Back to NEET PYQs
NEET 2025 May 04 Question Paper with Solutions
All 180 questions from the NEET 2025 (May 04) 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 2025Chemistry PYQs 2025Biology PYQs 2025
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctMechanical Properties of Fluids
Consider a water tank shown in the figure. It has one wall at and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density , the liquid surface makes angle () with the x-axis at . If y(x) is the height of the surface then the equation for y(x) is:
(take , g is the acceleration due to gravity)
(take , g is the acceleration due to gravity)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The excess pressure across a curved liquid surface is related to the surface tension and the local curvature of the surface. Equating this excess pressure to the hydrostatic pressure at depth yields the differential equation for the surface profile.
Step 1:The radius of curvature at a point on the surface determines the curvature.
Step 2:With the small-angle condition , the denominator reduces to one.
Step 3:The excess pressure due to surface tension balances the hydrostatic pressure at depth .
Final answer:
Q2Single correctRay Optics and Optical Instruments
A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2125
Approach:
The total magnification of a compound microscope is the product of the linear magnification of the objective and the angular magnification of the eyepiece.
Step 1:The objective contributes a factor equal to the tube length divided by its focal length.
Step 2:The eyepiece contributes a factor equal to the distance of distinct vision divided by its focal length.
Step 3:Multiplying the two factors gives the total magnification.
Final answer: 125
Q3Single correctMoving Charges and Magnetism
An electron (mass kg and charge C) moving with speed (c = speed of light) is injected into a magnetic field of magnitude T perpendicular to its direction of motion. We wish to apply an uniform electric field together with the magnetic field so that the electron does not deflect from its path. Then (speed of light m )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 is perpendicular to and its magnitude is V
Approach:
For an undeflected charged particle, the electric force must cancel the magnetic force. This balance fixes both the direction of the electric field and its magnitude through the velocity selector condition.
Step 1:Zero net force requires the electric force to oppose the magnetic force.
Step 2:The required field magnitude equals the product of speed and magnetic flux density.
Step 3:Evaluating the product gives the field magnitude.
Final answer: is perpendicular to and its magnitude is V
Q4Single correctLaws of Motion
There are two inclined surfaces of equal length (L) and same angle of inclination with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction () between the object and the rough surface is close to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 40.75
Approach:
The sliding time on an incline depends on the net acceleration. Taking the ratio of times for the smooth and rough surfaces relates the accelerations, which then yields the coefficient of kinetic friction.
Step 1:For a fixed length, the sliding time varies inversely with the square root of acceleration.
Step 2:The rough-surface time is twice the smooth-surface time, so the ratio of times equals two.
Step 3:Squaring both sides and inserting where gives the friction coefficient.
Final answer: 0.75
Q5Single correctWork, Energy and Power
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If and are the forces applied by the breaks on cars A and B respectively, then the ratio of is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By the work-energy theorem, the braking force equals the kinetic energy dissipated divided by the stopping distance. Taking the ratio of forces uses the given energies and distances.
Step 1:The braking force is the magnitude of kinetic energy lost over the stopping distance.
Step 2:Forming the ratio of the two forces using their energies and distances.
Step 3:Simplifying the product gives the required ratio.
Final answer:
Q6Single correctCurrent Electricity
The current passing through the battery in the given circuit, is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 20.5 A
Approach:
The central portion of the network forms a balanced Wheatstone bridge, so the bridge arm carries no current and can be removed. The remaining series-parallel combination reduces to an equivalent resistance from which the battery current follows by Ohm's law.
Step 1:The bridge is balanced, so the parallel combination of its two branches gives the equivalent of the bridge portion.
Step 2:The redrawn circuit places the bridge equivalent in series with the remaining resistors.
Step 3:Evaluating the total resistance and applying Ohm's law to the 5 V source gives the battery current.
Final answer: 0.5 A
Q7Single correctSystems of Particles and Rotational Motion
A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity as shown in figure. If the string gets slack at some point P making an angle from the horizontal, then the ratio of the speed v of the bob at point P to its initial speed is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
At the point where the string goes slack, the tension vanishes and gravity supplies the centripetal force. Combining this condition with conservation of mechanical energy between the bottom and point gives the speed ratio.
Step 1:At point the string is slack, so the radial component of gravity equals the centripetal requirement.
Step 2:Conservation of mechanical energy between the start and point relates the two speeds and the height gained.
Step 3:Substituting and simplifying isolates the speed ratio.
Final answer:
Q8Single correctSemiconductor Electronics
The output (Y) of the given logic implementation is similar to the output of an/a ______ gate.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4NOR
Approach:
Each gate output is written in Boolean form and the expressions are combined stage by stage. Simplification using De Morgan's laws identifies the equivalent single gate.
Step 1:The first two intermediate outputs are the complements of the products of the inputs.
Step 2:The final gate produces the complement of the product of the intermediate outputs.
Step 3:Applying De Morgan's law and simplifying reduces the expression to a single complemented sum.
Final answer: NOR
Q9Single correctElectromagnetic Waves
The electric field in a plane electromagnetic wave is given by
V/m.
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
V/m.
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 T
Approach:
In a plane electromagnetic wave the electric and magnetic fields oscillate in phase, are mutually perpendicular and both perpendicular to the propagation direction. The magnetic field amplitude is the electric amplitude divided by the speed of light.
Step 1:The magnetic field shares the same phase and spatial-temporal dependence as the electric field, with amplitude reduced by the speed of light.
Step 2:Dividing the electric amplitude by the speed of light gives the magnetic amplitude.
Step 3:The wave travels along and the electric field is along , so the magnetic field lies along .
Final answer: T
Q10Single correctSystems of Particles and Rotational Motion
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 121 NS
Approach:
The speeds just before and just after impact follow from the drop and rebound heights. The impulse is the change in momentum, accounting for the reversal of velocity direction at the ground.
Step 1:The downward speed at impact follows from the drop height.
Step 2:The upward rebound speed follows from the rise height.
Step 3:Taking upward as positive, the impulse is the mass times the change in velocity across the collision.
Final answer: 21 NS
Q11Single correctElectromagnetic Induction
AB is a part of an electrical circuit (see figure). The potential difference , at the instant when current A and is increasing at a rate of 1 amp/second is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 410 volt
Approach:
Traversing the branch from to , the potential drops across the resistor and the inductor and changes across the battery are summed. The inductor contributes a term proportional to the rate of change of current.
Step 1:The current is 2 A and increases at one ampere per second through a 1 H inductor and a 2 ohm resistor with a 5 V source in the branch.
Step 2:Writing the potential change from to accumulates the inductor, source and resistor contributions.
Step 3:Substituting the numerical values gives the potential difference.
...
Final answer: 10 volt
Q12Single correctMoving Charges and Magnetism
A 2 amp current is flowing through two different small circular copper coils having radii 1 : 2. The ratio of their respective magnetic moments will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11 : 4
Approach:
The magnetic moment of a single-turn current loop is the current times its area. With equal currents, the ratio of moments equals the ratio of the squares of the radii.
Step 1:For a circular loop the magnetic moment is the current times the enclosed area.
Step 2:With equal currents, the moment ratio reduces to the ratio of the areas, hence the squares of the radii.
Step 3:Evaluating gives the moment ratio.
Final answer: 1 : 4
Q13Single correctRay Optics and Optical Instruments
In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. The power of the combination and the total magnification in comparison to the power () and magnification () for each lens will be, respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 34p and
Approach:
For thin lenses in contact, the powers add directly, while the linear magnifications multiply because each lens magnifies the image formed by the previous one.
Step 1:Powers of lenses in contact add, and with four identical lenses the total power is four times each.
Step 2:Magnifications multiply across successive lenses, and four identical magnifications give the fourth power.
Final answer: 4p and
Q14Single correctKinetic Theory of Gases
An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27C. The mass of the oxygen withdrawn from the cylinder is nearly equal to
[Given, J mo , and molecular mass of = 32, 1 atm pressure = N/m]
[Given, J mo , and molecular mass of = 32, 1 atm pressure = N/m]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.116 kg
Approach:
The number of moles remaining is found from the ideal gas equation using the final pressure, volume and temperature. The withdrawn mass is the difference in moles multiplied by the molar mass.
Step 1:The remaining moles follow from the final pressure, volume and temperature applied to the ideal gas law.
Step 2:Evaluating gives the moles of oxygen left in the cylinder.
Step 3:The withdrawn amount is the loss in moles times the molar mass of oxygen.
Final answer: 0.116 kg
Q15Single correctMotion in a Straight Line
In some appropriate units, time (t) and position (x) relation of a moving particle is given by . The acceleration of the particle is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Differentiating the given relation provides the velocity as the reciprocal of . Differentiating the velocity with respect to time, using the chain rule, gives the acceleration.
Step 1:Differentiating the position-time relation with respect to position gives the inverse of velocity.
Step 2:Differentiating velocity with respect to position yields the gradient needed for acceleration.
Step 3:Acceleration is the product of velocity and its derivative with respect to position, and the result is negative because the particle decelerates as grows.
Final answer:
Q16Single correctAlternating Current
To an ac power supply of 220 V at 50 Hz, a resistor of 20 , a capacitor of reactance 25 and an inductor of reactance 45 are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 27.8 A and 45
Approach:
The impedance of the series circuit combines the resistance with the net reactance. The current follows from the supply voltage divided by the impedance, and the phase angle from the ratio of net reactance to resistance.
Step 1:The net reactance is the difference of inductive and capacitive reactances, here 20 ohm, equal to the resistance.
Step 2:The current is the supply voltage divided by the impedance.
Step 3:The phase angle follows from the ratio of net reactance to resistance.
Final answer: 7.8 A and 45
Q17Single correctSystems of Particles and Rotational Motion
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4108 days
Approach:
In the absence of external torque, angular momentum is conserved. The moment of inertia of a uniform sphere scales with the square of its radius, so the rotation period scales accordingly.
Step 1:With no external torque, the product of moment of inertia and angular velocity stays constant.
Step 2:Cancelling common factors leaves the period proportional to the square of the radius.
Step 3:Substituting the original period gives the new rotation period.
Final answer: 108 days
Q18Single correctAtoms
A model for quantized motion of an electron in a uniform magnetic field states that the flux passing through the orbit of the electron is where is an integer, is Planck's constant and is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state ( is the mass of the electron)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The radius of the circular orbit is fixed by equating the magnetic Lorentz force to the centripetal requirement and applying the flux quantization condition. The magnetic moment is the current times the orbit area, which simplifies using the quantization rule.
Step 1:Balancing the magnetic force against the centripetal requirement fixes the speed in terms of the radius.
Step 2:The quantization of flux through the orbit fixes the radius.
Step 3:The magnetic moment as current times area simplifies with the flux condition, and the lowest state corresponds to one.
Final answer:
Q19Single correctThermal Properties of Matter
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is and that at the right junction is . The ratio is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
In steady state the rate of heat flow is the same through each rod. Equating the heat current across the first rod with that across the whole series, and again across the last section, yields the two junction temperatures and their ratio.
Step 1:The total thermal resistance is the sum of the three rod resistances, with the middle rod contributing twice as much as each side rod.
Step 2:Equating heat current through the first rod with that through the whole combination gives the first junction temperature.
Step 3:Equating heat current through the last section with the total gives the second junction temperature, then the ratio follows.
Final answer:
Q20Single correctElectrostatic Potential and Capacitance
The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant and with thickness and respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.
If , then value of is:
If , then value of is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12.66
Approach:
The capacitor with the two slabs and the remaining air gap behaves as three capacitors in series. Imposing that the net capacitance is twice the empty value, together with the relation between the dielectric constants, determines .
Step 1:The remaining air gap has thickness , so the three layers carry thicknesses , and .
Step 2:Setting the combined capacitance to twice the empty value and inserting gives an equation in .
Step 3:Solving the resulting relation yields the first dielectric constant.
Final answer: 2.66
Q21Single correctMotion in a Straight Line
Two cities and are connected by a regular bus service with a bus leaving in either direction every min. A girl is driving scooty with a speed of 60 km/h in the direction to notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period of the bus service and the speed (assumed constant) of the buses.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 415 min, 120 km/h
Approach:
The spacing between consecutive buses equals the bus speed times the headway. The time between buses passing the scooty depends on the relative velocity, giving two equations for the bus speed and the headway.
Step 1:Buses in the direction of motion overtake the scooty every 30 minutes, so the spacing equals the relative speed times this interval.
Step 2:Buses in the opposite direction pass every 10 minutes, giving a second relation.
Step 3:Equating the two right-hand sides solves for the bus speed and then the headway.
Final answer: 15 min, 120 km/h
Q22Single correctSystems of Particles and Rotational Motion
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60 with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take m/)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 N
Approach:
Translational equilibrium fixes the vertical normal force from the floor and relates the friction to the wall's normal force. Taking torques about the floor end determines the wall normal force, which equals the friction.
Step 1:Vertical equilibrium gives the floor normal force, and horizontal equilibrium equates the wall normal force to the friction.
Step 2:Taking torque about the floor end, the weight torque balances the wall normal torque, where is the angle with the wall.
Step 3:Inserting the values with the rod at 60 to the wall gives the friction force.
Final answer: N
Q23Single correctOscillations
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency and average amplitude A(t) of the system change with time t. Which one of the following options schematically depicts these changes correctly?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 rising with t and A(t) falling with t
Approach:
As sand leaks out, the oscillating mass decreases. The angular frequency depends inversely on the square root of mass, while conservation of mechanical energy with decreasing mass governs how the amplitude evolves.
Step 1:Sand leaks out, so the oscillating mass decreases while the spring constant stays fixed.
Step 2:Angular frequency varies inversely with the square root of the mass, so as the mass falls the average angular frequency rises.
Step 3:At any instant the box hangs in equilibrium where the weight balances the spring force, so the equilibrium extension falls as the mass falls, and the oscillation about that shrinking extension has a steadily smaller average amplitude.
Final answer: rising with t and A(t) falling with t
Q24Single correctPhysical World and Measurement
A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as and then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The time is expressed as a product of powers of surface tension, outlet area, gas density and radius. Equating dimensions on both sides fixes each exponent, and continuity at the outlet fixes the dependence of the efflux speed on the radius.
Step 1:The efflux speed follows from the surface-tension excess pressure driving the gas out of the outlet, which gives the speed exponent directly.
Step 2:Continuity at the outlet equates the rate of change of balloon volume to the volume swept through the outlet area.
Step 3:Separating the variables and integrating the radius from down to zero gives the emptying time.
Step 4:A dimensional check confirms the exponents, noting that dimensions alone leave and undetermined and that the continuity integration is what fixes them.
Final answer:
Q25Single correctPhysical World and Measurement
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 cm
Approach:
The least count follows from the difference of one main-scale and one vernier-scale division. The zero error displaces the reading, and the corrected diameter is the main-scale plus vernier contribution minus the zero error.
Step 1:From the relation between vernier and main divisions, one vernier division equals nine tenths of one main division.
Step 2:With the smallest main division equal to 0.1 cm, the least count is one tenth of it.
cm
Step 3:With the jaws closed the vernier zero sits at cm on the main scale, so the instrument reads 0.1 cm too high and the zero error is positive.
cm
Step 4:The observed reading is the main-scale reading plus eight least counts, and subtracting the positive zero error gives the corrected diameter.
cm
Final answer: cm
Q26Single correctElectromagnetic Induction and Alternating Currents
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
Approach:
A uniformly increasing surface charge density gives a constant displacement current between the plates, which produces a circular magnetic field analogous to that of a current-carrying wire. The field grows with radius inside the plate region and falls outside, peaking at the plate edge.
Step 1:With the charge density rising at a constant rate, the rate of change of charge is constant, so the displacement current between the plates is constant.
Step 2:The constant displacement current behaves like the conduction current of a straight cylindrical wire, generating a circular magnetic field around the axis.
inside, outside
Step 3:The magnetic field therefore is non-zero everywhere and attains its maximum on the cylindrical surface joining the edges of the plates.
at
Final answer: Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
Q27Single correctOptics
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Reflected light is completely polarized and the angle of reflection is close to
Approach:
Brewster's law gives the polarizing angle from the tangent of the refractive index. At this angle the reflected ray is completely plane polarized, while the transmitted ray is only partially polarized.
Step 1:The polarizing angle is the inverse tangent of the refractive index.
Step 2:Evaluating the inverse tangent gives the Brewster angle, which equals the angle of reflection.
Step 3:At the polarizing angle the reflected beam is perfectly plane polarized while the transmitted beam remains partially polarized.
Final answer: Reflected light is completely polarized and the angle of reflection is close to
Q28Single correctElectrostatics
Two identical charged conducting spheres and have their centres separated by a certain distance. Charge on each sphere is and the force of repulsion between them is . A third identical uncharged conducting sphere is brought in contact with sphere first and then with and finally removed from both. New force of repulsion between spheres and (Radii of and are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Touching the uncharged sphere to A halves A's charge. Touching the same sphere to B then averages its charge with B's charge. The new force follows from the product of the resulting charges.
Step 1:The original force corresponds to two equal charges separated by the fixed distance.
Step 2:After contact with A, the charge on A becomes half its initial value, shared equally with the neutral sphere.
Step 3:The neutral sphere carrying half-charge then touches B; their total charge of three-halves q is shared equally, giving each three-quarters q.
Step 4:The new force uses the final charges on A and B.
Final answer:
Q29Single correctKinetic Theory of Gases
A container has two chambers of volumes litres and litres separated by a partition made of a thermal insulator. The chambers contain and moles of ideal gas at pressures atm and atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 atm
Approach:
Since temperature is unchanged across the insulating partition, the product of pressure and volume is conserved. The final pressure follows from equating the total initial pressure-volume product to the final pressure times the combined volume.
Step 1:The total pressure-volume product before mixing equals that of the combined chamber after mixing.
Step 2:Substituting the given values gives the total product and the combined volume.
Step 3:Dividing by the combined volume gives the equilibrium pressure.
atm
Final answer: atm
Q30Single correctAtoms and Nuclei
A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the orbit and the particle's speed v in the orbit depend on n as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A constant central force provides the centripetal force, fixing the speed-radius relation. Bohr's quantization of angular momentum gives a second relation, and solving the two yields the dependence of radius and speed on the quantum number.
Step 1:Equating the constant force to the centripetal requirement shows the product of mass and squared speed over radius is constant.
Step 2:Bohr's quantization condition gives the product of speed and radius proportional to the quantum number.
Step 3:Combining the two relations gives the radius proportional to the two-thirds power of the quantum number and the speed proportional to its one-third power.
Final answer:
Q31Single correctGravitation
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 earth days
Approach:
Kepler's third law relates the square of the orbital period to the cube of the orbital radius. The ratio of the two radii fixes the ratio of the periods, giving Mercury's year from the Martian year.
Step 1:The Martian orbital radius is four times that of Mercury.
Step 2:Applying Kepler's law, the ratio of periods is the three-halves power of the radius ratio.
Step 3:Mercury's year is the Martian year divided by eight.
days
Final answer: earth days
Q32Single correctGravitation
A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 N
Approach:
The acceleration due to gravity at a height falls off as the inverse square of the distance from the earth's centre. The new weight is the surface weight scaled by the square of the ratio of radii.
Step 1:The ratio of the weight at height to the surface weight equals the square of the ratio of the radii, with the height equal to one-third the radius.
Step 2:Simplifying the squared ratio gives nine-sixteenths.
Step 3:Scaling the surface weight by this factor gives the weight at the given height.
N
Final answer: N
Q33Single correctCurrent Electricity
A wire of resistance is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Cutting the wire into eight equal pieces gives each piece one-eighth of the original resistance. Four such pieces in parallel form a set, and two such sets in series give the final value.
Step 1:Each of the eight equal pieces has one-eighth of the original resistance.
Step 2:Four such pieces in parallel give a quarter of a piece's resistance.
Step 3:Two identical parallel sets in series add together.
Final answer:
Q34Single correctDual Nature of Radiation and Matter
De-Broglie wavelength of an electron orbiting in the state of hydrogen atom is close to (Given Bohr radius nm)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 nm
Approach:
The radius of the n=2 orbit follows from the Bohr radius scaled by the square of the quantum number. The de Broglie condition states that the orbit circumference equals an integer number of wavelengths, giving the wavelength directly.
Step 1:For the second orbit the radius is the Bohr radius times four.
nm
Step 2:The de Broglie standing-wave condition gives the wavelength as the circumference divided by the quantum number.
Step 3:Evaluating the product gives the wavelength.
nm
Final answer: nm
Q35Single correctElectrostatics
An electric dipole with dipole moment C m is aligned with the direction of a uniform electric field of magnitude N/C. The dipole is then rotated through an angle of with respect to the electric field. The change in the potential energy of the dipole is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 J
Approach:
The potential energy of a dipole in a uniform field varies as the negative cosine of the angle with the field. The change in energy is the difference between the final and initial orientations.
Step 1:The change in potential energy is the difference of the final and initial energies, with initial alignment at zero degrees and final at sixty degrees.
Step 2:Substituting the dipole moment and field strength gives the magnitude of the change.
Final answer: J
Q36Single correctCurrent Electricity
A constant voltage of 50 V is maintained between the points and of the circuit shown in the figure. The current through the branch of the circuit is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 A
Approach:
The network reduces to two parallel branches between A and B. The total current from the source splits between the branches, and the junction rule at the node connecting them gives the current through CD.
Step 1:The branches are a one-ohm with two-ohm series and a three-ohm with four-ohm series, taken in parallel between the terminals.
Step 2:The total current drawn from the cell follows from the source voltage and the equivalent resistance.
A
Step 3:The current divides between the two parallel branches in inverse proportion to their resistances, giving the branch currents.
Step 4:The bottom branch carries 16 A into the node and 8 A out, so the junction rule at C gives the current through CD.
Final answer: A
Q37Single correctDual Nature of Radiation and Matter
A photon and an electron (mass m) have the same energy E. The ratio of their de Broglie wavelengths is: (c is the speed of light)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The photon wavelength is set by its energy through Planck's relation, while the electron wavelength comes from its momentum in terms of its kinetic energy. The ratio of the two wavelengths simplifies to a compact expression.
Step 1:The photon wavelength follows from Planck's energy relation.
Step 2:The electron momentum in terms of its kinetic energy gives its de Broglie wavelength.
Step 3:Dividing the photon wavelength by the electron wavelength gives the required ratio.
Final answer:
Q38Single correctDual Nature of Radiation and Matter
Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A only
Approach:
The photoelectric current is proportional to the intensity of the incident light and is independent of its frequency once emission occurs. Only the graph showing a straight line of current against intensity is correct.
Step 1:The photoelectric current rises in direct proportion to the intensity of the incident light.
Step 2:Once above threshold, the saturation current does not depend on frequency, so only graph A correctly represents the variation.
Final answer: A only
Q39Single correctSystem of Particles and Rotational Motion
A sphere of radius is cut from a larger solid sphere of radius as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the -axis is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The moment of inertia of the whole sphere about a diameter is known. The mass of the smaller sphere follows from uniform density and the volume ratio. The parallel axis theorem gives the smaller sphere's inertia about the Y-axis, and subtraction gives that of the remaining part.
Step 1:The whole sphere of radius 2R about the Y-axis through its diameter has its moment of inertia in terms of the total mass.
Step 2:With uniform density, the smaller sphere of radius R has one-eighth the mass of the whole sphere.
Step 3:The parallel axis theorem about the Y-axis, with the smaller sphere centred at distance R, gives its moment of inertia.
Step 4:The remaining part's inertia is the whole minus the smaller sphere; the required ratio is the smaller to the remaining.
Final answer:
Q40Single correctSemiconductor Electronics
A full wave rectifier circuit with diodes () and () is shown in the figure. If input supply voltage volt, then at msec
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 is reverse biased, is forward biased
Approach:
The instantaneous phase of the input determines its sign at the given time. A negative half cycle reverse biases one diode and forward biases the other in a full-wave rectifier.
Step 1:The angular frequency of the supply gives a period of one-fiftieth of a second.
s
Step 2:The given time of 15 milliseconds corresponds to three-quarters of the period.
Step 3:Three-quarters of a period lies in the negative half cycle of the input, so the first diode is reverse biased and the second is forward biased.
at
Final answer: is reverse biased, is forward biased
Q41Single correctThermodynamics
Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius and , respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio is equal to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By the first law, with equal heat and equal change in internal energy, the work done by each gas is equal. At constant pressure the work is the pressure times the swept volume, equating the area times displacement for the two pistons.
Step 1:Equal heat and equal internal-energy change make the work done by the two gases equal.
Step 2:At constant common pressure the equal work gives equal swept volumes, so the piston area times displacement is the same for both.
Step 3:Solving for the radius ratio using the displacements of 16 cm and 9 cm gives the result.
Final answer:
Q42Single correctPhysical World and Measurement
A physical quantity P is related to four observations a, b, c and d as follows:
The percentage errors of measurement in a, b, c and d are 1%, 3%, 2% and 4% respectively. The percentage error in the quantity P is
The percentage errors of measurement in a, b, c and d are 1%, 3%, 2% and 4% respectively. The percentage error in the quantity P is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
For a product of powers the maximum fractional error adds the fractional errors of each factor weighted by the magnitude of its exponent. The total percentage error is the weighted sum.
Step 1:The relation expresses the maximum percentage error as a weighted sum of the individual percentage errors.
Step 2:Substituting the given percentage errors with the respective weights.
Step 3:Adding the terms gives the total percentage error.
Final answer:
Q43Single correctOptics
The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at from the polarization axis of one of the polaroids, is ( is the intensity of polarised light after passing through the first polaroid)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Malus's law is applied twice: first across the angle to the inserted polaroid, then across the complementary angle to the final crossed polaroid. The product of the two cosine-squared factors gives the transmitted intensity.
Step 1:The intensity after the inserted polaroid, oriented at half of forty-five degrees, follows from Malus's law.
Step 2:The final crossed polaroid makes the complementary angle with the inserted one, giving a further cosine-squared factor.
Step 3:Using the double-angle identity the product reduces to one-eighth of the initial intensity.
Final answer:
Q44Single correctOscillations and Waves
Two identical point masses P and Q, suspended from two separate massless springs of spring constants and , respectively, oscillate vertically. If their maximum speeds are the same, the ratio of the amplitude of mass Q to the amplitude of mass P is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The maximum speed of a simple harmonic oscillator is the product of amplitude and angular frequency. With equal masses and equal maximum speeds, the amplitude ratio is the inverse ratio of the angular frequencies, which depend on the spring constants.
Step 1:Equal maximum speeds make the product of amplitude and angular frequency equal for both masses.
Step 2:The masses are identical, so each angular frequency is the square root of its own spring constant divided by the common mass.
Step 3:The common mass cancels, leaving the amplitude of Q relative to P as the square root of over .
Final answer:
Q45Single correctOscillations and Waves
A pipe open at both ends has a fundamental frequency in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
An open pipe has a fundamental wavelength of twice its length. When dipped to half its length it behaves as a closed pipe of that half length, whose fundamental is a quarter wavelength. Comparing the two gives the new frequency.
Step 1:The open pipe of length L has its fundamental frequency set by twice its length.
Step 2:Once dipped to half its length, the pipe acts as a closed pipe of length half of L, with the fundamental set by four times this length.
Step 3:The two expressions are identical, so the fundamental frequency is unchanged.
Final answer:
Chemistry44 questions
Q46Single correctStructure of Atom
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes and transitions, respectively, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The wavelength of absorbed light is inverse to the energy gap, which follows the Rydberg expression for the difference of inverse-square principal quantum numbers. The ratio of the two wavelengths equals the inverse ratio of the corresponding energy gaps.
Step 1:For the transition from level 2 to level 3, the energy gap is the difference of the two level energies.
Step 2:For the transition from level 4 to level 6, the energy gap is computed the same way.
Step 3:Dividing the first wavelength by the second cancels the common factors and yields the required ratio.
Final answer:
Q47Single correctThe s-Block Elements
Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are not the same.
C. Ar, K, Cl, C, and are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below :
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are not the same.
C. Ar, K, Cl, C, and are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2C and E only
Approach:
Each statement is checked against established periodic trends in melting point, electronegativity, isoelectronic species, ionization enthalpy, and atomic radius.
Step 1:Statement A is false: both Ga and Cs have low melting points (Ga melts near 303 K and Cs near 302 K).
Step 2:Statement B is false: on the Pauling scale the electronegativity values of N and Cl are the same, both 3.0.
Step 3:Statement C is true: Ar, K, Cl, C and each carry 18 electrons and are isoelectronic.
Step 4:Statement D is false: the first ionization enthalpy of Mg exceeds that of Al because removal of an electron from Mg breaks into a stable filled 3s pair, so the order is not as stated.
Step 5:Statement E is true: atomic radius rises down a group, so Cs (262 pm) is larger than Rb (244 pm) and Li (152 pm).
Final answer: C and E only
Q48Single correctSome Basic Concepts of Chemistry
Choose the correct answer from the options given below :
| List-I (Ion) | List-II (Group Number in Cation Analysis) |
|---|---|
| A.. | I.. Group-I |
| B.. | II.. Group-III |
| C.. | III.. Group-IV |
| D.. | IV.. Group-VI |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-IV, C-I, D-II
Approach:
Each cation is placed in the analytical group corresponding to its characteristic precipitating reagent in qualitative analysis.
Step 1:C precipitates as sulphide in ammoniacal medium, placing it in Group-IV.
Step 2:M is identified in the last analytical stage, Group-VI.
Step 3:P forms an insoluble chloride and is detected in Group-I.
Step 4:A precipitates as its hydroxide in Group-III.
Step 5:Reading those analytical groups off the printed List-II labels, where I is Group-I, II is Group-III, III is Group-IV and IV is Group-VI, converts the assignments into the lettered pairing.
Final answer: A-III, B-IV, C-I, D-II
Q49Single correctHaloalkanes and Haloarenes
Predict the major product 'P' in the following sequence of reactions:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C and C on adjacent ring carbons (drawn structure, option 1)
Approach:
The sequence proceeds by anti-Markovnikov radical addition of HBr, nucleophilic substitution by cyanide, and reduction of the resulting nitrile to a primary amine.
Step 1:HBr with benzoyl peroxide adds across the terminal alkene by the anti-Markovnikov pathway, placing bromine on the terminal carbon to give a primary alkyl bromide.
Step 2:KCN displaces bromide by nucleophilic substitution, extending the chain by one carbon as a nitrile.
Step 3:Sodium amalgam in ethanol reduces the nitrile to a primary amine.
Final answer: C and C on adjacent ring carbons (drawn structure, option 1)
Q50Single correctStructure of Atom
Energy and radius of first Bohr orbit of He and L are
[Given = 2.18 1 J, = 52.9 pm]
[Given = 2.18 1 J, = 52.9 pm]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Bohr energy scales as the square of nuclear charge while the orbital radius scales inversely with nuclear charge. Each value is obtained by inserting Z = 2 for He and Z = 3 for L into the first-orbit expressions.
Step 1:For He (Z = 2, n = 1) the energy uses Z squared equal to 4.
Step 2:The radius for He uses Z = 2 in the denominator.
Step 3:For L (Z = 3, n = 1) the energy uses Z squared equal to 9.
Step 4:The radius for L uses Z = 3 in the denominator.
Final answer:
Q51Single correctCoordination Compounds
Which of the following are paramagnetic?
A.
B.
C.
D.
E.
Choose the correct answer from the options given below :
A.
B.
C.
D.
E.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A and D only
Approach:
The magnetic nature of each nickel complex is decided by counting unpaired d-electrons, which follows from the oxidation state of nickel and the field strength of the ligands.
Step 1: has N (3) with weak chloride ligands giving s hybridisation and two unpaired electrons, so it is paramagnetic.
Step 2: has Ni in the zero oxidation state (3) with s hybridisation and no unpaired electrons, so it is diamagnetic.
Step 3: has N with strong-field cyanide forcing ds hybridisation and zero unpaired electrons, so it is diamagnetic.
Step 4: has N with weak-field water giving s hybridisation and two unpaired electrons, so it is paramagnetic.
Step 5: has Ni(0) (3 4) with s hybridisation and zero unpaired electrons, so it is diamagnetic.
Final answer: A and D only
Q52Single correctThe p-Block Elements (Group 15)
Given below are two statements :
Statement I : Like nitrogen that can form ammonia, arsenic can form arsine.
Statement II : Antimony cannot form antimony pentoxide.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Like nitrogen that can form ammonia, arsenic can form arsine.
Statement II : Antimony cannot form antimony pentoxide.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect
Approach:
Each statement is judged against the known hydride- and oxide-forming behaviour of group 15 elements.
Step 1:All elements of group 15 form hydrides of E type, so nitrogen forms ammonia and arsenic forms arsine , making Statement I correct.
Step 2:All elements of group 15 form two types of oxides, the trioxide and the pentoxide , so antimony does form antimony pentoxide , making Statement II incorrect.
Final answer: Statement I is correct but Statement II is incorrect
Q53Single correctClassification of Elements and Periodicity
Which among the following electronic configurations belong to main group elements?
A.
B.
C.
D.
E.
Choose the correct answer from the option given below :
A.
B.
C.
D.
E.
Choose the correct answer from the option given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and C only
Approach:
The block of each configuration is read from its differentiating electron, and only s-block and p-block species count as main group elements.
Step 1: ends in an s subshell, identifying sodium, an s-block main group element.
Step 2: is a d-block configuration for vanadium, a transition element, not a main group element.
Step 3: ends in a p subshell, identifying iodine, a p-block main group element.
Step 4: is a d-block configuration for copper, a transition element.
Step 5: is thorium, an actinoid of the f-block, not a main group element.
Final answer: A and C only
Q54Single correctSome Basic Concepts of Chemistry
Dalton's Atomic theory could not explain which of the following?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Law of gaseous volume
Approach:
Dalton's theory accounts for the mass-based laws of chemical combination but not for the volume relationships of reacting gases.
Step 1:Dalton's theory explains the law of conservation of mass, the law of constant proportion, and the law of multiple proportion, all of which concern masses of combining elements.
Step 2:The law of gaseous volumes, which relates the volumes of reacting and product gases, lies outside Dalton's theory and is explained by Gay-Lussac and Avogadro.
Final answer: Law of gaseous volume
Q55Single correctRedox Reactions
Consider the following compounds :
, and
The oxidation state of the underlined elements in them are, respectively,
, and
The oxidation state of the underlined elements in them are, respectively,
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1+1, 1, and +6
Approach:
The oxidation state of each indicated element is fixed using the known states of the partner atoms and the overall neutrality of the compound.
Step 1:In , potassium as an alkali metal always shows the +1 oxidation state.
Step 2:In , the peroxide linkage gives each oxygen the 1 oxidation state.
Step 3:In , with hydrogen +1 and oxygen 2, sulphur carries the +6 oxidation state.
Final answer: +1, 1, and +6
Q56Single correctChemical Kinetics
If the half-life () for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 410 minutes
Approach:
For a first order reaction the time for a given fractional completion is found from the integrated rate law, and 99.9% completion corresponds to ten half-lives.
Step 1:For 99.9% completion the remaining fraction is one thousandth, so the logarithm term equals 3.
Step 2:Expressing this in terms of the half-life shows the required time is ten half-lives.
Step 3:With a one minute half-life the time for 99.9% completion is ten minutes.
Final answer: 10 minutes
Q57Single correctCoordination Compounds
The correct order of the wavelength of light absorbed by the following complexes is,
A.
B.
C.
D.
Choose the correct answer from the options given below :
A.
B.
C.
D.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B < A < D < C
Approach:
The wavelength of absorbed light is inversely related to the crystal field splitting, which itself is inverse to the ligand field strength, so stronger fields absorb shorter wavelengths.
Step 1:The strong cyanide field gives the largest splitting, so absorbs the shortest wavelength near 310 nm.
Step 2:Ammonia is a weaker field than cyanide, so absorbs near 475 nm.
Step 3: with aqua ligands absorbs near 498 nm.
Step 4: has the weakest effective field here and absorbs the longest wavelength near 600 nm.
Step 5:Arranging by increasing absorbed wavelength gives the order with C the largest and B the smallest.
Final answer: B < A < D < C
Q58Single correctOrganic Chemistry: Some Basic Principles
Which one of the following compounds can exist as cis-trans isomers?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 41, 2-Dimethylcyclohexane
Approach:
Cis-trans isomerism requires restricted rotation about a double bond or ring with two different groups on each of the two restricted-rotation centres.
Step 1:Pent-1-ene has a terminal double bond carrying two identical hydrogens on the end carbon, so it shows no cis-trans isomerism.
Step 2:2-Methylhex-2-ene carries two methyl groups on the same double-bond carbon, removing the requirement for cis-trans isomerism.
Step 3:1,1-Dimethylcyclopropane has both methyl groups on one ring carbon, so it has no cis-trans isomers.
Step 4:1,2-Dimethylcyclohexane has one methyl on each of two adjacent ring carbons, allowing cis and trans ring arrangements.
Final answer: 1, 2-Dimethylcyclohexane
Q59Single correctIonic Equilibrium
Phosphoric acid ionizes in three steps with their ionization constant values , , and , respectively, while K is the overall ionization constant. Which of the following statements are true?
A.
B. is a stronger acid than and
C.
D.
Choose the correct answer from the options given below :
A.
B. is a stronger acid than and
C.
D.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, B and C only
Approach:
The three successive ionizations of phosphoric acid are written out, their constants compared, and the overall constant related to the stepwise constants by multiplication.
Step 1:The first ionization has the largest constant near .
Step 2:The second ionization has a smaller constant near .
Step 3:The third ionization has the smallest constant near , confirming and that is the strongest of the three species (statements B and C true).
Step 4:Because the overall constant is the product of the stepwise constants, taking logarithms gives the sum of the individual log constants, so statement A is true while D, an arithmetic mean relation, is false.
Final answer: A, B and C only
Q60Single correctHydrocarbons
Which one of the following reactions does NOT give benzene as the product?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Benzenediazonium chloride warmed with water (option 4) gives phenol, and HCl, not benzene
Approach:
Each route is checked against the known laboratory preparations of benzene. Three of the four reactions are standard benzene syntheses; the fourth converts a diazonium salt to phenol, so it is the reaction that does not yield benzene.
Step 1:Heating sodium benzoate with sodalime (NaOH and CaO) removes the carboxylate group by decarboxylation.
Step 2:Passing n-hexane over Mo2O3 at 773 K and 10-20 atm causes aromatisation (dehydrocyclisation) of the straight chain into a six-membered aromatic ring.
Step 3:Three molecules of ethyne undergo cyclic polymerisation in a red hot iron tube at 873 K.
Step 4:Warming benzenediazonium chloride with water replaces the diazonium group with a hydroxyl group, releasing nitrogen and hydrogen chloride.
Final answer: Benzenediazonium chloride warmed with water (option 4) gives phenol, and HCl, not benzene
Q61Single correctElectrochemistry
If the molar conductivity () of a 0.050 mol solution of a monobasic weak acid is 90 S c mo, its extent (degree) of dissociation will be
[Assume S c mo and S c mo.]
[Assume S c mo and S c mo.]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.225
Approach:
The degree of dissociation of a weak electrolyte is the ratio of its molar conductivity to its limiting molar conductivity, where the latter is the sum of the cation and anion limiting values.
Step 1:The limiting molar conductivity is the sum of the ionic limiting conductivities.
Step 2:Dividing the measured molar conductivity by the limiting value gives the degree of dissociation.
Final answer: 0.225
Q62Single correctChemical Bonding and Molecular Structure
Given below are two statements :
Statement I : A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II : As bond order increases, the bond length increases.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II : As bond order increases, the bond length increases.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both Statement I and Statement II are false
Approach:
Each statement is tested against molecular orbital theory, where a positive bond order implies stability and bond order varies inversely with bond length.
Step 1:A positive bond order indicates a stable molecule, while a negative or zero bond order indicates an unstable molecule, so Statement I describing a bond-order-zero molecule as stable is false.
Step 2:Bond length falls as bond order rises, because more shared electron density pulls the nuclei closer, so the claim that bond length increases with bond order is false.
Final answer: Both Statement I and Statement II are false
Q63Single correctCoordination Compounds
Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Conductance depends on the number of ions a complex produces in solution; complexes that release no free ions have only the charge on the complex unit, which here is zero, giving minimum conductance.
Step 1:The conductance of a complex depends on the ions produced on dissolution; if no ions are released, the complex unit carries no net charge and conducts least.
Step 2:In the cobalt(III) centre is surrounded by three neutral ammonia ligands and three chloride ligands inside the coordination sphere, so the complex unit is neutral and releases no ions on dissolution.
carries no charge
Step 3:In the two chlorides also sit inside the coordination sphere and balance the metal charge, so that unit is likewise neutral and non-ionising, whereas and release three and one chloride ions respectively and conduct far better.
Final answer:
Q64Single correctThe p-Block Elements (Group 18)
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. | I.. sd; linear |
| B.. | II.. s; pyramidal |
| C.. | III.. s; distorted octahedral |
| D.. | IV.. s; square pyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B-I, C-IV, D-III
Approach:
The hybridisation and shape of each xenon compound are deduced from the number of bonding pairs and lone pairs around the central xenon atom.
Step 1: has three bond pairs and one lone pair, giving s hybridisation and a pyramidal shape.
Step 2: has two bond pairs and three lone pairs, giving sd hybridisation and a linear shape.
Step 3: has five bond pairs and one lone pair, giving s hybridisation and a square pyramidal shape.
Step 4: has six bond pairs and one lone pair, giving s hybridisation and a distorted octahedral shape.
Final answer: A-II, B-I, C-IV, D-III
Q65Single correctThermodynamics
C(s) + 2(g) C(g); H = 74.8 kJ mo. Which of the following diagrams gives an accurate representation of the above reaction? [R reactants; P products]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Energy profile with the product level P lying 74.8 kJ mo below the reactant level R (option 1)
Approach:
A negative reaction enthalpy marks an exothermic reaction whose products lie at lower energy than the reactants by the magnitude of the enthalpy change.
Step 1:The reaction enthalpy is kJ mo, a negative value, so the reaction is exothermic and releases energy.
Step 2:In an exothermic profile the product level lies below the reactant level by the enthalpy magnitude, which is the diagram of option 1.
Final answer: Energy profile with the product level P lying 74.8 kJ mo below the reactant level R (option 1)
Q66Single correctSurface Chemistry
Choose the correct answer from the options given below:
| List-I (Example) | List-II (Type of Solution) |
|---|---|
| A.. Humidity | I.. Solid in solid |
| B.. Alloys | II.. Liquid in gas |
| C.. Amalgams | III.. Solid in gas |
| D.. Smoke | IV.. Liquid in solid |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-II, B-I, C-IV, D-III
Approach:
Each example is classified as a solution by identifying its dispersed phase and dispersion medium.
Step 1:Humidity is liquid water dispersed in gaseous air, a liquid in gas.
Step 2:An alloy is a solid metal dissolved in another solid metal, a solid in solid.
Step 3:An amalgam is mercury, a liquid metal, dissolved in a solid metal, a liquid in solid.
Step 4:Smoke is fine solid particles dispersed in gas, a solid in gas.
Final answer: A-II, B-I, C-IV, D-III
Q67Single correctAmines
The correct order of decreasing basic strength of the given amines is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
Approach:
Basic strength is judged from the p value, with a higher basicity for aliphatic amines than aromatic amines, and the order ranked accordingly.
Step 1:A higher value of p corresponds to lower basicity, so basicity is compared inversely with p.
Step 2:Aliphatic amines are stronger bases than aromatic amines because aromatic lone-pair delocalisation lowers availability of the nitrogen lone pair.
Step 3:Ranking the four amines gives N-ethylethanamine the strongest, then ethanamine, then N-methylaniline, then benzenamine the weakest.
Final answer: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
Q68Single correctSome Basic Concepts of Chemistry
Among the following, choose the ones with equal number of atoms.
A. 212 g of N(s) [molar mass = 106 g]
B. 248 g of NO(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of (g) [molar mass = 2 g]
E. 220 g of C(g) [molar mass = 44 g]
Choose the correct answer from the options given below :
A. 212 g of N(s) [molar mass = 106 g]
B. 248 g of NO(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of (g) [molar mass = 2 g]
E. 220 g of C(g) [molar mass = 44 g]
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B, and D only
Approach:
The number of atoms in each sample is the moles of the substance multiplied by the atomicity and by Avogadro's number, and samples giving the same product are grouped.
Step 1:For N, 212 g is 2 mol and each formula unit has 6 atoms, giving 12 atoms.
Step 2:For NO, 248 g is 4 mol and each formula unit has 3 atoms, giving 12 atoms.
Step 3:For NaOH, 240 g is 6 mol and each formula unit has 3 atoms, giving 18 atoms.
Step 4:For , 12 g is 6 mol and each molecule has 2 atoms, giving 12 atoms.
Step 5:For C, 220 g is 5 mol and each molecule has 3 atoms, giving 15 atoms; the samples A, B and D all give 12 atoms.
Final answer: A, B, and D only
Q69Single correctBiomolecules
Match List-I with List-II. Choose the correct answer from the options given below:
| List-I (Name of Vitamin) | List-II (Deficiency disease) |
|---|---|
| A. Vitamin | I. Cheilosis |
| B. Vitamin D | II. Convulsions |
| C. Vitamin | III. Rickets |
| D. Vitamin | IV. Pernicious anaemia |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-III, C-I, D-II
Approach:
Each vitamin is paired with the disease arising from its deficiency, based on standard nutritional biochemistry.
Step 1:Vitamin B12 deficiency impairs red blood cell maturation, producing pernicious anaemia, so A matches IV.
Step 2:Vitamin D deficiency causes defective bone mineralisation in children, producing rickets, so B matches III.
Step 3:Vitamin B2 (riboflavin) deficiency causes cracking at the corners of the mouth, known as cheilosis, so C matches I.
Step 4:Vitamin B6 (pyridoxine) deficiency leads to neurological symptoms such as convulsions, so D matches II.
Final answer: A-IV, B-III, C-I, D-II
Q70Single correctAldehydes, Ketones and Carboxylic Acids
The correct order of decreasing acidity of the following aliphatic acids is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Acidity of a carboxylic acid increases as the electron density on the carboxylate is reduced; electron-donating alkyl groups destabilise the carboxylate and lower acidity.
Step 1:Alkyl groups release electron density through the positive inductive effect, intensifying the negative charge on the carboxylate and weakening the acid.
Step 2:Formic acid carries no alkyl group and is the strongest, while the three methyl groups of trimethylacetic acid make it the weakest.
Step 3:Arranging by increasing alkyl substitution gives the decreasing acidity order.
Final answer:
Q71Single correctThe d- and f-Block Elements
Given below are two statements :
Statement I : Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II : The number of unpaired electrons in a ion (Z = 24) is the same as that of a ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below :
Statement I : Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II : The number of unpaired electrons in a ion (Z = 24) is the same as that of a ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is true but Statement II is false
Approach:
Statement I is judged against the definition of ferromagnetism, and Statement II is tested by counting the unpaired electrons in each ion.
Step 1:Ferromagnetic substances are strongly attracted in an applied field, which represents an extreme form of paramagnetism, so Statement I is true.
Step 2:The chromium(II) ion has the configuration with four unpaired electrons in its 3d subshell.
Step 3:The neodymium(III) ion has three unpaired electrons in its 4f subshell.
Step 4:Since the two ions carry different numbers of unpaired electrons, Statement II is false.
Final answer: Statement I is true but Statement II is false
Q72Single correctOrganic Chemistry - Some Basic Principles and Techniques
Match List I with List II Choose the correct answer from the options given below :
| List-I (Mixture) | List-II (Method of separation) |
|---|---|
| A. | I. Distillation under reduced pressure |
| B. Crude oil in petroleum industry | II. Steam distillation |
| C. Glycerol from spent-lye | III. Fractional distillation |
| D. Aniline - water | IV. Simple distillation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-III, C-I, D-II
Approach:
Each mixture is paired with the purification technique suited to the boiling-point difference and thermal stability of its components.
Step 1:Chloroform and aniline differ widely in boiling point and are separated by simple distillation, so A matches IV.
Step 2:Crude oil contains many components of close boiling points that are resolved by fractional distillation, so B matches III.
Step 3:Glycerol decomposes near its boiling point and is recovered by distillation under reduced pressure, so C matches I.
Step 4:Aniline is steam-volatile and immiscible with water, so the aniline-water mixture is separated by steam distillation, giving D matches II.
Final answer: A-IV, B-III, C-I, D-II
Q73Single correctChemical Kinetics
For the reaction , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given : ]
for the reaction at 1000 K is
[Given : ]
for the reaction at 1000 K is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.033
Approach:
The equilibrium constant in concentration terms equals the ratio of forward to backward rate constants, and conversion to the pressure constant uses the change in moles of gas.
Step 1:The concentration equilibrium constant is the forward rate constant divided by the backward rate constant, which is the reciprocal of 2500.
Step 2:The reaction converts one mole of gas into two, so the change in gaseous moles is one.
Step 3:Substituting the values into the pressure-constant relation gives the result.
Final answer: 0.033
Q74Single correctAmines
Given below are two statements :
Statement I : Benzene diazonium salt is prepared by the reaction of aniline with nitrous acid at K. It decomposes easily in the dry state.
Statement II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Benzene diazonium salt is prepared by the reaction of aniline with nitrous acid at K. It decomposes easily in the dry state.
Statement II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both Statement I and Statement II are correct
Approach:
Each statement is checked against the standard preparation and properties of benzene diazonium chloride.
Step 1:Benzene diazonium chloride forms when aniline reacts with nitrous acid, generated in situ from sodium nitrite and hydrochloric acid, at 273-278 K, and the salt decomposes readily when dry, so Statement I is correct.
Step 2:Direct iodination of the benzene ring is difficult, so iodobenzene is obtained by treating the diazonium salt with potassium iodide, which makes Statement II correct.
Final answer: Both Statement I and Statement II are correct
Q75Single correctHydrocarbons
How many products (including stereoisomers) are expected from monochlorination of the following compound?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 46
Approach:
Each distinct hydrogen environment in 2-methylbutane gives a monochloro product, and products carrying a new stereocentre are counted with both configurations.
Step 1:Substitution at a primary methyl carbon of the isopropyl group gives 1-chloro-2-methylbutane, a single product with one isomer.
Step 2:Substitution at the tertiary carbon bearing the methyl branches gives 2-chloro-2-methylbutane as one product.
Step 3:Substitution at the internal methylene carbon creates a chiral centre, yielding two enantiomers.
Step 4:Substitution at the terminal ethyl carbon creates a chiral centre, yielding two further enantiomers.
Step 5:Summing the products across all positions gives the total.
Final answer: 6
Q76Single correctOrganic Chemistry - Some Basic Principles and Techniques
Among the given compounds I-III, the correct order of bond dissociation energy of C-H bond marked with * is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1II > I > III
Approach:
The strength of a C-H bond rises with the s-character of the carbon orbital forming it, since greater s-character draws the bonding electrons closer to the nucleus.
Step 1:In compound I the marked carbon of the benzene ring is sp2 hybridised, giving an intermediate s-character.
Step 2:In compound II the marked carbon is sp hybridised, giving the highest s-character and the strongest C-H bond.
Step 3:In compound III the marked carbon is sp3 hybridised, giving the lowest s-character and the weakest C-H bond.
Step 4:Ordering by decreasing s-character gives the order of bond dissociation energy.
Final answer: II > I > III
Q77Single correctHydrocarbons
Which one of the following compounds does not decolourize bromine water?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cyclohexane the drawn plain hexagon with no ring unsaturation (option 1)
Approach:
Bromine water is decolourised by unsaturated compounds through addition and by activated aromatic rings through electrophilic substitution; a saturated compound lacking such reactivity leaves the colour unchanged.
Step 1:Styrene contains a carbon-carbon double bond that adds bromine, discharging the colour.
Step 2:Phenol and aniline carry strongly activating groups that promote ring bromination, discharging the colour and forming substituted products.
Step 3:Cyclohexane is fully saturated with no activated ring and does not react with bromine water, so it retains the orange colour.
Final answer: Cyclohexane the drawn plain hexagon with no ring unsaturation (option 1)
Q78Single correctAldehydes, Ketones and Carboxylic Acids
The major product of the following reaction is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Benzylic tertiary alcohol (C, OH) with the nitrile converted to a methyl ketone (drawn structure, option 2)
Approach:
Excess methylmagnesium bromide attacks both the carbonyl of the ketone and the nitrile carbon; aqueous workup then converts each intermediate to its final functional group.
Step 1:The Grignard reagent adds a methyl group to the aromatic ketone, and acidic workup gives a tertiary alcohol at that carbon.
Step 2:A second equivalent of the Grignard reagent adds to the nitrile carbon to form a magnesium-bound imine intermediate.
Step 3:Acidic hydrolysis of this imine salt yields a methyl ketone in place of the original nitrile group.
Final answer: Benzylic tertiary alcohol (C, OH) with the nitrile converted to a methyl ketone (drawn structure, option 2)
Q79Single correctSolutions
Which of the following aqueous solution will exhibit highest boiling point?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.01M
Approach:
Boiling-point elevation depends on the total concentration of solute particles, given by the product of the van't Hoff factor and the molality, so the solution with the largest product boils highest.
Step 1:Urea is a non-electrolyte, so its particle product is one times 0.01, equal to 0.01.
Step 2:Potassium nitrate dissociates into two ions, giving a particle product of two times 0.01, equal to 0.02.
Step 3:Sodium sulphate dissociates into three ions, giving a particle product of three times 0.01, equal to 0.03.
Step 4:Glucose is a non-electrolyte, so its particle product is one times 0.015, equal to 0.015.
Step 5:The sodium sulphate solution has the largest particle product, so it shows the greatest elevation and the highest boiling point.
Final answer: 0.01M
Q80Single correctThe d- and f-Block Elements
Match List-I with List-II. Choose the correct answer from the options given below :
| List-I (Process) | List-II (Catalyst used) |
|---|---|
| A. Haber process | I. Fe catalyst |
| B. Wacker oxidation | II. |
| C. Wilkinson catalyst | III. |
| D. Ziegler catalyst | IV. with |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-I, B-II, C-III, D-IV
Approach:
Each named process is paired with the catalyst characteristically used to carry it out.
Step 1:The Haber process for ammonia synthesis uses an iron catalyst, so A matches I.
Step 2:Wacker oxidation of ethene to ethanal uses palladium(II) chloride, so B matches II.
Step 3:The Wilkinson catalyst is the rhodium complex with triphenylphosphine ligands, so C matches III.
Step 4:The Ziegler catalyst combines titanium tetrachloride with trimethylaluminium, so D matches IV.
Final answer: A-I, B-II, C-III, D-IV
Q81Single correctSolutions
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The solution shows negative deviation.
Approach:
The vapour pressure expected for an ideal solution is computed from Raoult's law and compared with the observed value; an observed pressure below the ideal one indicates negative deviation.
Step 1:The mole fractions are one-third for X and two-thirds for Y from the five and ten moles taken.
Step 2:Applying Raoult's law gives the ideal total pressure.
Step 3:The observed pressure of 70 torr is below the calculated 73 torr, so the solution shows negative deviation.
Final answer: The solution shows negative deviation.
Q82Single correctBiomolecules
Sugar "X"
A. Is found in honey
B. Is a keto sugar
C. exists in and - anomeric forms.
D. Is laevorotatory.
"X" is :
A. Is found in honey
B. Is a keto sugar
C. exists in and - anomeric forms.
D. Is laevorotatory.
"X" is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2D-Fructose
Approach:
The four properties listed are matched against the known characteristics of the candidate sugars to identify the one that satisfies all of them.
Step 1:The sugar occurs in honey and bears a keto group, which points to fructose rather than the aldose glucose.
Step 2:Fructose forms a five-membered furanose ring that exists in alpha and beta anomeric forms.
Step 3:Fructose rotates plane-polarised light to the left and is laevorotatory, satisfying the final property.
Final answer: D-Fructose
Q83Single correctAldehydes, Ketones and Carboxylic Acids
Identify the suitable reagent for the following conversion.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(i) , (ii)
Approach:
Partial reduction of an ester to an aldehyde requires a hindered hydride that stops at the aldehyde stage rather than reducing all the way to the alcohol.
Step 1:Lithium aluminium hydride and sodium borohydride reduce an ester past the aldehyde stage, so neither stops at the aldehyde.
Step 2:Diisobutylaluminium hydride is a bulky reagent that adds one hydride to the ester and is then hydrolysed to the aldehyde, giving the desired partial reduction.
Final answer: (i) , (ii)
Q85Single correctThermodynamics
The standard heat of formation, in kcal/mol of is :
[Given : standard heat of formation of ion (aq) = kcal/mol, standard heat of crystallisation of = kcal/mol, standard heat of formation of = kcal/mol]
[Given : standard heat of formation of ion (aq) = kcal/mol, standard heat of crystallisation of = kcal/mol, standard heat of formation of = kcal/mol]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The target reaction forming the barium ion is built by combining the given formation and crystallisation equations through Hess's law.
Step 1:The formation of the sulphate ion releases 216 kcal/mol, taken as equation one.
Step 2:The crystallisation forming solid barium sulphate from its aqueous ions releases 4.5 kcal/mol, taken as equation two.
Step 3:The formation of solid barium sulphate from its elements releases 349 kcal/mol, taken as equation three.
Step 4:Subtracting equations one and two from equation three isolates the formation of the barium ion and gives its heat of formation.
Final answer:
Q86Single correctOrganic Chemistry - Some Basic Principles and Techniques
Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 310
Approach:
All cyclic ethers of the given formula are enumerated by ring size and substitution pattern, with each stereocentre or geometric arrangement counted separately.
Step 1:An oxetane ring bearing an ethyl group and one bearing a single methyl group at two distinct positions, where one position is chiral, together contribute four isomers.
Step 2:A four-membered ring with two methyl groups gives cis and trans forms, the latter resolving into enantiomers, adding further isomers.
Step 3:Three-membered epoxide rings bearing the remaining carbon arrangements, including those with chiral or meso character, complete the count.
Step 4:Summing the isomers from every ring system and counting each stereoisomer gives the total.
Final answer: 10
Q87Single correctChemical Bonding and Molecular Structure
Identify the correct orders against the property mentioned
A. - dipole moment
B. - number of lone pairs on central atom
C. - bond length
D. - bond enthalpy
Choose the correct answer from the options given below:
A. - dipole moment
B. - number of lone pairs on central atom
C. - bond length
D. - bond enthalpy
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, D only
Approach:
Each ordering is tested against the relevant molecular property, and only the statements giving the correct trend are selected.
Step 1:The dipole moments fall in the order water, ammonia, chloroform with values of 1.85, 1.47 and 1.04 debye, so A is correct.
Step 2:Xenon tetrafluoride has two lone pairs, xenon trioxide has one and xenon difluoride has three, so the stated order is wrong and B is incorrect.
Step 3:The N-O bond is longer than the C-H bond, so the stated order placing C-H longer is wrong and C is incorrect.
Step 4:Nitrogen with a triple bond exceeds oxygen with a double bond, which exceeds hydrogen, so the bond enthalpy order in D is correct.
Final answer: A, D only
Q88Single correctEquilibrium
Higher yield of NO in can be obtained at
[ of the reaction = kJ mo]
A. Higher temperature
B. Lower temperature
C. Higher concentration of
D. Higher concentration of
Choose the correct answer from the options given below :
[ of the reaction = kJ mo]
A. Higher temperature
B. Lower temperature
C. Higher concentration of
D. Higher concentration of
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, C, D only
Approach:
The conditions favouring more product are found by applying Le Chatelier's principle to the endothermic, reactant-concentration-dependent equilibrium.
Step 1:The positive reaction enthalpy marks the forward reaction as endothermic, so a higher temperature shifts the equilibrium toward the product and raises the yield of NO.
Step 2:Raising the concentration of either nitrogen or oxygen drives the equilibrium toward the product, increasing the yield of NO.
Step 3:A lower temperature would shift the endothermic reaction backward and reduce the yield, so that condition is excluded.
Final answer: A, C, D only
Q89Single correctChemical Kinetics
If the rate constant of a reaction is , how much time does it take for concentration of the reactant to get reduced to ?
[Given: ]
[Given: ]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 169.3 s
Approach:
The first-order integrated rate law relates the time to the rate constant and the ratio of initial to final concentration through a logarithm.
Step 1:Substituting the rate constant and the initial and final concentrations into the rate law sets up the expression.
Step 2:The concentration ratio of 7.2 to 0.9 equals 8, whose logarithm is three times the logarithm of 2.
Step 3:Evaluating the product gives the time taken.
Final answer: 69.3 s
Q90Single correctOrganic Chemistry - Some Basic Principles and Techniques
Which one of the following reactions does NOT belong to "Lassaigne's test"?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Lassaigne's test detects nitrogen, sulphur and halogens by fusing the compound with sodium; reactions outside this fusion sequence do not belong to the test.
Step 1:Fusion with sodium converts nitrogen and carbon to sodium cyanide, which is part of the test.
Step 2:Fusion with sodium converts sulphur to sodium sulphide, which is part of the test.
Step 3:Fusion with sodium converts a halogen to a sodium halide, which is part of the test.
Step 4:The reduction of copper oxide by carbon is the basis of carbon estimation, not the sodium fusion test, so it does not belong to Lassaigne's test.
Final answer:
Biology89 questions
Q91Single correctRespiration in Plants
The complex II of mitochondrial electron transport chain is also known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Succinate dehydrogenase
Approach:
The mitochondrial electron transport chain consists of four protein complexes, each identified by its enzymatic activity.
Step 1:Complex I is NADH dehydrogenase, Complex III is cytochrome bc1, and Complex IV is cytochrome c oxidase.
Step 2:Complex II carries electrons from succinate to ubiquinone and is named succinate dehydrogenase.
Final answer: Succinate dehydrogenase
Q92Single correctBiotechnology: Principles and Processes
Polymerase chain reaction (PCR) amplifies DNA following the equation.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Each PCR cycle doubles the number of target DNA copies, producing exponential amplification.
Step 1:In one PCR cycle the template strands are denatured, primers anneal, and extension produces two copies from one.
Step 2:After n cycles the amplification factor follows the exponential relation, where n is the number of cycles.
Final answer:
Q93Single correctReproductive Health
What are the potential drawbacks in adoption of the IVF method?
A. High fatality risk to mother
B. Expensive instruments and reagents
C. Husband/wife necessary for being donors
D. Less adoption of orphans
E. Not available in India
F. Possibility that the early embryo does not survive
Choose the correct answer from the options given below:
A. High fatality risk to mother
B. Expensive instruments and reagents
C. Husband/wife necessary for being donors
D. Less adoption of orphans
E. Not available in India
F. Possibility that the early embryo does not survive
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B, D, F only
Approach:
The genuine drawbacks of in vitro fertilisation are evaluated statement by statement.
Step 1:Statements B, D and F are correct: IVF requires expensive instruments and reagents, it reduces adoption of orphans, and the early embryo may fail to survive.
Step 2:Statements A, C and E are incorrect. A husband or wife is not necessary for being donors, and IVF is available in India.
Final answer: B, D, F only
Q94Single correctStructural Organisation in Animals
What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Vena cava
Approach:
The vessel returning deoxygenated blood from the body tissues to the frog heart is identified from the circulatory plan.
Step 1:The frog heart is a muscular structure with three chambers and receives deoxygenated blood from the body parts through the major veins called vena cava.
Step 2:The aorta and pulmonary vein carry oxygenated blood, while the pulmonary artery carries deoxygenated blood from the heart towards the lungs.
Final answer: Vena cava
Q95Single correctThe Living World
Which one of the following statements refers to Reductionist Biology?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Physico-chemical approach to study and understand living organisms
Approach:
Reductionist biology explains living phenomena in terms of their fundamental physical and chemical components.
Step 1:The reductionist approach studies and understands living organisms through their physico-chemical interactions.
Step 2:Physiological, purely chemical, and behavioural approaches do not define reductionist biology.
Final answer: Physico-chemical approach to study and understand living organisms
Q96Single correctMolecular Basis of Inheritance
Given below are two statements :
Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both statement I and statement II are correct
Approach:
Each statement is evaluated against the RNA-world hypothesis and the molecular properties of RNA and DNA.
Step 1:Statement I describes the RNA world: RNA is regarded as the first genetic material that carried out essential life processes, functioning both as an information carrier and as a catalyst (ribozyme) for biochemical reactions. The presence of the 2'-OH group makes RNA chemically reactive and therefore unstable.
Step 2:Statement II describes the evolution of DNA from RNA as a more stable genetic material. The two complementary strands of the double helix allow damage in one strand to be corrected against the other, providing a repair mechanism that resists change.
Step 3:Both statements are correct, so the most appropriate option is the one stating that statement I and statement II are both correct.
Final answer: Both statement I and statement II are correct
Q97Single correctOrganisms and Populations
Epiphytes that are growing on a mango branch is an example of which of the following?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Commensalism
Approach:
The interaction type is determined by the effect on each species when an epiphyte grows on a host branch.
Step 1:Commensalism is the interaction in which one species benefits and the other is neither harmed nor benefited.
Step 2:An epiphyte growing on a mango branch gains support without harming the tree, which defines commensalism.
Final answer: Commensalism
Q98Single correctCell: The Unit of Life
From the statements given below choose the correct option :
A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S.
B. Each ribosome has two sub-units.
C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S.
D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S.
E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.
A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S.
B. Each ribosome has two sub-units.
C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S.
D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S.
E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B, C are true
Approach:
Each statement on ribosome size and subunit composition is checked against the structural facts.
Step 1:The eukaryotic ribosomes are 80S and prokaryotic are 70S, and each ribosome has two sub-units, so A and B are true.
Step 2:The two sub-units of the 80S ribosome are 60S and 40S, while those of 70S are 50S and 30S, so C is true.
Final answer: A, B, C are true
Q99Single correctBiodiversity and Conservation
Which one of the following is an example of ex-situ conservation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Zoos and botanical gardens
Approach:
Ex-situ conservation protects components of biodiversity outside their natural habitats.
Step 1:Zoological parks, botanical gardens and wildlife safari parks are examples of ex-situ conservation.
Step 2:Sacred groves, biosphere reserves, national parks and wildlife sanctuaries are examples of in-situ conservation.
Final answer: Zoos and botanical gardens
Q100Single correctEcosystem
Given below are two statements:
Statement I : The primary source of energy in an ecosystem is solar energy.
Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP).
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I : The primary source of energy in an ecosystem is solar energy.
Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP).
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but statement II is incorrect
Approach:
Each statement is tested against the definitions of energy source and primary productivity in ecosystems.
Step 1:The primary source of energy in an ecosystem is solar energy, so Statement I is correct.
Step 2:Gross primary productivity, not net primary productivity, is the rate of production of organic matter during photosynthesis, so Statement II is incorrect.
Final answer: Statement I is correct but statement II is incorrect
Q101Single correctBreathing and Exchange of Gases
Match List-I with List-II.
| List-I | List-II |
|---|---|
| A. Emphysema | I. Rapid spasms in muscle due to low C in body fluid |
| B. Angina Pectoris | II. Damaged alveolar walls and decreased respiratory surface |
| C. Glomerulonephritis | III. Acute chest pain when not enough oxygen is reaching to heart muscle |
| D. Tetany | IV. Inflammation of glomeruli of kidney |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-IV, D-I
Approach:
Each disorder in List-I is paired with its defining feature in List-II.
Step 1:Emphysema involves damaged alveolar walls and decreased respiratory surface, matching II, and Angina Pectoris is acute chest pain when not enough oxygen reaches the heart muscle, matching III.
Step 2:Glomerulonephritis is inflammation of glomeruli of kidney, matching IV, and Tetany is rapid spasms in muscle due to low calcium in body fluid, matching I.
Final answer: A-II, B-III, C-IV, D-I
Q102Single correctSexual Reproduction in Flowering Plants
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar.
Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers.
In the light of the above statements, choose the correct answer from the options given below:
Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar.
Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both A and R are true but R is NOT the correct explanation of A
Approach:
The truth of the assertion and reason on abiotic pollination is assessed along with their logical link.
Step 1:Wind and water pollinated flowers are not very colourful and do not produce nectar because they do not need to attract insects, so the assertion is true.
Step 2:The reason that such flowers produce enormous amounts of pollen grains is also true, but it does not explain why they lack colour and nectar, so it is not the correct explanation.
Final answer: Both and are true but is the correct explanation of
Q103Single correctMicrobes in Human Welfare
Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Beer
Approach:
Alcoholic beverages are classified by whether distillation follows fermentation.
Step 1:Wine and beer are produced without distillation of the fermented broth.
Step 2:Whisky, brandy and rum are produced by distillation of the fermented broth, so beer is the non-distilled beverage.
Final answer: Beer
Q104Single correctMorphology of Flowering Plants
Given below are two statements:
Statement I: In a floral formula stands for zygomorphic nature of the flower, and stands for inferior ovary.
Statement II: In a floral formula stands for actinomorphic nature of the flower and stands for superior ovary.
In the light of the above statements, choose the correct answer from the options given below:
Statement I: In a floral formula stands for zygomorphic nature of the flower, and stands for inferior ovary.
Statement II: In a floral formula stands for actinomorphic nature of the flower and stands for superior ovary.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct
Approach:
The floral formula symbols for symmetry and ovary position are checked against standard conventions.
Step 1:The floral symbol for the circle with a cross is used for actinomorphic nature of the flower, while a half sphere or bracket denotes zygomorphic flowers.
Step 2:The symbol G represents the gynoecium; a line above G denotes inferior ovary, while a line below G denotes superior ovary, so Statement I is incorrect and Statement II is correct.
Final answer: Statement I is incorrect but Statement II is correct
Q105Single correctMicrobes in Human Welfare
Streptokinase produced by bacterium Streptococcus is used for
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Removing clots from blood vessels
Approach:
The clinical role of streptokinase is identified from its action on blood clots.
Step 1:Streptokinase, produced by the bacterium Streptococcus and modified by genetic engineering, is used as a clot buster for removing clots from blood vessels of patients who have undergone myocardial infarction leading to heart attack.
Step 2:Curd production is done by Lactobacillus and ethanol production is done by Saccharomyces, so these are not roles of streptokinase.
Final answer: Removing clots from blood vessels
Q106Single correctMolecular Basis of Inheritance
Which chromosome in the human genome has the highest number of genes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Chromosome 1
Approach:
The chromosome carrying the maximum number of genes is recalled from the Human Genome Project findings.
Step 1:In the human genome, Chromosome 1 has the highest number of genes, about 2968.
Step 2:The Y chromosome has the fewest genes, so Chromosome 1 holds the maximum.
Final answer: Chromosome 1
Q107Single correctStructural Organisation in Animals
Which of the following statement is correct about location of the male frog copulatory pad?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4First digit of the fore limb
Approach:
The site of the copulatory pad, a male frog secondary sexual character, is located on the limb digits.
Step 1:In male frogs, the copulatory pad is present on the first digit of the forelimbs.
Step 2:These pads are absent in female frogs, confirming the forelimb first digit as the location.
Final answer: First digit of the fore limb
Q108Single correctPlant Growth and Development
Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Cytokinin
Approach:
The hormone that delays leaf senescence by promoting nutrient mobilisation is identified.
Step 1:Cytokinins help overcome apical dominance and promote nutrient mobilisation.
Step 2:This nutrient mobilisation by cytokinin helps in the delay of leaf senescence.
Final answer: Cytokinin
Q109Single correctAnimal Kingdom
While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelome of that animal?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pseudocoelomate
Approach:
The coelom type is inferred from the distribution of mesoderm around the body cavity.
Step 1:In pseudocoelomates, the body cavity is not entirely lined with mesoderm; instead, mesodermal tissue is present along the body wall but not towards the gut.
Step 2:Schizocoelomates have a coelom or body cavity that develops from a split in the mesoderm with the middle germ layer of the embryo, in acoelomates the coelom is absent, and spongocoel is a central cavity found in sponges.
Final answer: Pseudocoelome
Q110Single correctHuman Reproduction
Match List - I with List - II.
| List - I | List - II |
|---|---|
| A. Head | i. Enzymes |
| B. Middle piece | ii. Sperm motility |
| C. Acrosome | iii. Energy |
| D. Tail | iv. Genetic material |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-iv, B-iii, C-i, D-ii
Approach:
Each region of the sperm in List-I is matched with its primary function in List-II.
Step 1:The sperm head contains an elongated nucleus which possesses the genetic material, and the middle piece possesses numerous mitochondria which produce energy for movement.
Step 2:The acrosome is a cap-like structure filled with enzymes that help in fertilisation of the ovum, and the tail facilitates sperm motility essential for fertilisation.
Final answer: A-iv, B-iii, C-i, D-ii
Q111Single correctPlant Kingdom
Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence.
A. Prothallus stage
B. Meiosis in spore mother cells
C. Fertilisation
D. Formation of archegonia and antheridia in gametophyte.
E. Transfer of antherozoids to the archegonia in presence of water.
Choose the correct answer from the options given below:
A. Prothallus stage
B. Meiosis in spore mother cells
C. Fertilisation
D. Formation of archegonia and antheridia in gametophyte.
E. Transfer of antherozoids to the archegonia in presence of water.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1B, A, D, E, C
Approach:
The pteridophyte life cycle stages are ordered from spore formation through fertilisation.
Step 1:Meiosis in spore mother cells precedes the prothallus stage, which is followed by formation of archegonia and antheridia in the gametophyte.
Step 2:Transfer of antherozoids to the archegonia in presence of water then leads to fertilisation, giving the sequence B, A, D, E, C.
Final answer: B, A, D, E, C
Q112Single correctBody Fluids and Circulation
Cardiac activities of the heart are regulated by:
A. Nodal tissue
B. A special neural centre in the medulla oblongata
C. Adrenal medullary hormones
D. Adrenal cortical hormones
Choose the correct answer from the options given below :
A. Nodal tissue
B. A special neural centre in the medulla oblongata
C. Adrenal medullary hormones
D. Adrenal cortical hormones
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B and C Only
Approach:
The factors that regulate cardiac activity are evaluated for their role in controlling heart function.
Step 1:Normal cardiac activities of the heart are regulated intrinsically by specialised muscles, the nodal tissue, while a special neural centre in the medulla oblongata moderates cardiac function through the autonomic nervous system.
Step 2:Adrenal medullary hormones can also increase cardiac output, so A, B and C regulate cardiac activity, whereas adrenal cortical hormones do not.
Final answer: A, B and C Only
Q113Single correctMicrobes in Human Welfare
Which of following organisms cannot fix nitrogen?
A. Azotobacter
B. Oscillatoria
C. Anabaena
D. Volvox
E. Nostoc
Choose the correct answer from the options given below:
A. Azotobacter
B. Oscillatoria
C. Anabaena
D. Volvox
E. Nostoc
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2D only
Approach:
Each listed organism is checked for its nitrogen-fixing capability.
Step 1:Azotobacter, Oscillatoria, Anabaena and Nostoc can all fix atmospheric nitrogen.
Step 2:Volvox is a colonial green alga that cannot fix nitrogen, so D is the organism that cannot fix nitrogen.
Final answer: D only
Q114Single correctMolecular Basis of Inheritance
Given below are two statements :
Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA.
Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA.
Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but statement II is correct
Approach:
Each statement is evaluated against the established roles of tRNA, rRNA and the RNA interference pathway.
Step 1:Transfer RNAs read the codons of mRNA during translation and ribosomal RNA forms part of the ribosome that binds and moves along mRNA. Both interact with mRNA, so statement I is incorrect.
Step 2:RNA interference is a cellular defence mechanism present in all eukaryotic organisms in which complementary RNA silences specific mRNA. Statement II is correct.
Step 3:Statement I is incorrect and statement II is correct, which matches option 4.
Final answer: Statement I is incorrect but statement II is correct
Q116Single correctBiotechnology and its Applications
Which of the following genetically engineered organism was used by Eli Lilly to prepare human insulin?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Bacterium
Approach:
The organism is identified from the documented method used by Eli Lilly to manufacture human insulin in 1983.
Step 1:In 1983 Eli Lilly prepared two DNA sequences corresponding to the A and B chains of human insulin.
Step 2:These sequences were introduced into plasmids of Escherichia coli, a gram-negative bacterium, to produce the insulin chains.
Step 3:The host organism is a bacterium, matching option 1.
Final answer: Bacterium
Q117Single correctBiomolecules
Name the class of enzyme that usually catalyze the following reaction :
Where, a group other than hydrogen
a substrate
another substrate
Where, a group other than hydrogen
a substrate
another substrate
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Transferase
Approach:
The reaction type is matched to the enzyme class defined by the operation it performs.
Step 1:The reaction moves a group G, which is not hydrogen, from one substrate to another.
Step 2:Enzymes catalysing the transfer of a group other than hydrogen between substrates belong to the class transferases.
Step 3:Hydrolases cleave bonds with water, lyases add or remove groups to form double bonds without hydrolysis, and ligases join two molecules using ATP, none of which describes this group transfer.
Final answer: Transferase
Q118Single correctAnatomy of Flowering Plants
Find the statement that is NOT correct with regard to the structure of monocot stem.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hypodermis is parenchymatous.
Approach:
Each described feature is compared with the known anatomy of a typical monocot stem.
Step 1:In a monocot stem the hypodermis is sclerenchymatous, not parenchymatous, so option 1 states an incorrect feature.
Step 2:The vascular bundles in a monocot stem are scattered and are conjoint and closed, and phloem parenchyma is absent, so options 2, 3 and 4 are correct statements.
Step 3:The statement that is not correct is the parenchymatous hypodermis, matching option 1.
Final answer: Hypodermis is parenchymatous.
Q119Single correctPlant Kingdom
The correct sequence of events in the life cycle of bryophytes is
A. Fusion of antherozoid with egg.
B. Attachment of gametophyte to substratum.
C. Reduction division to produce haploid spores.
D. Formation of sporophyte.
E. Release of antherozoids into water.
Choose the correct answer from the options given below :
A. Fusion of antherozoid with egg.
B. Attachment of gametophyte to substratum.
C. Reduction division to produce haploid spores.
D. Formation of sporophyte.
E. Release of antherozoids into water.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3B, E, A, D, C
Approach:
The events are arranged in the order they occur during the bryophyte life cycle.
Step 1:The gametophyte first attaches to the substratum (B), then antherozoids are released into water (E).
Step 2:Fusion of antherozoid with the egg (A) follows, producing the zygote that forms the sporophyte (D).
Step 3:Reduction division in the sporophyte produces haploid spores (C). The order is B, E, A, D, C, matching option 3.
Final answer: B, E, A, D, C
Q120Single correctBiotechnology and its Applications
Which are correct:
A. Computed tomography and magnetic resonance imaging detect cancers of internal organs.
B. Chemotherapeutics drugs are used to kill non-cancerous cells.
C. -interferon activate the cancer patients' immune system and helps in destroying the tumour.
D. Chemotherapeutic drugs are biological response modifiers.
E. In the case of leukaemia blood cell counts are decreased.
Choose the correct answer from the options given below:
A. Computed tomography and magnetic resonance imaging detect cancers of internal organs.
B. Chemotherapeutics drugs are used to kill non-cancerous cells.
C. -interferon activate the cancer patients' immune system and helps in destroying the tumour.
D. Chemotherapeutic drugs are biological response modifiers.
E. In the case of leukaemia blood cell counts are decreased.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A and C only
Approach:
Each statement is judged against established facts of cancer detection and treatment.
Step 1:Computed tomography and magnetic resonance imaging are used to detect cancers of internal organs, so statement A is correct.
Step 2:Alpha-interferon activates the immune system of cancer patients and helps destroy the tumour, so statement C is correct.
Step 3:Chemotherapeutic drugs kill cancerous cells, not non-cancerous cells, so B is incorrect. Alpha-interferons, not chemotherapeutic drugs, are biological response modifiers, so D is incorrect. In leukaemia blood cell counts are increased, so E is incorrect. Only A and C are correct, matching option 4.
Final answer: A and C only
Q121Single correctCell: The Unit of Life
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A. Centromere | I. Mitochondrion |
| B. Cilium | II. Cell division |
| C. Cristae | III. Cell movement |
| D. Cell membrane | IV. Phospholipid Bilayer |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-II, B-III, C-I, D-IV
Approach:
Each structure in List-I is matched to its associated function or location in List-II.
Step 1:The centromere helps in cell division, so A pairs with II.
Step 2:The cilium helps in cell movement, so B pairs with III, and cristae are the finger-like structures of the mitochondrion, so C pairs with I.
Step 3:The cell membrane is a phospholipid bilayer, so D pairs with IV. The combination A-II, B-III, C-I, D-IV matches option 4.
Final answer: A-II, B-III, C-I, D-IV
Q122Single correctPhotosynthesis in Higher Plants
Choose the option with all correct matches.
| List-I | List-II |
|---|---|
| A. Chlorophyll a | I. Yellow-green |
| B. Chlorophyll b | II. Yellow |
| C. Xanthophylls | III. Blue-green |
| D. Carotenoids | IV. Yellow to Yellow-orange |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-I, C-II, D-IV
Approach:
Each photosynthetic pigment is matched to its characteristic colour on a chromatogram.
Step 1:Chlorophyll a appears bright or blue-green, so A pairs with III, and chlorophyll b appears yellow-green, so B pairs with I.
Step 2:Xanthophylls appear yellow, so C pairs with II, and carotenoids appear yellow to yellow-orange, so D pairs with IV.
Step 3:The combination A-III, B-I, C-II, D-IV matches option 2.
Final answer: A-III, B-I, C-II, D-IV
Q123Single correctHuman Reproduction
Find the correct statement :
(A) In human pregnancy, the major organ systems are formed at the end of 12 weeks.
(B) In human pregnancy the major organ systems are formed at the end of 8 weeks.
(C) In human pregnancy heart is formed after one month of gestation.
(D) In human pregnancy, limbs and digits develop by the end of second month.
(E) In human pregnancy the appearance of hair is usually observed in the fifth month.
Choose the correct answer from the options given below :
(A) In human pregnancy, the major organ systems are formed at the end of 12 weeks.
(B) In human pregnancy the major organ systems are formed at the end of 8 weeks.
(C) In human pregnancy heart is formed after one month of gestation.
(D) In human pregnancy, limbs and digits develop by the end of second month.
(E) In human pregnancy the appearance of hair is usually observed in the fifth month.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A, C, D and E only
Approach:
Each statement is checked against the NCERT timeline of human embryonic and foetal development.
Step 1:Statement A is correct: by the end of 12 weeks (first trimester) most of the major organ systems are formed. Statement B is therefore incorrect, since the major organ systems are not complete by the end of 8 weeks.
Step 2:Statement C is correct: the heart is formed after the first month of gestation. Statement D is correct: limbs and digits develop by the end of the second month.
Step 3:Statement E is correct: the appearance of hair on the head is usually observed by the fifth month. The combination of correct statements is A, C, D and E.
Final answer: A, C, D and E only
Q124Single correctSexual Reproduction in Flowering Plants
In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Aleurone layer
Approach:
The named layer is identified from the structure of a monocot cereal seed.
Step 1:In monocot seeds the outer covering of the endosperm is a separate proteinaceous layer that lies next to the embryo.
Step 2:This protein-rich layer is the aleurone layer.
Step 3:The layer separating the embryo by a protein-rich covering is the aleurone layer, matching option 4.
Final answer: Aleurone layer
Q125Single correctExcretory Products and their Elimination
Which of the following diagrams is correct with regard to the proximal (P) and distal (D) tubule of the Nephron.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Proximal tubule secreting H and N while reabsorbing HCO, NaCl and O; distal tubule secreting K and H (drawn diagram, option 2)
Approach:
The diagram is selected by matching the directions of secretion and reabsorption in the proximal and distal tubules.
Step 1:During urine formation the tubular cells secrete hydrogen ions, potassium ions and ammonia into the filtrate, maintaining the ionic and acid-base balance of body fluids.
Step 2:The proximal convoluted tubule selectively secretes hydrogen ions, ammonia and potassium ions into the filtrate.
Step 3:The distal convoluted tubule is capable of reabsorption of bicarbonate and selective secretion of hydrogen ions, potassium ions and ammonia. The diagram matching these directions is option 2.
Final answer: Proximal tubule secreting H and N while reabsorbing HCO, NaCl and O; distal tubule secreting K and H (drawn diagram, option 2)
Q126Single correctBiotechnology: Principles and Processes
Identify the part of a bio-reactor which is used as a foam breaker from the given figure.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C
Approach:
The labelled parts of the bioreactor are identified from their positions in the figure.
Step 1:In the bioreactor figure, B is the motor, A is the flat-bladed impeller and D supplies sterile air.
Step 2:The part labelled C is the foam breaker, positioned above the impeller.
Step 3:The foam breaker corresponds to part C, matching option 4.
Final answer: C
Q127Single correctSexual Reproduction in Flowering Plants
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled.
Reason (R) : The egg apparatus has 2 polar nuclei.
In the light of the above statements, choose the correct answer from the options given below :
Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled.
Reason (R) : The egg apparatus has 2 polar nuclei.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A is true but R is false
Approach:
The assertion and reason are each tested against the structure of a mature angiosperm embryo sac.
Step 1:A typical mature angiosperm embryo sac is 7-celled and 8-nucleate, so the assertion is true.
Step 2:The polar nuclei are situated in the large central cell, not in the egg apparatus. The egg apparatus consists of three cells grouped at the micropylar end, so the reason is false.
Step 3:The assertion is true while the reason is false, matching option 3.
Final answer: A is true but R is false
Q128Single correctCell: The Unit of Life
A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Mesosome
Approach:
The structure is identified from its membranous nature and its functions in a prokaryotic cell.
Step 1:The mesosome is a membranous extension of the plasma membrane in a bacterial cell.
Step 2:It helps in cell wall formation, DNA replication and contains enzymes for respiration.
Step 3:The described structure is the mesosome, matching option 1.
Final answer: Mesosome
Q129Single correctMolecular Basis of Inheritance
Which of the following are the post-transcriptional events in an eukaryotic cell?
A. Transport of pre-mRNA to cytoplasm prior to splicing.
B. Removal of introns and joining of exons.
C. Addition of methyl group at 5' end of hnRNA.
D. Addition of adenine residues at 3' end of hnRNA.
E. Base pairing of two complementary RNAs.
Choose the correct answer from the options given below :
A. Transport of pre-mRNA to cytoplasm prior to splicing.
B. Removal of introns and joining of exons.
C. Addition of methyl group at 5' end of hnRNA.
D. Addition of adenine residues at 3' end of hnRNA.
E. Base pairing of two complementary RNAs.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, C, D only
Approach:
Each listed event is classified as transcriptional or post-transcriptional.
Step 1:Transport of pre-mRNA to the cytoplasm prior to splicing is part of transcription, so A is excluded.
Step 2:Post-transcriptional processing of the primary transcript into functional mRNA involves capping by addition of a methyl group at the 5' end (C), tailing by addition of adenine residues at the 3' end (D), and splicing by removal of introns and joining of exons (B).
Step 3:Base pairing of two complementary RNAs is not an event of post-transcription, so E is excluded. The post-transcriptional events are B, C and D, matching option 2.
Final answer: B, C, D only
Q130Single correctPrinciples of Inheritance and Variation
What is the pattern of inheritance for polygenic trait?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Non-mendelian inheritance pattern
Approach:
The inheritance pattern is deduced from the genetic basis of polygenic traits.
Step 1:Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes.
Step 2:When human disorders are determined by mutation in a single gene, they are transmitted to the offspring as per the Mendelian principle.
Step 3:A polygenic trait, controlled by many genes, shows a non-Mendelian inheritance pattern, matching option 2.
Final answer: Non-mendelian inheritance pattern
Q131Single correctBiomolecules
Which one of the following enzymes contains 'Haem' as the prosthetic group?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Catalase
Approach:
The enzyme is identified by which one uses haem as its tightly bound prosthetic group.
Step 1:In peroxidase and catalase, which catalyse the breakdown of hydrogen peroxide into water and oxygen, haem is the prosthetic group and is part of the active site of the enzyme.
Step 2:Zinc is the cofactor in carbonic anhydrase, and RuBisCo is the most abundant protein in the biosphere; succinate dehydrogenase is a different oxidoreductase.
Step 3:The enzyme containing haem as the prosthetic group is catalase, matching option 4.
Final answer: Catalase
Q132Single correctBiological Classification
Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization.
A. Multicellular heterotrophs with cell wall made of chitin.
B. Heterotrophs with tissue/organ/organ system level of body organization.
C. Prokaryotes with cell wall made of polysaccharides and amino acids.
D. Eukaryotic autotrophs with tissue/organ level of body organization.
E. Eukaryotes with cellular body organization.
Choose the correct answer from the options given below :
A. Multicellular heterotrophs with cell wall made of chitin.
B. Heterotrophs with tissue/organ/organ system level of body organization.
C. Prokaryotes with cell wall made of polysaccharides and amino acids.
D. Eukaryotic autotrophs with tissue/organ level of body organization.
E. Eukaryotes with cellular body organization.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2C, E, A, D, B
Approach:
The five characteristics are arranged from the simplest to the most complex body organisation across Whittaker's kingdoms.
Step 1:Kingdom Monera comprises prokaryotes with cell wall made of polysaccharides and amino acids, the simplest organisation, so C comes first, followed by Protista, the unicellular eukaryotes with cellular organisation, so E comes second.
Step 2:Kingdom Fungi comprises multicellular heterotrophs with cell wall made of chitin, so A is third, then Plantae, eukaryotic autotrophs with tissue or organ level body organisation, so D is fourth.
Step 3:Kingdom Animalia comprises heterotrophs with tissue, organ and organ system level of body organisation, the most complex, so B is last. The sequence is C, E, A, D, B, matching option 2.
Final answer: C, E, A, D, B
Q133Single correctOrganisms and Populations
Who is known as the father of Ecology in India?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ramdeo Misra
Approach:
The figure is identified from the recognised title in Indian ecology.
Step 1:Ramdeo Misra is recognised for foundational contributions to ecological research and education in India.
Step 2:Ramdeo Misra is known as the father of Ecology in India.
Step 3:The correct choice is Ramdeo Misra, matching option 2.
Final answer: Ramdeo Misra
Q134Single correctMolecular Basis of Inheritance
Choose the correct answer from the options given below:
| List-I | List-II |
|---|---|
| A. Alfred Hershey and Martha Chase | I. Streptococcus pneumoniae |
| B. Euchromatin | II. Densely packed and dark-stained |
| C. Frederick Griffith | III. Loosely packed and light-stained |
| D. Heterochromatin | IV. DNA as genetic material confirmation |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-IV, B-III, C-I, D-II
Approach:
Each item in List-I is matched to its associated discovery or property in List-II.
Step 1:Alfred Hershey and Martha Chase gave unequivocal proof that DNA is the genetic material, pairing A with the confirmation of DNA as genetic material.
Step 2:Euchromatin is the lightly stained region with loosely packed chromatin fibre, and heterochromatin is the darkly stained region with tightly packed chromatin fibre, pairing B with loosely packed light-stained and D with densely packed dark-stained.
Step 3:Frederick Griffith performed his transformation experiments on different strains of Streptococcus pneumoniae, pairing C with Streptococcus pneumoniae and completing the set as A-IV, B-III, C-I, D-II.
Final answer: A-IV, B-III, C-I, D-II
Q135Single correctCell Cycle and Cell Division
Neoplastic characteristics of cells refer to :
A. A mass of proliferating cell
B. Rapid growth of cells
C. Invasion and damage to the surrounding tissue
D. Those confined to original location
Choose the correct answer from the options given below:
A. A mass of proliferating cell
B. Rapid growth of cells
C. Invasion and damage to the surrounding tissue
D. Those confined to original location
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A, B, C only
Approach:
Each characteristic is tested against the defining features of neoplastic, cancerous cells.
Step 1:A neoplasm is any abnormal growth of tissue, so neoplastic cells form a mass of proliferating cells (A) and show rapid growth of cells (B).
Step 2:Cancerous neoplasms are invasive and cause invasion and damage to the surrounding tissue (C), so C is included.
Step 3:Cells confined to their original location (D) describe benign tumours, not the malignant neoplastic characteristics, so D is excluded. The correct set is A, B and C, matching option 2.
Final answer: A, B, C only
Q136Single correctBiotechnology: Principles and Processes
Given below are two statements :
Statement I : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA.
Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA.
Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both statement I and statement II are correct
Approach:
Each statement is evaluated against the principle of agarose gel electrophoresis and its use in recombinant DNA technology.
Step 1:Restriction-digested DNA fragments separated on an agarose gel can be cut out, eluted from the gel piece, and joined to a cloning vector, so they serve as material for constructing recombinant DNA.
Step 2:In gel electrophoresis the DNA, being negatively charged, moves from the wells at the cathode towards the anode; smaller fragments migrate farther and lie nearer the anode, while larger fragments stay closer to the wells.
Step 3:Both statements describe valid features of the technique.
Final answer: Both statement I and statement II are correct
Q137Single correctMolecular Basis of Inheritance
Choose the option with all correct matches.
| List-I | List-II |
|---|---|
| A.. Adenosine | I.. Nitrogen base |
| B.. Adenylic acid | II.. Nucleotide |
| C.. Adenine | III.. Nucleoside |
| D.. Alanine | IV.. Amino acid |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A-III, B-II, C-I, D-IV
Approach:
Each entry of List I is classified by its chemical composition and matched to the correct biochemical category in List II.
Step 1:Adenosine is a nucleoside, being composed of the nitrogen base adenine joined to a sugar.
Step 2:Adenylic acid is a nucleotide, being composed of a nitrogen base, a sugar and a phosphate group.
Step 3:Adenine is a purine nitrogen base.
Step 4:Alanine is an amino acid that carries a methyl group as its side chain.
Final answer: A-III, B-II, C-I, D-IV
Q138Single correctHuman Reproduction
Consider the following :
A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis.
B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females.
C. The first polar body is associated with the formation of the primary oocyte.
D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.
Choose the correct answer from the options given below:
A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis.
B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females.
C. The first polar body is associated with the formation of the primary oocyte.
D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A and B are true
Approach:
Each statement on human gametogenesis and the menstrual cycle is checked against established reproductive physiology.
Step 1:In the female, meiosis I begins in foetal life and is arrested, so the reductive division starts earlier than in the male where it begins at puberty.
Step 2:In males the two meiotic divisions proceed continuously, so the interval between meiosis I and meiosis II is much shorter than in females where the oocyte is arrested for years.
Step 3:The first polar body is formed during the conversion of the secondary oocyte; the primary oocyte is formed before the first meiotic division.
Step 4:The LH surge triggers ovulation. Disintegration of the endometrium and menstrual bleeding follow the fall in progesterone during the luteal phase, not the LH surge.
Final answer: A and B are true
Q139Single correctAnimal Kingdom
All living members of the class Cyclostomata are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ectoparasite
Approach:
The mode of life of living cyclostomes is recalled from the characters of the class.
Step 1:Living members of Cyclostomata, such as lampreys, attach to the body of fishes and feed on their blood and tissues, living as external parasites.
Final answer: Ectoparasite
Q140Single correctCell - The Unit of Life
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell.
Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus.
In the light of the above statements, choose the correct answer from the options given below :
Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell.
Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus.
In the light of the above statements, choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are true and R is the correct explanation of A
Approach:
The truth of the Assertion and the Reason is established from NCERT, then it is determined whether the Reason explains the mechanism stated in the Assertion.
Step 1:The Assertion is correct: the primary function of the Golgi apparatus is to package materials synthesised by the endoplasmic reticulum and to deliver them to intracellular targets and to the exterior of the cell.
Step 2:The Reason is correct: vesicles carrying endoplasmic-reticulum products fuse with the cis (forming) face of the Golgi, are chemically modified within the cisternae, and are dispatched from the trans (maturing) face.
Step 3:The cis-to-trans transit described in the Reason is the exact mechanism by which the Golgi packages, modifies and delivers endoplasmic-reticulum products, so the Reason is the correct explanation of the Assertion.
Final answer: Both A and R are true and R is the correct explanation of A
Q141Single correctSexual Reproduction in Flowering Plants
Choose the option with all correct matches.
| List I | List II |
|---|---|
| A.. Scutellum | I.. Persistent nucellus |
| B.. Non-albuminous seed | II.. Cotyledon of Monocot seed |
| C.. Epiblast | III.. Groundnut |
| D.. Perisperm | IV.. Rudimentary cotyledon |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-II, B- III, C-IV, D-I
Approach:
Each seed structure in List I is matched with its correct description or example in List II.
Step 1:Scutellum is the single shield-shaped cotyledon of a monocot seed.
Step 2:Groundnut seed is non-albuminous, lacking residual endosperm at maturity.
Step 3:Epiblast is a rudimentary cotyledon present in monocot seeds.
Step 4:Perisperm is persistent nucellus retained in the mature seed.
Final answer: A-II, B- III, C-IV, D-I
Q142Single correctAnimal Kingdom
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate.
Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults.
In the light of the above statements, choose the correct answer from the options given below:
Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate.
Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults.
In the light of the above statements, choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both (A) and (R) are true and (R) is the correct explanation of (A)
Approach:
The assertion and reason on chordate and vertebrate relationship are tested for truth and for an explanatory link.
Step 1:All vertebrates are chordates, but chordates also include subphyla without a vertebral column, so not all chordates are vertebrates; the assertion is correct.
Step 2:Members of subphylum Vertebrata possess a notochord during the embryonic period that is replaced by a cartilaginous or bony vertebral column in the adult; the reason is correct.
Step 3:The defining vertebral column that separates vertebrates from other chordates explains why all vertebrates are chordates while all chordates are not vertebrates, so the reason explains the assertion.
Final answer: Both (A) and (R) are true and (R) is the correct explanation of (A)
Q143Single correctHuman Health and Disease
Identify the statement that is NOT correct.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Antigen binding site is located at C-terminal region of antibody molecules.
Approach:
Each statement about antibody structure is checked, and the incorrect one is identified.
Step 1:An antibody molecule has four peptide chains, two small light chains and two longer heavy chains, represented as , so statement (1) is correct.
Step 2:The heavy and light chains are held together by disulfide bonds, so statement (2) is correct.
Step 3:The antigen binding site lies in the variable region at the N-terminal end of the chains, not at the C-terminal region, so statement (3) is the one that is not correct.
Step 4:The constant regions of the heavy and light chains lie at the C-terminus, so statement (4) is correct.
Final answer: Antigen binding site is located at C-terminal region of antibody molecules.
Q144Single correctBiotechnology and its Applications
Silencing of specific mRNA is possible via RNAi because of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Complementary dsRNA
Approach:
The molecular basis of RNA interference is recalled to find the agent that silences a specific mRNA.
Step 1:RNA interference occurs in all eukaryotic organisms as a defence mechanism in which a double-stranded RNA complementary to the target mRNA binds it and prevents its translation, thereby silencing that specific mRNA.
Final answer: Complementary dsRNA
Q145Single correctPrinciples of Inheritance and Variation
Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Phenotypic ratio - 9 : 3 : 3 : 1
Approach:
A Mendelian dihybrid cross with two independently assorting gene pairs is analysed for its F2 phenotypic ratio.
Step 1:A cross between pure round yellow seeded pea plants (RRYY) and wrinkled green seeded plants (rryy) gives an F1 in which yellow colour is dominant over green and round seed shape is dominant over wrinkled seed shape.
Step 2:Selfing the dihybrid F1 with independent assortment of the two gene pairs distributes the four phenotype classes in the ratio 9 : 3 : 3 : 1.
Final answer: Phenotypic ratio - 9 : 3 : 3 : 1
Q146Single correctMolecular Basis of Inheritance
Histones are enriched with -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Lysine & Arginine
Approach:
The amino acid composition of histone proteins is recalled to identify the residues they are rich in.
Step 1:In eukaryotes, packaging of DNA is complex and involves a set of positively charged basic proteins called histones, which carry a positive charge.
Step 2:Histones are organised to form a unit of eight molecules called the histone octamer and are rich in the basic amino acid residues lysine and arginine, which provide the positive charge.
Final answer: Lysine & Arginine
Q147Single correctHuman Reproduction
The first menstruation is called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Menarche
Approach:
The term for the first menstrual flow at puberty is recalled and distinguished from related reproductive terms.
Step 1:The first menstruation begins at puberty and is called menarche.
Step 2:Ovulation is the release of the secondary oocyte from the mature Graafian follicle, and menopause is the cessation of menstrual cycles around fifty years of age, so neither names the first menstruation.
Final answer: Menarche
Q148Single correctChemical Coordination and Integration
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. Heart | I.. Erythropoietin |
| B.. Kidney | II.. Aldosterone |
| C.. Gastro-intestinal tract | III.. Atrial natriuretic factor |
| D.. Adrenal Cortex | IV.. Secretin |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A-III, B-I, C-IV, D-II
Approach:
Each organ in List-I is matched with the hormone or factor it secretes in List-II.
Step 1:The heart secretes atrial natriuretic factor.
Step 2:The kidney secretes erythropoietin.
Step 3:The gastro-intestinal tract secretes secretin.
Step 4:The adrenal cortex secretes aldosterone.
Final answer: A-III, B-I, C-IV, D-II
Q149Single correctBiomolecules
The protein portion of an enzyme is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Apoenzyme
Approach:
The term for the protein component of an enzyme bound to a cofactor is recalled.
Step 1:In many enzymes a non-protein constituent called a cofactor binds to the enzyme to make it catalytically active.
Step 2:In these instances the protein portion of the enzyme is called the apoenzyme.
Step 3:Prosthetic groups and coenzymes are organic cofactors and metal ions are inorganic cofactors, so none of them is the protein part of the enzyme.
Final answer: Apoenzyme
Q150Single correctEcosystem
Which of the following is the unit of productivity of an Ecosystem?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The dimensional unit of ecosystem productivity, which is a rate per unit area, is identified.
Step 1:The rate of biomass production is called productivity.
Step 2:It is expressed per unit area per unit time, in terms of or , to compare the productivity of different ecosystems.
Final answer:
Q151Single correctEvolution
Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Analogy, convergent
Approach:
The structural origin and function of sweet potato and potato are compared to classify them as homologous or analogous and to name the evolutionary pattern.
Step 1:Sweet potato is a modified root while potato is a modified stem, yet both store food and perform the same function, so they are analogous structures that are not anatomically similar though they perform similar functions.
Step 2:Analogous structures arise from convergent evolution, in which different structures come to perform similar functions.
Final answer: Analogy, convergent
Q152Single correctPrinciples of Inheritance and Variation
With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in generation.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11/4
Approach:
The pedigree is read to establish the mode of inheritance and the genotypes of the F2 parents, then the proportion of carrier, disease-free offspring in F3 is computed.
Step 1:In the F2 generation a carrier female and a non-affected normal male produce an affected male child, which indicates that the disorder is sex-linked recessive.
Step 2:The consanguineous mating between the carrier female and the normal male yields offspring , , and XY in equal proportions.
Step 3:Out of the four offspring only one is a disease-free carrier, giving a probability of one in four.
Final answer: 1/4
Q153Single correctSexual Reproduction in Flowering Plants
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.
Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.
Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both A and R are true and R is the correct explanation of A
Approach:
The truth of the Assertion and Reason is established from the NCERT description of the tapetum, then the explanatory link is assessed.
Step 1:The Assertion is correct: the innermost wall layer of the microsporangium, the tapetum, is made of cells with dense cytoplasm that generally contain more than one nucleus.
Step 2:The Reason states that the multinucleate condition of the tapetum increases the efficiency of nourishing the developing cells within the anther. The dense, multinucleate tapetal cytoplasm provides the metabolic capacity for this nutritive function.
Step 3:The nutritive role attributed to the multinucleate tapetum in the Reason accounts for the dense-cytoplasm, multinucleate condition stated in the Assertion, so the Reason explains the Assertion.
Final answer: Both A and R are true and R is the correct explanation of A
Q154Single correctSexual Reproduction in Flowering Plants
How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 31 Meiosis and 3 Mitosis
Approach:
The sequence of divisions from the megaspore mother cell to the mature embryo sac is counted.
Step 1:The megaspore mother cell undergoes one meiotic division to form the functional megaspore.
Step 2:Development of a mature female gametophyte, the embryo sac, from the functional megaspore then requires three mitotic divisions to give the eight-nucleate stage.
Final answer: 1 Meiosis and 3 Mitosis
Q155Single correctMorphology of Flowering Plants
Which of the following is an example of a zygomorphic flower?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Pea
Approach:
The definition of a zygomorphic flower is applied to the listed plants.
Step 1:Zygomorphic flowers can be divided into two equal halves by only a single vertical plane and show bilateral symmetry.
Step 2:Pea possesses zygomorphic flowers, whereas chilli, petunia and datura possess actinomorphic flowers.
Final answer: Pea
Q156Single correctHuman Health and Disease
After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s) / tissue(s) like
A. thymus
B. bone marrow
C. spleen
D. lymph nodes
E. Peyer's patches
Choose the correct answer from the options given below
A. thymus
B. bone marrow
C. spleen
D. lymph nodes
E. Peyer's patches
Choose the correct answer from the options given below
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4C, D, E only
Approach:
The given organs are sorted into primary and secondary lymphoid organs to find the destinations of mature lymphocytes.
Step 1:The primary lymphoid organs are the bone marrow and the thymus, where immature lymphocytes differentiate into antigen-sensitive lymphocytes.
Step 2:After maturation the lymphocytes migrate to the secondary lymphoid organs, namely the spleen, lymph nodes, Peyer's patches of the small intestine, and the appendix.
Step 3:These secondary lymphoid organs provide the sites where lymphocytes meet the antigen, so the spleen, lymph nodes and Peyer’s patches are the organs asked for.
Final answer: C, D, E only
Q157Single correctOrganisms and Populations
Given below are two statements :
Statement I : Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it.
Statement II : Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it.
Statement II : Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both statement I and statement II are incorrect
Approach:
Each statement on the fig and fig wasp relationship is checked against the biology of fig pollination and the nature of the fig fruit.
Step 1:The fig is a vegetarian fruit as it is only pollinated by wasp, so statement I is incorrect.
Step 2:The fig tree and fig wasp show mutualism in which both species are benefitted, but statement II is not correct as the fig inflorescence and flower get pollinated by the fig wasp rather than the fruit, so statement II is incorrect.
Final answer: Statement I and statement II are incorrect
Q158Single correctCell Cycle and Cell Division
What is the main function of the spindle fibers during mitosis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1To separate the chromosomes
Approach:
The role of spindle fibres during mitosis is recalled from the mechanism of chromosome movement.
Step 1:During mitosis, spindle fibres get attached to the kinetochores of the chromosomes and help in the separation of the chromosomes towards opposite poles.
Final answer: To separate the chromosomes
Q159Single correctPlant Kingdom
Which one of the following is the characteristic feature of gymnosperms?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Seeds are naked
Approach:
Gymnosperms are seed plants whose ovules are not enclosed within an ovary wall, so the seeds remain exposed.
Step 1:The term gymnosperm (Greek gymnos, naked; sperma, seed) describes plants in which the ovules are borne on the surface of megasporophylls rather than inside a closed ovary.
Step 2:Since the ovules are not enclosed, the seeds that develop after fertilisation are also exposed and uncovered.
Final answer: Seeds are naked
Q160Single correctChemical Coordination and Integration
Consider the following statements regarding function of adrenal medullary hormones :
(A) It causes pupilary constriction.
(B) It is a hyperglycemic hormone.
(C) It causes piloerection.
(D) It increases strength of heart contraction.
Choose the correct answer from the options given below :
(A) It causes pupilary constriction.
(B) It is a hyperglycemic hormone.
(C) It causes piloerection.
(D) It increases strength of heart contraction.
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2B, C and D only
Approach:
The adrenal medulla secretes adrenaline and noradrenaline, catecholamines that mediate emergency (fight-or-flight) responses.
Step 1:These hormones dilate the pupil rather than constrict it, so statement A is incorrect.
Step 2:They stimulate breakdown of glycogen, raising blood glucose, which makes them hyperglycemic; statement B is correct.
Step 3:They cause piloerection (raising of hairs) and increase the strength of heart contraction, so statements C and D are correct.
Final answer: B, C and D only
Q161Single correctChemical Coordination and Integration
Why can't insulin be given orally to diabetic patients?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2It will be digested in Gastro-Intestinal (GI) tract
Approach:
Insulin is a peptide (protein) hormone, and proteins taken by mouth are broken down by digestive enzymes.
Step 1:Insulin is a proteinaceous molecule made of amino acid chains.
Step 2:Proteases in the gastro-intestinal tract hydrolyse the peptide bonds of an orally administered protein, destroying its activity before absorption.
Final answer: It will be digested in Gastro-Intestinal (GI) tract
Q162Single correctPlant Kingdom
Choose the option with all correct matches.
| List-I | List-II |
|---|---|
| A.. Pteridophyte | I.. |
| B.. Bryophyte | II.. |
| C.. Angiosperm | III.. |
| D.. Gymnosperm | IV.. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-IV, B-III, C-I, D-II
Approach:
Each plant group is paired with its representative genus from the plant kingdom.
Step 1:Salvinia is an aquatic fern, hence a pteridophyte, giving A-IV.
Step 2:Polytrichum is a moss, hence a bryophyte, giving B-III.
Step 3:Salvia is a flowering plant, hence an angiosperm, giving C-I; Ginkgo is a naked-seeded plant, hence a gymnosperm, giving D-II.
Final answer: A-IV, B-III, C-I, D-II
Q163Single correctMolecular Basis of Inheritance
Who proposed that the genetic code for amino acids should be made up of three nucleotides?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1George Gamow
Approach:
The proposal that a triplet of bases codes for one amino acid is credited to a physicist who reasoned from the number of bases and amino acids.
Step 1:George Gamow, a physicist, argued that a code of three nucleotides is needed because four bases taken three at a time give sixty-four combinations, enough to specify twenty amino acids.
Final answer: George Gamow
Q164Single correctBiodiversity and Conservation
Choose the option with all correct matches.
| List I | List II |
|---|---|
| A.. The Evil Quartet | I.. Cryopreservation |
| B.. Ex situ conservation | II.. Alien species invasion |
| C.. Lantana\ camara | III.. Causes of biodiversity losses |
| D.. Dodo | IV.. Extinction |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A-III, B-I, C-II, D-IV
Approach:
Each item is paired with the conservation or biodiversity concept it represents.
Step 1:The Evil Quartet refers to the four major causes of biodiversity losses, giving A-III.
Step 2:Ex situ conservation includes techniques such as cryopreservation, giving B-I.
Step 3:Lantana camara is an example of alien species invasion, giving C-II; the Dodo is an example of extinction, giving D-IV.
Final answer: A-III, B-I, C-II, D-IV
Q165Single correctChemical Coordination and Integration
Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Anti-diuretic hormone (ADH)
Approach:
The posterior pituitary stores and releases hormones that are produced in the hypothalamus, while the anterior pituitary synthesises its own hormones.
Step 1:The neurohypophysis (posterior pituitary) stores and releases oxytocin and vasopressin, both manufactured by hypothalamic neurons and transported down axons.
Step 2:The pars distalis (anterior pituitary) itself produces FSH, LH and ACTH, so these are not hypothalamic in origin.
Final answer: Anti-diuretic hormone (ADH)
Q166Single correctAnimal Kingdom
Role of the water vascular system in Echinoderms is :
A. Respiration and Locomotion
B. Excretion and Locomotion
C. Capture and transport of food
D. Digestion and Respiration
E. Digestion and Excretion
Choose the correct answer from the options given below :
A. Respiration and Locomotion
B. Excretion and Locomotion
C. Capture and transport of food
D. Digestion and Respiration
E. Digestion and Excretion
Choose the correct answer from the options given below :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and C Only
Approach:
The water vascular system is a distinctive feature of echinoderms that serves several functions.
Step 1:The water vascular system assists in locomotion, capture and transport of food, and respiration, so statements A and C are correct.
Step 2:An excretory system is absent in echinoderms; excretion occurs through the general body surface, so statements involving excretion or digestion are incorrect.
Final answer: A and C Only
Q167Single correctHuman Health and Disease
Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Innate Immunity
Approach:
Immunity present from birth that provides general, non-specific protection is classified as innate immunity.
Step 1:Innate immunity is non-specific and present at birth; it provides different types of barriers to the entry of foreign agents into the body.
Step 2:Acquired immunity is pathogen-specific and develops after exposure; humoral and cell-mediated immunity are branches of acquired immunity.
Final answer: Innate Immunity
Q168Single correctPlant Kingdom
In bryophytes, the gemmae help in which one of the following?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Asexual reproduction
Approach:
Gemmae are specialized structures of bryophytes used for vegetative propagation.
Step 1:Gemmae are green, multicellular, asexual buds that develop in small receptacles called gemma cups.
Step 2:Each gemma detaches and grows into a new plant, accomplishing asexual reproduction.
Final answer: Asexual reproduction
Q169Single correctBody Fluids and Circulation
In frog, the Renal portal system is a special venous connection that acts to link :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Kidney and lower part of body
Approach:
A portal system carries blood from one capillary bed to another; the renal portal system is defined by the organs it connects.
Step 1:In frogs, a special venous connection exists between the kidney and the lower parts of the body.
Step 2:By contrast, the connection between the liver and the intestine forms the hepatic portal system.
Final answer: Kidney and lower part of body
Q170Single correctEcosystem
Given below are two statements:
Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers.
Statement II: Ecosystems are exempted from 2nd law of thermodynamics.
In the light of the above statements, choose the most appropriate answer from the options given below:
Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers.
Statement II: Ecosystems are exempted from 2nd law of thermodynamics.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Statement I is correct but statement II is incorrect
Approach:
Each statement is evaluated against the principles of energy flow and thermodynamics in ecosystems.
Step 1:The Sun is the only source of energy for all ecosystems on Earth except the deep-sea hydrothermal ecosystem, and this energy flow is unidirectional from the Sun to producers and then to consumers, so Statement A is correct.
Step 2:Ecosystems are not exempted from the second law of thermodynamics; they require a constant supply of energy to synthesise the molecules needed to counteract the universal tendency towards increasing disorderliness, so Statement B is incorrect.
Final answer: Statement I is correct but statement II is incorrect
Q171Single correctPhotosynthesis in Higher Plants
Which of the following statements about RuBisCO is true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4It catalyzes the carboxylation of RuBP
Approach:
RuBisCO is the key enzyme of the Calvin cycle, and its role is identified from its catalytic action.
Step 1:Carboxylation is the most crucial step of the Calvin cycle, in which carbon dioxide is fixed to the acceptor molecule RuBP by the enzyme ribulose bisphosphate carboxylase.
Step 2:The enzyme also has oxygenase activity and a higher affinity for carbon dioxide than for oxygen, so the statement about higher oxygen affinity is false.
Final answer: It catalyses the carboxylation of RuBP
Q172Single correctBiotechnology - Principles and Processes
Which of the following enzyme(s) are NOT essential for gene cloning?
A. Restriction enzymes
B. DNA ligase
C. DNA mutase
D. DNA recombinase
E. DNA polymerase
Choose the correct answer from the options given below:
A. Restriction enzymes
B. DNA ligase
C. DNA mutase
D. DNA recombinase
E. DNA polymerase
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C and D only
Approach:
Gene cloning isolates and replicates a specific gene, and only certain enzymes are required for this process.
Step 1:Gene cloning is a process where a specific gene or DNA sequence is isolated and replicated, creating multiple identical copies.
Step 2:Restriction enzymes, DNA ligase and DNA polymerase are primarily used in gene cloning, whereas DNA mutase and DNA recombinase are not essential.
Final answer: C and D only
Q173Single correctPlant Growth and Development
Read the following statements on plant growth and development.
(A) Parthenocarpy can be induced by auxins.
(B) Plant growth regulators can be involved in promotion as well as inhibition of growth.
(C) Dedifferentiation is a pre-requisite for re-differentiation.
(D) Abscisic acid is a plant growth promoter.
(E) Apical dominance promotes the growth of lateral buds.
Choose the option with all correct statements.
(A) Parthenocarpy can be induced by auxins.
(B) Plant growth regulators can be involved in promotion as well as inhibition of growth.
(C) Dedifferentiation is a pre-requisite for re-differentiation.
(D) Abscisic acid is a plant growth promoter.
(E) Apical dominance promotes the growth of lateral buds.
Choose the option with all correct statements.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A, B, C only
Approach:
Each statement is checked against the established roles of plant growth regulators and developmental processes.
Step 1:Auxins induce parthenocarpy, plant growth regulators both promote and inhibit growth, and dedifferentiation precedes re-differentiation, so statements A, B and C are correct.
Step 2:Abscisic acid is a plant growth inhibitor and an inhibitor of plant metabolism, and apical dominance suppresses rather than promotes the growth of lateral buds, so statements D and E are incorrect.
Final answer: A, B, C only
Q174Single correctMolecular Basis of Inheritance
Which factor is important for termination of transcription?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 (rho)
Approach:
In prokaryotic transcription, distinct accessory factors govern initiation and termination by the RNA polymerase.
Step 1:In prokaryotes the RNA polymerase alone catalyses elongation but associates transiently with an initiation factor, sigma, and a termination factor, rho.
Step 2:The rho factor is therefore responsible for terminating transcription.
Final answer: (rho)
Q175Single correctBreathing and Exchange of Gases
Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs.
Choose the correct answer from the following :
Choose the correct answer from the following :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The statement is false for water but true for land
Approach:
The respiratory surfaces used by frogs differ between aquatic and terrestrial conditions.
Step 1:In water, frogs respire through the skin alone and not through the buccal cavity, so they undergo cutaneous respiration only; the statement is false for water.
Step 2:On land, the buccal cavity, skin and lungs all act as respiratory organs, giving buccopharyngeal, cutaneous and pulmonary respiration, so the statement is true for land.
Final answer: The statement is false for water but true for land
Q176Single correctHuman Reproduction
Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2They are fraternal twins.
Approach:
The sex composition of the twins reveals whether they arose from one or two zygotes.
Step 1:Fraternal or dizygotic twins develop from two separate fertilized eggs and may be of different sexes.
Step 2:Since the twins are a boy and a girl, they cannot be monozygotic, which would be of the same sex; this indicates they are fraternal twins.
Final answer: They are fraternal twins.
Q177Single correctMicrobes in Human Welfare
Which of the following microbes is NOT involved in the preparation of household products?
A. Aspergillus niger
B. Lactobacillus
C. Trichoderma polysporum
D. Saccharomyces cerevisiae
E. Propionibacterium sharmanii
Choose the correct answer from the options given below:
A. Aspergillus niger
B. Lactobacillus
C. Trichoderma polysporum
D. Saccharomyces cerevisiae
E. Propionibacterium sharmanii
Choose the correct answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A and C only
Approach:
Each microbe is checked for whether it is used in making common household products such as curd, bread and toddy.
Step 1:Lactobacillus is used for production of curd, Saccharomyces cerevisiae for fermentation of palm sap to obtain toddy, and Propionibacterium sharmanii for production of Swiss cheese, so these are household-product microbes.
Step 2:Aspergillus niger is used for the commercial production of citric acid and Trichoderma polysporum for production of cyclosporin A and as a biocontrol agent, so A and C are industrial rather than household microbes.
Final answer: A and C only
Q178Single correctChemical Coordination and Integration
Choose the correct answer from the options given below :
| List-I | List-II |
|---|---|
| A.. Progesterone | I.. Pars intermedia |
| B.. Relaxin | II.. Ovary |
| C.. Melanocyte stimulating hormone | III.. Adrenal Medulla |
| D.. Catecholamines | IV.. Corpus luteum |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A-IV, B-II, C-I, D-III
Approach:
Each hormone is paired with the gland or tissue that secretes it.
Step 1:Progesterone is a steroidal hormone secreted by the corpus luteum, giving A-IV.
Step 2:Relaxin is a proteinaceous hormone secreted by the ovaries in the later stage of pregnancy, giving B-II.
Step 3:Melanocyte stimulating hormone is released by the pars intermedia, giving C-I, and catecholamines are released by the adrenal medulla, giving D-III.
Final answer: A-IV, B-II, C-I, D-III
Q179Single correctBiotechnology - Principles and Processes
The blue and white selectable markers have been developed which differentiate recombinant colonies from non-recombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate.
Given below are two statements about this method.
Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies.
Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies.
In the light of the above statements, choose the most appropriate answer from the options given below:
Given below are two statements about this method.
Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies.
Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Statement I is incorrect but Statement II is correct
Approach:
The blue-white screening relies on insertional inactivation of the enzyme beta-galactosidase.
Step 1:When a recombinant DNA is inserted within the coding sequence of the enzyme beta-galactosidase, the gene for synthesis of this enzyme is inactivated; presence of insert results in insertional inactivation of the beta-galactosidase gene and the colonies do not produce any colour, so they are identified as recombinant colonies.
Step 2:Non-recombinant transformants retain a functional enzyme and produce blue colour in the presence of the chromogenic substrate, so Statement I is incorrect.
Final answer: Statement I is incorrect but Statement II is correct
Q180Single correctOrganisms and Populations
Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Logistic growth incorporates the carrying capacity into the exponential growth model.
Step 1:Logistic growth is described by the Verhulst-Pearl logistic growth equation, in which the intrinsic rate of increase is modulated by the term that depends on carrying capacity.
Step 2:Here N is population size, r the intrinsic rate of natural increase and K the carrying capacity, which gives the sigmoid growth curve.
Final answer:
Frequently Asked Questions
How many questions are in the NEET 2025 May 04 paper?
The NEET 2025 May 04 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 2025 May 04 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 2025 May 04 paper as a timed mock test?
Yes. With a free NEETnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.
Solved this paper? Calculate your NEET score · most important chapters · formula sheets · all free tools.