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NEET 2016 May 01 Question Paper with Solutions
All 179 questions from the NEET 2016 (May 01) paper — Physics (45), Chemistry (44) 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 2016Chemistry PYQs 2016Biology PYQs 2016
- Questions
- 179
- Physics
- 45
- Chemistry
- 44
- Biology
- 90
Physics45 questions
Q1Single correctSystems of Particles and Rotational Motion
From a disc of radius R and mass M, a circular hole diameter R, whose rim passes through the centre is cut. What the moment of inertia of the remaining part of the disc about perpendicular axis & passing through the centre?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Compute the moment of inertia of the full disc about its central perpendicular axis, then subtract the contribution of the removed smaller disc using the parallel axis theorem.
Step 1:Mass of the full disc gives surface density . The removed disc has radius , so its mass is .
Step 2:Moment of inertia of full disc about central axis.
Step 3:Moment of inertia of removed disc about its own centre plus shift of to the main centre.
Step 4:Subtracting the hole contribution from the full disc gives the remaining moment of inertia.
Final answer:
Q2Single correctMoving Charges and Magnetism
A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I. the net force on the loop will be:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The net force on the loop is the difference between the forces on the two sides parallel to the wire, located at distances L/2 and 3L/2.
Step 1:The near side AB lies at distance from the wire and the far side CD at distance . Forces on the two sides perpendicular to the wire cancel.
Step 2:Force on near side.
Step 3:Force on far side.
Step 4:Net force is the difference of the two.
Final answer:
Q3Single correctMagnetism and Matter
The magnetic suscepetibility is negative for
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1diamagnetic material only
Approach:
Classify materials by the sign of magnetic susceptibility.
Step 1:Diamagnetic materials oppose the applied field, giving a small negative magnetization, so .
Step 2:Paramagnetic and ferromagnetic materials align with the field, giving positive susceptibility.
Step 3:Therefore only diamagnetic materials have negative susceptibility.
Final answer: diamagnetic material only
Q4Single correctWaves
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a a speed of 15 m. Then , the frequency of sound that the observer hears in the echo reflected from the cliff is :
(Take velocity of sound in air = 330 m)
(Take velocity of sound in air = 330 m)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3838 Hz
Approach:
The cliff receives sound from a source approaching it, then reflects that frequency back to a stationary observer.
Step 1:The siren moves toward the cliff at m/s, so the cliff (acting as detector) receives a raised frequency.
Step 2:The cliff reflects this frequency unchanged toward the stationary observer.
Final answer: 838 Hz
Q5Single correctElectrostatic Potential and Capacitance
A capacitor of 2 F is charged as shown in the diagram. When the switch S is turn to position 2, the percentage of its stored energy dissipated is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 480%
Approach:
Charge is conserved when the charged 2 microfarad capacitor is connected across the uncharged 8 microfarad capacitor; compare initial and final stored energies.
Step 1:Initial energy is stored on the 2 microfarad capacitor with charge at voltage .
Step 2:On switching, charge redistributes over total capacitance at common voltage .
Step 3:Final energy on the combined .
Step 4:Fraction dissipated equals the energy lost divided by initial energy.
Final answer: 80%
Q6Single correctWave Optics
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 3 when light of wavelength 5000 is incident on the slit. The first secondary maximum is observed at an angle of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Use the first minimum condition to find the slit width, then apply the first secondary maximum condition.
Step 1:First minimum at 30 degrees gives the slit width.
Step 2:First secondary maximum occurs where the path difference is .
Step 3:Therefore the angle is the inverse sine of three-fourths.
Final answer:
Q7Single correctGravitation
At what height from the surface of earth the gravitation potential and the value of g are J k and 6.0 m respectively? Take the radius of earth as 6400 km:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12600 km
Approach:
Take the ratio of gravitational potential to g at the height to eliminate GM and solve for the distance from the centre.
Step 1:Dividing the magnitude of potential by g gives the distance from the centre.
Step 2:Substituting the given values.
Step 3:Height above the surface is the distance from the centre minus the earth radius.
Final answer: 2600 km
Q8Single correctElectromagnetic Waves
Out of the following options which one can be used to produce a propagating electromagnetic wave?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4An accelerating charge
Approach:
Identify the source condition for radiating electromagnetic waves.
Step 1:A stationary charge produces only a static electric field and a charge at constant velocity produces steady fields, neither radiates.
Step 2:An accelerating charge produces time-varying electric and magnetic fields that sustain each other and propagate.
Final answer: An accelerating charge
Q9Single correctElectrostatic Potential and Capacitance
Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d l) apart because of their mutual repulsion. The charges begin to leak from the both the sphere at a constant rate. As a result, the sphere approach each other with a velocity . Then varies as a function of the distance x between the spheres as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Balance Coulomb repulsion against the restoring component of gravity for small separation, then differentiate the separation with respect to time.
Step 1:For small separation the equilibrium gives .
Step 2:Charge leaks at constant rate, so is constant. Differentiating with time.
Step 3:Since the rate of change of charge is constant, the velocity scales inversely with .
Final answer:
Q10Single correctWaves
A uniform rope of length L and mass hangs vertically from a rigid support. A block of mass is attached to the free end of the rope. A transverse pulse of wavelength is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is . The ratio is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Wave speed depends on tension, which differs at the bottom and top of the rope; frequency stays constant so wavelength scales with speed.
Step 1:At the lower end the tension equals the weight of the block.
Step 2:At the top the tension supports both block and rope.
Step 3:Frequency is unchanged, so the wavelength ratio equals the speed ratio, which is the square root of the tension ratio.
Final answer:
Q11Single correctThermodynamics
A refrigerator works between 4 C and 30 C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is:
(Take 1 cal = 4.2 joules)
(Take 1 cal = 4.2 joules)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3236.5 W
Approach:
Use the Carnot coefficient of performance to relate the heat extracted to the work input per second.
Step 1:Convert temperatures to kelvin: K, K.
Step 2:Heat removed per second in joules.
Step 3:Work input per second equals heat removed divided by COP.
Final answer: 236.5 W
Q12Single correctWaves
An air column, closed at one end and open at the other, resonates with a turning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same turning fork is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3150 cm
Approach:
A closed pipe resonates at odd multiples of the fundamental length, so the next resonance occurs at three times the smallest length.
Step 1:Smallest resonance corresponds to a quarter wavelength.
Step 2:Next resonance is three quarters of a wavelength.
Final answer: 150 cm
Q13Single correctSemiconductor Electronics
Consider the junction diode as ideal. The value of current flowing through AB is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 A
Approach:
Check the bias of the ideal diode, then apply Ohm's law across the resistor with the net potential difference.
Step 1:Terminal A is at V and B at V, so the diode (anode toward A) is forward biased and behaves as a short.
Step 2:The full 10 V appears across the 1 kohm resistor.
Final answer: A
Q14Single correctCurrent Electricity
The charge flowing through a resistance R varies with time t as Q = at b, where a and b are positive constants. The total heat produced in R is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Differentiate the charge to get current, find when current ceases, then integrate the Joule heating over that interval.
Step 1:Current is the derivative of charge.
Step 2:Current falls to zero at the upper time limit.
Step 3:Integrate from 0 to .
Final answer:
Q15Single correctThermal Properties of Matter
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is . At wavelength 500 nm is and that at 1000 nm is . Wien's constant, b = 2.88 1 nmK. Which of the following is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Find the wavelength of peak emission from Wien's law, then compare where the given wavelengths fall on the emission curve.
Step 1:Peak wavelength from Wien's law.
Step 2:The stated 500 nm is exactly the peak wavelength, so is the largest emission of the three; the curve falls away on both sides of the peak.
Step 3:A black body radiates at every wavelength, so neither nor is zero. Since 250 nm lies off the peak on the steep short-wavelength side, its emission is well below the peak value.
Final answer:
Q16Single correctThermal Properties of Matter
Coefficient of linear expansion of brass and steel rods are and . Lengths of brass and steel rods are and respectively. If ( ) is maintained same at all temperatures, which one of the following relations holds good?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
For the difference of lengths to stay constant, each rod must expand by the same amount for any temperature change.
Step 1:Difference of lengths constant means the change in each rod's length is equal.
Step 2:Setting the two expansions equal for the same temperature change.
Step 3:Therefore the products of length and expansion coefficient are equal.
Final answer:
Q17Single correctSemiconductor Electronics
A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 is connected in the collector circuit and the voltage drop across it is 0.8 V. Of the current amplification factor is 0.96 and the input resistance of the circuit is 192 , the voltage gain and the power gain of the amplifier will respectively be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 14, 3.84
Approach:
Compute the current gain beta from alpha, then voltage gain as beta times resistance ratio, and power gain as beta times voltage gain.
Step 1:Current gain from alpha.
Step 2:Voltage gain.
Step 3:Using and power gain .
Final answer: 4, 3.84
Q18Single correctWave Optics
The intensity of the maximum in a Young's double slit experiment is . Distance between two slits is d = 5, where is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Find the path difference at the point in front of one slit, convert to phase difference, then use the intensity formula.
Step 1:The point in front of one slit is at . Path difference.
Step 2:Phase difference.
Step 3:Intensity for this phase.
Final answer:
Q19Single correctSystems of Particles and Rotational Motion
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad . Its net acceleration in m at the end of 2.0 s is approximately:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 18.0
Approach:
Compute the tangential and centripetal accelerations of a rim point at t = 2 s, then combine them as the magnitude of the net acceleration.
Step 1:Tangential acceleration of a rim point.
Step 2:Angular velocity at 2 s and centripetal acceleration.
Step 3:Net acceleration is the resultant of tangential and centripetal parts.
Final answer: 8.0
Q20Single correctDual Nature of Radiation and Matter
An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Express the de Broglie wavelength of the electron from its energy and the photon wavelength from its energy, then take the ratio.
Step 1:Electron wavelength uses momentum .
Step 2:Photon wavelength from .
Step 3:Ratio of electron to photon wavelength.
Final answer:
Q21Single correctSystems of Particles and Rotational Motion
A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and lengths. Which one of the two objects gets to the bottom of the plane first?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Sphere
Approach:
Compare the accelerations of rolling bodies, which depend on the ratio of moment of inertia to mass times radius squared.
Step 1:For a disk , for a solid sphere .
Step 2:Smaller inertia factor gives larger acceleration, so the sphere accelerates faster and is independent of mass.
Final answer: Sphere
Q22Single correctRay Optics and Optical Instruments
The angle of incidence for a ray light at a refracting surface of a prism is 4. The angle of prism is 6. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
At minimum deviation use the relation between incidence angle, prism angle, and deviation, then apply the prism refractive index formula.
Step 1:At minimum deviation the incidence angle gives the deviation.
Step 2:Refractive index from the prism formula.
Final answer:
Q23Single correctAtoms
When an - particle of mass 'm' moving with velocity '' bombards on a heavy nucleus of charge 'Ze', its distance of closet approach from the nucleus depends on m as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Equate the initial kinetic energy of the alpha particle to the electrostatic potential energy at the distance of closest approach.
Step 1:At the closest approach all kinetic energy converts to potential energy.
Step 2:Solving for the closest distance.
Final answer:
Q24Single correctLaws of Motion / Circular Motion
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to J by the end of the second revolution after the beginning of the motion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Use the work-energy theorem over two revolutions to relate tangential force to the kinetic energy gained, then divide by mass.
Step 1:Distance covered in two revolutions of radius r = 6.4 cm = 0.064 m.
Step 2:Tangential force from the kinetic energy gained equals the work done by it.
Step 3:Divide tangential force by the mass m = 10 g = 0.01 kg.
Final answer:
Q25Single correctKinetic Theory of Gases
The molecules of a given mass of a gas have r.m.s. velocity of at C and pressure. When the temperature and pressure of the gas respectively, C and , the r.m.s. velocity of its molecules in is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
RMS velocity depends only on temperature for a given gas; scale the given value by the square root of the absolute-temperature ratio.
Step 1:Convert temperatures to kelvin.
Step 2:RMS speed is independent of pressure and scales with the square root of absolute temperature.
Step 3:Substituting the initial value gives the stated form.
Final answer:
Q26Single correctMoving Charges and Magnetism
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distances a/2 and 2a respectively, from the axis of the wire is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply Ampere's law inside (field proportional to r) and outside (field proportional to 1/r) the wire, then form the ratio.
Step 1:At r = a/2 (inside), the field is proportional to r.
Step 2:At r = 2a (outside), the field is proportional to 1/r.
Step 3:Form the ratio of the two equal fields.
Final answer:
Q27Single correctMotion in a Plane
A particle moves so that its position vector is given by . Where is a constant. Which of the following is true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Velocity is perpendicular to and acceleration is directed towards the origin.
Approach:
Differentiate the position vector to obtain velocity and acceleration, then compare their directions with the position vector.
Step 1:Differentiate the position vector once.
Step 2:The dot product of r and v is zero, so velocity is perpendicular to r.
Step 3:Differentiate again; acceleration is opposite to the position vector.
Final answer: Velocity is perpendicular to and acceleration is directed towards the origin.
Q28Single correctWork, Energy and Power
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Apply the minimum-speed condition at the top of the loop and conserve energy between the bottom and the top.
Step 1:At the top the minimum speed satisfies gravity providing centripetal force.
Step 2:Conserve mechanical energy from bottom to top, a height of 2R.
Step 3:Take the square root for the entry speed.
Final answer:
Q29Single correctDual Nature of Radiation and Matter
When a metallic surface is illuminated with radiation of wavelength , the stopping potential is V. If the same surface is illuminated with radiation of wavelength , the stopping potential is . The threshold wavelength for the metallic surface is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Write Einstein's photoelectric equation for both wavelengths and eliminate the work function and stopping potential.
Step 1:Write the equation for wavelength λ with stopping potential V.
Step 2:Write the equation for wavelength 2λ with stopping potential V/4.
Step 3:Multiply equation (2) by 4 and subtract equation (1) to eliminate V.
Final answer:
Q30Single correctThermodynamics
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Compressing the gas through adiabatic process will require more work to be done
Approach:
Compare the steepness of the adiabatic and isothermal curves on a P-V diagram for the same volume compression.
Step 1:An adiabatic curve is steeper than the isothermal curve since γ > 1.
Step 2:For compression to half the volume the area under the adiabatic curve exceeds that under the isothermal curve.
Step 3:Therefore more work is required for the adiabatic compression.
Final answer: Compressing the gas through adiabatic process will require more work to be done
Q31Single correctCurrent Electricity
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Balance length is proportional to the net emf; form and solve the two equations for the sum and difference of emfs.
Step 1:Sum of emfs corresponds to 50 cm and difference to 10 cm.
Step 2:Take the ratio of the two balance lengths.
Step 3:Solve for the ratio of the two emfs.
Final answer:
Q32Single correctRay Optics and Optical Instruments
A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 cm
Approach:
Find the image distance from the objective using the lens formula, then add the eyepiece focal length for normal adjustment.
Step 1:For the objective, u = -200 cm and f = 40 cm.
Step 2:The image from the objective lies 50 cm beyond it; the eyepiece adds its focal length.
Step 3:Total separation of the lenses.
Final answer: cm
Q33Single correctMechanical Properties of Fluids
Two non-mixing liquids of densities and are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL in the denser liquids. The density d is equal to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Balance the weight of the cylinder against the buoyant forces from the portions immersed in each liquid.
Step 1:Length pL is in the denser liquid (density nρ) and the remaining (1-p)L is in the lighter liquid (density ρ).
Step 2:Cancel L and A and expand.
Step 3:Write the result in the option form.
Final answer:
Q34Single correctSemiconductor Electronics
To get output 1 for the following circuit, the correct choice for the input is:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The shown circuit is an OR gate (inputs A, B) feeding an AND gate with input C; output is 1 only when (A OR B) AND C are true.
Step 1:The first gate gives A OR B; the second ANDs that with C.
Step 2:Output is 1 only if (A+B) = 1 and C = 1.
Step 3:Setting C = 1 and A = 1 (with B = 0) meets both requirements at once.
Final answer:
Q35Single correctThermal Properties of Matter
A piece of ice falls from a height h so that it melts completely. Only one - quarter of the heat produced is absorbed by the ice and all the energy of ice gets converted into heat during its fall. The value of h is: [Latent heat of ice is J/kg and g = 10 N/kg]
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 km
Approach:
Equate one quarter of the gravitational potential energy converted to heat with the latent heat needed to melt the ice.
Step 1:Only one quarter of mgh is absorbed by the ice to melt it.
Step 2:Solve for h.
Step 3:Convert to kilometres.
Final answer: km
Q36Single correctGravitation
The ratio of escape velocity of earth to the escape velocity at a planet whose radius and mean density are twice as that of earth is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Express escape velocity in terms of radius and density, then take the ratio with both quantities doubled.
Step 1:Escape velocity is proportional to R times the square root of density.
Step 2:For the planet, R and ρ are both doubled.
Step 3:Therefore the ratio of earth to planet escape velocity.
Final answer:
Q37Single correctMotion in a Plane
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Set the magnitudes of the sum and difference equal and solve for the cosine of the angle between the vectors.
Step 1:Equate the squared magnitudes of sum and difference.
Step 2:This forces the cosine of the angle to be zero.
Step 3:Therefore the angle is a right angle.
Final answer:
Q38Single correctAtoms
Given the value of Rydberg constant is , the wave number of the last line of the Balmer series in hydrogen spectrum will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The last line of the Balmer series corresponds to the transition from infinity to n = 2; apply the Rydberg formula.
Step 1:For the last (series-limit) Balmer line, n1 = 2 and n2 → ∞.
Step 2:Substitute the Rydberg constant.
Step 3:Express in the option form.
Final answer:
Q39Single correctWork, Energy and Power
A body of mass 1 kg begins to move under the action of a time dependent force N, where i and j are unit along x and y axis. What power will be developed by the force at the time t?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 W
Approach:
Integrate the force to obtain velocity (mass 1 kg), then compute power as the dot product of force and velocity.
Step 1:Integrate each component of the force with mass 1 kg to get velocity.
Step 2:Compute the dot product of force and velocity.
Step 3:Write the power expression.
Final answer: W
Q40Single correctAlternating Current
An inductor 20 mH, a capacitor 50 F and a resistor 40 are connected in series across a source of emf . The power loss in A.C circuit is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 W
Approach:
Compute the reactances at ω = 340, find the impedance, the rms current, and then the average power dissipated in the resistor.
Step 1:Find the inductive and capacitive reactances at ω = 340 rad/s.
Step 2:Compute the impedance.
Step 3:Find the rms current and the power dissipated.
Final answer: W
Q41Single correctMotion in a Straight Line
If the velocity of a particle is , where A and B are constants, then the distance travelled by it between 1 and 2s is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Integrate the velocity with respect to time from t = 1 s to t = 2 s to obtain the distance travelled.
Step 1:Integrate the velocity expression.
Step 2:Substitute the limits t = 2 and t = 1.
Step 3:Simplify the result.
Final answer:
Q42Single correctElectromagnetic Induction
A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is Wb. The self-inductance of the solenoid is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4H
Approach:
Use the relation between total flux linkage, self-inductance, and current.
Step 1:Total flux linkage equals number of turns times flux per turn.
Step 2:Divide by the current.
Step 3:State the self-inductance.
Final answer: H
Q43Single correctAlternating Current
A small signal voltage is applied across an ideal capacitor C:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Over a full cycle the capacitor C does not consume any energy from the voltage source.
Approach:
Analyse the phase relationship and the average power for a purely capacitive AC circuit.
Step 1:In a pure capacitor the current leads the voltage by 90 degrees.
Step 2:Average power over a cycle uses the power factor cosφ.
Step 3:Zero average power means no net energy is consumed over a full cycle.
Final answer: Over a full cycle the capacitor C does not consume any energy from the voltage source.
Q44Single correctRay Optics and Optical Instruments
Match the corresponding entries of column 1 with column 2 [where m is the magnification produced by the mirror]
| Column 1 | Column 2 |
|---|---|
| (A). | (a). Convex mirror |
| (B). | (b). Concave mirror |
| (C). | (c). Real image |
| (D). | (d). Virtual image |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1A b and c; B b and c; C b and d; D a and d
Approach:
Use the sign convention for mirror magnification to classify each value by mirror type and image nature.
Step 1:Negative magnification corresponds to a real image formed by a concave mirror.
Step 2:Positive magnification corresponds to a virtual image; m = +2 from a concave mirror and m = +1/2 from a convex mirror.
Step 3:A negative magnification means a real, inverted image, which a mirror forms only when it is concave; a positive magnification means a virtual, erect image. Magnitude greater than one enlarges, less than one diminishes.
Final answer: A b and c; B b and c; C b and d; D a and d
Q45Single correctLaws of Motion
A car is negotiating a curved road of radius R. The road is banked at an angle . The coefficient of friction between the tyres of the car and the road is . The maximum safe velocity on this road is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Resolve forces along and perpendicular to a banked road with friction acting down the incline at maximum speed, then solve for v.
Step 1:At maximum speed friction acts down the slope; resolve the normal and friction forces.
Step 2:Vertical balance of forces.
Step 3:Divide the equations and solve for v.
Final answer:
Chemistry44 questions
Q46Single correctChemical Bonding and Molecular Structure
Consider the molecules , and . Which of the given statements is false?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The bond angle in is larger than the bond angle in .
Approach:
Compare bond angles using VSEPR theory based on the number of lone pairs on the central atom.
Step 1:Methane has no lone pair, giving an ideal tetrahedral geometry.
Step 2:Ammonia has one lone pair, reducing the bond angle.
Step 3:Water has two lone pairs, reducing the bond angle further.
Step 4:The bond angle order is water less than ammonia less than methane. Thus the statement that water's angle is larger than methane's angle is false.
Final answer: The bond angle in is larger than the bond angle in .
Q47Single correctHydrocarbons
In the reaction
X and Y are :
X and Y are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1X = 1 – Butyne; Y = 3 – Hexyne
Approach:
Terminal alkyne acetylide is alkylated by an alkyl halide; track the position of the triple bond after each alkylation.
Step 1:Sodium amide deprotonates the terminal alkyne to form the acetylide ion.
Step 2:The acetylide attacks ethyl bromide, attaching an ethyl group to give X, a terminal alkyne.
Step 3:Repeating deprotonation and alkylation of the terminal carbon of X attaches a second ethyl group.
Step 4:The triple bond ends up internal and symmetric, giving 3-hexyne.
Final answer: X = 1 – Butyne; Y = 3 – Hexyne
Q48Single correctThe p-Block Elements
Among the following the correct order of acidity is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Acid strength of oxoacids increases with the number of oxygen atoms bonded to the central atom due to greater stabilization of the conjugate base.
Step 1:More terminal oxygen atoms increase the positive oxidation state of chlorine and stabilize the conjugate base by charge delocalization.
Step 2:Oxidation state of chlorine rises from +1 to +7 across the series.
Step 3:Greater conjugate base stability corresponds to stronger acidity.
Final answer:
Q49Single correctChemical Kinetics
The rate of first order reaction is at 10 seconds and at 20 seconds after intiation of the reaction. The half – life period of the reaction is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
For a first order reaction the rate is proportional to concentration, so the ratio of rates equals the ratio of concentrations; extract the rate constant and compute the half-life.
Step 1:Rate is proportional to concentration, so the ratio of rates at the two times equals the ratio of concentrations over the 10 second interval.
Step 2:Apply the first order relation over the 10 second interval to find the rate constant.
Step 3:Compute the half-life from the rate constant.
Final answer:
Q50Single correctSurface Chemistry
Which one of the following characteristics is associated with adsorption?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2, and all are negative
Approach:
Analyze the signs of the thermodynamic functions for a spontaneous adsorption process.
Step 1:Adsorption is exothermic, so the enthalpy change is negative.
Step 2:Gas molecules lose freedom on the surface, decreasing disorder, so the entropy change is negative.
Step 3:For spontaneity the Gibbs energy change must be negative; with negative enthalpy dominating over the positive entropy term, the process proceeds spontaneously.
Final answer: , and all are negative
Q51Single correctClassification of Elements and Periodicity
In which of the following options the order of arrangement does not agree with the variation of property indicated against it?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 (increasing first ionisation enthalpy)
Approach:
Check each listed order against the actual periodic trend to find arrangements that do not match the stated property.
Step 1:Isoelectronic species shrink with increasing nuclear charge, so true ionic size order is the reverse of option 1, making option 1 a correct match for increasing size in the given direction.
Step 2:First ionisation enthalpy of nitrogen exceeds that of oxygen due to the stable half-filled p subshell, so the strict increasing order B less than C less than N less than O does not hold.
Step 3:Electron gain enthalpy is most negative for chlorine, not fluorine, because the small fluorine atom concentrates inter-electronic repulsion. So the I < Br < Cl < F ordering also fails to describe its stated property.
in magnitude
Step 4:Metallic radius does increase down the group, so Li < Na < K < Rb agrees with its stated property and is not an answer. Two of the four orderings therefore fail, which is why this question carries two accepted answers.
Final answer: (increasing first ionisation enthalpy)
Q52Single correctThe s-Block Elements
Which of the following statements is false?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 ions are not important in maintaining the regular beating of the heart
Approach:
Evaluate each biological role statement of magnesium and calcium ions for correctness.
Step 1:Magnesium ions complex with ATP and are central to chlorophyll, so statements 1 and 4 are true.
Step 2:Calcium ions are essential for blood clotting, so statement 2 is true.
Step 3:Calcium ions are in fact important for the regular beating of the heart, so the statement denying this is false.
Final answer: ions are not important in maintaining the regular beating of the heart
Q53Single correctHydrogen
Which of the following statements about hydrogen is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hydrogen has three isotopes of which tritium is the most common
Approach:
Identify the incorrect statement about hydrogen and its species.
Step 1:Hydrogen has three isotopes, but protium is the most abundant, not tritium; statement 1 is incorrect.
Step 2:Dihydrogen does act as a reducing agent — it reduces metal oxides such as CuO to the metal — so the claim that it does not is also incorrect.
Step 3:Two of the four statements are therefore false, which is why this question carries two accepted answers. Hydrogen does form H+ salts and the hydronium ion is the solvated form of the proton, so those two statements stand.
Final answer: Hydrogen has three isotopes of which tritium is the most common
Q54Single correctAldehydes, Ketones and Carboxylic Acids
The correct statement regarding a carbonyl compound with a hydrogen atom on its alphacarbon is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4A carbonyl compound with a hydrogen atom on its alpha – carbon rapidly equilibrates with its corresponding enol and this process is known as keto – enol tautomerism.
Approach:
Identify the correct name and behaviour for the equilibrium of a carbonyl compound bearing an alpha hydrogen.
Step 1:An alpha hydrogen on a carbonyl compound allows interconversion between the keto and enol forms.
Step 2:This rapid interconversion is the keto-enol tautomerism, so option 4 is correct.
Final answer: A carbonyl compound with a hydrogen atom on its alpha – carbon rapidly equilibrates with its corresponding enol and this process is known as keto – enol tautomerism.
Q55Single correctEquilibrium
MY and two nearly insoluble salts, have the same values of at room temperature. Which statement would be true in regard of MY and ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The molar solubility of MY in water is less than that of
Approach:
Express the molar solubility of each salt in terms of its solubility product and compare.
Step 1:For the 1:1 salt MY the solubility equals the square root of the solubility product.
Step 2:For the 1:3 salt NY3 the solubility is the fourth root of the product divided by 27.
Step 3:Comparing the two values, the molar solubility of MY is smaller than that of NY3.
Final answer: The molar solubility of MY in water is less than that of
Q56Single correctBiomolecules
In a protein molecule various amino acids are linked together by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Peptide bond
Approach:
Recall the type of covalent bond connecting amino acids in proteins.
Step 1:Amino acids join through the reaction of the carboxyl group of one with the amino group of the next, releasing water.
Step 2:This amide linkage between amino acids is the peptide bond.
Final answer: Peptide bond
Q57Single correctPolymers
Natural rubber has:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1All cis – configuration
Approach:
Recall the geometric configuration of the double bonds in natural rubber.
Step 1:Natural rubber is cis-1,4-polyisoprene with all double bonds in the cis arrangement.
Step 2:This cis configuration is responsible for its elasticity, so option 1 is correct.
Final answer: All cis – configuration
Q58Single correctGeneral Principles of Isolation of Elements
Match items of Column I with the item of Column II and assign the correct code :
| Column I | Column II |
|---|---|
| (a). Cyanide process | (i). Ultrapure Ge |
| (b). Froth floatation process | (ii). Dressing of ZnS |
| (c). Electrolytic reduction | (iii). Extraction of Al |
| (d). Zone refining | (iv). Extraction of Au |
| (v). Purification of Ni |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Approach:
Match each metallurgical operation in Column I with its application in Column II.
Step 1:Cyanide process is used for the extraction of gold.
Step 2:Froth floatation is used for concentrating sulphide ore such as ZnS.
Step 3:Electrolytic reduction is used for the extraction of aluminium.
Step 4:Zone refining yields ultrapure germanium.
Final answer: (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Q59Single correctThe p-Block Elements
Which one of the following statements is correct when is passed through acidified solution?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Green is formed
Approach:
Determine the redox outcome when sulphur dioxide reacts with acidified potassium dichromate.
Step 1:Sulphur dioxide is a reducing agent and reduces orange dichromate to green chromium(III).
Step 2:The product formed is green chromium(III) sulphate.
Final answer: Green is formed
Q60Single correctThe d- and f-Block Elements
The electronic configurations of Eu (Atomic No.63), Gd (Atomic No.64 and Tb (Atomic No.65) are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4, and
Approach:
Write the ground state configurations of the three lanthanoids using the stability of half-filled f subshells.
Step 1:Europium attains a stable half-filled f subshell.
Step 2:Gadolinium keeps the half-filled f set and places one electron in 5d.
Step 3:Terbium fills additional f orbitals to give the f9 configuration.
Final answer: , and
Q61Single correctStructure of Atom
Two electrons occupying the same orbital are distinguished by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Spin quantum number
Approach:
Apply the Pauli exclusion principle to determine which quantum number differs for two electrons in one orbital.
Step 1:Two electrons in the same orbital share the same principal, azimuthal and magnetic quantum numbers.
Step 2:By the Pauli exclusion principle they must differ in spin quantum number.
Final answer: Spin quantum number
Q63Single correctAlcohols, Phenols and Ethers
Which of the following reagents would distinguish cis-cyclopenta-1, 2-diol from the trans-isomer?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Acetone
Approach:
Find a reagent whose reaction depends on the relative spatial arrangement of the two hydroxyl groups.
Step 1:A cis-1,2-diol has both hydroxyl groups on the same face, allowing formation of a cyclic acetonide with acetone.
Step 2:The trans-diol cannot form the five-membered cyclic acetal, so acetone distinguishes the isomers.
Final answer: Acetone
Q64Single correctThermodynamics
The correct thermodynamic conditions for the spontaneous reaction at all temperature is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 and
Approach:
Apply the Gibbs energy equation to find the signs of enthalpy and entropy that guarantee spontaneity at every temperature.
Step 1:Spontaneity requires the Gibbs energy change to be negative at all temperatures.
Step 2:A negative enthalpy makes the first term negative and a positive entropy makes the second term contribute negatively at every temperature.
Step 3:These signs ensure spontaneity independent of temperature.
Final answer: and
Q65Single correctThe Solid State
Lithium has a bcc structure. Its density is and its atomic mass is . Calculate the edge length of a unit cell of Lithium metal. ()
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Use the density relation for a cubic unit cell with two atoms per bcc cell to solve for the edge length.
Step 1:A bcc cell contains two atoms, and the density is expressed in consistent units.
Step 2:Rearrange to solve for the cube of the edge length.
Step 3:Take the cube root and convert to picometres.
Final answer:
Q66Single correctThe p-Block Elements
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Rank the halogens by bond dissociation enthalpy, noting the anomalously low value for fluorine.
Step 1:Chlorine has the highest bond dissociation enthalpy among the halogens, followed by bromine.
Step 2:Fluorine has an abnormally low value due to strong lone pair repulsion in the small F2 molecule, placing it above only iodine.
Step 3:Combining gives the overall order.
Final answer:
Q67Single correctChemistry in Everyday Life
Which of the following is an analgesic ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Novalgin
Approach:
Classify each listed drug by its therapeutic action.
Step 1:Penicillin, streptomycin and chloromycetin are antibiotics.
Step 2:Novalgin is a pain reliever and therefore the analgesic.
Final answer: Novalgin
Q68Single correctStates of Matter
Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Apply Graham's law to relate the rates of effusion of hydrogen and oxygen through the same pinhole.
Step 1:The ratio of effusion rates depends on the inverse square root of the molar masses.
Step 2:In the same time interval the amount of each gas effused is proportional to its rate, so oxygen escapes one quarter as much as hydrogen.
Step 3:When one half of the hydrogen escapes, the oxygen fraction is one quarter of one half.
Final answer:
Q69Single correctHaloalkanes and Haloarenes / Aromatic Compounds
Consider the nitration of benzene using mixed conc. and . If a large amount of is added to the mixture, the rate of nitration will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2slower
Approach:
Identify the active electrophile in nitration and the effect of added bisulphate on its concentration.
Step 1:The nitrating electrophile is the nitronium ion, generated as the conjugate acid of nitric acid is dehydrated.
Step 2:Adding a large amount of bisulphate increases the concentration of a product of this equilibrium.
Step 3:Shifting the equilibrium backward lowers the nitronium ion concentration available for attack on benzene.
Final answer: slower
Q70Single correctChemical Bonding and Molecular Structure
Predict the correct order among the following:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1lone pair lone pair lone pair bond pair bond pair bond pair
Approach:
Compare electron-pair repulsions by how localized each pair is relative to the central atom.
Step 1:A lone pair is held by only one nucleus, so it spreads closer to the central atom and repels more strongly than a bond pair, which is shared between two nuclei.
Step 2:Ranking pairwise repulsions by the number of lone pairs involved gives the standard VSEPR order.
Step 3:A lone pair is held by one nucleus only, so it spreads wider than a bond pair held by two; repulsion therefore falls off in the order lone-lone, lone-bond, bond-bond.
Final answer: lone pair lone pair lone pair bond pair bond pair bond pair
Q71Single correctThe p-Block / s-Block Elements
The product obtained as a result of a reaction of nitrogen with is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option None of the four — the product is calcium cyanamide,
Approach:
Recall the industrial reaction of calcium carbide with nitrogen and assess whether the listed products match the actual product.
Step 1:Calcium carbide reacts with nitrogen at high temperature to form calcium cyanamide and carbon.
Step 2:The mixture of calcium cyanamide and carbon produced this way is the fertiliser nitrolim, made by the Frank-Caro process.
Step 3:Note that calcium cyanamide carries the CN2(2-) ion, not the cyanide ion CN(-), so none of Ca(CN)2, CaCN, CaCN3 or Ca2CN describes it.
Final answer: Calcium cyanamide, (formed together with carbon)
Q72Single correctSolutions / Thermodynamics
Consider the following liquid vapour equilibrium.
Liquid Vapour
Which of the following relations is correct?
Liquid Vapour
Which of the following relations is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Recall the differential form of the Clausius-Clapeyron equation governing vapour pressure with temperature.
Step 1:For a liquid-vapour equilibrium the variation of vapour pressure with temperature follows the Clausius-Clapeyron relation.
Step 2:The enthalpy of vaporization is positive, so the slope is positive, ruling out the negative-sign forms.
Step 3:The correct differential is first order in T in the derivative and has in the denominator.
Final answer:
Q73Single correctThe p-Block Elements / Chemical Bonding
Match the compounds given in column I with the hybridization and shape given in column II and mark the correct option. Code:
| Column I | Column II |
|---|---|
| a. | i. distorted octahedral |
| b. | ii. square planar |
| c. | iii. pyramidal |
| d. | iv. square pyramidal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)(i) (b)(iii) (c)(iv) (d)(ii)
Approach:
Assign hybridization and molecular shape to each xenon compound using VSEPR, then match to column II.
Step 1:Xenon hexafluoride has seven electron domains (six bond pairs and one lone pair) and adopts a distorted octahedral shape.
Step 2:Xenon trioxide has three bond pairs and one lone pair, giving a pyramidal shape.
Step 3:Xenon oxytetrafluoride has five bond pairs and one lone pair, giving a square pyramidal shape.
Step 4:Xenon tetrafluoride has four bond pairs and two lone pairs, giving a square planar shape.
Final answer: (a)(i) (b)(iii) (c)(iv) (d)(ii)
Q74Single correctCoordination Compounds / d- and f-Block Elements
Which of the following has longest bond length? (Free bond length in CO is .)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Relate the C-O bond length in metal carbonyls to the extent of metal to ligand back-bonding, which depends on electron density on the metal.
Step 1:Greater electron density on the metal increases back-donation into the antibonding pi orbital of CO, weakening and lengthening the C-O bond.
Step 2:Compare net charge among the carbonyls: the doubly negative iron complex carries the most electron density on the metal.
Step 3:Strongest back-bonding produces the most weakened, hence longest, C-O bond.
Final answer:
Q75Single correctElectrochemistry
The pressure of required to make the potential of electrode zero in pure water at 298 K is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 atm
Approach:
Apply the Nernst equation to the hydrogen electrode in pure water and solve for the pressure that makes the potential zero.
Step 1:In pure water at 298 K the hydrogen ion concentration is fixed by neutrality.
Step 2:Set the electrode potential equal to zero in the Nernst expression.
Step 3:A zero logarithm forces the ratio to unity, so the pressure equals the square of the hydrogen ion concentration.
Final answer: atm
Q76Single correctChemical Kinetics
The addition of a catalyst during a chemical reaction alters which of the following quantities?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Activation Energy
Approach:
Recall how a catalyst influences the reaction pathway and which thermodynamic versus kinetic quantities it affects.
Step 1:A catalyst provides an alternative pathway with a lower energy barrier for the reaction.
Step 2:State functions such as entropy, internal energy and enthalpy depend only on initial and final states, which a catalyst leaves unchanged.
Step 3:The only quantity altered is the activation energy.
Final answer: Activation Energy
Q77Single correctThe Solid State
The ionic radii of and ions are m and m. The coordination number of each ion in AB is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 16
Approach:
Compute the radius ratio of the cation to the anion and assign the coordination number from the radius ratio ranges.
Step 1:Form the ratio of the cation radius to the anion radius.
Step 2:Compare the ratio with the standard radius ratio ranges.
Step 3:This range corresponds to octahedral coordination.
Final answer: 6
Q78Single correctThe p-Block Elements
Which is the correct statement for the given acids?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Phosphinic acid is a monoprotic acid while Phosphonic acid is a diprotic acid.
Approach:
Count the ionizable P-OH hydrogens in each oxoacid, since only hydrogens on oxygen are acidic.
Step 1:Phosphinic (hypophosphorous) acid has one P-OH group and two P-H bonds.
Step 2:Phosphonic (phosphorous) acid has two P-OH groups and one P-H bond.
Step 3:Phosphinic acid, H3PO2, has two P-H bonds and only one P-OH, so it releases a single proton; phosphonic acid, H3PO3, has one P-H and two P-OH, so it releases two. Only the hydrogens bonded to oxygen are ionisable.
Final answer: Phosphinic acid is a monoprotic acid while Phosphonic acid is a diprotic acid.
Q79Single correctSurface Chemistry
Fog is colloidal solution of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Liquid in gas
Approach:
Identify the dispersed phase and the dispersion medium of fog from its physical composition.
Step 1:Fog consists of tiny liquid water droplets suspended in air.
Step 2:The surrounding medium carrying the droplets is air, a gas.
Step 3:A liquid dispersed in a gas is classified as an aerosol of the liquid-in-gas type.
Final answer: Liquid in gas
Q80Single correctSolutions
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at C. (Given, Vapour Pressure Data at C, benzene kPa, toluene kPa
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The vapour will contain a higher percentage of benzene.
Approach:
Use Raoult's law to find each partial pressure and compare them, since the more volatile component is richer in the vapour.
Step 1:For a 1 : 1 molar mixture each mole fraction is one half.
Step 2:Compute partial pressures from Raoult's law.
Step 3:The component with the higher partial pressure is enriched in the vapour phase.
Final answer: The vapour will contain a higher percentage of benzene.
Q81Single correctOrganic Chemistry - Some Basic Principles / Hydrocarbons
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
Approach:
Compare the relative stabilities of the staggered and eclipsed conformations of ethane based on torsional strain.
Step 1:In the staggered conformation the C-H bonds on adjacent carbons are as far apart as possible, minimizing electron repulsion.
Step 2:In the eclipsed conformation the C-H bonds align, increasing electron repulsion and introducing torsional strain.
Step 3:The strain-free staggered conformation is therefore the more stable one.
Final answer: The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
Q82Single correctAlcohols, Phenols and Ethers / Haloalkanes
The reaction shown below can be classified as:

(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Williamson ether synthesis reaction
Approach:
Identify the reaction type from the conversion of an alcohol to an ether through an alkoxide and an alkyl halide.
Step 1:Sodium hydride deprotonates the alcohol to generate the corresponding alkoxide ion.
Step 2:The alkoxide attacks methyl iodide in a nucleophilic substitution, forming a new C-O bond.
Step 3:Formation of an ether from an alkoxide and an alkyl halide defines this named reaction.
Final answer: Williamson ether synthesis reaction
Q83Single correctAldehydes, Ketones and Carboxylic Acids / Amines
The product formed by the reaction of an aldehyde with a primary amine is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Schiff base
Approach:
Recall the condensation product formed when an aldehyde reacts with a primary amine.
Step 1:A primary amine adds to the carbonyl carbon of the aldehyde to give a carbinolamine intermediate.
Step 2:Loss of water from the intermediate produces a carbon-nitrogen double bond.
Step 3:An imine derived from a primary amine and a carbonyl compound is called a Schiff base.
Final answer: Schiff base
Q84Single correctOrganic Chemistry - Some Basic Principles
Which of the following biphenyl is optically active?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Dibromo- and diiodo-substituted biphenyl
Approach:
Biphenyls become optically active (atropisomerism) when bulky groups at all four ortho positions restrict rotation and the molecule lacks a plane of symmetry on each ring.
Step 1:Restricted rotation about the central bond requires bulky substituents at all ortho positions of both rings.
Step 2:Optical activity also requires that neither ring carries two identical ortho groups arranged symmetrically about the bond axis.
Step 3:The biphenyl bearing bromine and iodine at the four ortho positions in an unsymmetrical arrangement satisfies both conditions.
Final answer: The biphenyl bearing bromine at 2 and 2' and iodine at 6 and 6'
Q85Single correctHaloalkanes and Haloarenes / Hydrocarbons
For the following reactions:
(a)
Reactions (b) and (c) are shown below, in that order.
Which of the following statements is correct?
(a)
Reactions (b) and (c) are shown below, in that order.
Which of the following statements is correct?

(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a) is elimination, (b) is substitution and (c) is addition reaction.
Approach:
Classify each transformation by comparing the bonds broken and formed: loss of two groups to form a double bond is elimination, replacement of one group is substitution, and adding across a double bond is addition.
Step 1:Reaction (a) removes a hydrogen and a halogen from adjacent carbons to form a carbon-carbon double bond.
Step 2:Reaction (b) replaces the bromine of the alkyl bromide with a hydroxyl group, releasing KBr.
Step 3:Reaction (c) adds bromine across the carbon-carbon double bond of the alkene.
Final answer: (a) is elimination, (b) is substitution and (c) is addition reaction.
Q86Single correctSolutions
At C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If , the boiling point of this solution will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1C
Approach:
Find the molality from the relative lowering of vapour pressure, then apply the boiling point elevation formula.
Step 1:At 100 C the pure water vapour pressure is 760 mm, so compute the mole fraction of solute from the lowering.
Step 2:For a dilute solution the molality approximates the mole fraction of solute times moles of water per kilogram, giving close to one molal here.
Step 3:The vapour pressure drop from 760 mm to 732 mm fixes the mole fraction of solute, which gives a molality near 2 mol/kg; multiplying by Kb = 0.52 raises the boiling point by about one degree.
Final answer: C
Q87Single correctBiomolecules
The correct statement regarding RNA and DNA, respectively is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The sugar component in RNA is ribose and the sugar component in DNA is 2'- deoxyribose.
Approach:
Recall the pentose sugar present in each nucleic acid.
Step 1:RNA contains the pentose sugar ribose, which carries a hydroxyl group at the 2' position.
Step 2:DNA contains 2'-deoxyribose, lacking the hydroxyl group at the 2' position.
Step 3:RNA carries ribose, which has an OH at C-2'; DNA carries 2'-deoxyribose, where that OH is replaced by H. Arabinose is an epimer of ribose and appears in neither nucleic acid.
Final answer: The sugar component in RNA is ribose and the sugar component in DNA is 2'- deoxyribose.
Q88Single correctAmines / Organic Chemistry - Some Basic Principles
The correct statement regarding the basicity of arylamines is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring electron system.
Approach:
Compare the availability of the nitrogen lone pair in arylamines versus alkylamines to judge relative basicity.
Step 1:In arylamines the nitrogen lone pair conjugates with the aromatic ring, spreading the electron density into the ring.
Step 2:Reduced availability of the lone pair lowers the tendency of the nitrogen to accept a proton.
Step 3:In alkylamines the lone pair stays localized on nitrogen, making them more basic by comparison.
Final answer: Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring electron system.
Q89Single correctBiomolecules
Which one given below is a non-reducing sugar?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Sucrose
Approach:
A sugar is non-reducing only if both anomeric carbons are engaged in the glycosidic linkage, leaving no free reducing group.
Step 1:Glucose, maltose and lactose each retain a free anomeric (hemiacetal) carbon that can open to an aldehyde.
Step 2:In sucrose the glycosidic bond joins the anomeric carbon of glucose to the anomeric carbon of fructose, so neither is free.
Step 3:With both anomeric carbons locked, sucrose cannot reduce Fehling's or Tollens' reagent.
Final answer: Sucrose
Q90Single correctOrganic Chemistry - Some Basic Principles
The pair of electron in the given carbanion, , is present in which of the following orbitals?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Determine the hybridization of the carbanion carbon from the triple bond, then place the lone pair in the appropriate hybrid orbital.
Step 1:The carbanion carbon is bonded through a triple bond, so it has two sigma frameworks worth of regions around it characteristic of sp hybridization.
Step 2:An sp hybridized carbon forms two sp hybrid orbitals and retains two unhybridized p orbitals used for the pi bonds.
Step 3:The non-bonding electron pair of the carbanion therefore resides in an sp orbital.
Final answer:
Biology90 questions
Q91Single correctOrganisms and Populations
Gause's principle of competitive exclusion states that:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3No two species can occupy the same niche indefinitely for the same limiting resources.
Approach:
Apply Gause's competitive exclusion principle from ecology.
Step 1:Gause's competitive exclusion principle states that two species competing for the same limiting resource cannot coexist; one is eventually eliminated.
Final answer: No two species can occupy the same niche indefinitely for the same limiting resources.
Q92Single correctBiomolecules
The two polypeptides of human insulin are linked together by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Disulphide bridges
Approach:
Recall structure of mature human insulin.
Step 1:Human insulin consists of A and B chains joined by inter-chain disulphide bridges between cysteine residues.
Final answer: Disulphide bridges
Q93Single correctSexual Reproduction in Flowering Plants
The coconut water from tender coconut represents:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Free nuclear endosperm
Approach:
Recall endosperm development in coconut.
Step 1:Coconut water is the liquid free-nuclear stage of endosperm formed before cellularisation; the white kernel is the cellular endosperm.
Final answer: Free nuclear endosperm
Q94Single correctBiological Classification
Which of the following statements is wrong for viroids?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Their RNA is of high molecular weight
Approach:
Recall properties of viroids.
Step 1:Viroids are free RNA without a protein coat, smaller than viruses, and cause infections; their RNA is of low molecular weight, so the high molecular weight statement is wrong.
Final answer: Their RNA is of high molecular weight
Q95Single correctAnimal Kingdom
Which of the following features is not present in the Phylum-Arthropoda?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Parapodia
Approach:
Recall arthropod characteristics.
Step 1:Arthropods have a chitinous exoskeleton, metameric segmentation and jointed appendages. Parapodia are lateral appendages found in Annelida (Polychaeta), not Arthropoda.
Final answer: Parapodia
Q96Single correctPrinciples of Inheritance and Variation
Which of the following most appropriately describes haemophilia?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2X-linked recessive gene disorder
Approach:
Recall the genetic basis of haemophilia.
Step 1:Haemophilia is a sex-linked recessive disorder; the defective gene lies on the X chromosome, so X-linked recessive is the most precise description.
Final answer: X-linked recessive gene disorder
Q97Single correctPhotosynthesis in Higher Plants
Emerson's enhancement effect and Red drop have been instrumental in the discovery of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Two photosystems operating simultaneously
Approach:
Recall significance of Emerson's experiments.
Step 1:The red drop and the enhancement effect demonstrated that maximum photosynthesis requires two light reactions, leading to the concept of two photosystems (PS I and PS II) operating together.
Final answer: Two photosystems operating simultaneously
Q98Single correctMineral Nutrition
In which of the following, all three are macronutrients?
(1)
(2)
(3)
(4)
SolutionAnswer: Option None of the four — every set contains at least one micronutrient
Approach:
Classify the listed elements into macro- and micronutrients.
Step 1:Macronutrients are needed in concentrations above about 10 mmol per kg of dry matter: C, H, O, N, P, K, Ca, Mg and S. Micronutrients are needed in trace amounts: Fe, Mn, Cu, Zn, B, Mo, Cl and Ni.
Step 2:Testing each set: boron, zinc and manganese are all micronutrients; iron, copper and molybdenum are all micronutrients; molybdenum and manganese are micronutrients while only magnesium is a macronutrient; and in the last set nitrogen and phosphorus are macronutrients but nickel is a micronutrient.
Final answer: No listed set contains three macronutrients; the closest, nitrogen and phosphorus with nickel, fails because nickel is a micronutrient.
Q99Single correctBreathing and Exchange of Gases
Name the chronic respiratory disorder caused mainly by cigarette smoking:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Emphysema
Approach:
Recall respiratory disorders linked to smoking.
Step 1:Emphysema is a chronic disorder where alveolar walls are damaged and the respiratory surface decreases; cigarette smoking is a major cause.
Final answer: Emphysema
Q100Single correctStrategies for Enhancement in Food Production
A system of rotating crops with legume or grass pasture to improve soil structure and fertility is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ley farming
Approach:
Recall farming system definitions.
Step 1:Ley farming rotates arable crops with legume or grass pasture leys, improving soil structure and fertility through nitrogen fixation and organic matter.
Final answer: Ley farming
Q101Single correctCell - The Unit of Life
Mitochondria and chloroplast are:
(a) semi-autonomous organelles.
(b) formed by division of pre-existing organelles and they contain DNA but lack protein synthesizing machinery.
Which one of the following options is correct?
(a) semi-autonomous organelles.
(b) formed by division of pre-existing organelles and they contain DNA but lack protein synthesizing machinery.
Which one of the following options is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(a) is true but (b) is false.
Approach:
Evaluate each statement about mitochondria and chloroplast.
Step 1:Both organelles are semi-autonomous and divide by fission, so (a) is true. They contain DNA AND their own ribosomes (protein synthesizing machinery), so the claim that they lack it makes (b) false.
Final answer: (a) is true but (b) is false.
Q102Single correctReproductive Health
In context of Amniocentesis, which of the following statement is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4It can be used for detection of Cleft palate.
Approach:
Recall uses and limits of amniocentesis.
Step 1:Amniocentesis analyses foetal cells/chromosomes to detect chromosomal and metabolic disorders and sex. Cleft palate is a morphological defect not detectable by this chromosomal test, making that statement incorrect.
Final answer: It can be used for detection of Cleft palate.
Q103Single correctPhotosynthesis in Higher Plants
In a chloroplast the highest number of protons are found in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Lumen of thylakoids
Approach:
Recall proton accumulation during photosynthesis.
Step 1:During the light reaction, protons are pumped from the stroma into the thylakoid lumen and also generated there by water splitting, so the lumen has the highest proton concentration.
Final answer: Lumen of thylakoids
Q104Single correctNeural Control and Coordination
Photosensitive compound in human eye is made up of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Opsin and Retinal
Approach:
Recall the composition of visual pigment.
Step 1:The photosensitive pigment rhodopsin is composed of the protein opsin combined with the aldehyde of vitamin A, retinal.
Final answer: Opsin and Retinal
Q105Single correctCell Cycle and Cell Division
Spindle fibres attach on to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Kinetochore of the chromosome
Approach:
Recall site of spindle fibre attachment.
Step 1:Spindle fibres attach to the kinetochore, a disc-shaped protein structure on the centromere of the chromosome.
Final answer: Kinetochore of the chromosome
Q106Single correctBiodiversity and Conservation
Which is the National Aquatic Animal of India?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2River dolphin
Approach:
Recall India's national aquatic animal.
Step 1:The Ganges river dolphin is designated the National Aquatic Animal of India.
Final answer: River dolphin
Q107Single correctMolecular Basis of Inheritance
Which of the following is required as inducer(s) for the expression of Lac operon?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3lactose
Approach:
Recall regulation of the lac operon.
Step 1:Lactose acts as the inducer of the lac operon; it binds the repressor and inactivates it, allowing transcription of the structural genes.
Final answer: lactose
Q108Single correctChemical Coordination and Integration
Which of the following pairs of hormones are not antagonistic (having opposite effects) to each other?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Relaxin - Inhibin
Approach:
Compare the action of each hormone pair.
Step 1:Parathormone/calcitonin oppose on calcium, insulin/glucagon oppose on glucose, aldosterone/ANF oppose on sodium-water balance. Relaxin and inhibin have unrelated, non-opposing roles, so they are not antagonistic.
Final answer: Relaxin - Inhibin
Q109Single correctCell - The Unit of Life
Microtubules are the constituents of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Spindle fibres, Centrioles and Cilia
Approach:
Identify structures built from microtubules.
Step 1:Microtubules form spindle fibres, centrioles, cilia and flagella. Peroxisomes, chromatin and nucleosomes are not microtubular, so only option 2 lists three microtubular structures.
Final answer: Spindle fibres, Centrioles and Cilia
Q110Single correctMolecular Basis of Inheritance
A complex of ribosomes attached to a single strand of RNA is known as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Polysome
Approach:
Recall translation machinery terminology.
Step 1:Several ribosomes translating a single mRNA simultaneously form a polysome (polyribosome).
Final answer: Polysome
Q111Single correctHuman Reproduction
Fertilization in humans is practically feasible only if:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the fallopian tube.
Approach:
Recall site and timing of human fertilization.
Step 1:Fertilization occurs at the ampullary-isthmic junction of the fallopian tube and requires the ovum and sperms to reach this site simultaneously.
Final answer: the ovum and sperms are transported simultaneously to ampullary – isthmic junction of the fallopian tube.
Q112Single correctBreathing and Exchange of Gases
Asthma may be attributed to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2allergic reaction of the mast cells in the lungs
Approach:
Recall the cause of asthma.
Step 1:Asthma results from an allergic (hypersensitivity) response in which mast cells in the lungs release mediators causing bronchial inflammation and difficulty in breathing.
Final answer: allergic reaction of the mast cells in the lungs
Q113Single correctPlant Growth and Development
The Avena curvature is used for bioassay of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3IAA
Approach:
Recall the Avena curvature bioassay.
Step 1:The Avena coleoptile curvature test is the classical bioassay for auxin (indole-3-acetic acid, IAA), where the degree of curvature is proportional to auxin concentration.
Final answer: IAA
Q114Single correctMorphology of Flowering Plants
The standard petal of a papilionaceous corolla is also called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Vexillum
Approach:
Recall the floral structure of a papilionaceous (butterfly-shaped) corolla found in Fabaceae.
Step 1:A papilionaceous corolla has five petals: one large posterior petal, two lateral petals, and two anterior fused petals.
Step 2:The largest posterior petal is the standard, technically termed vexillum; the lateral wings are alae and the fused anterior keel is carina.
Final answer: Vexillum
Q115Single correctMorphology of Flowering Plants
Tricarpellary, syncarpous gynoecium is found in flowers of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Liliaceae
Approach:
Identify the family whose gynoecium is tricarpellary and syncarpous.
Step 1:A tricarpellary gynoecium has three carpels, and syncarpous means the carpels are fused.
Step 2:In Liliaceae the gynoecium is tricarpellary and syncarpous with a superior ovary, whereas Solanaceae is bicarpellary, Fabaceae monocarpellary, and Poaceae monocarpellary.
Final answer: Liliaceae
Q116Single correctBiological Classification
One of the major components of cell wall of most fungi is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Chitin
Approach:
Recall the chemical composition of the fungal cell wall.
Step 1:Fungal cell walls are composed mainly of chitin, a polymer of N-acetylglucosamine.
Step 2:Peptidoglycan is found in bacteria and cellulose/hemicellulose in plant cell walls, ruling out the other options.
Final answer: Chitin
Q117Single correctHuman Reproduction
Select the incorrect statement :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3LH and FSH decrease gradually during the follicular phase.
Approach:
Evaluate each statement on gonadotropin function and identify the false one.
Step 1:FSH acts on Sertoli cells aiding spermiogenesis, LH triggers ovulation, and LH stimulates Leydig cells to secrete androgens; these are correct.
Step 2:During the follicular phase LH and FSH levels gradually increase toward the mid-cycle surge rather than decrease, making this statement incorrect.
Final answer: LH and FSH decrease gradually during the follicular phase.
Q118Single correctCell Cycle and Cell Division
In meiosis crossing over is initiated at:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pachytene
Approach:
Recall the sub-stages of prophase I and when crossing over begins.
Step 1:Prophase I proceeds through leptotene, zygotene, pachytene, diplotene, and diakinesis.
Step 2:Synapsis completes by zygotene; crossing over between non-sister chromatids of bivalents is initiated at pachytene with recombination nodules.
Final answer: Pachytene
Q119Single correctPrinciples of Inheritance and Variation
A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the plants were selfed the resulting genotypes were in the ratio of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 11 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
Approach:
Apply Mendel's monohybrid cross and read out the genotypic ratio.
Step 1:Crossing TT with tt gives all Tt F1 plants.
Step 2:Selfing Tt yields genotypes 1 TT : 2 Tt : 1 tt, that is tall homozygous : tall heterozygous : dwarf.
Final answer: 1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
Q120Single correctBiodiversity and Conservation
Which of the following is the most important cause of animals and plants being driven to extinction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Habitat loss and fragmentation
Approach:
Recall the major causes of biodiversity loss and their relative importance.
Step 1:The four major causes of biodiversity loss are habitat loss and fragmentation, over-exploitation, alien species invasion, and co-extinctions.
Step 2:Habitat loss and fragmentation is the most important cause driving species to extinction.
Final answer: Habitat loss and fragmentation
Q121Single correctEcosystem
Which one of the following is a characteristic feature of cropland ecosystem?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Least genetic diversity
Approach:
Identify the trait that distinguishes a cultivated cropland from natural ecosystems.
Step 1:Croplands are monocultures of a single high-yielding variety, so they show very low genetic diversity.
Step 2:Soil organisms and weeds are present, and managed croplands do not undergo natural ecological succession, eliminating the other options.
Final answer: Least genetic diversity
Q122Single correctHuman Reproduction
Changes in GnRH pulse frequency in females is controlled by circulating levels of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1estrogen and progesterone
Approach:
Recall the feedback regulators of hypothalamic GnRH pulses in females.
Step 1:GnRH secretion from the hypothalamus is pulsatile and is modulated by ovarian steroid feedback.
Step 2:The circulating ovarian steroids estrogen and progesterone control the changes in GnRH pulse frequency.
Final answer: estrogen and progesterone
Q123Single correctBiotechnology - Principles and Processes
Which of the following is not a feature of the plasmids?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Single-stranded
Approach:
Recall the structural and functional features of bacterial plasmids.
Step 1:Plasmids are extrachromosomal, circular, double-stranded DNA molecules that replicate independently and can be transferred between cells.
Step 2:Plasmids are double-stranded, not single-stranded, so single-stranded is not a feature.
Final answer: Single-stranded
Q124Single correctAnimal Kingdom
Which of the following features is not present in Periplaneta\ americana?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Indeterminate and radial cleavage during embryonic development
Approach:
Recall the developmental and structural features of the cockroach as an arthropod.
Step 1:Periplaneta americana is an arthropod with a schizocoelous body cavity, a chitinous exoskeleton of N-acetylglucosamine, and a metamerically segmented body.
Step 2:Arthropods are protostomes showing determinate and spiral cleavage, so indeterminate and radial cleavage (a deuterostome trait) is absent.
Final answer: Indeterminate and radial cleavage during embryonic development
Q125Single correctHuman Health and Disease
In higher vertebrates, the immune system can distinguish self-cells and non-self. If this property is lost due to genetic abnormality and it attacks self-cells, then it leads to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Auto-immune disease
Approach:
Relate loss of self/non-self discrimination to the resulting immune disorder.
Step 1:Normally the immune system tolerates self-antigens and reacts only to foreign antigens.
Step 2:When this discrimination fails and the body attacks its own cells, the condition is an auto-immune disease.
Final answer: Auto-immune disease
Q126Single correctPrinciples of Inheritance and Variation
Match Column I with Column II
| Column I | Column II |
|---|---|
| (a). Dominance | (i). Many genes govern a single character |
| (b). Codominance | (ii). In a heterozygous organism only one allele expresses itself |
| (c). Pleiotropy | (iii). In a heterozygous organism both alleles express themselves fully |
| (d). Polygenic inheritance | (iv). A single gene influences many characters |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Approach:
Match each genetic term with its correct definition.
Step 1:Dominance means in a heterozygote only one allele expresses, matching (ii); codominance means both alleles express fully, matching (iii).
Step 2:Pleiotropy is a single gene influencing many characters, matching (iv); polygenic inheritance is many genes governing a single character, matching (i).
Final answer: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Q127Single correctEnvironmental Issues
Joint Forest Management Concept was introduced in India during:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 31980s
Approach:
Recall the historical introduction of Joint Forest Management in India.
Step 1:Joint Forest Management (JFM) involves working with local communities to protect and manage forests.
Step 2:The JFM concept was introduced in India during the 1980s.
Final answer: 1980s
Q128Single correctPrinciples of Inheritance and Variation
Pick out the correct statements:
(a) Haemophilia is a sex-linked recessive disease.
(b) Down's syndrome is due to aneuploidy.
(c) Phenylketonuria is an autosomal recessive gene disorder.
(d) Sickle cell anaemia is an X-linked recessive gene disorder.
(a) Haemophilia is a sex-linked recessive disease.
(b) Down's syndrome is due to aneuploidy.
(c) Phenylketonuria is an autosomal recessive gene disorder.
(d) Sickle cell anaemia is an X-linked recessive gene disorder.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a), (b) and (c) are correct.
Approach:
Assess each genetic statement and select the combination of correct ones.
Step 1:Haemophilia is sex-linked recessive, Down's syndrome results from aneuploidy (trisomy 21), and phenylketonuria is an autosomal recessive disorder; all three are correct.
Step 2:Sickle cell anaemia is an autosomal recessive disorder, not X-linked, so (d) is wrong.
Final answer: (a), (b) and (c) are correct.
Q129Single correctBiological Classification
Which one of the following statements is wrong?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Eubacteria are also called false bacteria.
Approach:
Evaluate each alternate name and find the incorrect one.
Step 1:Cyanobacteria are blue-green algae, golden algae are desmids, and phycomycetes are algal fungi; these are correct.
Step 2:Eubacteria means true bacteria, not false bacteria, so this statement is wrong.
Final answer: Eubacteria are also called false bacteria.
Q130Single correctSexual Reproduction in Flowering Plants
Proximal end of the filament of stamen is attached to the:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Thalamus or petal
Approach:
Recall the structure of a stamen and the attachment of its filament.
Step 1:A stamen consists of a filament and an anther; the distal end of the filament bears the anther.
Step 2:The proximal (basal) end of the filament is attached to the thalamus or the petal of the flower.
Final answer: Thalamus or petal
Q131Single correctReproductive Health
Which of the following approaches does not give the defined action of contraceptive?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Vasectomy - prevents spermatogenesis
Approach:
Match each contraceptive with its actual mechanism and find the mismatch.
Step 1:Barrier methods prevent fertilization, IUDs increase phagocytosis and suppress sperm function, and hormonal contraceptives prevent ovulation and sperm entry; these are correctly defined.
Step 2:Vasectomy blocks the vas deferens so sperm do not reach the semen, but spermatogenesis continues in the testis; therefore this defined action is wrong.
Final answer: Vasectomy - prevents spermatogenesis
Q132Single correctBiotechnology - Principles and Processes
The taq polymerase enzyme is obtained from:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Thermus\ aquaticus
Approach:
Recall the source organism of the thermostable Taq DNA polymerase.
Step 1:Taq polymerase is a heat-stable DNA polymerase used in PCR.
Step 2:It is isolated from the thermophilic bacterium Thermus aquaticus, from which it derives its name.
Final answer: Thermus\ aquaticus
Q133Single correctHuman Reproduction
Identify the correct statement on 'inhibin':
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Is produced by granulose cells in ovary and inhibits the secretion of FSH.
Approach:
Recall the source and target of the hormone inhibin.
Step 1:Inhibin is produced by granulosa cells of the ovary (and by Sertoli cells in testes).
Step 2:Inhibin selectively suppresses the secretion of FSH from the anterior pituitary, identifying statement 2 as correct.
Final answer: Is produced by granulose cells in ovary and inhibits the secretion of FSH.
Q134Single correctBiotechnology and its Applications
Which part of the tobacco plant is infected by Meloidogyne\ incognita?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Root
Approach:
Recall the part of the tobacco plant attacked by the root-knot nematode.
Step 1:Meloidogyne incognita is a root-knot nematode that infests the roots of tobacco plants and reduces yield.
Step 2:RNA interference targeting nematode genes is used to develop tobacco resistant to this root infection.
Final answer: Root
Q135Single correctHuman Health and Disease
Antivenom injection contains preformed antibodies while polio drops that are administered into the body contain:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Attenuated pathogens
Approach:
Distinguish passive immunisation (antivenom) from active immunisation (polio vaccine).
Step 1:Antivenom supplies preformed antibodies, an example of passive immunisation.
Step 2:Oral polio drops contain attenuated (weakened) pathogens that stimulate the body to produce its own antibodies, an example of active immunisation.
Final answer: Attenuated pathogens
Q136Single correctCell: The Unit of Life
Which one of the following cell organelles is enclosed by a single membrane?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Lysosomes
Approach:
Classify each organelle by the number of bounding membranes.
Step 1:Mitochondria, chloroplasts and nuclei are each bounded by a double (two) membrane envelope.
Step 2:Lysosomes are vesicles budded from the Golgi apparatus and are bounded by a single membrane.
Final answer: Lysosomes
Q137Single correctLocomotion and Movement
Lack of relaxation between successive stimuli in sustained muscle contraction is known as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Tetanus
Approach:
Match the physiological term to a sustained contraction without relaxation.
Step 1:When stimuli arrive in rapid succession the muscle cannot relax between them and the twitches fuse into a smooth, maintained contraction.
Step 2:This summated, sustained state of contraction without relaxation is termed tetanus.
Final answer: Tetanus
Q138Single correctMorphology of Flowering Plants
Which of the following is not a stem modification?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pitcher of
Approach:
Identify which structure originates from a leaf rather than a stem.
Step 1:Thorns of citrus, tendrils of cucumber and the flattened phylloclade of Opuntia are all axillary or apical stem modifications.
Step 2:The pitcher of Nepenthes is a modification of the leaf lamina into an insect-trapping pitcher, so it is not a stem modification.
Final answer: Pitcher of
Q139Single correctCell: The Unit of Life
Water soluble pigments found in plant cell vacuoles are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Anthocyanins
Approach:
Distinguish water-soluble vacuolar pigments from lipid-soluble plastid pigments.
Step 1:Chlorophylls, xanthophylls and carotenoids are fat-soluble pigments located within plastids.
Step 2:Anthocyanins are water-soluble flavonoid pigments dissolved in the cell sap of the vacuole, giving red, purple and blue colours.
Final answer: Anthocyanins
Q140Single correctPlant Kingdom
Select the correct statement:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 is one of the tallest trees
Approach:
Evaluate each statement against gymnosperm biology.
Step 1:Gymnosperms are exclusively heterosporous, so statement 1 is wrong; Salvinia is a pteridophyte, so statement 2 is wrong.
Step 2:Gymnosperm leaves bear thick cuticle and sunken stomata adapting them to extremes, so statement 4 is wrong; Sequoia (giant redwood) is among the tallest trees.
Final answer: is one of the tallest trees
Q141Single correctBiotechnology: Principles and Processes
Which of the following is not required for any of the techniques of DNA fingerprinting available at present?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Zinc finger analysis
Approach:
Identify the tool that is not part of DNA fingerprinting.
Step 1:PCR amplifies DNA, restriction enzymes cut DNA at specific sites, and DNA-DNA hybridization with labelled probes detects VNTR sequences in fingerprinting.
Step 2:Zinc finger analysis relates to DNA-binding protein motifs and is not a technique used in DNA fingerprinting.
Final answer: Zinc finger analysis
Q142Single correctStructural Organisation in Animals
Which type of tissue correctly matches with its location?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Smooth muscle — Wall of intestine
Approach:
Match each tissue to its correct anatomical location.
Step 1:Tendons are dense regular connective tissue not areolar; the tip of nose has hyaline/elastic cartilage not transitional epithelium; the stomach lining is columnar not cuboidal epithelium.
Step 2:Smooth (visceral) muscle forms the wall of the intestine, which is the correct match.
Final answer: Smooth muscle — Wall of intestine
Q143Single correctPhotosynthesis in Higher Plants
A plant in your garden avoids photorespiratory losses, has improved water use efficiency, shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilisation. In which of the following physiological groups would you assign this plant?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Map the listed adaptive traits to a photosynthetic group.
Step 1:Avoidance of photorespiration through CO2 concentration in bundle sheath cells, high water-use and nitrogen-use efficiency, and high optimum temperature are hallmark features.
Step 2:These traits collectively define C4 plants.
Final answer:
Q144Single correctEvolution
Which of the following structures is homologous to the wing of a bird?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Flipper of Whale
Approach:
Apply the definition of homology, common origin with different function.
Step 1:Shark dorsal fin and moth wing share no common forelimb origin; the rabbit hind limb is a hind, not fore, limb.
Step 2:The whale flipper and the bird wing are both modified pentadactyl forelimbs built on the same skeletal plan, making them homologous.
Final answer: Flipper of Whale
Q145Single correctAnimal Kingdom
Which of the following characteristic features always holds true for the corresponding group of animals?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cartilaginous endoskeleton — Chondrichthyes
Approach:
Test each feature for being universally true of the stated group.
Step 1:Some mammals (monotremes) are oviparous; many chordates such as agnathans lack jaws; crocodiles among reptiles have a 4-chambered heart, so options 2, 3, 4 are not always true.
Step 2:All members of Chondrichthyes possess a cartilaginous endoskeleton without exception.
Final answer: Cartilaginous endoskeleton — Chondrichthyes
Q146Single correctCell Cycle and Cell Division
Which of the following statements is not true for cancer cells in relation to mutations?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Mutations inhibit production of telomerase.
Approach:
Identify the statement that contradicts cancer cell biology.
Step 1:Cancer arises when proto-oncogenes are activated, cell-cycle control is lost, and telomerase activity is enhanced, which permits unlimited division.
Step 2:Cancer cells switch telomerase production on rather than inhibiting it, so the statement that mutations inhibit telomerase production is not true.
Final answer: Mutations inhibit production of telomerase.
Q147Single correctChemical Coordination and Integration
The amino acid Tryptophan is the precursor for the synthesis of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Melatonin and Serotonin
Approach:
Trace the biosynthetic origin of each listed hormone pair.
Step 1:Thyroid hormones derive from tyrosine, while estrogen, progesterone, cortisol and cortisone are steroids derived from cholesterol.
Step 2:Tryptophan is hydroxylated and decarboxylated to serotonin, which is further converted to melatonin in the pineal gland.
Final answer: Melatonin and Serotonin
Q148Single correctOrigin and Evolution of Life
Following are the two statements regarding the origin of life :
(a) The earliest organisms that appeared on the earth were non-green and presumably anaerobes.
(b) The first autotrophic organisms were the chemoautotrophs that never released oxygen.
Of the above statements which one of the following options is correct?
(a) The earliest organisms that appeared on the earth were non-green and presumably anaerobes.
(b) The first autotrophic organisms were the chemoautotrophs that never released oxygen.
Of the above statements which one of the following options is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Both (a) and (b) are correct.
Approach:
Assess the historical accuracy of each statement about life's origin.
Step 1:The first cells arose in an oxygen-free reducing atmosphere, so they were non-green anaerobes, making statement (a) correct.
Step 2:The earliest autotrophs were chemoautotrophs that obtained energy from chemical oxidation and did not evolve oxygen, so statement (b) is also correct.
Final answer: Both (a) and (b) are correct.
Q149Single correctBody Fluids and Circulation
Reduction in pH of blood will:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3decrease the affinity of hemoglobin with oxygen.
Approach:
Apply the Bohr effect to a fall in blood pH.
Step 1:A reduction in pH (rise in H+ and CO2) shifts the oxygen dissociation curve to the right.
Step 2:A rightward shift lowers haemoglobin's affinity for oxygen, promoting unloading in tissues.
Final answer: decrease the affinity of hemoglobin with oxygen.
Q150Single correctEvolution
Analogous structures are a result of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Convergent evolution
Approach:
Link analogy to the underlying evolutionary process.
Step 1:Analogous structures perform similar functions but differ in origin, arising in unrelated lineages exposed to similar environments.
Step 2:This independent acquisition of similar features is the outcome of convergent evolution.
Final answer: Convergent evolution
Q151Single correctBiotechnology: Principles and Processes
Which of the following is a restriction endonuclease?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Hind II
Approach:
Identify the enzyme that cleaves DNA at specific recognition sequences.
Step 1:Protease degrades proteins, DNase I cuts DNA non-specifically, and RNase digests RNA, so none are sequence-specific DNA cutters.
Step 2:Hind II is a restriction endonuclease that cleaves DNA at a specific palindromic recognition sequence.
Final answer: Hind II
Q152Single correctEcosystem
The term ecosystem was coined by:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2A.G. Tansley
Approach:
Attribute the coining of the term ecosystem to its author.
Step 1:Haeckel coined ecology, Odum popularised modern ecosystem ecology, and Warming pioneered plant ecology.
Step 2:A.G. Tansley introduced the term ecosystem in 1935.
Final answer: A.G. Tansley
Q153Single correctBiomolecules
Which one of the following statements is wrong?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Glycine is a sulphur containing amino acid.
Approach:
Find the false biochemical statement among the four.
Step 1:Sucrose is indeed a disaccharide, cellulose a polysaccharide, and uracil a pyrimidine base, so those statements are correct.
Step 2:Glycine has the side chain of a single hydrogen and contains no sulphur; the sulphur-containing amino acids are cysteine and methionine.
Final answer: Glycine is a sulphur containing amino acid.
Q154Single correctPlant Kingdom
In bryophytes and pteridophytes, transport of male gametes requires:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Water
Approach:
Recall the fertilisation medium of lower plants.
Step 1:Bryophytes and pteridophytes produce flagellated, motile male gametes (antherozoids).
Step 2:These swimming gametes require a film of water to reach the archegonium, so water is essential for their transport.
Final answer: Water
Q155Single correctOrganisms and Populations
When does the growth rate of a population following the logistic model equal zero? The logistic model is given as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1when N/K is exactly one.
Approach:
Determine the condition that makes the logistic growth rate vanish.
Step 1:The growth rate becomes zero when the bracketed term equals zero.
Step 2:Thus the rate is exactly zero when N/K equals one, that is when N equals the carrying capacity exactly.
Final answer: when N/K is exactly one.
Q156Single correctSexual Reproduction in Flowering Plants
Which one of the following statements is not true?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Tapetum helps in the dehiscence of anther
Approach:
Identify the incorrect statement about pollen and anther.
Step 1:Exine is built of resistant sporopollenin, certain pollens cause allergies such as Parthenium, and cryopreserved pollen aids breeding, so those statements are correct.
Step 2:The tapetum nourishes developing pollen grains; dehiscence of the anther is brought about by the endothecium, so statement 1 is not true.
Final answer: Tapetum helps in the dehiscence of anther
Q157Single correctEcosystem
Which of the following would appear as the pioneer organisms on bare rocks?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Lichens
Approach:
Recall the first colonisers in primary succession on rock.
Step 1:Primary succession on bare rock (xerarch) begins with organisms able to grow on dry rock and secrete acids that weather it.
Step 2:Lichens are the pioneer organisms that colonise bare rock, preparing soil for mosses and higher plants.
Final answer: Lichens
Q158Single correctMolecular Basis of Inheritance
Which one of the following is the starter codon?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1AUG
Approach:
Distinguish the initiation codon from termination codons.
Step 1:UGA, UAA and UAG are the three stop codons that terminate translation.
Step 2:AUG codes for methionine and serves as the initiation (start) codon for translation.
Final answer: AUG
Q159Single correctAnimal Kingdom
Which one of the following characteristics is not shared by birds and mammals?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Viviparity
Approach:
Compare reproductive and other organ-system traits between Aves and Mammalia.
Step 1:Both birds and mammals possess a bony (ossified) endoskeleton, respire through lungs and are warm blooded (homeothermic).
Step 2:Birds are oviparous (egg-laying), whereas viviparity (giving birth to live young) is characteristic of most mammals; hence it is not common to both.
Final answer: Viviparity
Q160Single correctBiological Classification
Nomenclature is governed by certain universal rules.
Which one of the following is contrary to the rules of nomenclature?
Which one of the following is contrary to the rules of nomenclature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Biological names can be written in any language
Approach:
Recall the universal rules of binomial nomenclature.
Step 1:Biological names are derived from Latin (or latinised), written in italics, with the genus name first and the specific epithet second; when handwritten they are underlined.
Step 2:Names are not written in any arbitrary language; they follow Latin conventions, so this statement violates the rules.
Final answer: Biological names can be written in any language
Q161Single correctBody Fluids and Circulation
Blood pressure in the pulmonary artery is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3more than that in the pulmonary vein.
Approach:
Compare pressures across the pulmonary circuit.
Step 1:Blood is driven by the right ventricle into the pulmonary artery, then through the lung capillaries, and returns at low pressure via the pulmonary vein.
Step 2:Therefore the pulmonary artery carries blood at a higher pressure than the pulmonary vein, though much lower than the systemic aorta/carotid.
Final answer: more than that in the pulmonary vein.
Q162Single correctMorphology of Flowering Plants
Cotyledon of maize grain is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4scutellum
Approach:
Identify the modified single cotyledon of a monocot grain.
Step 1:In maize (a monocot), the single shield-shaped cotyledon that lies against the endosperm and absorbs nutrients is the scutellum.
Step 2:Plumule and radicle are protected by coleoptile and coleorhiza respectively, which are sheaths, not the cotyledon.
Final answer: scutellum
Q163Single correctDigestion and Absorption
In the stomach, gastric acid is secreted by the :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2parietal cells
Approach:
Match the secretory product to the gastric gland cell type.
Step 1:Parietal (oxyntic) cells of the gastric glands secrete hydrochloric acid and intrinsic factor.
Step 2:Peptic (chief/zymogen) cells secrete pepsinogen, while gastrin is a hormone from G-cells; ‘acidic cells’ is not a defined type.
Final answer: parietal cells
Q164Single correctEnvironmental Issues
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancers?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ozone
Approach:
Link atmospheric gas depletion to UV exposure and skin cancer.
Step 1:Stratospheric ozone absorbs harmful ultraviolet (UV-B) radiation from sunlight.
Step 2:Depletion of the ozone layer increases UV-B reaching the surface, raising the incidence of skin cancers.
Final answer: Ozone
Q165Single correctBiological Classification
Chrysophytes, Euglenoids, Dinoflagellates and Slime moulds are included in the kingdom:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Protista
Approach:
Classify the listed groups within Whittaker's five-kingdom scheme.
Step 1:Chrysophytes, euglenoids, dinoflagellates and slime moulds are all eukaryotic single-celled or simple eukaryotic organisms.
Step 2:In the five-kingdom classification these are placed in Kingdom Protista.
Final answer: Protista
Q166Single correctTransport in Plants
Water vapour comes out from the plant leaf through the stomatal opening. Through the same stomatal opening carbon dioxide diffuses into the plant during photosynthesis. Reason out the above statements using one of following options:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Both processes can happen together because the diffusion coefficient of water and C is different.
Approach:
Reason simultaneous transpiration and CO2 uptake through the same pore.
Step 1:Through an open stoma, water vapour diffuses outward (transpiration) while CO2 diffuses inward for photosynthesis at the same time.
Step 2:This is possible because water vapour and CO2 have different diffusion coefficients, allowing both to move along their own concentration gradients through the same opening.
Final answer: Both processes can happen together because the diffusion coefficient of water and C is different.
Q167Single correctBody Fluids and Circulation
In mammals, which blood vessel would normally carry largest amount of urea?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Hepatic Vein
Approach:
Trace where urea is produced and how it enters the bloodstream.
Step 1:Urea is synthesised in the liver via the ornithine cycle; blood leaving the liver carries the freshly produced urea.
Step 2:The hepatic vein drains the liver, so it carries the largest amount of urea; the renal vein carries blood after the kidneys have removed urea (lowest urea).
Final answer: Hepatic Vein
Q168Single correctSexual Reproduction in Flowering Plants
Seed formation without fertilization in flowering plants involves the process of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Apomixis
Approach:
Identify the asexual seed-forming mechanism in angiosperms.
Step 1:Apomixis is the formation of seeds without fertilization, producing offspring genetically identical to the parent.
Step 2:Sporulation, budding and somatic hybridization are unrelated to apomictic seed formation.
Final answer: Apomixis
Q169Single correctMicrobes in Human Welfare
Which of the following is wrongly matched in the given table?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Clostridium butylicum — Lipase — Removal of oil stains
Approach:
Verify each microbe-product-application triplet against known facts.
Step 1:Trichoderma polysporum yields cyclosporin A (immunosuppressant), Monascus purpureus yields statins (cholesterol lowering), and Streptococcus yields streptokinase (clot dissolving) — all correct.
Step 2:Lipase used to remove oily stains in detergents is obtained from fungi/bacteria; pairing it with Clostridium butylicum is incorrect (lipase is the wrong source/match here).
Final answer: Clostridium butylicum — Lipase — Removal of oil stains
Q170Single correctPrinciples of Inheritance and Variation
In a testcross involving dihybrid flies, more parental-type offspring were produced than the recombinant-type offspring. This indicates:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The two genes are linked and present on the same chromosome.
Approach:
Interpret an excess of parental over recombinant types in a testcross.
Step 1:If two genes assorted independently, parental and recombinant offspring would appear in equal proportions.
Step 2:A higher frequency of parental types means the alleles tend to stay together, indicating the genes are linked on the same chromosome.
Final answer: The two genes are linked and present on the same chromosome.
Q171Single correctLocomotion and Movement
It is much easier for a small animal to run uphill than for a large animal, because:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Smaller animals have a higher metabolic rate
Approach:
Relate body size to metabolic rate and uphill locomotion.
Step 1:Smaller animals have a larger surface-area-to-volume ratio and a correspondingly higher mass-specific metabolic rate.
Step 2:This higher metabolic capacity allows them to generate the extra power needed to run uphill more easily than large animals.
Final answer: Smaller animals have a higher metabolic rate
Q172Single correctCell Cycle and Cell Division
Which of the following is not a characteristic feature during mitosis in somatic cells?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Synapsis
Approach:
Distinguish mitotic events from meiosis-specific events.
Step 1:Spindle fibre formation, disappearance of the nucleolus and chromosome movement all occur during mitosis.
Step 2:Synapsis (pairing of homologous chromosomes) occurs only during prophase I of meiosis, not in mitosis.
Final answer: Synapsis
Q173Single correctSexual Reproduction in Flowering Plants
Which of the following statements is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1pollen grains of many species can germinate on the stigma of a flower, but only one pollen tube of the same species grows into the style.
Approach:
Identify the incorrect statement on pollen-pistil interaction.
Step 1:Statements 2, 3 and 4 are factually correct descriptions of pollen robbers, chemical regulation of pollen growth, and reptile pollinators.
Step 2:Pollen-pistil interaction allows the pistil to recognise compatible (same-species) pollen and reject incompatible pollen; the statement that pollen of many species germinate but only one same-species tube grows misrepresents this selective recognition.
Final answer: pollen grains of many species can germinate on the stigma of a flower, but only one pollen tube of the same species grows into the style.
Q174Single correctAnatomy of Flowering Plants
Specialised epidermal cells surrounding the guard cells are called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Subsidiary cells
Approach:
Name the epidermal cells associated with guard cells.
Step 1:The stomatal apparatus consists of guard cells and the surrounding specialised epidermal cells.
Step 2:These specialised epidermal cells around the guard cells are termed subsidiary cells.
Final answer: Subsidiary cells
Q175Single correctDigestion and Absorption
Which of the following guards the opening of hepatopancreatic duct into the duodenum?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Sphincter of Oddi
Approach:
Identify the sphincter at the hepatopancreatic duct opening.
Step 1:The common bile and pancreatic ducts join to form the hepatopancreatic duct, which opens into the duodenum.
Step 2:This opening is guarded by the sphincter of Oddi (hepatopancreatic sphincter).
Final answer: Sphincter of Oddi
Q176Single correctMorphology of Flowering Plants
Stems modified into flat green organs performing the functions of leaves are known as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Phylloclades
Approach:
Name the flattened green photosynthetic stem modification.
Step 1:In xerophytes, stems become flattened, green and photosynthetic, taking over the role of leaves.
Step 2:These flattened green stems of unlimited growth performing leaf functions are called phylloclades.
Final answer: Phylloclades
Q177Single correctMicrobes in Human Welfare
The primitive prokaryotes responsible for the production of biogas from the dung of ruminant animals, include:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Methanogens
Approach:
Identify the archaebacteria producing biogas in ruminant dung.
Step 1:Biogas (methane-rich) is produced by microbes present in the rumen and dung of cattle.
Step 2:These primitive prokaryotes are methanogens (archaebacteria), which generate methane and hence biogas.
Final answer: Methanogens
Q178Single correctEnvironmental Issues
A river with an inflow of domestic sewage rich in organic waste may result in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Death of fish due to lack of oxygen.
Approach:
Trace the effect of organic sewage on dissolved oxygen and fish.
Step 1:Organic-rich sewage raises the biochemical oxygen demand as microbes decompose the waste, consuming dissolved oxygen.
Step 2:The resulting oxygen depletion suffocates aquatic animals, leading to death of fish.
Final answer: Death of fish due to lack of oxygen.
Q179Single correctCell Cycle and Cell Division
A cell at telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Polyploidy
Approach:
Determine the outcome of failed cytokinesis after chromosome doubling.
Step 1:Absence of cell plate formation means cytokinesis fails, so the two complete chromosome sets remain within one cell.
Step 2:Possessing more than two complete sets of chromosomes is termed polyploidy.
Final answer: Polyploidy
Q180Single correctBiomolecules
A typical fat molecule is made up of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2One glycerol and three fatty acid molecules
Approach:
Recall the structure of a triglyceride (typical fat).
Step 1:A typical fat is a triglyceride, formed by esterification of glycerol with fatty acids.
Step 2:Each of glycerol's three hydroxyl groups bonds with one fatty acid, giving one glycerol and three fatty acid molecules.
Final answer: One glycerol and three fatty acid molecules
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