Loop is a non-medicated IUD; CuT, Cu7 and Multiload 375 are copper-releasing IUDs; Vaults is not hormone-releasing.
Final answer: Progestasert, LNG-20
Q151Single correctEvolution
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.16(AA); 0.48(Aa); 0.36(aa)
Approach:
Apply the Hardy-Weinberg genotype frequencies with given allele frequencies.
Step 1:Let p be the frequency of dominant allele A; p = 0.4, so the recessive allele frequency q = 1 - 0.4 = 0.6.
Step 2:Frequency of homozygous dominant individuals (AA) is p squared.
Step 3:Frequency of heterozygous individuals (Aa) is 2pq.
Step 4:Frequency of homozygous recessive individuals (aa) is q squared.
Final answer: 0.16(AA); 0.48(Aa); 0.36(aa)
Q152Single correctBiotechnology and its Applications
Which of the following is true for Golden rice?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1It is Vitamin A enriched, with a gene from daffodil
Approach:
Recall the defining property of Golden rice.
Step 1:Golden rice is enriched in vitamin A (beta-carotene), with a gene derived from daffodil that enables provitamin-A biosynthesis in the rice endosperm.
Final answer: It is Vitamin A enriched, with a gene from daffodil
Q153Single correctSexual Reproduction in Flowering Plants
Pinus seeds cannot germinate and establish without fungal association. This is because :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2it has obligate association with mycorrhizae.
Approach:
Explain why Pinus seeds need a fungus to germinate and establish.
Step 1:The fungus associated with the roots of Pinus increases mineral and water absorption, helping the plant by increasing surface area and converting nutrient-rich soil into usable forms.
Step 2:Because this mycorrhizal association is obligate for Pinus seed germination and establishment, the seeds cannot establish without the fungal partner.
Final answer: it has obligate association with mycorrhizae.
Q154Single correctMolecular Basis of Inheritance
Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Genetic code is nearly universal
Approach:
Identify the property of the genetic code that permits a human gene to be expressed in bacteria.
Step 1:In recombinant DNA technology bacteria are able to produce human insulin because the genetic code is nearly universal: the same codons specify the same amino acids in nearly all organisms.
Final answer: Genetic code is nearly universal
Q155Single correctHuman Health and Disease
Which of the following sexually transmitted diseases is not completely curable?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Genital herpes
Approach:
Identify the STI that cannot be completely cured.
Step 1:Genital herpes is caused by herpes simplex virus. As the disease is viral, there is no complete cure for type-II herpes simplex virus and therefore the disease caused, genital herpes, is not completely curable.
Step 2:Hepatitis-B and HIV are also non-curable, but of the four diseases named here genital herpes is the one with no complete cure.
Final answer: Genital herpes
Q156Single correctBiological Classification
Which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Infective constituent in viruses is the protein coat.
Approach:
Identify the incorrect statement about viruses, viroids and prions.
Step 1:Viroids lack a protein coat (true); viruses are obligate intracellular parasites (true); prions are abnormally folded proteins (true).
Step 2:The infective constituent in viruses is the nucleic acid (DNA or RNA), not the protein coat. Hence the statement that the infective constituent is the protein coat is incorrect.
Final answer: Infective constituent in viruses is the protein coat.
Q157Single correctAnimal Kingdom
Match the following organisms with their respective characteristics. Select the correct option from the following :
| List - I | List - II |
|---|---|
| (a). | (i). Flame cells |
| (b). | (ii). Comb plates |
| (c). | (iii). Radula |
| (d). | (iv). Malpighian tubules |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Approach:
Assign each organism its characteristic structure, then pair the columns.
Step 1:Pila is a mollusc whose mouth contains a file-like rasping organ for feeding, called the radula, so Pila pairs with radula (iii).
Step 2:Bombyx is an arthropod; in arthropods excretion takes place through malpighian tubules, so Bombyx pairs with malpighian tubules (iv).
Step 3:Pleurobrachia is a ctenophore whose body bears eight external rows of ciliated comb plates that help in locomotion, so it pairs with comb plates (ii).
Step 4:Taenia is a platyhelminth specialised cells called flame cells help in osmoregulation and excretion, so it pairs with flame cells (i).
Final answer: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Q158Single correctBiotechnology and its Applications
Expressed Sequence Tags (ESTs) refers to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Genes expressed as RNA
Approach:
Recall the definition of Expressed Sequence Tags in the Human Genome Project.
Step 1:Expressed Sequence Tags (ESTs) are DNA sequences corresponding to genes that are expressed as mRNA, identifying the transcribed (expressed) genes.
Step 2:This expression-based approach was one of the methods used in the Human Genome Project, focusing on genes expressed as RNA.
Final answer: Genes expressed as RNA
Q159Single correctStrategies for Enhancement in Food Production
Which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Yeasts have filamentous bodies with long thread-like hyphae.
Approach:
Test each statement against fungal biology and identify the one that does not hold.
Step 1:Morels and truffles are ascomycete fungi prized as edible delicacies, so statement 1 is a true biological fact.
Step 2:Claviceps purpurea (ergot) produces ergot alkaloids, including lysergic acid derivatives related to LSD, so statement 2 is also a true fact.
Step 3:Conidia form exogenously on conidiophores while ascospores form endogenously within asci, so statement 3 is true.
Step 4:Yeasts are unicellular fungi that reproduce by budding and have no true filamentous mycelium of long thread-like hyphae, so this is the incorrect statement.
Final answer: Yeasts have filamentous bodies with long thread-like hyphae.
Q160Single correctBiological Classification
Match Column - I with Column - II. Choose the correct answer from the option given below.
| Column - I | Column - II |
|---|---|
| (a). Saprophyte | (i). Symbiotic association of fungi with plant roots |
| (b). Parasite | (ii). Decomposition of dead organic materials |
| (c). Lichens | (iii). Living on living plants or animals |
| (d). Mycorrhiza | (iv). Symbiotic association of algae and fungi |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Each nutritional mode in Column - I is paired with its defining ecological description in Column - II.
Step 1:Saprophytes obtain nutrition from dead and decaying organic matter, driving decomposition.
Step 2:Parasites derive nourishment while residing on or in living host plants or animals.
Step 3:Lichens are a symbiotic association between an alga and a fungus.
Step 4:Mycorrhiza is a symbiotic association of fungi with the roots of higher plants.
Step 5:Combining the pairings gives saprophyte with decomposition of dead organic material, parasite with living on living plants or animals, lichens with the symbiosis of algae and fungi, and mycorrhiza with the symbiosis of fungi and plant roots.
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q161Single correctChemical Coordination and Integration
Which of the following glucose transporters is insulin-dependent?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4GLUT IV
Approach:
Identify the glucose transporter isoform whose membrane translocation is stimulated by insulin.
Step 1:GLUT I, II and III mediate basal glucose uptake and operate independently of insulin in tissues such as erythrocytes, liver and neurons.
Step 2:GLUT IV resides in intracellular vesicles of muscle and adipose tissue and is recruited to the plasma membrane upon insulin signalling, making its glucose uptake insulin-dependent.
Final answer: GLUT IV
Q162Single correctHuman Health and Disease
Which of the following immune responses is responsible for rejection of kidney graft?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cell-mediated immune response
Approach:
Identify the immune mechanism that drives rejection of a transplanted organ.
Step 1:Graft rejection arises because the recipient distinguishes self from non-self antigens on the donor tissue.
Step 2:T lymphocytes recognise the foreign MHC molecules and mount a cell-mediated response that destroys the graft.
Final answer: Cell-mediated immune response
Q163Single correctExcretory Products and their Elimination
Use of an artificial kidney during hemodialysis may result in :
(a) Nitrogenous waste build-up in the body
(b) Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
(d) Reduced RBC production
Which of the following options is the most appropriate?
(a) Nitrogenous waste build-up in the body
(b) Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
(d) Reduced RBC production
Which of the following options is the most appropriate?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(c) and (d) are correct
Approach:
Distinguish the functions that haemodialysis performs from the kidney functions it cannot replace.
Step 1:Haemodialysis removes nitrogenous wastes and excess potassium from the blood, so statements (a) and (b) do not represent consequences of its use.
Step 2:The healthy kidney activates vitamin D for calcium absorption and secretes erythropoietin; the artificial kidney performs neither, so calcium absorption from the gut is reduced and RBC production declines.
Final answer: (c) and (d) are correct
Q164Single correctSensory Perception
Which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cornea consists of dense matrix of collagen and is the most sensitive portion of the eye.
Approach:
Compare each description of the cornea against its anatomy.
Step 1:The cornea is the transparent anterior layer of the eye built from a dense, regularly arranged matrix of collagen fibres.
Step 2:It is avascular and richly supplied with sensory nerve endings, which makes it the most sensitive part of the eye; descriptions calling it proteinaceous, elastin-based or highly vascularised misstate its composition and blood supply.
Final answer: Cornea consists of dense matrix of collagen and is the most sensitive portion of the eye.
Q165Single correctPrinciples of Inheritance and Variation
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Alfred Sturtevant
Approach:
Recall the scientist who used recombination frequency to map gene distances.
Step 1:Alfred Sturtevant, a student in Morgan's laboratory, used the frequency of recombination between linked genes to determine their relative positions and construct the first genetic map.
Final answer: Alfred Sturtevant
Q166Single correctMolecular Basis of Inheritance
Match the following genes of the Lac operon with their respective products : Select the correct option.
| Column - I | Column - II |
|---|---|
| a. i gene | i. β-galactosidase |
| b. z gene | ii. Permease |
| c. a gene | iii. Repressor |
| d. y gene | iv. Transacetylase |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Approach:
Assign each lac operon gene to the protein it encodes.
Step 1:The i gene encodes the repressor protein.
Step 2:The z gene encodes β-galactosidase.
Step 3:The a gene encodes transacetylase.
Step 4:The y gene encodes permease.
Final answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Q167Single correctPlant Growth and Development
It takes very long time for pineapple plants to produce flowers. Which combination of hormones can be applied to artificially induce flowering in pineapple plants throughout the year to increase yield?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Auxin and Ethylene
Approach:
Recall the hormones that promote flowering in pineapple.
Step 1:Application of auxin and ethylene induces synchronised flowering and fruit set in pineapple irrespective of season, allowing year-round production.
Final answer: Auxin and Ethylene
Q168Single correctDigestion and Absorption
Identify the cells whose secretion protects the lining of gastro-intestinal tract from various enzymes.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Goblet Cells
Approach:
Identify the cell type secreting protective mucus in the gut lining.
Step 1:Goblet cells of the intestinal and gastric mucosa secrete mucus and bicarbonates that coat and shield the epithelium from digestive enzymes and acid.
Final answer: Goblet Cells
Q169Single correctMicrobes in Human Welfare
Which of the following can be used as a biocontrol agent in the treatment of plant disease?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Recall the organism used as a biological control agent against plant pathogens.
Step 1:The free-living fungus Trichoderma is an effective biocontrol agent that suppresses several soil-borne plant pathogens.
Final answer:
Q170Single correctAnatomy of Flowering Plants
Phloem in gymnosperms lacks :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both sieve tubes and companion cells
Approach:
Recall the phloem elements absent in gymnosperms.
Step 1:Gymnosperm phloem contains sieve cells and albuminous cells rather than the sieve tubes and companion cells found in angiosperms.
Final answer: Both sieve tubes and companion cells
Q171Single correctHuman Reproduction
Extrusion of second polar body from egg nucleus occurs :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1after entry of sperm but before fertilization
Approach:
The timing of the second meiotic division of the oocyte relative to sperm entry and fertilisation determines when the second polar body is released.
Step 1:The secondary oocyte is ovulated arrested in metaphase of meiosis II and remains a haploid secondary oocyte until stimulated.
Step 2:Entry of the sperm into the secondary oocyte triggers completion of the arrested second meiotic division.
Step 3:Completion of meiosis II expels the second polar body, producing the mature ovum nucleus, which then fuses with the sperm nucleus to complete fertilisation.
Step 4:Extrusion therefore takes place after the sperm has entered but before fertilization is complete.
Final answer: after entry of sperm but before fertilization
Q172Single correctMolecular Basis of Inheritance
Under which of the following conditions will there be no change in the reading frame of following mRNA?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Deletion of GGU from , and positions
Approach:
Determine which mutation preserves the triplet reading frame of the mRNA.
Step 1:Inserting or deleting a single nucleotide shifts every downstream codon, so both single-nucleotide changes cause a frameshift.
Step 2:Inserting two nucleotides also breaks the triplet grouping, so that change shifts the frame as well.
Step 3:Deleting three contiguous nucleotides (GGU) removes one whole codon and leaves the downstream reading frame intact.
Final answer: Deletion of GGU from , and positions
Q173Single correctCell - The Unit of Life
The concept of “ - ” regarding cell division was first proposed by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Rudolf Virchow
Approach:
Recall the scientist who proposed that cells arise from pre-existing cells.
Step 1:Rudolf Virchow proposed the concept Omnis cellula-e cellula, stating that new cells arise from the division of pre-existing cells.
Final answer: Rudolf Virchow
Q174Single correctMicrobes in Human Welfare
What triggers activation of protoxin to active Bt toxin of in boll worm?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Alkaline pH of gut
Approach:
Identify the gut condition that converts the inactive Bt protoxin into active toxin.
Step 1:The Bt protoxin crystal is ingested by the insect larva and dissolves in the alkaline pH of the midgut, releasing the active toxin that perforates the gut epithelium.
Final answer: Alkaline pH of gut
Q175Single correctHuman Health and Disease
Identify the correct pair representing the causative agent of typhoid fever and the confirmatory test for typhoid.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 / Widal test
Approach:
Match the typhoid pathogen with its confirmatory diagnostic test.
Step 1:Typhoid fever is caused by Salmonella typhi.
Step 2:The Widal test, an antigen-antibody agglutination assay, confirms typhoid, so the correct pair is Salmonella typhi / Widal test.
Final answer: / Widal test
Q176Single correctPrinciples of Inheritance and Variation
What is the genetic disorder in which an individual has an overall masculine development gynaecomastia, and is sterile ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Klinefelter's syndrome
Approach:
Identify the chromosomal disorder fitting masculine development with gynaecomastia and sterility.
Step 1:Klinefelter's syndrome results from an extra X chromosome (44 + XXY, 47). Affected individuals show overall masculine development together with gynaecomastia and are sterile.
Final answer: Klinefelter's syndrome
Q177Single correctEnvironmental Issues
Polyblend, a fine powder of recycled modified plastic, has proved to be a good material for :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Construction of roads
Approach:
Recall the practical application of polyblend.
Step 1:Polyblend, a fine powder of recycled modified plastic, is mixed with bitumen to improve the durability of road surfaces, making it useful in road construction.
Final answer: Construction of roads
Q178Single correctEnvironmental Issues
Which of these following methods is the most suitable for disposal of nuclear waste?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Bury the waste within rocks deep below the Earth's surface
Approach:
Identify the recommended method for safe nuclear waste disposal.
Step 1:Nuclear waste should first be stored in suitably shielded containers and then buried deep within stable rock formations below the Earth's surface, about 500 metres deep, to isolate the radiation.
Final answer: Bury the waste within rocks deep below the Earth's surface
Q179Single correctChemical Coordination and Integration
Match the following hormones with the respective disease Select the correct option.
| Column - I | Column - II |
|---|---|
| a. Insulin | i. Addison's disease |
| b. Thyroxin | ii. Diabetes insipidus |
| c. Corticoids | iii. Acromegaly |
| d. Growth Hormone | iv. Goitre |
| v. Diabetes mellitus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a)-(v), (b)-(iv), (c)-(i), (d)-(iii)
Approach:
Link each hormone disorder to the disease it produces.
Step 1:Insulin deficiency leads to diabetes mellitus.
Step 2:Hyposecretion or hypersecretion of thyroxine can be associated with enlargement of the thyroid called goitre.
Step 3:Deficiency of corticoids (glucocorticoid and mineralocorticoid) leads to Addison's disease.
Step 4:Growth hormone hypersecretion in adults leads to acromegaly.
Final answer: (a)-(v), (b)-(iv), (c)-(i), (d)-(iii)
Q180Single correctLocomotion and Movement
Select the correct option.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4There are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs.
Approach:
Evaluate each statement about the rib cage against human skeletal anatomy.
Step 1:The first seven pairs are true (vertebrosternal) ribs attached dorsally to the thoracic vertebrae and ventrally to the sternum through hyaline cartilage.
Step 2:The 8th, 9th and 10th pairs are vertebrochondral (false) ribs that join the seventh rib's cartilage instead of articulating directly with the sternum.
Step 3:The last two pairs (11th and 12th) are floating (vertebral) ribs with no ventral attachment, giving seven vertebrosternal, three vertebrochondral and two vertebral pairs.
Final answer: There are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs.
Frequently Asked Questions
How many questions are in the NEET 2019 May 05 paper?
The NEET 2019 May 05 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 2019 May 05 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 2019 May 05 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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