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![Three bicyclo[2.2.1] structures. I is a bridged bicyclic ketone whose carbonyl sits at the one-carbon bridge. II is a bicyclic ketone with the carbonyl on the ring and two phenyl groups on the adjacent carbon. III is the same bicyclic ketone with the carbonyl on the ring and an unsubstituted adjacent carbon.](/api/qna-image?path=QnA%2F0b631c94-c7ee-48c9-b195-b99ec63eda67%2F5ef15551-aed6-4f93-a8d5-a219b7a5d19b%2F2cbdf665-15ab-4b56-a08c-af446b298807%2F2cbdf665-15ab-4b56-a08c-af446b298807%2Fimages%2FQ88_bicyclic_ketones_v3.webp)
NEET 2016 Jul 24 Question Paper with Solutions
All 180 questions from the NEET 2016 (Jul 24) paper — Physics (45), Chemistry (45) and Biology (90) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.Physics PYQs 2016Chemistry PYQs 2016Biology PYQs 2016
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctUnits and Measurements
Planck's constant , speed of light in vacuum and Newton's gravitational constant are three fundamental constants. Which of the following combinations of these has the dimension of length?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Form a product of powers of h, c and G whose net dimensions reduce to length [L], by writing each constant in [M], [L], [T] and matching exponents.
Step 1:Assume length L equals h to the power x, c to the power y and G to the power z, and write the combined dimensions.
Step 2:Match the exponents of M, L and T on both sides.
Step 3:Substituting x = z = 1/2 and y = -3/2 gives the required combination.
Final answer:
Q2Single correctMotion in a Straight Line
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by and . At what time do the cars have the same velocity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Differentiate each position function with respect to time to obtain the velocities, then equate them and solve for t.
Step 1:Differentiate the position of P to get its velocity.
Step 2:Differentiate the position of Q to get its velocity.
Step 3:Equate the two velocities and solve for t.
Final answer:
Q3Single correctMotion in a Plane
In the given figure, represents the total acceleration of a particle moving in the clockwise direction in a circle of radius m at a given instant of time. The speed of the particle is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Resolve the total acceleration into its centripetal (radial) component, which the 30 degree angle fixes as a cos or sin of the total, then equate that radial part to v squared over R and solve for speed.
Step 1:The total acceleration makes 30 degrees with the radius, so its component along the radius (centripetal direction) is a times sin of 30 degrees.
Step 2:Equate this radial component to the centripetal acceleration expression.
Step 3:Substitute R = 2.5 m and take the square root.
Final answer:
Q4Single correctLaws of Motion
A rigid ball of mass m strikes a rigid wall at and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Impulse equals the change in momentum. With elastic reflection the component of velocity parallel to the wall is unchanged, while the normal component reverses, so only the normal momentum contributes to the impulse.
Step 1:The ball strikes at 60 degrees, so the component of velocity normal to the wall is V cos of 60 degrees, which reverses on reflection.
Step 2:The change in normal momentum has magnitude twice the incoming normal momentum, since the component reverses with equal magnitude.
Final answer:
Q5Single correctWork, Energy and Power
A bullet of mass 10 g moving horizontally with a velocity of strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the block's launch speed from energy conservation over the 10 cm rise, then apply linear momentum conservation to the bullet-block collision to obtain the emerging bullet speed.
Step 1:Determine the block's speed just after impact from the height it rises, using g = 10 m per second squared and h = 0.1 m.
Step 2:Apply momentum conservation with bullet mass 0.01 kg, initial bullet speed 400 m/s, block mass 2 kg.
Step 3:Solve for the emerging bullet speed.
Final answer:
Q6Single correctWork, Energy and Power
Two identical balls A and B having velocities of 0.5 m/s and – 0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a one-dimensional elastic collision between two equal masses the particles simply exchange their velocities, so determine the new velocities of B and A by swapping the initial values.
Step 1:In a head-on elastic collision of identical masses the velocities are interchanged.
Step 2:The question asks for B then A, so report the velocity of B first followed by A.
Final answer:
Q7Single correctWork, Energy and Power
A particle moves from a point to when a force of N is applied. How much work has been done by the force ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Work for a constant force equals the dot product of the force with the displacement, where the displacement is the difference of final and initial position vectors.
Step 1:Form the displacement vector as the difference of the final and initial position vectors.
Step 2:Take the dot product of the force with the displacement.
Final answer:
Q8Single correctSystem of Particles and Rotational Motion
Two rotating bodies A and B of masses m and 2m with moments of inertia and have equal kinetic energy of rotation. If and be their respective angular momenta, then –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Express rotational kinetic energy in terms of angular momentum and moment of inertia, then compare the two angular momenta using the equal-energy condition and the inequality between the moments of inertia.
Step 1:Write the equal kinetic energies in terms of angular momentum and moment of inertia.
Step 2:Form the ratio of angular momenta from this relation.
Step 3:Since the moment of inertia of B exceeds that of A, the ratio is less than one, so is smaller than .
Final answer:
Q9Single correctSystem of Particles and Rotational Motion
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Write each rotational kinetic energy using its moment of inertia and angular speed, insert the standard moments of inertia for a solid sphere and a solid cylinder, and use the doubled angular speed of the cylinder.
Step 1:Write the ratio of energies in terms of moments of inertia and the square of angular speeds, with the cylinder at twice the sphere's angular speed.
Step 2:Simplify the numerical ratio.
Final answer:
Q10Single correctSystem of Particles and Rotational Motion
A light rod of length l has two masses and attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Locate the centre of mass along the rod, find the distance of each mass from it, then sum the products of mass and distance squared to obtain the moment of inertia.
Step 1:Balance moments about the centre of mass to locate the distances of each mass from it.
Step 2:Substitute these distances into the sum of mass times distance squared.
Step 3:Cancel the common factor of the total mass.
Final answer:
Q11Single correctGravitation
Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Graph: g rises linearly from O to R, then falls as
Approach:
Compare the behaviour of g inside the earth, where it grows in direct proportion to distance from the centre, with its behaviour outside, where it falls off as the inverse square of the distance, and pick the graph matching both regimes.
Step 1:Inside the earth gravity is proportional to the distance from the centre, so it increases linearly from zero at the centre up to its surface value at r = R.
Step 2:Outside the earth gravity falls off as the inverse square of the distance from the centre.
Step 3:A curve that rises straight to r = R and then decays as one over r squared reproduces both regimes.
Final answer: Graph: g rises linearly from O to R, then falls as
Q12Single correctGravitation
A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of , the value of acceleration due to gravity at the earth's surface, is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Use the standard expression for the total mechanical energy of a circular orbit at radius R plus h, then replace the product of G and the earth's mass using the surface gravity relation.
Step 1:Write the total energy of a satellite in a circular orbit of radius R plus h.
Step 2:Substitute the product of G and earth's mass with surface gravity times radius squared.
Final answer:
Q13Single correctMechanical Properties of Fluids
A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is J, the value of the surface tension of the liquid is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Relate the work done in extending a liquid film to the change in its surface area times the surface tension, remembering a film has two surfaces, then solve for the surface tension.
Step 1:Compute the change in geometric area between the final and initial rectangles.
Step 2:A film has two free surfaces, so the work equals surface tension times twice the change in area; solve for the surface tension.
Step 3:Evaluate the fraction.
Final answer:
Q14Single correctMechanical Properties of Fluids
Three liquids of densities and (with ), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact and obey –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
From the capillary rise formula with equal height, surface tension and radius, the cosine of the contact angle must vary in direct proportion to density; translate that proportionality into an ordering of the angles.
Step 1:With h, T and r fixed across the three capillaries, the cosine of the contact angle is directly proportional to the density.
Step 2:Since density of the first liquid is greatest, its cosine is largest, so its angle is smallest; the ordering of cosines reverses for the angles.
Step 3:Positive cosines keep all three contact angles below a right angle, so they increase in the order theta1 < theta2 < theta3, all under pi/2.
Final answer:
Q15Single correctThermal Properties of Matter
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at C, while the other one is at C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2more than
Approach:
Equate the heat released by the hotter body to the heat gained by the colder body, noting that the hotter body has a larger heat capacity because capacity rises with temperature, which biases the equilibrium toward the hot side.
Step 1:The hotter body sits at a higher temperature, so its heat capacity exceeds that of the colder body throughout the process.
Step 2:Balancing heat lost against heat gained, the larger capacity of the hot body releases more heat per degree, so the common temperature settles above the simple midpoint.
Final answer: more than
Q16Single correctThermal Properties of Matter
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Apply Newton's law of cooling in its approximate finite-difference form over each 10-minute interval, equating the rate of fall to a constant times the excess of the average temperature over the surroundings, then chain the two intervals.
Step 1:Write the cooling law for the first interval where the body falls from 3T to 2T in ten minutes.
Step 2:Write the same law for the second interval where the body cools from 2T to an unknown final temperature.
Step 3:Divide the two relations to remove the constant k and the factor of ten, then solve for the final temperature.
Final answer:
Q17Single correctThermodynamics
One mole of an ideal monatomic gas undergoes a process described by the equation constant. The heat capacity of the gas during this process is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Treat the relation as a polytropic process with index 3 and use the molar heat capacity formula for a polytropic process, inserting the monatomic value of the constant-volume heat capacity.
Step 1:Identify the polytropic index from the process equation, which has P times V cubed constant.
Step 2:Substitute the monatomic constant-volume capacity and the index into the polytropic heat capacity formula.
Step 3:Combine the terms.
Final answer:
Q18Single correctThermodynamics
The temperature inside a refrigerator is C and the room temperature is C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For an ideal refrigerator the heat delivered to the room per unit work is the ratio of the hot-reservoir absolute temperature to the difference of the two absolute temperatures; convert the Celsius temperatures to kelvin and simplify.
Step 1:Express the heat delivered to the room per joule of work using absolute temperatures of room and interior.
Step 2:Convert to Celsius; the constant 273 cancels in the temperature difference but remains in the numerator.
Final answer:
Q19Single correctKinetic Theory
A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Write the ideal gas law in terms of number of molecules and Boltzmann's constant to get the number density, then multiply by the mass per molecule to obtain the mass density.
Step 1:Rearrange the molecular ideal gas law to obtain the number of molecules per unit volume.
Step 2:Multiply the number density by the mass of a single molecule to obtain the mass density.
Final answer:
Q20Single correctOscillations
A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Write the spring-mass period for both masses, take the ratio of the two periods to cancel the spring constant, then solve the resulting equation for the original mass.
Step 1:Write the periods for mass m and for mass m plus one, with 3 s and 5 s respectively.
Step 2:Divide the first by the second to remove the spring constant and the constant factor, then square.
Step 3:Solve the linear equation for m.
Final answer:
Q21Single correctWaves
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Express the second overtone frequency of an open pipe and the first overtone frequency of a closed pipe in terms of their lengths, equate the two frequencies and solve for the open-pipe length.
Step 1:The second overtone of an open pipe is the third harmonic, and the first overtone of a closed pipe is its third harmonic; write both frequencies.
Step 2:Cancel the common factors and solve for the open-pipe length.
Final answer:
Q22Single correctWaves
Three sound waves of equal amplitudes have frequencies . They superimpose to give beats. The number of beats produced per second will be –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The audible beat frequency among the three equally spaced frequencies corresponds to the largest frequency difference present, since pairwise differences of one combine to a dominant difference of two between the extreme frequencies.
Step 1:Identify the extreme frequencies among the three.
Step 2:Take the difference of the extreme frequencies to obtain the maximum beat frequency.
Final answer:
Q23Single correctElectric Charges and Fields
An electric dipole is placed at an angle of with an electric field intensity N/C. It experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is –
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Use the torque on a dipole as the product of dipole moment, field strength and the sine of the angle, write the dipole moment as charge times length, and solve for the charge.
Step 1:Express the torque in terms of charge, dipole length, field and angle, then isolate the charge.
Step 2:Substitute torque 4 N m, length 0.02 m, field 2e5 N/C and sine of 30 degrees equal to one half.
Step 3:Evaluate the expression.
Final answer:
Q24Single correctElectrostatic Potential and Capacitance
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants , , and as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Treat the top three dielectrics (each of width A/3 and thickness d/2) as three capacitors in parallel, then combine that parallel group in series with the bottom dielectric (area A, thickness d/2). Equate the net capacitance to a single equivalent dielectric filling area A and separation d.
Step 1:Each top dielectric spans area A/3 and thickness d/2, so the parallel group of three gives capacitance C1.
Step 2:The bottom dielectric spans area A and thickness d/2.
Step 3:C1 and C2 are in series; combine and set equal to the single equivalent capacitance with constant k.
Step 4:Cancel common factors to obtain the relation for k.
Final answer:
Q25Single correctCurrent Electricity
The potential difference () between the points A and B in the given figure is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Travel from A to B along the branch, accounting for the IR drop across each resistor and the EMF of the battery. A current of 2 A flows from A toward B through a 2 Ω resistor, a 3 V source, and a 1 Ω resistor.
Step 1:Across the 2 Ω resistor the potential falls by I times R with I = 2 A.
Step 2:Crossing the 3 V source from + to - lowers the potential by 3 V.
Step 3:Across the 1 Ω resistor the potential falls by another I times R.
Step 4:Rearrange to find the required potential difference.
Final answer:
Q26Single correctCurrent Electricity
A filament bulb (500 W, 100 V) is to be used in 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the bulb resistance from its rated power and voltage, find the operating current, then size R so that the remaining 130 V (out of 230 V) drops across it.
Step 1:Bulb resistance from rated values.
Step 2:Operating current of the bulb at 100 V.
Step 3:The series resistor must drop the extra 130 V at the same current.
Final answer:
Q27Single correctMoving Charges and Magnetism
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Keep the total wire length fixed. For one turn of radius R the field is known; for n turns the same length forms a smaller radius R/n, and the field scales with both the number of turns and the inverse radius.
Step 1:Single turn of radius R gives the reference field.
Step 2:Fixed wire length wound into n turns gives radius r = R/n.
Step 3:Field of the n-turn coil substituting r = R/n.
Final answer:
Q28Single correctMagnetism and Matter
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by is W. Now the torque required to keep the magnet in this new position is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Express the work to rotate the dipole from alignment through 60° and the torque holding it at 60° in terms of the dipole moment M and field B, then take the ratio.
Step 1:Work done rotating from 0° to 60°.
Step 2:Torque required to hold the magnet at 60°.
Step 3:Substitute MB = 2W.
Final answer:
Q29Single correctMoving Charges and Magnetism
An electron is moving in a circular path under the influence of a transverse magnetic field of T. If the value of e/m is C/kg, the frequency of revolution of the electron is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Use the cyclotron frequency relation, which depends only on the charge-to-mass ratio and the magnetic field, and substitute the given values.
Step 1:Write the cyclotron frequency in terms of e/m and B.
Step 2:Substitute e/m = 1.76e11 C/kg and B = 3.57e-2 T.
Step 3:Convert to a named unit.
Final answer:
Q30Single correctAlternating Current
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Better tuning corresponds to a sharper resonance, that is the highest quality factor Q. Q increases with L and decreases with R and C, so the combination with smallest R, largest L and smallest C is best.
Step 1:Sharp tuning demands a high quality factor.
Step 2:The quality factor is maximised by the smallest R, the largest L and the smallest C.
Final answer:
Q31Single correctElectromagnetic Induction
A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate . Loop 1 of radius encloses the region r and Loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 in loop 1 and zero in loop 2
Approach:
Apply Faraday's law to the flux enclosed by each loop. Only the area where the field exists (radius r) contributes to the flux through loop 1; loop 2 encloses no field region so its flux stays zero.
Step 1:Loop 1 encloses the field region of area πr², so its flux equals B times πr².
Step 2:Loop 2 lies entirely outside the field region, so it encloses no changing flux.
Final answer: in loop 1 and zero in loop 2
Q32Single correctAlternating Current
The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The power factor equals the ratio of the resistive voltage to the total applied voltage, where the total voltage combines the resistive component with the net reactive component (inductive minus capacitive).
Step 1:Compute the net reactive voltage.
Step 2:Total applied voltage from the resistive and net reactive parts.
Step 3:Power factor is the ratio of resistive voltage to total voltage.
Final answer:
Q33Single correctAlternating Current
A 100 resistance and a capacitor of 100 reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Find the circuit impedance from the equal resistance and capacitive reactance, obtain the rms current from the source voltage, then convert to the peak (maximum) value.
Step 1:Impedance with R = 100 Ω and = 100 Ω.
Step 2:Rms current from the 220 V source.
Step 3:Peak displacement current equals √2 times the rms current.
Final answer:
Q34Single correctRay Optics and Optical Instruments
Two identical glass () equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (). The focal length of the combination is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Each equiconvex glass lens has focal length f; relate its surface radius to f. The water trapped between them forms a biconcave lens. Add the powers of glass-water-glass in series.
Step 1:For the equiconvex glass lens (R and -R) the lensmaker relation gives R = f.
Step 2:The water region is a biconcave lens with radii -R and R.
Step 3:Combine the two glass lenses (each f) and the water lens in contact.
Final answer:
Q35Single correctRay Optics and Optical Instruments
An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Convert each apparent depth to its real depth using the refractive index, then add the two real depths to get the full slab thickness.
Step 1:Real depth of the bubble from the first surface.
Step 2:Real depth of the bubble from the opposite face.
Step 3:Total thickness is the sum of the two real depths.
Final answer:
Q36Single correctWave Optics
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Write maximum and minimum intensities in terms of the source amplitudes, take their ratio with the amplitude ratio set by the intensity ratio n, and simplify.
Step 1:With intensity ratio n, take I1 = n I2 so the amplitude ratio is √n to 1.
Step 2:Form the visibility ratio and expand the squares.
Step 3:Simplify.
Final answer:
Q37Single correctRay Optics and Optical Instruments
A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2concave, diopter
Approach:
To shift the far point from 400 cm to infinity, the corrective lens must form a virtual image at the existing far point (400 cm) for an object at infinity. Apply the lens equation with the far point as the image distance.
Step 1:Object at infinity, image at the far point -4 m (virtual, same side).
Step 2:A negative power corresponds to a diverging concave lens.
Final answer: concave, diopter
Q38Single correctWave Optics
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The first diffraction minimum of a single slit lies at a position given by the wavelength times the focal length divided by the slit width. Substitute the given values consistently in cm.
Step 1:Write the position of the first dark band.
Step 2:Substitute λ = 5e-5 cm, f = 60 cm, d = 0.02 cm.
Step 3:Evaluate the expression.
Final answer:
Q39Single correctDual Nature of Radiation and Matter
Electrons of mass m with de-Broglie wavelength fall on the target in an X-ray tube. The cutoff wavelength () of the emitted X-ray is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Equate the kinetic energy of the incident electron (expressed via its de-Broglie wavelength) to the maximum X-ray photon energy at the cutoff wavelength, then solve for the cutoff wavelength.
Step 1:Cutoff photon energy equals the electron kinetic energy.
Step 2:Express the kinetic energy through the de-Broglie wavelength.
Step 3:Combine and solve for the cutoff wavelength.
Final answer:
Q40Single correctDual Nature of Radiation and Matter
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Find the work function from the first illumination, then use it to find the maximum kinetic energy at 6 eV; the stopping potential of A relative to C is the negative of that energy in volts.
Step 1:Work function from the 5 eV illumination giving 2 eV electrons.
Step 2:Maximum kinetic energy with 6 eV photons.
Step 3:Stopping potential of A relative to C must be -3 V to repel these electrons.
Final answer:
Q41Single correctAtoms
If an electron in a hydrogen atom jumps from the orbit to the orbit, it emits a photon of wavelength . When it jumps from the orbit to the orbit, the corresponding wavelength of the photon will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply the Rydberg formula for both transitions and take the ratio of wavelengths, so the Rydberg constant cancels and only the energy-level terms remain.
Step 1:Transition 3 to 2.
Step 2:Transition 4 to 3.
Step 3:Take the ratio to eliminate R.
Final answer:
Q42Single correctNuclei
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
At 40% decay 60% of the sample remains; at 85% decay 15% remains. Use the exponential decay law to find the time interval between these two fractions, expressed through the half-life.
Step 1:Remaining fractions at the two instants.
Step 2:The interval depends on the ratio of remaining fractions.
Step 3:Evaluate.
Final answer:
Q43Single correctSemiconductor Electronics
For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 k, then the input signal voltage is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The voltage gain of a CE amplifier equals the current amplification factor times the ratio of collector to base resistance. Divide the output voltage by this gain to get the input signal voltage.
Step 1:Voltage gain from the given parameters.
Step 2:Input voltage from output divided by gain.
Final answer:
Q44Single correctSemiconductor Electronics
The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Determine the bias of each ideal diode. The diode in the branch with R2 is reverse biased and blocks current, while the other diode is forward biased, so current flows through R1 and the 2 Ω resistor in series across the 10 V source.
Step 1:Diode D1 is reverse biased and D2 is forward biased, so the R2 (3 Ω) branch carries no current.
Step 2:Effective circuit is the 2 Ω resistor R1 in series with the 2 Ω resistor R3 across 10 V.
Final answer:
Q45Single correctSemiconductor Electronics
What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1 ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Trace the two-gate logic for each input set. Gate P (NAND) acts on A and B; its output and C feed the second gate Q (NAND) to give Y.
Step 1:Inputs all 0: gate P output is 1, and the second gate gives Y = 1.
Step 2:Inputs all 1: gate P output is 0 (or carries forward), and the second gate gives Y = 0.
Final answer:
Chemistry45 questions
Q46Single correctChemical Bonding and Molecular Structure
Which one of the following compounds shows the presence of intramolecular hydrogen bond?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cellulose
Approach:
Identify which species can form a hydrogen bond within a single molecule (intramolecular) rather than between separate molecules.
Step 1:Hydrogen peroxide, HCN and acetic acid associate through intermolecular hydrogen bonds between separate molecules, not within one molecule.
Step 2:Cellulose is a polysaccharide in which the chain conformation is stabilised by hydrogen bonds between hydroxyl groups that are part of the same macromolecule, an intramolecular hydrogen bond parallel to the glycosidic linkage.
Final answer: Cellulose
Q47Single correctElectrochemistry
The molar conductivity of a solution of with electrolytic conductivity of at 298 K is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Relate molar conductivity to electrolytic (specific) conductivity and concentration, taking care of the unit conversion from to .
Step 1:Concentration in mol per litre equals the given molarity.
Step 2:Substituting the specific conductivity and concentration with the factor 1000 that converts per to per litre.
Step 3:Therefore the molar conductivity carries units of S per mol.
Final answer:
Q48Single correctSurface Chemistry
The decomposition of phosphine () on tungsten at low pressure is a first-order reaction. It is because the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1rate is proportional to the surface coverage
Approach:
Apply the Langmuir adsorption isotherm to the heterogeneous decomposition and examine the limiting form at low pressure.
Step 1:The rate of decomposition is proportional to the fraction of surface covered by phosphine.
Step 2:At low pressure the term is much smaller than unity, so the surface coverage becomes directly proportional to the pressure.
Step 3:Therefore the rate is proportional to the surface coverage, giving first-order behaviour.
Final answer: rate is proportional to the surface coverage
Q49Single correctSurface Chemistry
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of are given below :
I. (NaCl) = 52,
II. () = 0.69,
III. () = 0.22
The correct order of the their coagulating power is
I. (NaCl) = 52,
II. () = 0.69,
III. () = 0.22
The correct order of the their coagulating power is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3III > II > I
Approach:
Coagulating power is inversely related to the coagulation value; the smaller the coagulation value, the greater the coagulating power.
Step 1:List the coagulation values: NaCl 52, BaCl2 0.69, MgSO4 0.22.
Step 2:Taking the reciprocal order, the smallest coagulation value gives the largest coagulating power, so III exceeds II exceeds I.
Final answer: III > II > I
Q50Single correctElectrochemistry
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2110 minutes
Approach:
Use Faraday's law: the charge equals moles of electrons times Faraday constant, with two electrons required per mole of chlorine gas, then convert charge to time at the given current.
Step 1:Production of one mole of chlorine gas requires two moles of electrons.
Step 2:Charge required is moles of electrons times the Faraday constant.
Step 3:Time equals charge divided by current.
Step 4:A charge of 0.2 F at 3 A takes 6433 s, that is 107 minutes, so the electrolysis runs for very nearly 110 minutes.
Final answer: 110 minutes
Q51Single correctStructure of Atom
How many electrons can fit in the orbital for which and ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 12
Approach:
Distinguish between a subshell and a single orbital; the question asks for one orbital, which holds at most two electrons.
Step 1:The values n equal to 3 and l equal to 1 specify a 3p subshell.
Step 2:An individual orbital, defined by a single set of n, l and m, accommodates a maximum of two electrons by the Pauli exclusion principle.
Final answer: 2
Q52Single correctThermodynamics
For a sample of perfect gas when its pressure is changed isothermally from to , the entropy change is given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Start from the isothermal entropy change expressed in volumes and convert to pressures using Boyle's law for a perfect gas.
Step 1:For an isothermal process the entropy change in terms of volume is the standard logarithmic expression.
Step 2:At constant temperature pressure and volume are inversely related, so the volume ratio becomes the inverse pressure ratio.
Final answer:
Q53Single correctSolutions
The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 43
Approach:
Count the total number of ions produced on complete dissociation of the strong electrolyte; the van't Hoff factor equals that number for complete ionisation.
Step 1:Barium hydroxide dissociates completely into one barium ion and two hydroxide ions.
Step 2:With degree of dissociation equal to one for a strong electrolyte, the van't Hoff factor equals the number of ions.
Final answer: 3
Q54Single correctEquilibrium
The percentage of pyridine () that forms pyridinium ion () in a 0.10 M aqueous pyridine solution ( for ) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Compute the degree of dissociation of the weak base from the dissociation-constant expression, then express it as a percentage.
Step 1:Substituting the base dissociation constant and concentration.
Step 2:Evaluating the square root.
Step 3:Expressing the fraction as a percentage.
Final answer:
Q55Single correctSolid State
In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion () and fluoride ion () are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 38 and 4
Approach:
Apply the known coordination ratio of the fluorite structure, where the cation-to-anion stoichiometry fixes the ratio of coordination numbers.
Step 1:Calcium fluoride adopts the fluorite structure with formula CaF2.
Step 2:Each calcium ion is surrounded by eight fluoride ions while each fluoride ion is surrounded by four calcium ions, preserving the 8 : 4 ratio.
Final answer: 8 and 4
Q56Single correctElectrochemistry
If the for a given reaction has a negative value, which of the following gives the correct relationships for the values of and ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Connect standard cell potential to standard Gibbs energy and to the equilibrium constant through the two standard thermodynamic relations.
Step 1:A negative standard cell potential makes the standard Gibbs energy positive.
Step 2:A positive standard Gibbs energy forces the logarithm of the equilibrium constant to be negative, so the constant is less than one.
Final answer:
Q57Single correctSolutions
Which one of the following is incorrect for ideal solution ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Recall the defining thermodynamic conditions of an ideal solution and identify which listed statement contradicts spontaneous mixing.
Step 1:An ideal solution has zero enthalpy and zero internal-energy change of mixing and obeys Raoult's law exactly.
Step 2:Mixing increases entropy, so with zero enthalpy of mixing the Gibbs energy of mixing is negative rather than zero.
Final answer:
Q58Single correctEquilibrium
The solubility of AgCl (s) with solubility product in 0.1 M NaCl solution would be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Apply the common ion effect: the chloride from NaCl fixes the chloride concentration, so the silver ion solubility equals the solubility product divided by that chloride concentration.
Step 1:The chloride ion concentration is dominated by 0.1 M NaCl through the common ion effect.
Step 2:The solubility equals the silver ion concentration obtained from the solubility product.
Final answer:
Q59Single correctSome Basic Concepts of Chemistry
Suppose the elements X and Y combine to form two compounds and . When 0.1 mole of weighs 10 g and 0.05 mole of weighs 9 g, the atomic weights of X and Y are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 140, 30
Approach:
Convert the given masses and moles into molar masses of each compound, write two equations in the atomic weights, and solve the simultaneous equations.
Step 1:For XY2, 0.1 mole weighing 10 g gives a molar mass of 100, so the atomic weight of X plus twice that of Y equals 100.
Step 2:For X3Y2, 0.05 mole weighing 9 g gives a molar mass of 180, so three times the atomic weight of X plus twice that of Y equals 180.
Step 3:Subtracting the first equation from the second gives the atomic weight of X.
Step 4:Substituting back gives the atomic weight of Y.
Final answer: 40, 30
Q60Single correctElectrochemistry
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the total charge passed, then divide by the charge on a single electron to count the electrons.
Step 1:The total charge equals current times time.
Step 2:Dividing the charge by the electronic charge gives the number of electrons.
Final answer:
Q61Single correctThe p-Block Elements
Boric acid is an acid because its molecule
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3accepts from water releasing proton
Approach:
Recall that boric acid is a weak monobasic Lewis acid that does not donate its own proton but acts as an electron-pair acceptor toward water.
Step 1:Boric acid is electron deficient, so it accepts a lone pair of electrons from the hydroxide of water.
Step 2:This abstraction of hydroxide releases a proton into solution, accounting for the acidic character.
Final answer: accepts from water releasing proton
Q62Single correctThe p-Block Elements
is soluble in HF only in presence of KF. It is due to the formation of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Recognise that potassium fluoride supplies fluoride ions that combine with aluminium fluoride to form a soluble hexafluoroaluminate complex.
Step 1:Potassium fluoride provides fluoride ions which add to aluminium fluoride to give the soluble complex.
Step 2:The formation of the hexafluoroaluminate ion accounts for the dissolution that does not occur in HF alone.
Final answer:
Q63Single correctElectrochemistry
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4zinc has higher negative electrode potential than iron
Approach:
Compare the standard electrode potentials of zinc and iron to decide which metal acts as the sacrificial anode in galvanization.
Step 1:Zinc has a more negative standard electrode potential than iron, making it more readily oxidised.
Step 2:Therefore zinc corrodes preferentially and protects the iron beneath, so galvanization works only with zinc on iron and not the reverse.
Final answer: zinc has higher negative electrode potential than iron
Q64Single correctThe s-Block Elements
The suspension of slaked lime in water is known as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3milk of lime
Approach:
Recall the standard names of calcium hydroxide preparations to distinguish a suspension from a clear solution.
Step 1:Calcium hydroxide, slaked lime, is sparingly soluble in water; its dilute clear solution is limewater.
Step 2:A suspension of slaked lime in water is called milk of lime.
Final answer: milk of lime
Q65Single correctChemical Bonding and Molecular Structure
The hybridizations of atomic orbitals of nitrogen in , and respectively are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Determine the hybridisation of nitrogen in each species from the number of sigma bonds and lone pairs around it.
Step 1:In the nitronium ion the nitrogen forms two regions of electron density in a linear arrangement, giving sp hybridisation.
Step 2:In the nitrate ion the nitrogen has three regions in a trigonal planar arrangement, giving sp2 hybridisation.
Step 3:In the ammonium ion the nitrogen has four sigma bonds in a tetrahedral arrangement, giving sp3 hybridisation.
Final answer:
Q66Single correctThe p-Block Elements
Which of the following fluoro-compound is most likely to behave as a Lewis base ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
A Lewis base donates a lone pair, so identify the molecule whose central atom retains an available lone pair.
Step 1:Phosphorus trifluoride has a lone pair on phosphorus that can be donated to an electron-pair acceptor.
Step 2:Boron trifluoride is electron deficient and the carbon and silicon tetrafluorides have no lone pair on the central atom, so they cannot act as Lewis bases.
Final answer:
Q67Single correctChemical Bonding and Molecular Structure
Which of the following pairs of ions is isoelectronic and isostructural?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Two species are isoelectronic and isostructural when they have the same number of valence electrons and the same geometry; check each pair for matching electron count and shape.
Step 1:Carbonate and nitrate ions are both trigonal planar with sp2 central atoms and equal valence-electron counts.
Step 2:Chlorate and sulphite ions are both trigonal pyramidal with sp3 central atoms and equal valence-electron counts.
Step 3:Carbonate with nitrate (24 valence electrons, trigonal planar sp2) and chlorate with sulphite (26 valence electrons, pyramidal sp3) are each isoelectronic and isostructural pairs.
Final answer:
Q68Single correctThe s-Block Elements
In context with beryllium, which one of the following statements is incorrect ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Its salts rarely hydrolyse
Approach:
Evaluate each statement against the known chemistry of beryllium and identify the false one.
Step 1:Beryllium is passivated by nitric acid, forms the carbide Be2C, and has an electron-deficient polymeric hydride, so those statements are correct.
Step 2:Beryllium salts are readily hydrolysed owing to the small, highly polarising beryllium ion, so the claim that its salts rarely hydrolyse is incorrect.
Final answer: Its salts rarely hydrolyse
Q69Single correctThe p-Block Elements
Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reaction does not show oxidizing behaviour?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify the reaction where sulphur in sulphuric acid keeps its oxidation state, so no oxidation occurs.
Step 1:In a redox reaction the oxidation state of an element must change. Track sulphur, which is +6 in sulphuric acid.
Step 2:With Cu, S, and C the sulphur is reduced from +6 to +4 (forming sulphur dioxide), so these are oxidation reactions.
Step 3:With calcium fluoride the sulphur stays +6 throughout; this is a simple acid displacement giving hydrogen fluoride, with no change of oxidation state.
Final answer:
Q70Single correctCoordination Compounds
Which of the following pairs of d-orbitals will have electron density along the axes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Recall which of the five d-orbitals point their lobes directly along the Cartesian axes.
Step 1:The five d-orbitals split into two sets by lobe orientation: the set points along the axes and the g set points between the axes.
Step 2:The 2 lobe lies along the z-axis and 2-y2 lobes lie along the x and y axes, so both have electron density along the axes.
Final answer:
Q71Single correctChemical Bonding and Molecular Structure
The correct geometry and hybridization for are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Use VSEPR: count bond pairs and lone pairs on the central xenon atom of xenon tetrafluoride.
Step 1:Xenon contributes 8 valence electrons; four bonds to fluorine use four, leaving two lone pairs.
Step 2:Total of six electron domains gives sp3d2 hybridisation with an octahedral electron arrangement.
Step 3:The two lone pairs occupy opposite axial positions, leaving the four fluorine atoms in a plane.
Final answer:
Q72Single correctChemical Bonding and Molecular Structure
Among the following, which one is a wrong statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Evaluate each statement against known shapes and bonding, and pick the false one.
Step 1:Selenium tetrafluoride has one lone pair on a sp3d centre giving a see-saw shape, whereas methane is a regular tetrahedron.
Step 2:The statement claiming the two have the same shape is therefore incorrect, while the other three statements are correct.
Final answer:
Q73Single correctCoordination Compounds
The correct increasing order of trans-effect of the following species is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Order the ligands by their position in the experimentally established trans-effect series.
Step 1:The trans-effect series places strong pi-acceptors and strong sigma-donors highest, with ammonia among the weakest.
Step 2:Cyanide is the strongest trans-director, phenyl comes next, then bromide, and ammonia is weakest.
Final answer:
Q74Single correctThe d- and f-Block Elements
Which one of the following statements related to lanthanons is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3All the lanthanons are much more reactive than aluminium
Approach:
Assess each statement on lanthanoid behaviour and select the false one.
Step 1:Earlier lanthanoids resemble calcium in reactivity, and reactivity decreases across the series toward aluminium-like values, so the blanket claim that all are far more reactive than aluminium is false.
Step 2:Europium commonly shows +2, basicity falls with the lanthanoid contraction, and cerium(IV) is a standard oxidant, so the remaining statements are correct.
Final answer: All the lanthanons are much more reactive than aluminium
Q75Single correctCoordination Compounds
Jahn-Teller effect is not observed in high spin complexes of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Jahn-Teller distortion appears when the set is unequally occupied; find the configuration whose high-spin filling is symmetric.
Step 1:Distortion arises from uneven occupancy of the degenerate orbitals. For high-spin octahedral complexes test the population.
Step 2:High-spin d8 is with one electron in each of the two orbitals, an evenly filled set, so no distortion arises; d4, d7 and d9 all leave the pair unevenly occupied.
Final answer:
Q76Single correctHaloalkanes and Haloarenes
Which of the following can be used as the halide component for Friedel-Crafts reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Isopropyl chloride
Approach:
Friedel-Crafts alkylation needs a halide that forms a stable carbocation; rule out aryl and vinyl halides.
Step 1:Aryl halides (chlorobenzene, bromobenzene) and vinyl halides (chloroethene) have carbon-halogen bonds with partial double-bond character and cannot ionise to give carbocations.
Step 2:Isopropyl chloride ionises with aluminium chloride to a stable secondary carbocation that acts as the electrophile.
Final answer: Isopropyl chloride
Q77Single correctHydrocarbons
In which of the following molecules, all atoms are coplanar?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Biphenyl — two benzene rings joined by a single bond
Approach:
All atoms are coplanar only when every carbon is sp2 hybridised; find the structure with no sp3 carbon.
Step 1:Cyclohexane and bicyclohexyl are built from sp3 ring carbons whose tetrahedral geometry forces atoms out of one plane, and the tetracyano alkene carries sp3 methyl hydrogens.
Step 2:The structure built entirely on a benzene framework with only sp2 carbons keeps all atoms in a single plane.
Final answer: Biphenyl — two benzene rings joined by a single bond
Q78Single correctPolymers
Which one of the following structures represents nylon 6,6 polymer?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Recall that nylon-6,6 is the condensation polymer of hexamethylenediamine and adipic acid, joined by amide linkages.
Step 1:Adipic acid and hexamethylenediamine condense with loss of water to form repeating amide bonds.
Step 2:The repeat unit of nylon-6,6 is -CO-(CH2)4-CO-NH-(CH2)6-NH-, carrying two amide bonds per unit.
Final answer:
Q79Single correctHydrocarbons
In pyrrole (ring positions numbered with nitrogen at position 1) the electron density is maximum on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 42 and 5
Approach:
Use resonance of the nitrogen lone pair into the ring to find the carbons that gain negative charge.
Step 1:The nitrogen lone pair delocalises into the ring, placing negative charge on the alpha carbons adjacent to nitrogen.
Step 2:Electrophilic attack and highest electron density therefore occur at positions 2 and 5, which also give the most stable protonated intermediate.
Final answer: 2 and 5
Q80Single correctHaloalkanes and Haloarenes
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Test each compound for whether the HBr-addition/elimination sequence can generate the three-carbon alkene propene.
Step 1:Cyclopropane opens with HBr to a bromopropane and then eliminates to propene; the alcohol and the alkyl bromide eliminate directly to propene.
Step 2:Ketene adds HBr to an acyl bromide that on the workup gives an acid bromide route, not propene, so it cannot form the alkene.
Final answer:
Q81Single correctAmines / Nitrogen compounds
Which one of the following nitro compounds does not react with nitrous acid?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Reaction with nitrous acid requires an alpha hydrogen on the nitro-bearing carbon; find the compound lacking it.
Step 1:Primary and secondary nitroalkanes possess alpha hydrogens on the carbon carrying the nitro group and react with nitrous acid.
Step 2:The tertiary nitroalkane has no alpha hydrogen on the nitro carbon, so it gives no reaction with nitrous acid.
Final answer:
Q82Single correctBiomolecules
The central dogma of molecular genetics states that the genetics information flows from :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Recall the directional flow of genetic information in the central dogma.
Step 1:Genetic information is transcribed from DNA to RNA and then translated from RNA to protein.
Final answer:
Q83Single correctBiomolecules
The correct corresponding order of names of four aldoses with configuration given below, respectively, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4D-erythrose, D-threose, L-erythrose, L-threose
Approach:
Assign D or L from the lowest chiral centre and erythro or threo from the relative arrangement of the two hydroxyl groups in each Fischer projection.
Step 1:The configuration of the bottom-most chiral carbon sets D when its hydroxyl is on the right and L when on the left.
Step 2:Both hydroxyls on the same side gives erythro and on opposite sides gives threo; applying this in order yields D-erythrose, D-threose, L-erythrose, L-threose.
Final answer: D-erythrose, D-threose, L-erythrose, L-threose
Q84Single correctHydrocarbons
In the given reaction, benzene and cyclohexene are treated with HF at C, and the product P is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cyclohexylbenzene
Approach:
Treat the acid-catalysed addition of cyclohexene to benzene as a Friedel-Crafts type alkylation through a carbocation.
Step 1:Cyclohexene is protonated by hydrogen fluoride to a cyclohexyl carbocation.
Step 2:The carbocation alkylates benzene to give cyclohexylbenzene.
Final answer: Cyclohexylbenzene
Q85Single correctAmines / Nitrogen compounds
A given nitrogen-containing aromatic compound A reacts with , followed by to give an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C with the molecular formula . The structure of compound A is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Nitrobenzene,
Approach:
Work backward from the azo dye product through diazotisation and reduction to identify starting compound A.
Step 1:Coupling of a diazonium salt with phenol gives the orange azo dye of formula matching C, so B is benzenediazonium chloride.
Step 2:The diazonium salt forms from aniline with nitrous acid, and aniline itself comes from nitrobenzene by Sn/HCl reduction, so A is nitrobenzene.
Final answer: Nitrobenzene,
Q86Single correctHaloalkanes and Haloarenes
Consider the reaction
This reaction will be the fastest in
This reaction will be the fastest in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3N, N - dimethylformamide (DMF)
Approach:
Recognise the reaction as a bimolecular nucleophilic substitution and pick the solvent that best frees the nucleophile.
Step 1:The reaction is an SN2 substitution whose rate rises when the nucleophile is unhindered by solvation.
Step 2:Polar aprotic dimethylformamide solvates the sodium cation but not the cyanide, leaving a highly reactive naked nucleophile, unlike the protic alcohols and water.
Final answer: N, N - dimethylformamide (DMF)
Q87Single correctAldehydes, Ketones and Carboxylic Acids
The correct structure of the product A formed in the reaction of cyclohex-2-en-1-one with gas at 1 atmosphere over Pd/carbon in ethanol is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cyclohexanone — the saturated ring ketone
Approach:
Identify which functional group hydrogen over palladium reduces in an alpha,beta-unsaturated ketone.
Step 1:Hydrogen over palladium on carbon reduces the carbon-carbon double bond but leaves the carbonyl untouched.
Step 2:The alpha,beta-unsaturated ketone is therefore reduced to the saturated ketone.
Final answer: Cyclohexanone — the saturated ring ketone
Q88Single correctAldehydes, Ketones and Carboxylic Acids
Which among the given molecules can exhibit tautomerism?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1III only
Approach:
Tautomerism requires an alpha hydrogen on a carbon next to the carbonyl and a stable enol; test each molecule.
Step 1:Molecule I would need an enol that violates Bredt's rule at a bridgehead, so its enol cannot exist.
Step 2:Molecule II has no alpha hydrogen available for the shift, while molecule III has a suitable alpha hydrogen and a stable enol, so only III shows tautomerism.
Final answer: III only
Q89Single correctAldehydes, Ketones and Carboxylic Acids
The correct order of strengths of the carboxylic acids I, II and III shown below is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Rank the acids by the electron-withdrawing inductive effect of the ring substituents on the carboxyl group.
Step 1:Acidity rises as the negative inductive effect near the carboxyl group increases, stabilising the carboxylate.
Step 2:The molecule with the strongest electron-withdrawing ring oxygen near the carboxyl is most acidic, giving the order II greater than III greater than I.
Final answer:
Q90Single correctHydrocarbons
The compound that will react most readily with gaseous bromine has the formula
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Compare the molecules by the electron density of their multiple bonds, since bromine adds fastest to the most electron-rich unsaturation.
Step 1:Bromine adds to multiple bonds by electrophilic addition, which is fastest where the pi bond is most electron-rich.
Step 2:Propene carries an electron-donating methyl group that enriches its double bond more than ethene, while butane is saturated and ethyne is less reactive, so propene reacts most readily.
Final answer:
Biology90 questions
Q91Single correctBiological Classification
Which one of the following is for fungi ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2All fungi possess a purely cellulosic cell wall
Approach:
Evaluate each statement against the structural and nutritional characteristics of Kingdom Fungi.
Step 1:Fungi are eukaryotic organisms, so statement 1 holds.
Step 2:The fungal cell wall is composed mainly of chitin (and other polysaccharides), not pure cellulose, so the claim of a purely cellulosic cell wall is incorrect.
Step 3:Fungi lack chlorophyll and obtain nutrition by absorption, confirming heterotrophy; statements 3 and 4 hold.
Final answer: All fungi possess a purely cellulosic cell wall
Q92Single correctBiological Classification
Methanogens belong to -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Archaebacteria
Approach:
Classify methanogens within the prokaryotic groups based on their habitat and metabolism.
Step 1:Methanogens are prokaryotes that produce methane in anaerobic environments such as marshes and the gut of ruminants.
Step 2:Such organisms living in extreme habitats are placed within the archaebacteria.
Final answer: Archaebacteria
Q93Single correctBiological Classification
Select the statement -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The walls of diatoms are easily destructible
Approach:
Assess each claim about diatom cell walls, ecology and morphology.
Step 1:The cell walls of diatoms are made of silica, rendering them nearly indestructible and non-biodegradable; the statement calling them easily destructible is false.
Step 2:Accumulated silica walls form diatomaceous earth, diatoms are major oceanic producers, and they are microscopic planktonic floaters, so the remaining statements hold.
Final answer: The walls of diatoms are easily destructible
Q94Single correctThe Living World
The label of a herbarium sheet carry information on -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4height of the plant
Approach:
Recall the standard contents recorded on a herbarium label.
Step 1:A herbarium label records the date and place of collection, scientific and local names, and the collector's name.
Step 2:The height of the plant is not part of the recorded herbarium label information.
Final answer: height of the plant
Q95Single correctPlant Kingdom
Conifers are adapted to tolerate extreme environmental conditions because of -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3thick cuticle
Approach:
Identify the xerophytic adaptation that enables conifers to survive harsh climates.
Step 1:Conifers (gymnosperms) possess needle-like leaves, sunken stomata and a thick cuticle to reduce water loss.
Step 2:A thick cuticle restricts water loss, and that is the feature which lets conifers tolerate extremes.
Final answer: thick cuticle
Q96Single correctPlant Kingdom
Which one of the following statements is ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Algin is obtained from red algae, and carrageenan from brown algae
Approach:
Check the source organisms of algal products against the listed statements.
Step 1:Algin is a product of brown algae while carrageenan comes from red algae, so the statement reverses these sources and is false.
Step 2:Algae oxygenate water, agar comes from Gelidium and Gracilaria, and Laminaria and Sargassum are edible, so the other statements hold.
Final answer: Algin is obtained from red algae, and carrageenan from brown algae
Q97Single correctMorphology of Flowering Plants
The term 'polyadelphous' is related to -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2androecium
Approach:
Interpret the term polyadelphous in the context of floral whorls.
Step 1:Polyadelphous describes stamens united by their filaments into more than two bundles, a cohesion of the male reproductive whorl.
Step 2:The whorl of stamens constitutes the androecium.
Final answer: androecium
Q98Single correctMorphology of Flowering Plants
How many plants among Indigofera, Sesbania , Salvia, Allium, Aloe, mustard, groundnut, radish, gram and turnip have stamens with different lengths in their flowers ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Four
Approach:
Count plants whose flowers show stamens of unequal length (didynamous or tetradynamous conditions).
Step 1:Salvia shows didynamous stamens, while mustard, radish and turnip (Brassicaceae) show tetradynamous stamens of unequal length.
Step 2:These four plants therefore have stamens of differing lengths.
Final answer: Four
Q99Single correctMorphology of Flowering Plants
Radial symmetry is found in the flowers of -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Distinguish actinomorphic (radial) flowers from zygomorphic (bilateral) ones among the given genera.
Step 1:Trifolium, Pisum and Cassia bear zygomorphic flowers with bilateral symmetry.
Step 2:Brassica flowers are actinomorphic and can be divided into equal halves along multiple planes, showing radial symmetry.
Final answer:
Q100Single correctMorphology of Flowering Plants
Free-central placentation is found in -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Match each genus with its type of placentation.
Step 1:Free-central placentation has ovules borne on a central axis in a unilocular ovary without septa, as in Dianthus.
Step 2:Argemone and Brassica show parietal placentation and Citrus shows axile placentation.
Final answer:
Q101Single correctAnatomy of Flowering Plants
Cortex is the region found between -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1epidermis and stele
Approach:
Locate the cortex within the concentric tissue zones of a root or stem.
Step 1:Moving inward from the surface the tissues are epidermis, cortex, endodermis and then the stele.
Step 2:The cortex lies between the epidermis and the stele.
Final answer: epidermis and stele
Q102Single correctAnatomy of Flowering Plants
The balloon-shaped structures called tyloses -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3are extensions of xylem parenchyma cells into vessels
Approach:
Recall the origin and nature of tyloses during heartwood formation.
Step 1:Tyloses form when adjacent xylem parenchyma cells balloon out through pits into the lumen of vessels in the heartwood.
Step 2:They are therefore extensions of xylem parenchyma cells into the vessels, characterising heartwood rather than sapwood.
Final answer: are extensions of xylem parenchyma cells into vessels
Q103Single correctBiomolecules
A non-proteinaceous enzyme is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2ribozyme
Approach:
Identify which listed catalyst is not made of protein.
Step 1:Lysozyme, ligase and deoxyribonuclease are protein enzymes.
Step 2:A ribozyme is an RNA molecule with catalytic activity, making it non-proteinaceous.
Final answer: ribozyme
Q104Single correctCell - The Unit of Life
Select the -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Large central vacuoles – Animal cells
Approach:
Test each pairing of a feature with a cell type for correctness.
Step 1:Large central vacuoles are characteristic of plant cells, not animal cells, so this pairing is mismatched.
Step 2:Gas vacuoles in green bacteria, protists as eukaryotes and methanogens as prokaryotes are correct pairings.
Final answer: Large central vacuoles – Animal cells
Q105Single correctCell - The Unit of Life
Select the statement -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Pili and fimbriae are mainly involved in motility of bacterial cells
Approach:
Examine each statement about prokaryotic surface structures.
Step 1:Pili function in conjugation and fimbriae help attachment to surfaces; neither provides motility, so this statement is false.
Step 2:Peptidoglycan walls, absence of flagellated cyanobacterial cells and the wall-less Mycoplasma are all correct.
Final answer: Pili and fimbriae are mainly involved in motility of bacterial cells
Q106Single correctCell - The Unit of Life
A cell organelle containing hydrolytic enzymes is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Lysosome
Approach:
Recall which organelle stores digestive hydrolytic enzymes.
Step 1:Lysosomes are membrane-bound vesicles rich in hydrolytic enzymes active at acidic pH for intracellular digestion.
Step 2:Ribosomes synthesise protein, mesosomes are membrane infoldings, and microsomes are vesicle fragments, none storing hydrolases.
Final answer: Lysosome
Q107Single correctCell Cycle and Cell Division
During cell growth, DNA synthesis takes place in -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1S phase
Approach:
Identify the phase of interphase devoted to DNA replication.
Step 1:DNA replication occurs during the synthesis phase of interphase, doubling the genetic material.
Step 2:G1 and G2 are growth gaps and M phase is for division, none of which replicate DNA.
Final answer: S phase
Q108Single correctRespiration in Plants
Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Acetyl CoA
Approach:
Find the metabolite where catabolism of all three macromolecule classes converges.
Step 1:Fatty acids, sugars and amino acids are each ultimately converted to acetyl CoA before entering the citric acid cycle.
Step 2:Glucose-6-phosphate, fructose 1,6-bisphosphate and pyruvic acid lie only on the carbohydrate route.
Final answer: Acetyl CoA
Q109Single correctPlant Physiology - Transport
A few drops of sap were collected by cutting across a plant stem by a suitable method. The sap was tested chemically. Which one of the following test results indicates that it is phloem sap ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Alkaline
Approach:
Recall the chemical character of phloem sap carrying dissolved sugars.
Step 1:Phloem sap is a sucrose solution that is alkaline in nature.
Step 2:An acidic reaction, low refractive index or absence of sugar would not indicate phloem sap.
Final answer: Alkaline
Q110Single correctPlant Growth and Development
You are given a tissue with its potential for differentiation in an artificial culture. Which of the following pairs of hormones would you add to the medium to secure shoots as well as roots ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Auxin and cytokinin
Approach:
Select the hormone pair that drives root and shoot differentiation in tissue culture.
Step 1:Auxin promotes root initiation while cytokinin promotes shoot formation, and their balance directs organogenesis.
Step 2:Gibberellin and abscisic acid do not jointly induce roots and shoots.
Final answer: Auxin and cytokinin
Q111Single correctPlant Growth and Development
Phytochrome is a -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Chromoprotein
Approach:
Classify phytochrome by the nature of its protein-pigment association.
Step 1:Phytochrome consists of a protein bound to a light-absorbing pigment, structurally similar to phycobilins.
Step 2:A protein carrying a chromophore is termed a chromoprotein.
Final answer: Chromoprotein
Q112Single correctMineral Nutrition
Which is essential for the growth of root tip ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ca
Approach:
Determine which mineral element supports activity of the root apical meristem.
Step 1:Calcium is required for the activity of meristematic cells and is essential for elongation and growth at the root tip.
Step 2:Zinc, iron and manganese serve other roles and are not the element required for root tip growth.
Final answer: Ca
Q113Single correctPhotosynthesis in Higher Plants
The process which makes major difference between and plants is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Photorespiration
Approach:
Identify the process that distinguishes C3 from C4 photosynthesis.
Step 1:C4 plants concentrate carbon dioxide in bundle sheath cells, suppressing the oxygenase activity of RuBisCO.
Step 2:As a result photorespiration is essentially absent in C4 plants but significant in C3 plants, marking the major difference.
Final answer: Photorespiration
Q114Single correctMorphology of Flowering Plants
Which one of the following statements is correct ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3In potato, banana and ginger, the plantlets arise from the internodes present in the modified stem
Approach:
Identify the incorrect statement among descriptions of vegetative and asexual reproduction.
Step 1:In potato, banana and ginger the new shoots develop from buds present at the nodes (eyes) of the modified underground stem, not from the internodes.
Step 2:Clones arise from asexual reproduction, zoospores are motile asexual spores, and water hyacinth depletes dissolved oxygen, so the remaining statements are correct.
Final answer: In potato, banana and ginger, the plantlets arise from the internodes present in the modified stem
Q115Single correctSexual Reproduction in Flowering Plants
Which one of the following generates new genetic combinations leading to variation ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Sexual reproduction
Approach:
Determine which mode of reproduction shuffles genetic material to create new combinations.
Step 1:Sexual reproduction involves meiosis and fusion of gametes from two parents, which through crossing over and random fertilization yields new genetic combinations.
Step 2:Vegetative reproduction, parthenogenesis and nucellar polyembryony are asexual or apomictic and produce genetically identical offspring.
Final answer: Sexual reproduction
Q116Single correctAnatomy of Flowering Plants
Match Column-I with Column-II and select the correct option using the codes given below –
| Column-I | Column-II |
|---|---|
| (a). Pistils fused together | (i). Gametogenesis |
| (b). Formation of gametes | (ii). Pistillate |
| (c). Hyphae of higher Ascomycetes | (iii). Syncarpous |
| (d). Unisexual female flower | (iv). Dikaryotic |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4a-iii, b-i, c-iv, d-ii
Approach:
Match each botanical term in Column-I with its defining concept in Column-II.
Step 1:Pistils fused together describes a syncarpous gynoecium, so (a) pairs with (iii).
Step 2:Formation of gametes is gametogenesis, so (b) pairs with (i).
Step 3:Hyphae of higher Ascomycetes carry two nuclei per cell and are dikaryotic, so (c) pairs with (iv).
Step 4:A unisexual female flower bearing only pistils is pistillate, so (d) pairs with (ii).
Final answer: a-iii, b-i, c-iv, d-ii
Q117Single correctSexual Reproduction in Flowering Plants
In majority of angiosperms -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3reduction division occurs in the megaspore mother cells
Approach:
Identify the correct statement about embryo sac development in angiosperms.
Step 1:The diploid megaspore mother cell undergoes meiosis (reduction division) to form four haploid megaspores.
Step 2:The filiform apparatus belongs to synergids, antipodals are typically only three, and the central cell is large with two polar nuclei, making the other statements incorrect.
Final answer: reduction division occurs in the megaspore mother cells
Q118Single correctSexual Reproduction in Flowering Plants
Pollination in water hyacinth and water lily is brought about by the agency of -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2insects or wind
Approach:
Recall the pollinating agents for water hyacinth and water lily.
Step 1:Although they are aquatic plants, water hyacinth and water lily raise their flowers above the water surface and are pollinated by insects or wind rather than by water.
Final answer: insects or wind
Q119Single correctSexual Reproduction in Flowering Plants
The ovule of an angiosperm is technically equivalent to -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1megasporangium
Approach:
Relate the angiosperm ovule to its homologous reproductive structure.
Step 1:The ovule encloses the megaspore-producing tissue (nucellus) within integuments, so it is structurally equivalent to a megasporangium.
Final answer: megasporangium
Q120Single correctMolecular Basis of Inheritance
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Recall the organism Taylor used to demonstrate semiconservative chromosome replication.
Step 1:Taylor and coworkers used tritiated thymidine autoradiography on root tip cells of the broad bean Vicia faba to show that chromosomes replicate semiconservatively.
Final answer:
Q121Single correctPrinciples of Inheritance and Variation
The mechanism that causes a gene to move from one linkage group to another is called -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3translocation
Approach:
Identify the chromosomal change that shifts a gene to a different linkage group.
Step 1:A linkage group corresponds to a single chromosome. Translocation transfers a segment to a non-homologous chromosome, moving its genes into a new linkage group.
Step 2:Inversion and duplication remain on the same chromosome, and crossing over exchanges segments between homologues of the same group.
Final answer: translocation
Q122Single correctMolecular Basis of Inheritance
The equivalent of a structural gene is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2cistron
Approach:
Match the term that corresponds to a single structural gene.
Step 1:A cistron is the segment of DNA that codes for one polypeptide, which is the functional definition of a structural gene.
Step 2:A muton is the smallest mutable unit, a recon the smallest recombinable unit, and an operon a cluster of genes with regulatory regions.
Final answer: cistron
Q123Single correctPrinciples of Inheritance and Variation
A true breeding plant is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3near homozygous and produces offspring of its own kind
Approach:
Define a true breeding plant in terms of genotype and progeny.
Step 1:A true breeding line results from continued self-pollination, becoming nearly homozygous and yielding offspring with the same traits as the parent.
Step 2:A true breeding line need not be homozygous recessive, and it is not produced by crossing unrelated plants.
Final answer: near homozygous and produces offspring of its own kind
Q124Single correctMolecular Basis of Inheritance
Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 323 S rRNA
Approach:
Identify the bacterial rRNA that is both structural and catalytic.
Step 1:In bacteria the large 50S ribosomal subunit contains 23S rRNA, which both shapes the subunit and catalyses peptide bond formation, acting as a ribozyme.
Step 2:18S and 5.8S rRNAs occur in eukaryotes, and 5S rRNA is purely structural, so the others are incorrect for bacteria.
Final answer: 23 S rRNA
Q125Single correctBiotechnology and its Applications
Stirred-tank bioreactors have been designed for -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3availability of oxygen throughout the process
Approach:
Recall the purpose of stirring in a stirred-tank bioreactor.
Step 1:Stirring mixes the culture and disperses sparged air uniformly, ensuring oxygen is available to cells throughout the vessel.
Step 2:Purification and preservative addition are downstream steps, and bioreactors usually maintain aerobic, not anaerobic, conditions.
Final answer: availability of oxygen throughout the process
Q126Single correctBiotechnology - Principles and Processes
A foreign DNA and plasmid cut by the same restriction endonuclease can be joined to form a recombinant plasmid using
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ligase
Approach:
Identify the enzyme that joins DNA fragments to make a recombinant plasmid.
Step 1:DNA ligase seals the phosphodiester bonds between the complementary sticky ends of the foreign DNA and plasmid, forming the recombinant molecule.
Step 2:Eco RI cuts DNA, while Taq polymerase and Polymerase III synthesize new strands rather than joining fragments.
Final answer: Ligase
Q127Single correctBiotechnology - Principles and Processes
Which of the following is a component of downstream processing ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Expression
Approach:
Determine which listed step is not part of downstream processing.
Step 1:Downstream processing covers separation, purification and preservation or formulation of the product after the bioreactor stage.
Step 2:Expression of the gene occurs during the production phase inside the host, so it is not a downstream step.
Final answer: Expression
Q128Single correctBiotechnology - Principles and Processes
Which of the following restriction enzymes produces blunt ends ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 RV
Approach:
Identify the restriction enzyme that cuts to leave blunt ends.
Step 1:Eco RV recognizes the palindrome and cleaves both strands at the centre of its site, producing flush blunt ends.
Step 2:Sal I, Xho I and Hind III cut asymmetrically and leave overhanging sticky ends.
Final answer: RV
Q129Single correctBiotechnology and its Applications
Which kind of therapy was given in 1990 to a four-year-old girl with adenosine deaminase (ADA) deficiency ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gene therapy
Approach:
Recall the treatment used for the first ADA deficiency case in 1990.
Step 1:In 1990 the first clinical gene therapy was performed on a four-year-old girl with ADA deficiency by introducing a functional ADA gene into her lymphocytes.
Final answer: Gene therapy
Q130Single correctBiodiversity and Conservation
How many hot spots of biodiversity in the world have been identified till date by Norman Myers ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 334
Approach:
Recall the number of biodiversity hotspots identified to date.
Step 1:The originally proposed 25 biodiversity hotspots were later expanded, and the count recognized in the NCERT text is 34 hotspots worldwide.
Final answer: 34
Q131Single correctEcosystem
The primary producers of the deep-sea hydrothermal vent ecosystem are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2chemosynthetic bacteria
Approach:
Identify the primary producers where no sunlight reaches the deep sea vents.
Step 1:Deep-sea hydrothermal vents receive no light, so chemosynthetic bacteria oxidize inorganic chemicals to fix carbon and form the base of the food web.
Step 2:Green algae, blue-green algae and corals depend on light or other inputs and cannot serve as producers there.
Final answer: chemosynthetic bacteria
Q132Single correctOrganisms and Populations
Which of the following is for r-selected species ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Large number of progeny with small size
Approach:
Recall the reproductive strategy of r-selected species.
Step 1:r-selected species maximize reproductive rate by producing a large number of small offspring with little parental investment.
Step 2:Fewer but larger progeny characterise K-selected species.
Final answer: Large number of progeny with small size
Q133Single correctOrganisms and Populations
If '+' sign is assigned to beneficial interaction, '–' sign to detrimental and '0' sign to neutral interaction, then the population interaction represented by '+' '–' refers to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4parasitism
Approach:
Match the (+, –) interaction sign pattern to the correct ecological relationship.
Step 1:A (+ , –) pattern means one species benefits while the other is harmed, which defines parasitism (and predation).
Step 2:Mutualism is (+ , +), commensalism is (+ , 0) and amensalism is (– , 0), so they do not fit.
Final answer: parasitism
Q134Single correctBiodiversity and Conservation
Which of the following is matched ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3–Threat to biodiversity
Approach:
Find the pair that is correctly associated.
Step 1:Parthenium hysterophorus is an invasive alien weed that threatens native biodiversity, making this pairing correct.
Step 2:Aerenchyma belongs to hydrophytes not Opuntia, age pyramids relate to populations not biomes, and stratification is a community feature, so the rest are mismatched.
Final answer: –Threat to biodiversity
Q135Single correctBiodiversity and Conservation
Red List contains data or information on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3threatened species
Approach:
Recall the content of the IUCN Red List.
Step 1:The Red List is the IUCN inventory of the conservation status of species, documenting those that are threatened with extinction.
Step 2:The Red List is not limited to economically important plants, to trade products, or to marine vertebrates.
Final answer: threatened species
Q136Single correctHuman Health and Disease
Which of the following sets of diseases is caused by bacteria ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cholera and tetanus
Approach:
Identify the causative agents of each disease to find the pair that is purely bacterial.
Step 1:Cholera is caused by Vibrio cholerae, a bacterium.
Step 2:Tetanus is caused by Clostridium tetani, a bacterium.
Step 3:Smallpox, mumps, herpes and influenza are all viral diseases.
Final answer: Cholera and tetanus
Q137Single correctAnimal Kingdom
Match Column-I with Column-II for housefly classification and select the correct option using the codes given below :
| Column-I | Column-II |
|---|---|
| (a). Family | (i). Diptera |
| (b). Order | (ii). Arthropoda |
| (c). Class | (iii). Muscidae |
| (d). Phylum | (iv). Insecta |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1a-(iii), b-(i), c-(iv), d-(ii)
Approach:
Recall the systematic classification of the housefly Musca domestica and match each taxonomic rank to its correct name.
Step 1:Family of Musca domestica is Muscidae, matching a with (iii).
Step 2:Order is Diptera, matching b with (i).
Step 3:Class is Insecta, matching c with (iv).
Step 4:Phylum is Arthropoda, matching d with (ii).
Final answer: a-(iii), b-(i), c-(iv), d-(ii)
Q138Single correctAnimal Kingdom
Choose the correct statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2All cyclostomes do not possess jaws and paired fins.
Approach:
Test each statement against known characteristics of mammals, cyclostomes, reptiles and fishes.
Step 1:Cyclostomes such as Petromyzon and Myxine are jawless agnathans and lack paired fins.
Step 2:Egg-laying mammals (Ornithorhynchus, Echidna) are oviparous, so not all mammals are viviparous.
Step 3:Crocodiles among reptiles have a four-chambered heart, so not all reptiles have three chambers; only bony fishes possess an operculum, cartilaginous fishes do not.
Final answer: All cyclostomes do not possess jaws and paired fins.
Q139Single correctBiological Classification
Study the four statement (A-D) given below and select the two correct ones out of them :
(A) Definition of biological species was given by Ernst Mayr.
(B) Photoperiod does not affect reproduction in plants.
(C) Binomial nomenclature system was given by R.H. Whittaker.
(D) In unicellular organisms, reproduction is synonymous with growth.
The two correct statements are
(A) Definition of biological species was given by Ernst Mayr.
(B) Photoperiod does not affect reproduction in plants.
(C) Binomial nomenclature system was given by R.H. Whittaker.
(D) In unicellular organisms, reproduction is synonymous with growth.
The two correct statements are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3A and D
Approach:
Evaluate the truth of each of the four statements and pick the pair that is correct.
Step 1:The biological species concept was indeed defined by Ernst Mayr, so A is correct.
Step 2:In unicellular organisms the increase in cell number through division serves as both growth and reproduction, so D is correct.
Step 3:Photoperiod does affect reproduction (flowering) in plants, making B false; binomial nomenclature was given by Carolus Linnaeus, not Whittaker, making C false.
Final answer: A and D
Q140Single correctReproduction in Organisms
In male cockroaches, sperms are stored in which part of the reproductive system ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Seminal vesicles
Approach:
Recall the male reproductive anatomy of the cockroach and the site of sperm storage.
Step 1:In the male cockroach, sperms produced by the testes pass into the seminal vesicles where they are stored before release.
Step 2:Mushroom glands secrete accessory fluid, the vas deferens conducts sperm, and the testis produces sperm but does not store them.
Final answer: Seminal vesicles
Q141Single correctStructural Organisation in Animals
Smooth muscles are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1involuntary, fusiform, non-striated
Approach:
Recall the structural and functional features of smooth muscle.
Step 1:Smooth muscle cells are spindle (fusiform) shaped and uninucleate.
Step 2:They lack striations because myofilaments are diffusely arranged, and they are controlled by the autonomic nervous system, making them involuntary.
Final answer: involuntary, fusiform, non-striated
Q142Single correctCell Respiration
Oxidative phosphorylation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4formation of ATP by energy released from electrons removed during substrate oxidation
Approach:
Define oxidative phosphorylation and contrast it with substrate-level phosphorylation.
Step 1:Oxidative phosphorylation synthesises ATP using energy from electrons removed during the oxidation of substrates as they pass down the electron transport chain.
Step 2:Transfer of a phosphate group from a substrate to ADP is substrate-level phosphorylation, and the two remaining descriptions misstate the process.
Final answer: formation of ATP by energy released from electrons removed during substrate oxidation
Q143Single correctBiomolecules
Which of the following is the least likely to be involved in stabilizing the three—dimensional folding of most proteins ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Ester bonds
Approach:
Identify which listed interaction does not normally stabilise the three-dimensional folding of proteins.
Step 1:Hydrogen bonds, electrostatic (ionic) interactions and hydrophobic interactions all contribute to stabilising protein tertiary structure.
Step 2:Ester (printed as Easter) bonds are not a typical stabilising force in protein folding, so this is the least likely.
Final answer: Ester bonds
Q144Single correctEnzymes
Which of the following describes the given graph correctly ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Exothermic reaction with energy A in presence of enzyme and B in absence of enzyme
Approach:
Interpret the potential-energy versus reaction-progress graph to deduce the reaction type and the role of the enzyme.
Step 1:Enzymatic and biochemical reactions of this profile are exothermic, since the product energy lies below the substrate energy.
Step 2:The presence of an enzyme lowers the activation energy, so the smaller barrier A corresponds to the enzyme-catalysed path and the larger barrier B to the uncatalysed path.
Final answer: Exothermic reaction with energy A in presence of enzyme and B in absence of enzyme
Q145Single correctCell Cycle and Cell Division
When cell has stalled DNA replication fork, which checkpoint should be predominantly activated ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Relate a stalled DNA replication fork to the cell-cycle checkpoint that prevents progression into mitosis with incomplete or damaged DNA.
Step 1:A stalled replication fork signals incomplete or damaged DNA that must be resolved before mitosis.
Step 2:The G2/M checkpoint monitors DNA integrity at the end of G2 and blocks entry into mitosis until replication is complete, so it is predominantly activated.
Final answer:
Q146Single correctCell Cycle and Cell Division
Match the stages of meiosis in Column-I to their characteristic features in Column-II and select the correct option using the codes given below :
| Column-I | Column-II |
|---|---|
| (a). Pachytene | (i). Pairing of homologous chromosomes |
| (b). Metaphase I | (ii). Terminalization of chiasmata |
| (c). Diakinesis | (iii). Crossing-over takes place |
| (d). Zygotene | (iv). Chromosomes align at equatorial plate |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1a-(iii), b-(iv), c-(ii), d-(i)
Approach:
Recall the defining event of each meiotic sub-stage and match it to its characteristic feature.
Step 1:Crossing-over occurs during pachytene, matching a with (iii).
Step 2:At metaphase I chromosomes align at the equatorial plate, matching b with (iv).
Step 3:Chiasmata terminalise at diakinesis, matching c with (ii); homologous chromosomes pair during zygotene, matching d with (i).
Final answer: a-(iii), b-(iv), c-(ii), d-(i)
Q147Single correctDigestion and Absorption
Which hormones do stimulate the production of pancreatic juice and bicarbonate ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Cholecystokinin and secretin
Approach:
Identify the gut hormones that regulate pancreatic secretion.
Step 1:Cholecystokinin stimulates the secretion of pancreatic enzymes, while secretin stimulates the release of bicarbonate-rich pancreatic juice.
Step 2:These intestinal hormones act on the pancreas, whereas the other listed hormones regulate blood pressure or blood glucose instead.
Final answer: Cholecystokinin and secretin
Q148Single correctBreathing and Exchange of Gases
The partial pressure of oxygen in the alveoli of the lungs is -
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2more than that in the blood
Approach:
Compare the partial pressure of oxygen in alveolar air with that in deoxygenated blood to determine the direction of diffusion.
Step 1:Alveolar oxygen partial pressure is about 104 mm Hg while that in oxygenated/deoxygenated blood entering the alveolar capillaries is about 95 mm Hg.
Step 2:Since alveolar value exceeds the blood value, oxygen diffuses from alveoli into blood, so alveolar partial pressure is more than that in the blood.
Final answer: more than that in the blood
Q149Single correctNeural Control and Coordination
Choose the correct statement.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Photoreceptors in the human eye are depolarized during darkness and become hyperpolarized in response to the light stimulus.
Approach:
Assess each statement about sensory receptors for correctness.
Step 1:In darkness photoreceptors (rods and cones) are depolarized, and light triggers hyperpolarization.
Step 2:Nociceptors detect pain not pressure, Meissner's corpuscles are mechanoreceptors not thermoreceptors, and receptors do produce graded potentials, so the other statements are false.
Final answer: Photoreceptors in the human eye are depolarized during darkness and become hyperpolarized in response to the light stimulus.
Q150Single correctChemical Coordination and Integration
Graves' disease is caused due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2hypersecretion of thyroid gland
Approach:
Recall the endocrine basis of Graves' disease.
Step 1:Graves' disease is an autoimmune condition causing over-activity of the thyroid gland, that is hypersecretion of thyroxine (hyperthyroidism).
Step 2:Graves' disease involves the thyroid, not the adrenal gland.
Final answer: hypersecretion of thyroid gland
Q151Single correctLocomotion and Movement
Name the ion responsible for unmasking of active sites for myosin for cross-bridge activity during muscle contraction.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Calcium
Approach:
Recall the ion that triggers exposure of myosin-binding sites on actin during the sliding-filament mechanism.
Step 1:A neural signal releases calcium from the sarcoplasmic reticulum into the sarcoplasm.
Step 2:Calcium binds troponin, shifting tropomyosin to unmask the actin active sites so myosin heads can form cross-bridges.
Final answer: Calcium
Q152Single correctBody Fluids and Circulation
Name the blood cells, whose reduction in number can cause clotting disorder, leading to excessive loss of blood from the body.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Thrombocytes
Approach:
Identify the blood cell whose deficiency impairs clotting.
Step 1:Thrombocytes (platelets) are essential for blood clotting.
Step 2:A reduction in thrombocyte number causes a clotting disorder leading to excessive blood loss, whereas erythrocytes, leucocytes and neutrophils have other functions.
Final answer: Thrombocytes
Q153Single correctChemical Coordination and Integration
Name a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular glucose uptake and utilization.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Insulin
Approach:
Identify the peptide hormone that promotes cellular glucose uptake in liver and fat cells.
Step 1:Insulin is a peptide hormone that acts mainly on hepatocytes and adipocytes to enhance cellular uptake and utilization of glucose.
Step 2:Glucagon raises blood glucose, while secretin and gastrin regulate digestion rather than glucose uptake.
Final answer: Insulin
Q154Single correctLocomotion and Movement
Osteoporosis, an age-related disease of skeletal system, may occur due to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3decreased level of estrogen
Approach:
Recall the hormonal cause linked to age-related osteoporosis.
Step 1:Osteoporosis is an age-related bone disease more common in older women.
Step 2:A decreased level of estrogen, as occurs after menopause, reduces bone density and causes osteoporosis.
Final answer: decreased level of estrogen
Q155Single correctBody Fluids and Circulation
Serum differs from blood in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3lacking clotting factors
Approach:
Compare the composition of serum with that of whole blood/plasma.
Step 1:During clotting, blood clotting factors such as fibrinogen are consumed to form the clot.
Step 2:The remaining plasma after clotting is called serum, which therefore lacks clotting factors while still containing globulins, albumins and antibodies.
Final answer: lacking clotting factors
Q156Single correctBreathing and Exchange of Gases
Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2there is a negative intrapleural pressure pulling at the lung walls
Approach:
Explain why the lungs remain partially inflated between breaths.
Step 1:The intrapleural space carries a negative (sub-atmospheric) pressure relative to the lung interior.
Step 2:This negative pressure continually pulls on the lung walls, preventing collapse and keeping residual air in the lungs.
Final answer: there is a negative intrapleural pressure pulling at the lung walls
Q157Single correctChemical Coordination and Integration
The posterior pituitary gland is not a 'true' endocrine gland because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2it only stores and releases hormones
Approach:
Recall the functional nature of the posterior pituitary (pars nervosa).
Step 1:The posterior pituitary does not synthesise hormones of its own.
Step 2:It only stores and releases hormones such as ADH and oxytocin produced by the hypothalamus, so it is not a true endocrine gland.
Final answer: it only stores and releases hormones
Q158Single correctExcretory Products and their Elimination
The part of nephron involved in active reabsorption of sodium is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1distal convoluted tubule
Approach:
Identify the nephron segment that carries out active, regulated reabsorption of sodium.
Step 1:The distal convoluted tubule performs active reabsorption of sodium ions, often under hormonal control such as aldosterone.
Step 2:The proximal convoluted tubule reabsorbs the bulk of solutes, the descending limb is permeable mainly to water, and Bowman's capsule is the site of filtration, so the regulated active sodium reabsorption is attributed to the distal convoluted tubule.
Final answer: distal convoluted tubule
Q159Single correctReproductive Health
Which of the following is hormone releasing IUD ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1LNG-20
Approach:
Identify the intrauterine device that works by releasing a hormone rather than acting only physically or by releasing copper ions.
Step 1:Classify the listed IUDs by their mode of action.
Step 2:Recognise the progestasert-type device that releases a progestational hormone (levonorgestrel) into the uterine cavity.
Final answer: LNG-20
Q160Single correctReproductive Health
Which of the following is incorrect regarding vasectomy ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2No sperm occurs in epididymis
Approach:
Evaluate each statement about vasectomy and select the one that is biologically false.
Step 1:Recall that vasectomy cuts and ties the vas deferens, blocking sperm transport beyond that point.
Step 2:Determine where spermatogenesis continues after the procedure.
Step 3:Identify the incorrect statement.
Final answer: No sperm occurs in epididymis
Q161Single correctReproductive Health
Embryo with more than 16 blastomeres formed due to in vitro fertilization is transferred into
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1uterus
Approach:
Match the developmental stage of the embryo with the corresponding site of transfer in assisted reproductive technology.
Step 1:Distinguish ZIFT from IUT by the embryonic stage transferred.
Step 2:Apply the rule to an embryo with more than 16 blastomeres.
Final answer: uterus
Q162Single correctHuman Reproduction
Which of the following depicts the correct pathway of transport of sperms ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Rete testis Efferent ductules Epididymis Vas deferens
Approach:
Trace the anatomical route taken by sperm from the seminiferous tubules to the vas deferens.
Step 1:Recall the sequence of the male duct system following the testis.
Step 2:Continue the path beyond the epididymis.
Step 3:Assemble the ordered pathway.
Final answer: Rete testis Efferent ductules Epididymis Vas deferens
Q163Single correctHuman Reproduction
Match Column – I with Column – II and select the correct option using the codes given below :
| Column – I | Column – II |
|---|---|
| (a). Mons pubis | (i). Embryo formation |
| (b). Antrum | (ii). Sperm |
| (c). Trophectoderm | (iii). Female external genitalia |
| (d). Nebenkern | (iv). Graafian follicle |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2a-(iii), b-(iv), c-(i), d-(ii)
Approach:
Pair each structure in Column-I with its correct biological association in Column-II, then read off the code row carrying those pairs.
Step 1:Identify the mons pubis.
Step 2:Identify the antrum.
Step 3:Identify the trophectoderm.
Step 4:Identify the nebenkern.
Step 5:Read the four pairings off as a code.
Final answer: a-(iii), b-(iv), c-(i), d-(ii)
Q164Single correctHuman Reproduction
Several hormones like hCG, hPL, estrogen, progesterone are produced by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2placenta
Approach:
Identify the endocrine organ of pregnancy that secretes the listed group of hormones.
Step 1:Recall the hormones secreted during pregnancy.
Step 2:Attribute this combined secretion to the correct structure.
Final answer: placenta
Q165Single correctGenetics
If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10
Approach:
Set up the X-linked cross between a colour-blind father and a homozygous normal mother and read the sons' genotypes.
Step 1:Assign genotypes.
Step 2:Determine the gametes sons receive.
Step 3:Evaluate the colour-vision status of the sons.
Final answer: 0
Q166Single correctEvolution
Genetic drift operates in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1small isolated population
Approach:
Recall the population conditions under which random changes in allele frequency dominate.
Step 1:Define genetic drift.
Step 2:Identify where chance effects are strongest.
Final answer: small isolated population
Q167Single correctEvolution
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Recall the genotype-frequency terms in the expanded Hardy-Weinberg equation.
Step 1:Write the genotype frequencies.
Step 2:Identify the heterozygous term.
Final answer:
Q168Single correctEvolution
The chronological order of human evolution from early to the recent is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ramapithecus Australopithecus Homo habilis Homo erectus
Approach:
Arrange the listed hominins by their appearance in geological time from earliest to most recent.
Step 1:Place the oldest form first.
Step 2:Order the subsequent forms.
Final answer: Ramapithecus Australopithecus Homo habilis Homo erectus
Q169Single correctEvolution
Which of the following is the correct sequence of events in the origin of life ?
I. Formation of protobionts
II. Synthesis of organic monomers
III. Synthesis of organic polymers
IV. Formation of DNA-based genetic systems
I. Formation of protobionts
II. Synthesis of organic monomers
III. Synthesis of organic polymers
IV. Formation of DNA-based genetic systems
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3II, III, I, IV
Approach:
Order the chemical-evolution events from simple precursors to a self-replicating genetic system.
Step 1:Start with the simplest building blocks.
Step 2:Combine monomers and then organise them.
Step 3:Conclude with heredity.
Final answer: II, III, I, IV
Q170Single correctMolecular Basis of Inheritance
A molecule that can act as a genetic material must fulfill the traits given below, except
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3it should be unstable structurally and chemically
Approach:
Recall the required properties of genetic material and pick the statement that contradicts them.
Step 1:List the genuine requirements.
Step 2:Recall the stability requirement.
Step 3:Identify the exception.
Final answer: it should be unstable structurally and chemically
Q171Single correctMolecular Basis of Inheritance
DNA-dependent RNA polymerase catalyzes transcription on one strand of the DNA which is called the
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1template strand
Approach:
Identify the DNA strand that RNA polymerase reads to synthesise RNA.
Step 1:Distinguish the two strands during transcription.
Step 2:Name the strand used by RNA polymerase.
Final answer: template strand
Q172Single correctStrategies for Enhancement in Food Production
Interspecific hybridization is the mating of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2two different related species
Approach:
Define interspecific hybridization by the taxonomic level of the parents.
Step 1:Interpret the prefix inter-specific.
Step 2:Mating between two different but related species is interspecific hybridisation.
Final answer: two different related species
Q173Single correctHuman Health and Disease
Which of the following is correct regarding AIDS causative agent HIV ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2HIV is enveloped virus that containing two identical molecule of single-stranded RNA and two molecule of reverse transcriptase.
Approach:
Recall the structural composition of the HIV particle and choose the accurate description.
Step 1:Note HIV is an enveloped retrovirus.
Step 2:Recall the genome and enzyme content.
Step 3:Select the matching statement.
Final answer: HIV is enveloped virus that containing two identical molecule of single-stranded RNA and two molecule of reverse transcriptase.
Q174Single correctStrategies for Enhancement in Food Production
Among the following edible fishes, which one is a marine fish having rich source of omega-3 fatty acids ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Mackerel
Approach:
Pick the marine edible fish among the listed names known to be rich in omega-3 fatty acids.
Step 1:Classify the fishes by habitat.
Step 2:Identify the marine omega-3 source.
Final answer: Mackerel
Q175Single correctMicrobes in Human Welfare
Match Column-I with Column-II and select the correct option using the codes given below :
| Column-I | Column-II |
|---|---|
| (a). Citric acid | (i). Trichoderma |
| (b). Cyclosporin A | (ii). Clostridium |
| (c). Statins | (iii). Aspergillus |
| (d). Butyric acid | (iv). Monascus |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2a-(iii), b-(i), c-(iv), d-(ii)
Approach:
Pair each microbial product with its source organism and read the matching code.
Step 1:Source of citric acid.
Step 2:Source of cyclosporin A.
Step 3:Source of statins.
Step 4:Source of butyric acid.
Step 5:Match to the code.
Final answer: a-(iii), b-(i), c-(iv), d-(ii)
Q176Single correctEnvironmental Issues
Biochemical Oxygen Demand (BOD) may not be a good index for pollution for water bodies receiving effluents from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3petroleum industry
Approach:
Identify the effluent whose pollutants are not biodegradable, so BOD fails to reflect their pollution load.
Step 1:Recall what BOD measures.
Step 2:Consider petroleum effluents.
Step 3:Conclude where BOD is a poor index.
Final answer: petroleum industry
Q177Single correctOrganisms and Populations
The principle of competitive exclusion was stated by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2G. F. Gause
Approach:
Attribute the competitive exclusion principle to its proposer.
Step 1:Recall the originator of the principle.
Final answer: G. F. Gause
Q178Single correctBiodiversity and Conservation
Which of the following National Parks is home to the famous musk deer or hangul ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Dachigam National Park, Jammu & Kashmir
Approach:
Match the endangered musk deer / hangul to the national park that conserves it.
Step 1:Recall the habitat of the hangul and musk deer.
Step 2:Name the park.
Final answer: Dachigam National Park, Jammu & Kashmir
Q179Single correctEnvironmental Issues
A lake which is rich in organic waste may result in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4mortality of fish due to lack of oxygen
Approach:
Trace the effect of high organic load in a lake on dissolved oxygen and aquatic life.
Step 1:Relate organic waste to microbial activity.
Step 2:Determine the consequence for fish.
Final answer: mortality of fish due to lack of oxygen
Q180Single correctEnvironmental Issues
The highest DDT concentration in aquatic food chain shall occur in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2seagull
Approach:
Apply biomagnification to find the trophic level with the greatest accumulated DDT.
Step 1:State the biomagnification rule.
Step 2:Order the given organisms by trophic position.
Step 3:Identify the maximum.
Final answer: seagull
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