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NEET 2018 May 06 Question Paper with Solutions
All 180 questions from the NEET 2018 (May 06) 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 2018Chemistry PYQs 2018Biology PYQs 2018
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q1Single correctThermodynamics
The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The V-T graph is a straight line through the origin, so the process is isobaric (V/T constant means constant pressure). For an isobaric process the ratio of work done to heat absorbed equals R/Cp.
Step 1:A linear V vs T graph through the origin corresponds to constant pressure, an isobaric process.
Step 2:For a monatomic gas the molar heat capacity at constant pressure is Cp = (5/2)R.
Step 3:The ratio of work done to heat absorbed for an isobaric process is dW/dQ = nR dT / (nCp dT).
Final answer:
Q2Single correctWaves
The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 313.2 cm
Approach:
Equate the fundamental frequency of the open pipe to the third harmonic of the closed pipe and solve for the open pipe length.
Step 1:For a closed organ pipe of length l, the third harmonic frequency is 3v/(4l).
Step 2:For an open organ pipe of length l', the fundamental frequency is v/(2l').
Step 3:Equate the two frequencies and substitute l = 20 cm.
Final answer: 13.2 cm
Q3Single correctKinetic Theory of Gases
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = kg
Boltzmann's constant )
(Given :
Mass of oxygen molecule (m) = kg
Boltzmann's constant )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 K
Approach:
Set the rms speed equal to the escape speed and solve for temperature using the kinetic theory expression for rms speed.
Step 1:The escape speed from Earth's surface is taken as 11200 m/s.
Step 2:Set the rms speed equal to the escape speed.
Step 3:Solve for T using the given mass and Boltzmann constant.
Final answer: K
Q4Single correctThermodynamics
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 326.8%
Approach:
Apply the ideal (Carnot) efficiency formula with the sink at the freezing point and the source at the boiling point of water in kelvin.
Step 1:The sink temperature is the freezing point T2 = 273 K and the source temperature is the boiling point T1 = 373 K.
Step 2:Substitute into the Carnot efficiency expression.
Step 3:Evaluate the percentage efficiency.
Final answer: 26.8%
Q5Single correctCurrent Electricity
A carbon resistor of k is to be marked with rings of different colours for its identification. The colour code sequence will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Yellow Violet Orange Silver
Approach:
Express the resistance as a base value times a power of ten with a tolerance, then map the digits, multiplier, and tolerance to their standard resistor colour codes.
Step 1:Write the resistance as significant figures times a multiplier with tolerance.
Step 2:The first digit 4 is Yellow and the second digit 7 is Violet.
Step 3:The multiplier is Orange and the 10% tolerance is Silver.
Final answer: Yellow Violet Orange Silver
Q6Single correctCurrent Electricity
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes 10 I. The value of 'n' is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 310
Approach:
Write the series current and the parallel current in terms of E, R and n, then use the ratio of currents (10I/I = 10) to solve for n.
Step 1:In series the total resistance is nR plus the internal resistance R.
Step 2:In parallel the combination resistance is R/n plus the internal resistance R.
Step 3:Divide the two equations to eliminate E and R, giving 10 = (n+1)/((1/n)+1).
Final answer: 10
Q7Single correctCurrent Electricity
A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The horizontal straight line: I is constant and independent of n
Approach:
For n identical cells in series, both the total emf and the total internal resistance scale with n, so the short-circuit current is independent of n, giving a horizontal line.
Step 1:For n cells in series the total emf is n times epsilon and total internal resistance is n times r.
Step 2:On short-circuit the current is the total emf divided by the total internal resistance.
Step 3:The factor n cancels, so I equals epsilon over r and is constant for all n.
Final answer: I is independent of n, so the graph is a horizontal straight line
Q8Single correctWave Optics
Unpolarised light is incident from air on a plane surface of a material of refractive index ''. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Reflected light is polarised with its electric vector perpendicular to the plane of incidence
Approach:
When the reflected and refracted rays are perpendicular, the incidence angle is the Brewster angle and the reflected light is completely plane polarised.
Step 1:The condition that reflected and refracted rays are perpendicular defines the Brewster (polarising) angle.
Step 2:At the Brewster angle the reflected ray is fully plane polarised.
Step 3:The electric field of the reflected polarised light is perpendicular to the plane of incidence.
Final answer: Reflected light is polarised with its electric vector perpendicular to the plane of incidence
Q9Single correctWave Optics
In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is . To increase the fringe angular width to (with same and D) the separation between the slits needs to be changed to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 21.9 mm
Approach:
The angular fringe width is lambda over d. Form the ratio of the two angular widths to relate the new slit separation to the original one.
Step 1:The angular width of the fringes equals the wavelength divided by the slit separation.
Step 2:For the increased angular width the new separation d' satisfies a similar relation.
Step 3:Dividing the relations gives 0.20/0.21 = d'/2 mm, so d' = 1.9 mm.
Final answer: 1.9 mm
Q10Single correctRay Optics and Optical Instruments
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Large focal length and large diameter
Approach:
Angular magnification of a telescope rises with objective focal length, and angular resolution improves with objective diameter, so both quantities must be large.
Step 1:Angular magnification equals the objective focal length divided by the eyepiece focal length, so a large objective focal length increases magnification.
Step 2:The smallest resolvable angle decreases as the objective diameter increases, so a large diameter gives high resolution.
Step 3:Both requirements together call for a large focal length and large diameter objective.
Final answer: Large focal length and large diameter
Q11Single correctAtoms
The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
In a Bohr orbit the total energy equals the negative of the kinetic energy, so the ratio of kinetic energy to total energy follows directly.
Step 1:For an electron in a Bohr orbit, the total energy equals the negative of the kinetic energy.
Step 2:Form the ratio of kinetic energy to total energy.
Step 3:The ratio simplifies to 1 : -1.
Final answer:
Q12Single correctDual Nature of Radiation and Matter
An electron of mass m with an initial velocity () enters an electric field ( constant > 0) at . If is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the acceleration of the electron in the field, compute its velocity at time t, then write the de-Broglie wavelength as h over momentum and relate it to the initial wavelength.
Step 1:The field exerts a force on the electron giving an acceleration eE0/m directed along +i.
Step 2:The speed after time t is V0 plus the acceleration times t.
Step 3:Substitute into the de-Broglie relation and factor out the initial wavelength lambda0 = h/(mV0).
Final answer:
Q13Single correctNuclei
For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 320
Approach:
Determine how many nuclei remain after the disintegration, express that fraction as a power of one-half, and read off the number of half-lives to find the elapsed time.
Step 1:After 450 of the 600 nuclei disintegrate, the number remaining is 150.
Step 2:The remaining fraction 150/600 equals one-quarter, which is (1/2) raised to the power 2.
Step 3:Two half-lives of 10 minutes each give a total time of 20 minutes.
Final answer: 20
Q14Single correctDual Nature of Radiation and Matter
When the light of frequency (where is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is . When the frequency of the incident radiation is increased to , the maximum velocity of electrons emitted from the same plate is . The ratio of to is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Apply Einstein's photoelectric equation for both frequencies using the work function as h times the threshold frequency, then take the ratio of the resulting kinetic energies.
Step 1:For incident frequency 2*nu0 the kinetic energy gives h(2 nu0) = h nu0 + (1/2) m , leading to h nu0 = (1/2) m .
Step 2:For incident frequency 5*nu0 similarly h(5 nu0) = h nu0 + (1/2) m , leading to 4 h nu0 = (1/2) m .
Step 3:Divide the two relations to find / = 1/4, so v1 : v2 = 1 : 2.
Final answer:
Q15Single correctSemiconductor Electronics
In the circuit shown in the figure, the input voltage is 20 V, and . The values of , and are given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
With the junction voltages set to zero, the base current is the input voltage across the 500 k-ohm base resistor and the collector current is the supply voltage across the 4 k-ohm collector resistor; the ratio of collector to base current gives beta.
Step 1:With VBE = 0 the base current equals 20 V across the 500 k-ohm resistor.
Step 2:With VCE = 0 the collector current equals the 20 V supply across the 4 k-ohm resistor.
Step 3:The current gain is the ratio of collector current to base current.
Final answer:
Q16Single correctSemiconductor Electronics
In a p-n junction diode, change in temperature due to heating
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Affects the overall V - I characteristics of p-n junction
Approach:
Heating a diode increases the generation of electron-hole pairs throughout the device, altering both forward and reverse behaviour and hence the complete characteristic curve.
Step 1:Heating increases the number of thermally generated electron-hole pairs in the semiconductor.
Step 2:More carriers change the resistance under both forward and reverse bias conditions.
Step 3:Since both forward and reverse behaviour change, the entire V-I characteristic is affected.
Final answer: Affects the overall V - I characteristics of p-n junction
Q17Single correctSemiconductor Electronics
In the combination of the following gates the output Y can be written in terms of inputs A and B as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Trace the logic network from the inputs through the intermediate gates to the final OR gate, writing the Boolean expression at each stage to obtain the output Y.
Step 1:The upper branch produces the product of A with the complement of B.
Step 2:The lower branch produces the product of the complement of A with B.
Step 3:The final OR gate combines the two branches to give Y.
Final answer:
Q18Single correctElectromagnetic Waves
An em wave is propagating in a medium with a velocity . The instantaneous oscillating electric field of this em wave is along axis. Then the direction of oscillating magnetic field of the em wave will be along
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 direction
Approach:
The propagation direction of an em wave is along E cross B, so given the propagation along +x and E along +y, the magnetic field direction follows from the vector relation.
Step 1:The propagation direction is parallel to E cross B.
Step 2:Substitute E along +y and V along +x to find B.
Step 3:The vector that satisfies this is B along +z, since j-hat cross k-hat equals i-hat.
Final answer: direction
Q19Single correctRay Optics and Optical Instruments
The refractive index of the material of a prism is and the angle of the prism is . One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For the beam to retrace its path it must strike the silvered surface normally, so the refraction angle at the first surface equals the prism angle; applying Snell's law at the entry face gives the incidence angle.
Step 1:To retrace its path the ray must hit the silvered surface normally, so the refraction angle at the first face equals the prism angle of 30 degrees.
Step 2:Apply Snell's law at the entry face with mu = sqrt 2 and r = 30 degrees.
Step 3:Solve for the incidence angle.
Final answer:
Q20Single correctRay Optics and Optical Instruments
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 236 cm away from the mirror
Approach:
Use the mirror formula for the two object positions (40 cm and 20 cm from the mirror) to find the two image distances, then take their difference to find the image displacement.
Step 1:For the initial object at u1 = -40 cm with f = -15 cm, the mirror formula gives v1 = -24 cm.
Step 2:After displacement towards the mirror the object is at u2 = -20 cm, giving v2 = -60 cm.
Step 3:The image moves from 24 cm to 60 cm, a shift of 36 cm away from the mirror.
Final answer: 36 cm away from the mirror
Q21Single correctElectromagnetic Induction and Alternating Currents
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
(1)
(2)
(3)
(4)
SolutionAnswer: Option 413.89 H
Approach:
Use the expression for energy stored in an inductor and rearrange to solve for the inductance from the given energy and current.
Step 1:The magnetic potential energy of an inductor is half L times current squared.
Step 2:Substitute U = 25 mJ and I = 60 mA.
Step 3:Solve for the inductance.
Final answer: 13.89 H
Q22Single correctElectric Charges and Fields
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Smaller
Approach:
Both particles fall through the same distance under an electric force of equal magnitude; the time of fall depends on the mass, so comparing the masses of the electron and proton gives the comparison of times.
Step 1:Each particle experiences an electric force eE producing acceleration eE/m, and falls the same distance h from rest.
Step 2:Solving for the time of fall shows it is proportional to the square root of the mass.
Step 3:Since the electron has a much smaller mass than the proton, the electron takes a smaller time to fall.
Final answer: Smaller
Q23Single correctElectrostatic Potential and Capacitance
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Independent of the distance between the plates
Approach:
For an isolated capacitor the charge Q is fixed; the force on one plate equals the charge times the field due to the other plate, which depends only on Q and A and not on the separation.
Step 1:For an isolated capacitor the charge Q remains constant when the separation changes.
Step 2:The force on one plate equals the charge times the field produced by the other plate, over 2 A epsilon-zero.
Step 3:This expression contains no plate separation, so the force is independent of the distance.
Final answer: Independent of the distance between the plates
Q24Single correctWaves
A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of two successive resonances are produced at and of column length. If the frequency of the tuning fork is , the velocity of sound in air at is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Two successive resonances of an air column differ in length by half the wavelength. The wavelength is found from this difference, then the speed of sound follows from the wave relation with the given frequency.
Step 1:The gap between two successive resonance lengths equals half a wavelength.
Step 2:Apply the wave relation with the tuning fork frequency.
Final answer:
Q25Single correctOscillations
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is at a distance of from the mean position. The time period of oscillation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
In simple harmonic motion the magnitude of acceleration equals the angular frequency squared times the displacement. The angular frequency gives the time period directly.
Step 1:Relate the given acceleration magnitude and displacement to the angular frequency.
Step 2:Take the square root to find the angular frequency.
Step 3:Compute the time period from the angular frequency.
Final answer:
Q26Single correctMoving Charges and Magnetism
A metallic rod of mass per unit length is lying horizontally on a smooth inclined plane which makes an angle of with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Balance the component of gravity along the incline against the component of the horizontal magnetic force along the incline. With a vertical field and horizontal current the force is horizontal; resolving both along the incline gives the required current.
Step 1:Set the gravity component along the incline equal to the magnetic force component along the incline.
Step 2:Substitute the mass per unit length, g, and the field with tangent of thirty degrees.
Final answer:
Q27Single correctElectromagnetic Induction
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3The current source
Approach:
Identify the ultimate origin of the energy that becomes the gravitational potential energy of the rod. The magnetic field does no net work on charges and merely mediates the interaction; the energy is supplied by the source maintaining the current.
Step 1:The rod is lifted, so its gravitational potential energy increases and this energy must be supplied externally.
Step 2:A magnetic field does no work on moving charges, so the field itself cannot be the energy source; the current source that drives the electromagnet supplies the energy.
Final answer: The current source
Q28Single correctAlternating Current
An inductor of , a capacitor and a resistor are connected in series across a source of emf, . The power loss in the circuit is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Find the impedance from the resistance and the net reactance, then the average power dissipated equals the mean square emf divided by the impedance times the power factor, which reduces to the rms voltage squared times resistance over impedance squared.
Step 1:Compute the net reactance and impedance using omega equal to 314 rad/s.
Step 2:Apply the average power expression with peak emf 10 V.
Final answer:
Q29Single correctMoving Charges and Magnetism
Current sensitivity of a moving coil galvanometer is and its voltage sensitivity (angular deflection per unit voltage applied) is . The resistance of the galvanometer is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The galvanometer resistance equals the current sensitivity divided by the voltage sensitivity, since current sensitivity is deflection per unit current and voltage sensitivity is deflection per unit voltage.
Step 1:Express the resistance as the ratio of current sensitivity to voltage sensitivity, keeping consistent units.
Step 2:Simplify the ratio.
Final answer:
Q30Single correctSystems of Particles and Rotational Motion
A body initially at rest and sliding along a frictionless track from a height (as shown in the figure) just completes a vertical circle of diameter . The height is equal to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
For a body to just complete a vertical loop on a frictionless track, the speed at the top must satisfy the minimum condition. Energy conservation from the release height to the top of the loop, accounting for the rise to the top diameter, fixes h in terms of the diameter.
Step 1:At the top of the circle the minimum speed for just completing the loop is set by gravity providing the centripetal force, with R the radius equal to D over two.
Step 2:Apply energy conservation between the release point and the top of the loop, where the top is at height equal to the diameter.
Final answer:
Q31Single correctSystems of Particles and Rotational Motion
Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The work to bring a spinning body to rest equals its rotational kinetic energy, which for a common angular speed is proportional to the moment of inertia. Ordering the moments of inertia about the symmetry axis orders the required work.
Step 1:The work to stop each body equals its rotational kinetic energy, which scales with its moment of inertia at fixed angular speed.
Step 2:Compare the moments of inertia for the same M and R in the ratio two fifths, one half, and one.
Final answer:
Q32Single correctWork, Energy and Power
A moving block having mass , collides with another stationary block having mass . The lighter block comes to rest after collision. When the initial velocity of the lighter block is , then the value of coefficient of restitution (e) will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Use conservation of linear momentum to find the velocity of the heavier block after collision, since the lighter block stops. The coefficient of restitution is the ratio of relative separation speed to relative approach speed.
Step 1:Conserve momentum with the lighter block stopping after collision.
Step 2:Form the coefficient of restitution as the separation speed over the approach speed.
Final answer:
Q33Single correctLaws of Motion
Which one of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Coefficient of sliding friction has dimensions of length.
Approach:
Evaluate each statement against the established properties of friction. The coefficient of friction is the ratio of the friction force to the normal reaction, a ratio of two forces, hence dimensionless.
Step 1:The coefficient of sliding friction is the ratio of friction force to normal reaction.
Step 2:Since both numerator and denominator are forces, the coefficient is dimensionless, so the statement assigning it dimensions of length is incorrect.
Final answer: Coefficient of sliding friction has dimensions of length.
Q34Single correctElectric Charges and Fields
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field . Due to the force , its velocity increases from to in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between to seconds are respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
The field gives constant acceleration of magnitude 6 m/s squared. Track the velocity and displacement over three one-second intervals with the field reversed after the first second, then form average velocity as net displacement over time and average speed as total distance over time.
Step 1:The acceleration magnitude is 6 m/s squared from the velocity change in the first second. For t from 0 to 1 s the displacement is computed.
Step 2:For t from 1 to 2 s the field is reversed so the car decelerates from 6 m/s; displacement in this interval is found.
Step 3:For t from 2 to 3 s the velocity reverses; displacement is back toward the start.
Step 4:Total displacement gives average velocity; total distance gives average speed.
Final answer:
Q35Single correctLaws of Motion
A block of mass m is placed on a smooth inclined wedge ABC of inclination as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and for the block to remain stationary on the wedge is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Work in the non-inertial frame of the wedge, adding a pseudo force on the block opposite to the wedge acceleration. For the block to stay put on the smooth incline, the components of gravity and pseudo force along the incline must balance.
Step 1:In the wedge frame the pseudo force ma acts on the block; resolving the normal reaction, the horizontal balance gives the first relation and the vertical balance gives the second.
Step 2:Divide the two relations to eliminate the normal reaction.
Final answer:
Q36Single correctSystems of Particles and Rotational Motion
The moment of the force, at , about the point , is given by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The moment of a force about a point is the cross product of the position vector from that point to the point of application with the force vector. Form the position vector difference, then evaluate the determinant.
Step 1:Compute the position vector from the reference point to the point of application.
Step 2:Evaluate the cross product determinant of the position vector with the force.
Final answer:
Q37Single correctUnits and Measurements
A student measured the diameter of a small steel ball using a screw gauge of least count . The main scale reading is and zero of circular scale division coincides with divisions above the reference level. If screw gauge has a zero error of , the correct diameter of the ball is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The measured diameter is the main scale reading plus the circular scale reading times the least count. Subtracting the zero error (which is negative) corrects the reading.
Step 1:Add the main scale reading of 5 mm, equal to 0.5 cm, to the circular scale contribution of 25 times the least count, then subtract the negative zero error.
Step 2:Sum the contributions.
Final answer:
Q38Single correctSystems of Particles and Rotational Motion
A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Angular momentum
Approach:
In free space with no external torque, angular momentum is conserved. As the radius grows, the moment of inertia increases and the angular velocity decreases, while the product remains fixed.
Step 1:With no external torque acting, the rate of change of angular momentum is zero.
Step 2:Increasing the radius raises the moment of inertia and lowers the angular velocity, so kinetic energy, moment of inertia, and angular velocity all change while angular momentum stays constant.
Final answer: Angular momentum
Q39Single correctGravitation
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are , and , respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By Kepler's second law the planet moves fastest at perihelion (closest to the Sun) and slowest at aphelion (farthest). Identify A as perihelion and C as aphelion, with B at an intermediate distance, then order the kinetic energies by speed.
Step 1:Point A is perihelion and C is aphelion, with B between them, so the orbital speed is greatest at A and least at C.
Step 2:Kinetic energy increases with the square of speed, so it follows the same ordering.
Final answer:
Q40Single correctGravitation
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4'g' on the Earth will not change
Approach:
Surface gravity on Earth depends on the gravitational constant G and Earth's mass, not on the Sun's mass; scaling G changes g and every g-dependent effect.
Step 1:The Sun's mass does not enter the surface gravity on Earth; g is fixed by G and Earth's mass.
Step 2:Making the gravitational constant ten times larger scales g by the same factor.
Step 3:With larger g: the pendulum period decreases, effective weight rises so walking is harder, and raindrop terminal speed rises — statements (1), (2), (3) are correct.
Step 4:Statement (4) claims g stays unchanged, contradicting the increase; it is the statement that is not correct.
Final answer: 'g' on the Earth will not change
Q41Single correctSystems of Particles and Rotational Motion
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy as well as rotational kinetic energy simultaneously. The ratio for the sphere is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
For a rolling solid sphere, express both translational and rotational kinetic energies in terms of the translational speed using the moment of inertia and the rolling constraint, then form the requested ratio.
Step 1:Add the translational and rotational kinetic energies using the moment of inertia of a solid sphere and the rolling condition.
Step 2:Form the ratio of translational energy to total energy.
Final answer:
Q42Single correctMechanical Properties of Fluids
A small sphere of radius '' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
At terminal velocity the rate of heat production equals the power dissipated by the viscous drag, which is the drag force times the terminal velocity. Use Stokes' drag with the terminal velocity dependence on radius to find the power's dependence on r.
Step 1:The rate of heat production at terminal velocity is the Stokes drag force times the terminal velocity.
Step 2:Since the terminal velocity scales as the square of the radius, substitute its dependence.
Final answer:
Q43Single correctThermal Properties of Matter
The power radiated by a black body is P and it radiates maximum energy at wavelength, . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength , the power radiated by it becomes nP. The value of n is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Wien's law relates peak wavelength inversely to temperature, fixing the new temperature ratio. The Stefan-Boltzmann law makes power scale with the fourth power of temperature, giving the factor n.
Step 1:By Wien's law the temperatures are inversely proportional to the peak wavelengths.
Step 2:Apply the fourth power dependence of radiated power on temperature.
Final answer:
Q44Single correctMechanical Properties of Solids
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area . If the length of the first wire is increased by on applying a force F, how much force is needed to stretch the second wire by the same amount?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Equal volume with three times the area means one third the length for the second wire. Apply the Young's modulus relation to each wire for the same extension and take the ratio of the forces.
Step 1:For the first wire of area A and length l the extension relation is written.
Step 2:Equal volume gives the second wire length l over three for area three A; impose the same extension.
Step 3:Equate the two extensions and solve for the required force.
Final answer:
Q45Single correctThermodynamics
A sample of of water at and normal pressure requires of heat energy to convert to steam at . If the volume of the steam produced is , the change in internal energy of the sample, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
By the first law of thermodynamics, the change in internal energy equals the heat supplied minus the work done by the system during the expansion at constant pressure. Convert the supplied heat to joules and compute the pressure-volume work for the volume change.
Step 1:Convert the heat supplied to joules and subtract the pressure-volume work for the expansion from negligible liquid volume to the steam volume.
Step 2:Carry out the subtraction.
Final answer:
Chemistry45 questions
Q46Single correctThe p-Block Elements (Group 15)
The correct order of N-compounds in its decreasing order of oxidation states is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Assign the oxidation state of nitrogen in each species, then arrange in decreasing order.
Step 1:Nitrogen in nitric acid.
Step 2:Nitrogen in nitric oxide.
Step 3:Nitrogen in dinitrogen.
Step 4:Nitrogen in ammonium chloride.
Step 5:Arrange from highest to lowest oxidation state.
Final answer:
Q47Single correctThe p-Block Elements (Group 13)
Which one of the following elements is unable to form ion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Identify which Group 13 element lacks the orbitals required to expand its covalency to six.
Step 1:Boron belongs to the second period and has only 2s and 2p orbitals available in its valence shell, with no vacant d-orbitals.
Step 2:Without accessible d-orbitals, boron cannot extend its covalency beyond four, so a six-coordinate fluoride ion is not possible for it.
Step 3:The heavier elements Al, Ga and In possess vacant d-orbitals and can expand covalency to six.
Final answer:
Q48Single correctGeneral Principles and Processes of Isolation of Elements
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
A metal can reduce alumina only if its oxide formation line lies below that of aluminium oxide in the Ellingham diagram, i.e. the metal is more reactive than aluminium.
Step 1:Reduction of alumina requires a metal whose oxidation to oxide is more thermodynamically favourable than that of aluminium.
Step 2:Only magnesium is more reactive than aluminium; its oxide line lies below the aluminium line.
Step 3:Zn, Fe and Cu are less reactive than aluminium and cannot reduce alumina.
Final answer:
Q49Single correctThe p-Block Elements (Group 13)
The correct order of atomic radii in group 13 elements is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Compare the experimental atomic radii of the Group 13 elements; gallium is anomalously small due to poor shielding by the intervening 3d electrons.
Step 1:List the atomic radii in picometres.
Step 2:Gallium is smaller than aluminium because the poor shielding by the filled 3d subshell increases the effective nuclear charge on the outer electrons.
Step 3:Arrange all five values in increasing order.
Final answer:
Q50Single correctThe p-Block Elements (Group 17)
Which of the following statements is not true for halogens?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1All but fluorine show positive oxidation states
Approach:
Examine each statement for halogens and identify the false one. The question asks for the statement that is not true.
Step 1:Fluorine, being the most electronegative element, exhibits only the -1 and 0 oxidation states; the other halogens display positive oxidation states.
Step 2:Owing to its high electronegativity and small size, fluorine forms the oxoacid hypofluorous acid, HOF, in which fluorine carries a +1 oxidation state.
Step 3:Because fluorine attains a positive oxidation state in HOF, the claim that all halogens except fluorine show positive oxidation states is the untrue statement.
Final answer: All but fluorine show positive oxidation states
Q51Single correctChemical Bonding and Molecular Structure
In the structure of , the number of lone pair of electrons on central atom 'Cl' is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Two
Approach:
Apply VSEPR theory to count the bond pairs and lone pairs around the central chlorine atom in chlorine trifluoride.
Step 1:Chlorine, the central atom, contributes seven valence electrons.
Step 2:Three chlorine-fluorine bonds use three of these electrons in bonding.
Step 3:The remaining four electrons form two lone pairs on chlorine.
Final answer: Two
Q52Single correctHydrocarbons / Alcohols, Phenols and Ethers
Identify the major products P, Q and R in the following sequence of reactions:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Trace the Friedel-Crafts alkylation with carbocation rearrangement to give cumene, then apply the cumene (Hock) process to obtain phenol and acetone.
Step 1:Anhydrous aluminium chloride ionizes 1-chloropropane; the primary carbocation rearranges by a 1,2-hydride shift to the more stable secondary isopropyl carbocation, which alkylates benzene to give cumene (isopropylbenzene), P.
Step 2:Air oxidation of cumene at the benzylic position produces cumene hydroperoxide.
Step 3:Acid-catalysed cleavage of the hydroperoxide (Hock rearrangement) furnishes phenol Q and acetone R.
Final answer:
Q53Single correctBiomolecules / Amines
Which of the following compounds can form a zwitterion?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Glycine
Approach:
A zwitterion forms only when a molecule contains both an acidic and a basic group that can undergo internal proton transfer.
Step 1:Glycine possesses a carboxylic acid group and an amino group on the same molecule.
Step 2:The carboxyl proton transfers to the amino nitrogen, giving a dipolar ion bearing both a positive and a negative charge.
Step 3:Benzoic acid, acetanilide and aniline each lack the simultaneous free acidic and basic groups required for an internal salt.
Final answer: Glycine
Q54Single correctPolymers
Regarding cross-linked or network polymers, which of the following statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4They contain strong covalent bonds in their polymer chains.
Approach:
Recall the defining feature of cross-linked polymers and identify the statement that does not describe the cross-linking itself.
Step 1:Cross-linked or network polymers form from bi-functional and tri-functional monomers, with examples such as bakelite and melamine.
Step 2:Their characteristic feature is strong covalent cross-links joining various linear polymer chains together.
Step 3:Stating that the strong covalent bonds lie within the polymer chains misses the cross-linking between chains, so this statement is not related to cross-linking and is incorrect.
Final answer: They contain strong covalent bonds in their polymer chains.
Q55Single correctAmines
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4In acidic (strong) medium aniline is present as anilinium ion.
Approach:
Determine the directing behaviour of the species actually present when aniline is nitrated in strong acid.
Step 1:In strong acid, the lone pair on the amino nitrogen is protonated, converting aniline into the anilinium ion.
Step 2:The positively charged ammonium substituent is electron-withdrawing and therefore meta-directing.
Step 3:Consequently, alongside para (about 51%) and ortho (about 2%) products, a significant meta product (about 47%) is obtained, explaining the formation of m-nitroaniline.
Final answer: In acidic (strong) medium aniline is present as anilinium ion.
Q56Single correctBiomolecules
The difference between amylose and amylopectin is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Amylopectin have -linkage and -linkage
Approach:
Compare the glycosidic linkages present in the two components of starch.
Step 1:Both amylose and amylopectin are polymers of alpha-D-glucose.
Step 2:Amylose is a linear chain held together only by 1 to 4 alpha linkages.
Step 3:Amylopectin is branched, with 1 to 4 alpha linkages along the chains and 1 to 6 alpha linkages at the branch points.
Final answer: Amylopectin have -linkage and -linkage
Q57Single correctAldehydes, Ketones and Carboxylic Acids
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Determine the gases produced by sulphuric-acid dehydration of each acid, remove the CO2 absorbed by KOH, and weigh the remaining CO.
Step 1:Formic acid (molar mass 46) is dehydrated by concentrated sulphuric acid to carbon monoxide and water.
Step 2:Oxalic acid (molar mass 90) is dehydrated to carbon monoxide, carbon dioxide and water.
Step 3:KOH pellets absorb all the carbon dioxide, so only the carbon monoxide remains in the gaseous mixture.
Step 4:Convert the remaining moles of carbon monoxide to mass.
Final answer:
Q58Single correctThe s-Block Elements
Which of the following oxides is most acidic in nature?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Within Group 2, basic character of the oxide increases down the group, so the topmost element gives the least basic (most acidic) oxide.
Step 1:Down Group 2, the metallic and basic character of the oxides increases from BeO to BaO.
Step 2:Beryllium oxide is amphoteric, whereas the remaining oxides are basic.
Step 3:The least basic and amphoteric beryllium oxide is therefore the most acidic of the listed oxides.
Final answer:
Q59Single correctEnvironmental Chemistry / Oxides of Nitrogen
Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Recall which oxides of nitrogen are recognised atmospheric pollutants from natural and anthropogenic sources.
Step 1:Nitrous oxide, nitric oxide and nitrogen dioxide are released into the atmosphere from both natural and human activities and are common pollutants.
Step 2:Dinitrogen pentoxide is unstable under ambient atmospheric conditions and is not introduced as a common pollutant.
Final answer:
Q60Single correctAlcohols, Phenols and Ethers
The compound A on treatment with Na gives B, and with gives C. B and C react together to give diethyl ether. A, B and C are in the order
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify A, B and C from the described reactions, using the Williamson synthesis of diethyl ether as the key constraint.
Step 1:Ethanol reacts with sodium to give sodium ethoxide, identifying A as ethanol and B as sodium ethoxide.
Step 2:Ethanol with phosphorus pentachloride gives chloroethane, identifying C.
Step 3:Sodium ethoxide reacts with chloroethane by the Williamson synthesis to give diethyl ether, confirming B and C.
Final answer:
Q61Single correctHaloalkanes and Haloarenes / Hydrocarbons
The compound undergoes the following reactions:
The product 'C' is
The product 'C' is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3m-bromotoluene
Approach:
Follow the side-chain chlorination of toluene, the meta-directed ring bromination, and the reductive removal of the chlorines.
Step 1:Toluene (C7H8) undergoes free-radical side-chain chlorination with three equivalents of chlorine under heat to give (trichloromethyl)benzene, A.
Step 2:The strongly electron-withdrawing trichloromethyl group is meta-directing, so bromination with Br2/Fe places bromine at the meta position to give B.
Step 3:Reduction with Zn/HCl converts the trichloromethyl group back to a methyl group, giving m-bromotoluene as C.
Final answer: m-bromotoluene
Q62Single correctHydrocarbons / Haloalkanes
Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Work backward from the Wurtz product, which must be a gaseous hydrocarbon with fewer than four carbons, to identify the starting hydrocarbon that reacts by substitution.
Step 1:Substitution with bromine requires a saturated hydrocarbon; methane reacts by free-radical substitution to give bromomethane.
Step 2:The Wurtz reaction couples two bromomethane molecules to give ethane, a gaseous hydrocarbon with two carbon atoms.
Step 3:Ethane (C2) satisfies the requirement of a gaseous product with fewer than four carbon atoms, identifying A as methane.
Final answer:
Q63Single correctSome Basic Concepts / Chemical Bonding
Which of the following molecules represents the order of hybridisation from left to right atoms?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Assign hybridisation to each carbon from left to right, where a carbon in a double bond is sp2 and a carbon in a triple bond is sp.
Step 1:Hybridisation is governed by the number of sigma bonds plus lone pairs around each carbon; doubly bonded carbons are sp2 and triply bonded carbons are sp.
Step 2:In but-1-en-3-yne the first two carbons of the double bond are sp2 and the last two carbons of the triple bond are sp.
Step 3:This pattern matches the required order from left to right.
Final answer:
Q64Single correctOrganic Chemistry Some Basic Principles
Which of the following carbocations is expected to be most stable?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cyclohexadienyl (arenium) cation in which is meta to the carbon (the one bearing H and Y); the positive charge sits on the ring carbon ortho to that carbon and para to
Approach:
An arenium ion is most stable when the strongly electron-withdrawing nitro group is not located on a carbon that bears the positive charge, since adjacent positive charge and -NO2 destabilise the cation.
Step 1:The nitro group is a strong electron-withdrawing substituent that intensifies and destabilises a neighbouring positive charge.
Step 2:A cation is therefore most stable when the positive charge resides on carbons remote from the nitro group, avoiding direct or adjacent placement of charge next to -NO2.
Step 3:Among the drawn structures, the one in which the positive charge is farthest from the nitro group experiences the least destabilisation and is the most stable carbocation.
Final answer: Cyclohexadienyl (arenium) cation with the group meta to the carbon (bearing H and Y) and the positive charge located ortho/para to the carbon, i.e. positive charge away from the group (see figure).
Q65Single correctOrganic Chemistry Some Basic Principles
Which of the following is correct with respect to effect of the substituents? (R = alkyl)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The -I effect increases with the electronegativity of the atom attached, so order the substituents by the electronegativity of nitrogen, oxygen and fluorine.
Step 1:The inductive electron-withdrawing effect strengthens as the electronegativity of the directly bonded atom increases.
Step 2:Electronegativity rises across the period from nitrogen to oxygen to fluorine.
Step 3:Therefore the -I effect of the substituents increases in the order amino, alkoxy, fluoro.
Final answer:
Q66Single correctAldehydes, Ketones and Carboxylic Acids / Phenols
In the reaction shown below, the electrophile involved is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Dichlorocarbene
Approach:
Recall the mechanism of the Reimer-Tiemann reaction and identify the reactive electrophilic species generated from chloroform and base.
Step 1:Hydroxide ion abstracts a proton from chloroform to give the trichloromethyl carbanion.
Step 2:The carbanion loses a chloride ion to form the electron-deficient dichlorocarbene.
Step 3:Dichlorocarbene is the electrophile that attacks the phenoxide ring to give, after hydrolysis, salicylaldehyde.
Final answer: Dichlorocarbene
Q67Single correctAldehydes, Ketones and Carboxylic Acids
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Formation of intermolecular H-bonding
Approach:
Identify the intermolecular interaction responsible for the unusually high boiling points of carboxylic acids.
Step 1:Carboxylic acid molecules associate through intermolecular hydrogen bonding, commonly forming dimers.
Step 2:This association raises the energy required to separate the molecules into the vapour phase, increasing the boiling point above that of comparable aldehydes, ketones and alcohols.
Final answer: Formation of intermolecular H-bonding
Q68Single correctAldehydes, Ketones and Carboxylic Acids / Alcohols
Compound A, , is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell.
A and Y are respectively
A and Y are respectively
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 (1-phenylethanol) and
Approach:
A positive iodoform test requires a CH3CO- group or a secondary alcohol of the form CH3CH(OH)- that is oxidised to a methyl ketone; identify A and the reagent Y that generates NaOI.
Step 1:Sodium hypoiodite is generated from iodine and sodium hydroxide, so Y is iodine.
Step 2:1-Phenylethanol is a secondary alcohol of the type CH3CH(OH)-, which is oxidised by hypoiodite to the methyl ketone acetophenone.
Step 3:Acetophenone bears a methyl ketone group and undergoes the iodoform reaction to give the yellow iodoform precipitate and sodium benzoate.
Final answer: (1-phenylethanol) and
Q69Single correctCoordination Compounds
Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :
| Column I | Column II |
|---|---|
| a. | i. |
| b. | ii. |
| c. | iii. |
| d. | iv. |
| v. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a-iv, b-v, c-ii, d-i
Approach:
The spin-only magnetic moment is , where n is the number of unpaired electrons. Determine n for each ion from its d-electron configuration and match the resulting value.
Step 1:Co3+ is [Ar]3d6 with 4 unpaired electrons.
Step 2:Cr3+ is [Ar]3d3 with 3 unpaired electrons.
Step 3:Fe3+ is [Ar]3d5 with 5 unpaired electrons.
Step 4:Ni2+ is [Ar]3d8 with 2 unpaired electrons.
Final answer: a-iv, b-v, c-ii, d-i
Q70Single correctThe d- and f-Block Elements
Which one of the following ions exhibits d-d transition and paramagnetism as well?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A d-d transition and paramagnetism both require at least one unpaired d-electron on the central metal ion. Determine the d-electron count of the metal in each ion.
Step 1:In MnO4-, Mn is in the +7 state, giving [Ar] (d0); - and - have Cr in the +6 state, giving [Ar] (d0). These are d0, so they are diamagnetic and show no d-d transition.
Step 2:In -, Mn is in the +6 state, giving [Ar]3d1, which carries one unpaired electron.
Step 3:The presence of one d-electron permits a d-d electronic transition and makes the ion paramagnetic.
Final answer:
Q71Single correctCoordination Compounds
Iron carbonyl, is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mononuclear
Approach:
The nuclearity of a metal carbonyl equals the number of metal atoms present in one molecule of the complex.
Step 1:The formula Fe(CO)5 contains exactly one iron atom bonded to five carbonyl ligands.
Step 2:A carbonyl with a single metal atom is classified as mononuclear.
Final answer: Mononuclear
Q72Single correctCoordination Compounds
The type of isomerism shown by the complex is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Geometrical isomerism
Approach:
Identify the geometry and ligand arrangement of the octahedral complex to determine the form of stereoisomerism it can display.
Step 1:In [CoCl2(en)2], cobalt has a coordination number of six and the complex is octahedral.
Step 2:The two chloride ligands can occupy adjacent or opposite positions, producing cis and trans arrangements.
Step 3:The existence of cis and trans forms identifies geometrical isomerism.
Final answer: Geometrical isomerism
Q73Single correctCoordination Compounds
The geometry and magnetic behaviour of the complex are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Tetrahedral geometry and diamagnetic
Approach:
Determine the oxidation state of nickel, the effect of the strong-field CO ligand on its electrons, the resulting hybridisation, and hence the geometry and magnetic nature.
Step 1:In Ni(CO)4 nickel is in the zero oxidation state with configuration [Ar]3d8 4s2.
Step 2:CO is a strong field ligand and pairs the 4s electrons into the 3d orbitals, giving a 3d10 4s0 configuration with no unpaired electrons.
Step 3:The four CO ligands bond through sp3 hybridisation, producing a tetrahedral geometry; the absence of unpaired electrons makes the complex diamagnetic.
Final answer: Tetrahedral geometry and diamagnetic
Q74Single correctEquilibrium
Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :
a.
b.
c.
d.
pH of which one of them will be equal to 1?
a.
b.
c.
d.
pH of which one of them will be equal to 1?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4c
Approach:
For each mixture compute the milliequivalents of HCl and NaOH, find the excess acid, divide by the total volume to get the H+ concentration, and obtain the pH. pH equals 1 when the resulting H+ concentration is 0.1 M.
Step 1:For mixture c, meq of HCl is computed from 75 mL of M/5 acid.
Step 2:Meq of NaOH for mixture c from 25 mL of M/5 base.
Step 3:Excess acid is 15 - 5 = 10 meq in a total volume of 100 mL, giving the H+ concentration.
Step 4:The pH follows from the H+ concentration.
Final answer: c
Q75Single correctSurface Chemistry
On which of the following properties does the coagulating power of an ion depend?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Both magnitude and sign of the charge on the ion
Approach:
Recall the factors that govern the coagulating ability of an electrolyte ion on a charged colloidal sol.
Step 1:Coagulation occurs when an ion of charge opposite to that of the colloidal particles neutralises their surface charge, so the sign of the ion's charge is decisive.
Step 2:By the Hardy-Schulze rule, the greater the magnitude of the ionic charge, the stronger its coagulating power.
Step 3:Coagulating power therefore depends on both the magnitude and the sign of the charge on the ion.
Final answer: Both magnitude and sign of the charge on the ion
Q76Single correctStates of Matter
Given van der Waals constant for , , and are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The van der Waals constant 'a' measures the strength of intermolecular attraction; the larger the value of 'a', the more easily the gas liquefies.
Step 1:The constant 'a' signifies the magnitude of intermolecular attractive forces.
Step 2:Comparing the given values, NH3 has the highest 'a' (4.17) among the four gases.
Step 3:The gas with the largest 'a' is the easiest to liquefy.
Final answer:
Q77Single correctEquilibrium
The solubility of in water is at 298 K. The value of its solubility product will be
(Given molar mass of )
(Given molar mass of )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Convert the solubility in g/L into mol/L using the molar mass, then express Ksp in terms of the molar solubility for the 1:1 dissociation of BaSO4.
Step 1:Divide the solubility by the molar mass to obtain the molar solubility.
Step 2:BaSO4 dissociates into Ba2+ and - in a 1:1 ratio, so each ion concentration equals s.
Step 3:The solubility product is the square of the molar solubility.
Step 4:Evaluate the square.
Final answer:
Q78Single correctSome Basic Concepts of Chemistry
In which case is number of molecules of water maximum?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3 of water
Approach:
Convert each quantity into moles of water and multiply by Avogadro's number; the largest mole count corresponds to the maximum number of molecules.
Step 1:0.00224 L of vapour at STP gives moles equal to volume divided by 22.4 L.
Step 2:0.18 g of water gives moles equal to mass divided by molar mass 18.
Step 3:18 mL of water has a mass of 18 g (density 1 g/mL), which is 1 mole.
Step 4: mol of water gives NA molecules.
Final answer: of water
Q79Single correctChemical Kinetics
The correct difference between first and second order reactions is that
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The half-life of a first-order reaction does not depend on ; the half-life of a second-order reaction does depend on
Approach:
Compare the half-life expressions of first-order and second-order reactions to identify their dependence on the initial concentration.
Step 1:For a first-order reaction the half-life is independent of the initial concentration.
Step 2:For a second-order reaction the half-life is inversely proportional to the initial concentration.
Step 3:Therefore the distinguishing feature is the dependence of the second-order half-life on initial concentration.
Final answer: The half-life of a first-order reaction does not depend on ; the half-life of a second-order reaction does depend on
Q80Single correctThe s-Block Elements
Among , , , the order of ionic character is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The ionic character of a metal hydride increases as the metallic (electropositive) character of the metal increases down the group.
Step 1:Down group 2, atomic size increases and the metals become more electropositive, so metallic character rises in the order Be < Ca < Ba.
Step 2:Greater metallic character of the metal produces a more ionic hydride.
Final answer:
Q81Single correctRedox Reactions
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
Then the species undergoing disproportionation is
Then the species undergoing disproportionation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
A species disproportionates when the standard cell potential for its reduction to a lower state minus its formation from a higher state (i.e. ight - eft across it) is positive. Evaluate this for HBrO using the given potentials.
Step 1:HBrO can be reduced to Br2 with potential 1.595 V and is itself produced from BrO3- with potential 1.5 V.
Step 2:The cell potential for the disproportionation of HBrO is the difference of the two relevant potentials.
Step 3:Since the cell potential is positive (greater than zero), HBrO disproportionates.
Final answer:
Q82Single correctRedox Reactions
For the redox reaction
The correct coefficients of the reactants for the balanced equation are
The correct coefficients of the reactants for the balanced equation are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Balance the redox reaction by equating the total electrons gained by the reduction half (MnO4- to Mn2+) with the electrons lost in the oxidation half (- to CO2), then balance oxygen with water and hydrogen with H+.
Step 1:Each MnO4- gains 5 electrons going from Mn(+7) to Mn(+2); each - loses 2 electrons going to 2 CO2.
Step 2:Equalising electrons gives a mole ratio of MnO4- to - of 2 : 5.
Step 3:Balancing oxygen and hydrogen with water and H+ gives the full balanced equation.
Final answer:
Q83Single correctEquilibrium
Which one of the following conditions will favour maximum formation of the product in the reaction,
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Low temperature and high pressure
Approach:
Apply Le Chatelier's principle, considering the change in number of gaseous moles and the exothermic nature of the forward reaction.
Step 1:The forward reaction reduces the number of gaseous moles from two to one, so increasing pressure shifts the equilibrium toward the product.
Step 2:The forward reaction is exothermic, so lowering the temperature shifts the equilibrium toward the product.
Step 3:Maximum product formation is therefore achieved at low temperature and high pressure.
Final answer: Low temperature and high pressure
Q84Single correctChemical Kinetics
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Is doubled
Approach:
Use the half-life expression for a zero order reaction to determine how it scales with the initial concentration.
Step 1:For a zero order reaction the half-life is directly proportional to the initial concentration.
Step 2:Doubling the initial concentration therefore doubles the half-life.
Final answer: Is doubled
Q85Single correctThermodynamics
The bond dissociation energies of , and XY are in the ratio of . for the formation of XY is . The bond dissociation energy of will be
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Write the enthalpy of formation of XY as the bonds broken (half X2 and half Y2) minus the bond formed (XY), express each bond energy in terms of X, and solve for X.
Step 1:The formation reaction is half a mole of X2 plus half a mole of Y2 forming one mole of XY.
Step 2:With bond energies in the ratio X2 : Y2 : XY = 1 : 0.5 : 1, set X2 = X, Y2 = X/2 and XY = X, then apply the enthalpy relation.
Step 3:Simplify the expression for the enthalpy.
Step 4:Solve for the bond dissociation energy of X2.
Final answer:
Q86Single correctStates of Matter
The correction factor 'a' to the ideal gas equation corresponds to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Forces of attraction between the gas molecules
Approach:
Recall the physical meaning of the constant 'a' in the van der Waals equation of state for real gases.
Step 1:The pressure correction term contains 'a', which accounts for the intermolecular attractive forces that reduce the observed pressure of a real gas.
Step 2:The constant 'a' therefore signifies the magnitude of the intermolecular forces of attraction.
Final answer: Forces of attraction between the gas molecules
Q87Single correctChemical Bonding and Molecular Structure
Consider the following species :
and
Which one of these will have the highest bond order?
and
Which one of these will have the highest bond order?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Determine the number of electrons in each species, fill the molecular orbitals, and compute the bond order as half the difference between bonding and antibonding electrons.
Step 1:CN- has 14 electrons filling up to the sigma 2pz bonding orbital, giving the maximum bond order.
Step 2:Computing the bond order for CN- gives 3.
Step 3:CN (13 e, B.O. 2.5), CN+ (12 e, B.O. 2) and NO (15 e, B.O. 2.5) all have a lower bond order than CN-.
Final answer:
Q88Single correctClassification of Elements and Periodicity
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is , the simplest formula for this compound is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Determine the valency of element X from its electronic configuration and combine it with the valency of magnesium to obtain the simplest formula.
Step 1:The configuration 1s2 2s2 2p3 has five valence electrons, so X needs three electrons to complete its octet, giving a valency of 3 and a charge of -3.
Step 2:Magnesium has a valency of 2 and forms a +2 ion.
Step 3:Cross-combining the charges gives three magnesium ions for two X ions.
Final answer:
Q89Single correctThe Solid State
Iron exhibits bcc structure at room temperature. Above , it transforms to fcc structure. The ratio of density of iron at room temperature to that at (assuming molar mass and atomic radii of iron remains constant with temperature) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Express the density of each unit cell in terms of the number of atoms Z and the edge length a (related to the atomic radius r), then form the ratio of bcc to fcc density.
Step 1:For the bcc lattice, Z = 2 and the edge length relates to the radius as a = 4r/sqrt(3).
Step 2:For the fcc lattice, Z = 4 and the edge length relates to the radius as a = 2*sqrt(2)*r.
Step 3:Form the ratio of densities; molar mass and radius cancel, leaving Z and terms.
Step 4:Simplify the ratio.
Final answer:
Q90Single correctStructure of Atom
Which one is a wrong statement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1The electronic configuration of N atom is shown by the orbital diagram with the three 2p boxes filled as , , , i.e. the three 2p electrons are not all of parallel spin
Approach:
Evaluate each statement and identify the one that violates a fundamental rule of electron configuration.
Step 1:By Hund's rule of maximum multiplicity, the three 2p electrons of nitrogen occupy the three 2p orbitals singly with parallel spins, giving 1s2 2s2 2px1 2py1 2pz1.
Step 2:The orbital diagram shown in statement (1) pairs electrons in the 2p orbitals, which contradicts Hund's rule, so statement (1) is wrong.
Step 3:The remaining statements about quantum number designations, the orbital angular momentum of an s electron being zero, and m = 0 for dz2 are all correct.
Final answer: The electronic configuration of N atom is (shown as the orbital diagram with all three 2p orbitals doubly occupied)
Biology90 questions
Q91Single correctPhotosynthesis in Higher Plants
Oxygen is not produced during photosynthesis by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Green sulphur bacteria
Approach:
Identify the organism that performs anoxygenic photosynthesis and therefore releases no molecular oxygen.
Step 1:Green sulphur bacteria use hydrogen sulphide as the hydrogen-donor instead of water.
Step 2:, and all carry out oxygenic photosynthesis, splitting water and evolving oxygen.
Final answer: Green sulphur bacteria
Q92Single correctSexual Reproduction in Flowering Plants
Double fertilization is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Syngamy and triple fusion
Approach:
Recall the two fusion events that together constitute double fertilization in angiosperms.
Step 1:One male gamete fuses with the egg to form the zygote, a process called syngamy.
Step 2:The second male gamete fuses with the two polar nuclei (secondary nucleus), a triple fusion forming the primary endosperm nucleus.
Final answer: Syngamy and triple fusion
Q93Single correctOrganisms and Populations
Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Recall the classic obligate plant-pollinator mutualism involving a moth.
Step 1: and its specific moth pollinator depend completely on each other.
Step 2:Neither nor the moth can complete its life cycle in the absence of the partner.
Final answer:
Q94Single correctSexual Reproduction in Flowering Plants
Pollen grains can be stored for several years in liquid nitrogen having a temperature of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Recall the temperature of liquid nitrogen used for cryopreservation of pollen.
Step 1:Liquid nitrogen is maintained at .
Final answer:
Q95Single correctMineral Nutrition
Which of the following elements is responsible for maintaining turgor in cells?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Potassium
Approach:
Identify the mineral element that regulates cell turgidity.
Step 1:Potassium maintains the turgidity of cells and regulates stomatal opening and closing.
Final answer: Potassium
Q96Single correctRespiration in Plants
What is the role of NA in cellular respiration?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2It functions as an electron carrier.
Approach:
Recall the function of NA during cellular respiration.
Step 1:NA picks up hydrogen and electrons removed during the oxidation of respiratory substrates, becoming NADH.
Final answer: It functions as an electron carrier.
Q97Single correctMineral Nutrition
In which of the following forms is iron absorbed by plants?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ferric
Approach:
Recall the ionic form in which plants absorb iron, as per NCERT.
Step 1:According to NCERT, iron is absorbed by plants in the form of ferric ions, (preferably).
Final answer: Ferric
Q98Single correctBiotechnology: Principles and Processes
Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Retrovirus
Approach:
Identify the vector used to introduce DNA into human lymphocytes during gene therapy.
Step 1:Retrovirus is commonly used as the vector for introducing a DNA fragment in human lymphocytes.
Final answer: Retrovirus
Q99Single correctBiotechnology and its Applications
Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Biopiracy
Approach:
Recall the term for unauthorised use of bioresources by multinational companies.
Step 1:Biopiracy is the term used for the use of bioresources by multinational companies and other organisations without proper authorisation from the countries and people concerned, without compensatory payment.
Final answer: Biopiracy
Q100Single correctBiotechnology and its Applications
In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Genetic Engineering Appraisal Committee (GEAC)
Approach:
Identify the Indian regulatory body that decides the safety of GM organisms for public service.
Step 1:The Indian Government set up the organisation GEAC (Genetic Engineering Appraisal Committee), which makes decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.
Final answer: Genetic Engineering Appraisal Committee (GEAC)
Q101Single correctBiotechnology: Principles and Processes
The correct order of steps in Polymerase Chain Reaction (PCR) is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Denaturation, Annealing, Extension
Approach:
Recall the sequence of the three repeated steps of a PCR cycle.
Step 1:PCR amplifies copies of gene (or DNA) of interest in vitro through three steps in each cycle.
Final answer: Denaturation, Annealing, Extension
Q102Single correctMolecular Basis of Inheritance
Select the correct match
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ribozyme - Nucleic acid
Approach:
Examine each pair and identify the one that is correctly matched.
Step 1:Ribozyme is a catalytic RNA, that is, a nucleic acid that acts as an enzyme.
Final answer: Ribozyme - Nucleic acid
Q103Single correctBiotechnology and its Applications
A 'new' variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Basmati
Approach:
Recall the rice variety involved in the well-known patenting controversy in India.
Step 1:In 1997, an American company got patent rights on Basmati rice through the US patent and trademark office that was actually derived from Indian farmer's varieties.
Final answer: Basmati
Q104Single correctPrinciples of Inheritance and Variation
Which of the following pairs is wrongly matched?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Starch synthesis in pea : Multiple alleles
Approach:
Identify the wrongly matched pair among the four.
Step 1:Starch synthesis in pea is controlled by pleiotropic gene, not by multiple alleles.
Final answer: Starch synthesis in pea : Multiple alleles
Q105Single correctMolecular Basis of Inheritance
Select the correct statement
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Punnett square was developed by a British scientist
Approach:
Identify the only statement that is factually correct.
Step 1:Punnett square was developed by Reginald C. Punnett, a British geneticist.
Step 2:Franklin Stahl shared the semiconservative mode of replication work; transduction was discovered by Zinder and Lederberg; spliceosomes function in post-transcriptional change in eukaryotes, not translation.
Final answer: Punnett square was developed by a British scientist
Q106Single correctMolecular Basis of Inheritance
The experimental proof for semiconservative replication of DNA was first shown in a
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Bacterium
Approach:
Recall the organism in which Meselson and Stahl first demonstrated semiconservative replication.
Step 1:Semiconservative DNA replication was first shown in the bacterium Escherichia\ coli by Matthew Meselson and Franklin Stahl.
Final answer: Bacterium
Q107Single correctSexual Reproduction in Flowering Plants
Which of the following flowers only once in its life-time?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Bamboo species
Approach:
Identify the monocarpic plant that flowers only once in its lifetime.
Step 1:Bamboo species are monocarpic; they generally flower only once in their life-time after 50-100 years.
Step 2:Jackfruit, papaya and mango are polycarpic, that is, they produce flowers and fruits many times in their life-time.
Final answer: Bamboo species
Q108Single correctReproduction in Organisms
Offsets are produced by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mitotic divisions
Approach:
Recall the type of cell division by which offsets, a vegetative propagule, arise.
Step 1:Offset is a vegetative part of a plant, formed by mitosis.
Final answer: Mitotic divisions
Q109Single correctMolecular Basis of Inheritance
Select the correct match
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Francois Jacob and Jacques Monod - Lac operon
Approach:
Identify the correctly matched scientist-discovery pair.
Step 1:Francois Jacob and Jacques Monod proposed the model of gene regulation known as operon model or lac/lac operon.
Step 2:Alec Jeffreys developed DNA fingerprinting; Matthew Meselson and F. Stahl demonstrated semiconservative DNA replication in E. coli; Alfred Hershey and Martha Chase proved DNA as genetic material, not protein.
Final answer: Francois Jacob and Jacques Monod - Lac operon
Q110Single correctSexual Reproduction in Flowering Plants
Which of the following has proved helpful in preserving pollen as fossils?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Sporopollenin
Approach:
Identify the pollen wall component resistant to degradation that helps preserve pollen as fossils.
Step 1:Sporopollenin cannot be degraded by enzymes, strong acids and alkali; therefore it is helpful in preserving pollen as fossil.
Step 2:Pollenkitt helps in insect pollination; cellulosic intine is the inner sporoderm layer of pollen grain known as intine made up of cellulose and pectin; oil content does not preserve pollen.
Final answer: Sporopollenin
Q111Single correctOrganisms and Populations
Natality refers to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Birth rate
Approach:
Recall the population attribute denoted by natality.
Step 1:Natality refers to birth rate. Mortality is death rate, immigration is the number of individuals entering a habitat, and emigration is the number of individuals leaving the habitat.
Final answer: Birth rate
Q112Single correctEnvironmental Issues
World Ozone Day is celebrated on
(1)
(2)
(3)
(4)
SolutionAnswer: Option 116 September
Approach:
Recall the internationally observed date for World Ozone Day.
Step 1:World Ozone Day (International Day for the Preservation of the Ozone Layer) is observed on 16th September, marking the signing of the Montreal Protocol in 1987.
Step 2:The other listed dates correspond to different observances: 5th June is World Environment Day and 22nd April is Earth Day; neither is World Ozone Day.
Final answer: 16 September
Q113Single correctEnvironmental Issues
Which of the following is a secondary pollutant?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Distinguish a secondary pollutant, formed by reaction of primary pollutants, from primary pollutants themselves.
Step 1: (ozone) is a secondary pollutant, formed by the reaction of primary pollutants. CO is a quantitative pollutant, while and are primary pollutants.
Final answer:
Q114Single correctOrganisms and Populations
Niche is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4the functional role played by the organism where it lives
Approach:
Define the ecological concept of niche.
Step 1:A niche describes the total functional role of a species in its ecosystem, including its use of resources and interactions with other organisms.
Step 2:The physical space an organism occupies is its habitat, which is distinct from its niche.
Final answer: the functional role played by the organism where it lives
Q115Single correctEcosystem
What type of ecological pyramid would be obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Inverted pyramid of biomass
Approach:
Compare the biomass values across trophic levels to determine the pyramid shape.
Step 1:The data gives biomass (in g) per trophic level: producer 10 g, primary consumer 60 g, secondary consumer 120 g.
Step 2:Biomass increases from producer to top consumer, so the broad base is at the top and the apex at the bottom.
Step 3:This produces an inverted pyramid of biomass, characteristic of an aquatic ecosystem.
Final answer: Inverted pyramid of biomass
Q116Single correctEnvironmental Issues
In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Cl
Approach:
Identify the catalytic species responsible for stratospheric ozone breakdown.
Step 1:UV radiation acts on chlorofluorocarbons (CFCs) in the stratosphere, releasing chlorine atoms.
Step 2:Chlorine atoms react with ozone, degrading it to molecular oxygen, and chlorine is regenerated, acting catalytically.
Final answer: Cl
Q117Single correctBiomolecules
The two functional groups characteristic of sugars are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Carbonyl and hydroxyl
Approach:
Recall the defining functional groups of sugar molecules.
Step 1:Sugar is a common term used to denote carbohydrate.
Step 2:Carbohydrates are polyhydroxy aldehydes or ketones, meaning they bear carbonyl and hydroxyl groups.
Final answer: Carbonyl and hydroxyl
Q118Single correctBiological Classification
Which among the following is not a prokaryote?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Identify the eukaryotic organism among the listed microbes.
Step 1:Saccharomyces (yeast) is a unicellular fungus and is eukaryotic.
Step 2:Mycobacterium is a bacterium, while Oscillatoria and Nostoc are cyanobacteria, all prokaryotic.
Final answer:
Q119Single correctCell: The Unit of Life
The Golgi complex participates in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Formation of secretory vesicles
Approach:
Recall the principal function of the Golgi complex.
Step 1:The Golgi complex, after processing materials, packages secretory products into vesicles that bud from its trans-face.
Final answer: Formation of secretory vesicles
Q120Single correctPhotosynthesis in Higher Plants
Which of the following is not a product of light reaction of photosynthesis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2NADH
Approach:
Identify the molecule not formed during the light reaction.
Step 1:The light reaction produces ATP, NADPH and oxygen.
Step 2:NADH is a product of respiration, not of the photosynthetic light reaction.
Final answer: NADH
Q121Single correctCell: The Unit of Life
Which of the following is true for nucleolus?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4It is a site for active ribosomal RNA synthesis
Approach:
Recall the structure and function of the nucleolus.
Step 1:The nucleolus is a non membranous structure and is a site of r-RNA synthesis.
Step 2:It lacks a limiting membrane and does not form the spindle.
Final answer: It is a site for active ribosomal RNA synthesis
Q122Single correctAnatomy of Flowering Plants
Stomatal movement is not affected by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 concentration
Approach:
Identify the factor that does not govern opening and closing of stomata.
Step 1:Light, temperature and CO2 concentration affect opening and closing of stomata.
Step 2:Stomatal movement is not affected by O2 concentration.
Final answer: concentration
Q123Single correctCell Cycle and Cell Division
The stage during which separation of the paired homologous chromosomes begins is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Diplotene
Approach:
Identify the sub-stage of meiotic prophase I where homologues begin separating.
Step 1:Synaptonemal complex disintegrates and terminalisation begins at diplotene stage, where the synapsed homologous chromosomes start to separate while remaining linked at chiasmata.
Final answer: Diplotene
Q124Single correctAnatomy of Flowering Plants
Stomata in grass leaf are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Dumb-bell shaped
Approach:
Recall the shape of guard cells in grasses (monocots).
Step 1:Grass, being a monocot, has dumb-bell shaped guard cells surrounding its stomata.
Final answer: Dumb-bell shaped
Q125Single correctAnatomy of Flowering Plants
Secondary xylem and phloem in dicot stem are produced by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Vascular cambium
Approach:
Identify the meristem producing secondary vascular tissues in dicot stems.
Step 1:Vascular cambium is partially secondary; it forms secondary xylem towards the inside and secondary phloem towards the outside.
Step 2:It produces far more secondary xylem than secondary phloem.
Final answer: Vascular cambium
Q126Single correctAnatomy of Flowering Plants
Pneumatophores occur in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Halophytes
Approach:
Recall the plants that bear pneumatophores.
Step 1:Halophytes like mangroves have pneumatophores.
Step 2:These apogeotropic (negatively geotropic) roots have lenticels called pneumathodes to take up O2.
Final answer: Halophytes
Q127Single correctAnatomy of Flowering Plants
Casparian strips occur in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Endodermis
Approach:
Locate the tissue bearing Casparian strips in the root.
Step 1:Endodermis has casparian strip on radial and inner tangential wall.
Step 2:The strip is composed of suberin, which is rich in the endodermal walls.
Final answer: Endodermis
Q128Single correctAnatomy of Flowering Plants
Plants having little or no secondary growth are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Grasses
Approach:
Identify the group lacking appreciable secondary growth.
Step 1:Grasses are monocots and usually do not have secondary growth.
Step 2:Palm-like monocots show only anomalous secondary growth, but typical grasses lack it.
Final answer: Grasses
Q129Single correctMorphology of Flowering Plants
Sweet potato is a modified
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Adventitious root
Approach:
Classify the storage organ of sweet potato.
Step 1:Sweet potato is a modified adventitious root for storage of food.
Step 2:Rhizomes are underground modified stems, while a tap root is the primary root directly elongated from the radicle, neither of which describes sweet potato.
Final answer: Adventitious root
Q130Single correctPlant Kingdom
Which of the following statements is correct?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Ovules are not enclosed by ovary wall in gymnosperms
Approach:
Evaluate each statement and select the accurate one.
Step 1:Gymnosperms have naked ovules, so the ovules are not enclosed by an ovary wall.
Step 2:Horsetails are pteridophytes, both Selaginella and Salvinia are heterosporous, and Cycas is unbranched while Cedrus is branched, making the other statements incorrect.
Final answer: Ovules are not enclosed by ovary wall in gymnosperms
Q131Single correctBiological Classification
Select the wrong statement :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pseudopodia are locomotory and feeding structures in Sporozoans
Approach:
Identify the incorrect statement among the four.
Step 1:Pseudopodia are locomotory structures in sarcodines (Amoeboid protozoans), not in sporozoans, making this statement wrong.
Final answer: Pseudopodia are locomotory and feeding structures in Sporozoans
Q132Single correctBiological Classification
After karyogamy followed by meiosis, spores are produced exogenously in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Match exogenous spore production to the correct fungal genus.
Step 1:In Agaricus (a genus of basidiomycetes), basidiospores or meiospores are produced exogenously on the basidium.
Step 2:Alternaria produces asexual conidia, Neurospora (an ascomycete) makes meiospores endogenously inside the ascus, and Saccharomyces (unicellular ascomycetes) produces ascospores endogenously.
Final answer:
Q133Single correctThe Living World
Match the items given in Column I with those in Column II and select the correct option given below:
| Column I | Column II |
|---|---|
| a. Herbarium | i. It is a place having a collection of preserved plants and animals |
| b. Key | ii. A list that enumerates methodically all the species found in an area with brief description aiding identification |
| c. Museum | iii. Is a place where dried and pressed plant specimens mounted on sheets are kept |
| d. Catalogue | iv. A booklet containing a list of characters and their alternates which are helpful in identification of various taxa. |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4a-(iii), b-(iv), c-(i), d-(ii)
Approach:
Pair each taxonomic aid with its correct description.
Step 1:Herbarium stores dried and pressed plant specimens, matching (iii).
Step 2:Key is a booklet for identification of various taxa using contrasting characters, matching (iv).
Step 3:Museum holds preserved plant and animal specimens, matching (i); Catalogue is an alphabetical listing of species matching (ii).
Final answer: a-(iii), b-(iv), c-(i), d-(ii)
Q134Single correctSexual Reproduction in Flowering Plants
Winged pollen grains are present in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify the plant whose pollen bears wings.
Step 1:In Pinus, winged pollen grains are present, with two air bladders extended from the exine that form the wings to aid wind pollination.
Step 2:Pollen grains of Mustard, Cycas and Mango are not winged.
Final answer:
Q135Single correctPlant Kingdom
Which one is wrongly matched?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Uniflagellate gametes
Approach:
Find the incorrectly paired characteristic and organism.
Step 1:Polysiphonia is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated, so pairing it with uniflagellate gametes is wrong.
Step 2:Gemma cups do occur in Marchantia, brown algae do produce biflagellate zoospores, and Chlorella is unicellular.
Final answer: Uniflagellate gametes
Q136Single correctBreathing and Exchange of Gases
Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Inflammation of bronchioles; Decreased respiratory surface
Approach:
Identify the characteristic structural change in each disorder.
Step 1:Asthma is difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles.
Step 2:Emphysema is a chronic disorder in which alveolar walls are damaged, reducing the respiratory surface for gaseous exchange.
Final answer: Inflammation of bronchioles; Decreased respiratory surface
Q137Single correctBody Fluids and Circulation
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I | Column II |
|---|---|
| a. Tricuspid valve | i. Between left atrium and left ventricle |
| b. Bicuspid valve | ii. Between right ventricle and pulmonary artery |
| c. Semilunar valve | iii. Between right atrium and right ventricle |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3a-iii, b-i, c-ii
Approach:
Recall the anatomical position of each heart valve.
Step 1:The tricuspid valve lies between the right atrium and right ventricle.
Step 2:The bicuspid (mitral) valve lies between the left atrium and left ventricle.
Step 3:The semilunar valves guard the openings of the aorta and pulmonary artery, here matched to the pulmonary artery.
Final answer: a-iii, b-i, c-ii
Q138Single correctBreathing and Exchange of Gases
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I | Column II |
|---|---|
| a. Tidal volume | i. 2500 – 3000 mL |
| b. Inspiratory Reserve volume | ii. 1100 – 1200 mL |
| c. Expiratory Reserve volume | iii. 500 – 550 mL |
| d. Residual volume | iv. 1000 – 1100 mL |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2a-iii, b-i, c-iv, d-ii
Approach:
Recall standard values of respiratory volumes.
Step 1:Tidal volume is the volume of air inspired or expired during normal respiration, about 500 mL.
Step 2:Inspiratory reserve volume is the additional air a person can inspire by forceful inspiration, about 2500 – 3000 mL.
Step 3:Expiratory reserve volume is the additional air a person can expire by forceful expiration, about 1000 – 1100 mL.
Step 4:Residual volume is the volume of air remaining in the lungs even after forceful expiration, about 1100 – 1200 mL.
Final answer: a-iii, b-i, c-iv, d-ii
Q139Single correctNeural Control and Coordination
The transparent lens in the human eye is held in its place by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3ligaments attached to the ciliary body
Approach:
Recall the suspensory attachment of the eye lens.
Step 1:The lens in the human eye is held in place by suspensory ligaments attached to the ciliary body.
Final answer: ligaments attached to the ciliary body
Q140Single correctChemical Coordination and Integration
Which of the following is an amino acid derived hormone?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Epinephrine
Approach:
Identify the hormone derived from an amino acid.
Step 1:Epinephrine is derived from the amino acid tyrosine by removal of the carboxyl group, making it a catecholamine.
Step 2:Estradiol and estriol are steroid hormones, and ecdysone is also a steroid hormone of insects.
Final answer: Epinephrine
Q141Single correctChemical Coordination and Integration
Which of the following hormones can play a significant role in osteoporosis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Estrogen and Parathyroid hormone
Approach:
Identify hormones that regulate bone density.
Step 1:Estrogen promotes the activity of osteoblasts and inhibits osteoclasts; its decline in an ageing female causes osteoporosis.
Step 2:Parathyroid hormone causes excessive mobilisation of calcium from bone into blood, demineralisation that leads to osteoporosis.
Final answer: Estrogen and Parathyroid hormone
Q142Single correctNeural Control and Coordination
Which of the following structures or regions is incorrectly paired with its functions?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Limbic system : consists of fibre tracts that interconnect different regions of brain; controls movement.
Approach:
Identify the wrongly described structure.
Step 1:The limbic system is the emotional brain and controls sexual behaviour, expression of emotions and motivation, not bodily movements.
Step 2:The descriptions for hypothalamus, medulla oblongata and corpus callosum are all correct.
Final answer: Limbic system : consists of fibre tracts that interconnect different regions of brain; controls movement.
Q143Single correctHuman Reproduction
The amnion of mammalian embryo is derived from
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3ectoderm and mesoderm
Approach:
Recall the germ layer origin of extra-embryonic membranes.
Step 1:The extra-embryonic or foetal membranes are the amnion, chorion, allantois and yolk sac.
Step 2:The amnion is formed from mesoderm on the outer side and ectoderm on the inner side.
Final answer: ectoderm and mesoderm
Q144Single correctHuman Reproduction
Hormones secreted by the placenta to maintain pregnancy are
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1hCG, hPL, progestogens, estrogens
Approach:
List the hormones produced by the placenta.
Step 1:The placenta releases human chorionic gonadotropin (hCG), human placental lactogen (hPL), progestogens (progesterone) and estrogens to maintain pregnancy.
Final answer: hCG, hPL, progestogens, estrogens
Q145Single correctHuman Reproduction
The difference between spermiogenesis and spermiation is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules.
Approach:
Distinguish the two processes of sperm development.
Step 1:Spermiogenesis is the transformation of spermatids into spermatozoa.
Step 2:Spermiation is the release of the formed sperms from sertoli cells into the lumen of seminiferous tubules.
Final answer: In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules.
Q146Single correctReproductive Health
The contraceptive 'SAHELI'
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
Approach:
Recall the mechanism of the oral contraceptive Saheli.
Step 1:Saheli is the first non-steroidal, once-a-week oral contraceptive pill containing the drug centchroman.
Step 2:It functions by blocking estrogen receptors so the uterus does not allow implantation of the egg.
Final answer: blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
Q147Single correctBiological Classification
Ciliates differ from all other protozoans in
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4having two types of nuclei
Approach:
Identify the unique feature of ciliates.
Step 1:Ciliates differ from other protozoans in having two types of nuclei.
Step 2:For example, Paramoecium has two types of nuclei, a macronucleus and a micronucleus.
Final answer: having two types of nuclei
Q148Single correctAnimal Kingdom
Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Aves
Approach:
Recall which class has a crop and gizzard.
Step 1:The digestive tract of Aves has additional chambers, the crop for storage of food grains and the gizzard for crushing food grains.
Final answer: Aves
Q149Single correctAnimal Kingdom
Which of the following features is used to identify a male cockroach from a female cockroach?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Presence of caudal styles
Approach:
Identify the male-specific external structure in cockroach.
Step 1:Males bear a pair of short, thread like anal styles which are absent in females.
Final answer: Presence of caudal styles
Q150Single correctAnimal Kingdom
Which one of these animals is not a homeotherm?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Chelone
Approach:
Identify the poikilotherm among the listed animals.
Step 1:Homeotherms are animals that maintain a constant body temperature irrespective of surrounding temperature, including birds and mammals.
Step 2:Chelone (turtle) is a reptile, which is poikilothermic or cold blooded.
Final answer: Chelone
Q151Single correctAnimal Kingdom
Which of the following animals does not undergo metamorphosis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Earthworm
Approach:
Identify the animal with direct development.
Step 1:Metamorphosis refers to transformation of larva into adult; animals that perform metamorphosis are said to have indirect development.
Step 2:In earthworm development is direct, with no larval stage and hence no metamorphosis.
Final answer: Earthworm
Q152Single correctEcosystem
Which of the following organisms are known as chief producers in the oceans?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Diatoms
Approach:
Recall the primary producers of marine ecosystems.
Step 1:Diatoms are the chief producers of the ocean.
Final answer: Diatoms
Q153Single correctOrganisms and Populations
Which one of the following population interactions is widely used in medical science for the production of antibiotics?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Amensalism
Approach:
Identify the interaction underlying antibiotic production.
Step 1:Amensalism is an interaction (0, –) in which antibiotics or chemicals secreted by one microbial group, for example Penicillium, harm other microbes such as Staphylococcus.
Step 2:Penicillin secreted by the organism Penicillium is used as an antibiotic.
Final answer: Amensalism
Q154Single correctBiodiversity and Conservation
All of the following are included in 'ex-situ conservation' except
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Sacred groves
Approach:
Separate in-situ from ex-situ conservation methods.
Step 1:Sacred groves are an example of in-situ conservation, representing pristine forest patches protected by tribal groups.
Step 2:Botanical gardens, wildlife safari parks and seed banks are forms of ex-situ conservation.
Final answer: Sacred groves
Q155Single correctEnvironmental Issues
Match the items given in Column I with those in Column II and select the correct option given below :
| Column-I | Column-II |
|---|---|
| a. Eutrophication | i. UV-B radiation |
| b. Sanitary landfill | ii. Deforestation |
| c. Snow blindness | iii. Nutrient enrichment |
| d. Jhum cultivation | iv. Waste disposal |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1a-iii, b-iv, c-i, d-ii
Approach:
Match each environmental term with its cause or effect.
Step 1:Eutrophication is nutrient enrichment of a water body.
Step 2:Sanitary landfill is a method of solid waste disposal.
Step 3:Snow blindness is caused by UV-B radiation.
Step 4:Jhum cultivation is a shifting practice that causes deforestation.
Final answer: a-iii, b-iv, c-i, d-ii
Q156Single correctOrganisms and Populations
In a growing population of a country,
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3pre-reproductive individuals are more than the reproductive individuals.
Approach:
Interpret the age pyramid of a growing population.
Step 1:Whenever the pre-reproductive population of the younger age group is larger than the reproductive group, the population grows with an increasing population.
Final answer: pre-reproductive individuals are more than the reproductive individuals.
Q157Single correctHuman Health and Disease
Which part of poppy plant is used to obtain the drug "Smack"?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Latex
Approach:
Trace the source of smack within the poppy plant.
Step 1:Smack, also called brown sugar or heroin, is formed by acetylation of morphine and is obtained from the latex of the unripe capsule of the poppy plant.
Final answer: Latex
Q158Single correctMolecular Basis of Inheritance
All of the following are part of an operon except
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1an enhancer
Approach:
List the components of a prokaryotic operon.
Step 1:An operon consists of structural genes, an operator and a promoter.
Step 2:Enhancer sequences are present in eukaryotes, while the operon concept applies to prokaryotes, so an enhancer is not part of an operon.
Final answer: an enhancer
Q159Single correctPrinciples of Inheritance and Variation
A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Both sons and daughters
Approach:
Determine which offspring can receive a particular X chromosome from a mother.
Step 1:A female carries two X chromosomes and passes one X to every child, irrespective of the child's sex.
Step 2:Sons receive an X from the mother (and a Y from the father); daughters receive an X from the mother (and an X from the father). Hence the X chromosome bearing the condition can pass to either sex.
Final answer: Both sons and daughters
Q160Single correctEvolution
According to Hugo de Vries, the mechanism of evolution is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Saltation
Approach:
Recall Hugo de Vries' mutation theory and the term used for single large-step change.
Step 1:Hugo de Vries proposed that evolution proceeds through mutations, which are large and sudden rather than gradual.
Step 2:De Vries called this single-step large mutation responsible for speciation as saltation.
Final answer: Saltation
Q161Single correctMolecular Basis of Inheritance
AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3AGGUAUCGCAU
Approach:
Derive the mRNA from the coding strand by copying its sequence and replacing thymine with uracil.
Step 1:The mRNA has the same nucleotide sequence as the coding (sense) strand of the gene.
Step 2:In RNA, thymine is replaced by uracil. Replacing every T in AGGTATCGCAT with U gives AGGUAUCGCAU.
Final answer: AGGUAUCGCAU
Q162Single correctHuman Reproduction
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I | Column II |
|---|---|
| a. Proliferative Phase | i. Breakdown of endometrial lining |
| b. Secretory Phase | ii. Follicular Phase |
| c. Menstruation | iii. Luteal Phase |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1a-ii, b-iii, c-i
Approach:
Map each menstrual cycle phase to its corresponding description or alternate name.
Step 1:The proliferative phase coincides with the follicular phase of the ovarian cycle, during which the follicle matures.
Step 2:The secretory phase corresponds to the luteal phase, dominated by the corpus luteum.
Step 3:Menstruation involves breakdown of the endometrial lining.
Final answer: a-ii, b-iii, c-i
Q163Single correctExcretory Products and their Elimination
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I | Column II |
|---|---|
| a. Glycosuria | i. Accumulation of uric acid in joints |
| b. Gout | ii. Mass of crystallised salts within the kidney |
| c. Renal calculi | iii. Inflammation in glomeruli |
| d. Glomerular nephritis | iv. Presence of glucose in urine |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4a-iv, b-i, c-ii, d-iii
Approach:
Associate each excretory disorder with its defining feature.
Step 1:Glycosuria denotes the presence of glucose in urine.
Step 2:Gout arises from accumulation of uric acid in the joints.
Step 3:Renal calculi are masses of crystallised salts within the kidney.
Step 4:Glomerular nephritis is inflammation of the glomeruli.
Final answer: a-iv, b-i, c-ii, d-iii
Q164Single correctExcretory Products and their Elimination
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I (Function) | Column II (Part of Excretory system) |
|---|---|
| a. Ultrafiltration | i. Henle's loop |
| b. Concentration of urine | ii. Ureter |
| c. Transport of urine | iii. Urinary bladder |
| d. Storage of urine | iv. Malpighian corpuscle |
| v. Proximal convoluted tubule |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2a-iv, b-i, c-ii, d-iii
Approach:
Match each excretory function to the structure that performs it.
Step 1:Ultrafiltration occurs in the Malpighian corpuscle (glomerulus plus Bowman's capsule).
Step 2:Concentration of urine, by establishing the medullary osmotic gradient, is carried out by Henle's loop.
Step 3:Transport of urine from the kidney to the bladder occurs through the ureter.
Step 4:Storage of urine takes place in the urinary bladder.
Final answer: a-iv, b-i, c-ii, d-iii
Q165Single correctDigestion and Absorption
Which of the following gastric cells indirectly help in erythropoiesis?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Parietal cells
Approach:
Identify the gastric cell whose secretion enables vitamin B12 absorption, supporting red blood cell formation.
Step 1:Parietal (oxyntic) cells secrete intrinsic factor in addition to hydrochloric acid.
Step 2:Intrinsic factor is essential for absorption of vitamin B12, which is required for normal erythropoiesis; its deficiency causes pernicious anaemia.
Final answer: Parietal cells
Q166Single correctBody Fluids and Circulation
Match the items given in Column I with those in Column II and select the correct option given below :
| Column I | Column II |
|---|---|
| a. Fibrinogen | i. Osmotic balance |
| b. Globulin | ii. Blood clotting |
| c. Albumin | iii. Defence mechanism |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4a-ii, b-iii, c-i
Approach:
Pair each plasma protein with its principal physiological role.
Step 1:Fibrinogen is involved in coagulation, forming fibrin threads during blood clotting.
Step 2:Globulins, including immunoglobulins, participate in the defence mechanism.
Step 3:Albumin maintains osmotic balance of the blood.
Final answer: a-ii, b-iii, c-i
Q167Single correctHuman Health and Disease
Which of the following is an occupational respiratory disorder?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Silicosis
Approach:
Identify the disease caused by occupational exposure that affects the respiratory system.
Step 1:Silicosis results from inhalation of silica dust by workers involved in grinding or stone breaking industries.
Step 2:Long exposure leads to inflammation and fibrosis of the lungs, causing serious lung damage, marking it as an occupational respiratory disorder.
Final answer: Silicosis
Q168Single correctLocomotion and Movement
Calcium is important in skeletal muscle contraction because it
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Binds to troponin to remove the masking of active sites on actin for myosin.
Approach:
Recall the role of calcium ions in initiating the sliding filament mechanism.
Step 1:A signal for contraction raises the cytosolic Ca2+ level, releasing calcium from the sarcoplasm.
Step 2:Ca2+ binds to a subunit of troponin (troponin C), shifting the masking position so that the active sites on actin are exposed.
Step 3:Once the active site is exposed, the myosin head attaches and initiates contraction by sliding the actin over myosin.
Final answer: Binds to troponin to remove the masking of active sites on actin for myosin.
Q169Single correctNeural Control and Coordination
Nissl bodies are mainly composed of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Free ribosomes and RER
Approach:
Recall the composition and function of Nissl granules in neurons.
Step 1:Nissl granules occur in the cyton and extend into the dendrite, but are absent in the axon and the axon hillock.
Step 2:Each Nissl granule is composed of free ribosomes and rough endoplasmic reticulum and is responsible for protein synthesis.
Final answer: Free ribosomes and RER
Q170Single correctBreathing and Exchange of Gases / Cell Respiration
Which of these statements is incorrect?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Oxidative phosphorylation takes place in outer mitochondrial membrane
Approach:
Evaluate each statement on the site and conditions of respiratory pathways to find the false one.
Step 1:Glycolysis proceeds while NAD is available to accept hydrogen, occurs in the cytosol, and TCA cycle enzymes reside in the mitochondrial matrix; these statements are correct.
Step 2:Oxidative phosphorylation, driven by the electron transport chain and ATP synthase, takes place on the inner mitochondrial membrane, not the outer membrane.
Final answer: Oxidative phosphorylation takes place in outer mitochondrial membrane
Q171Single correctCell Cycle and Cell Division / Chromosome structure
Select the incorrect match :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Polytene chromosomes – Oocytes of amphibians
Approach:
Check each chromosome type against its correct association to locate the wrong pairing.
Step 1:Submetacentric chromosomes appear L-shaped, allosomes are sex chromosomes, and lampbrush chromosomes occur as diplotene bivalents; these matches are correct.
Step 2:Polytene chromosomes are found in the salivary glands of dipteran insects, not in amphibian oocytes; lampbrush chromosomes occur in amphibian oocytes. The polytene-oocyte pairing is wrong.
Final answer: Polytene chromosomes – Oocytes amphibians
Q172Single correctDigestion and Absorption
Which of the following terms describe human dentition?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Thecodont, Diphyodont, Heterodont
Approach:
Classify human teeth by attachment, number of sets, and type variety.
Step 1:Human teeth are embedded in sockets of the jaw bone, making the dentition thecodont.
Step 2:Two successive sets of teeth (temporary milk teeth and permanent teeth) make it diphyodont.
Step 3:Teeth are of different types: incisors, canines, premolars and molars, making it heterodont.
Final answer: Thecodont, Diphyodont, Heterodont
Q173Single correctCell: Structure and Functions
Which of the following events does not occur in rough endoplasmic reticulum?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Phospholipid synthesis
Approach:
Distinguish processes carried out by rough versus smooth endoplasmic reticulum.
Step 1:The rough endoplasmic reticulum, studded with ribosomes, carries out cleavage of the signal peptide, protein glycosylation and protein folding.
Step 2:Phospholipid synthesis is a function of the smooth endoplasmic reticulum, which lacks ribosomes and is involved in lipid synthesis.
Final answer: Phospholipid synthesis
Q174Single correctCell: Structure and Functions
Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Polysome
Approach:
Name the structure formed by several ribosomes translating one mRNA at once.
Step 1:Association of several ribosomes with a single mRNA leads to formation of polyribosomes, also called polysomes or ergasomes.
Step 2:This arrangement permits simultaneous synthesis of multiple copies of the same polypeptide from one mRNA.
Final answer: Polysome
Q175Single correctHuman Health and Disease
In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Elephantiasis
Approach:
Identify the mosquito-borne disease marked by chronic lymphatic inflammation.
Step 1:Elephantiasis (filariasis) is caused by the roundworm Wuchereria bancrofti and is transmitted by the Culex mosquito.
Step 2:The pathogen causes chronic inflammation of the lymphatic vessels, leading to gross swelling of the affected parts.
Final answer: Elephantiasis
Q176Single correctHuman Health and Disease
Which of the following is not an autoimmune disease?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Alzheimer's disease
Approach:
Separate the autoimmune disorders from the one neurodegenerative condition.
Step 1:Rheumatoid arthritis, psoriasis and vitiligo are autoimmune disorders in which the immune system attacks the body's own tissues.
Step 2:Alzheimer's disease is a neurodegenerative disorder linked to deficiency of the neurotransmitter acetylcholine, not an autoimmune reaction.
Final answer: Alzheimer's disease
Q177Single correctEvolution
Among the following sets of examples for divergent evolution, select the incorrect option :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Eye of octopus, bat and man
Approach:
Distinguish homologous (divergent) examples from a set that includes a convergent analogous structure.
Step 1:Divergent evolution gives rise to homologous structures (forelimbs, heart and brain of bat, man and cheetah) that develop along different directions from a common ancestral plan.
Step 2:The eye of octopus versus that of bat and man represents analogous organs formed by convergent evolution, so this set is incorrect for divergent evolution.
Final answer: Eye of octopus, bat and man
Q178Single correctBiology in Human Welfare / Microbes in Human Welfare
Conversion of milk to curd improves its nutritional value by increasing the amount of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Vitamin
Approach:
Recall how microbial fermentation of milk into curd enhances vitamin content.
Step 1:Lactic acid bacteria ferment milk into curd, and during this process curd becomes more nourishing than milk.
Step 2:The enrichment is due to an increased presence of vitamins, specifically vitamin B12.
Final answer: Vitamin
Q179Single correctEvolution
The similarity of bone structure in the forelimbs of many vertebrates is an example of
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Homology
Approach:
Classify the shared forelimb bone plan among vertebrates.
Step 1:In different vertebrates, the bones of the forelimbs share the same basic plan but their forelimbs are adapted in different ways to serve different functions.
Step 2:Structures with a common ancestral plan but different functions are homologous, illustrating homology.
Final answer: Homology
Q180Single correctPrinciples of Inheritance and Variation
Which of the following characteristics represent 'Inheritance of blood groups' in humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2a, b and c
Approach:
Identify the genetic phenomena illustrated by the inheritance of the ABO blood group system.
Step 1:Alleles IA and IB are each dominant over IO, illustrating a dominant-recessive relationship.
Step 2:When IA and IB occur together they are equally expressed, demonstrating codominance.
Step 3:Three different alleles IA, IB and IO control the trait, demonstrating multiple allelism.
&
Step 4:Incomplete dominance and polygenic inheritance are not features of ABO blood group inheritance, so a, b and c apply.
Final answer: a, b and c
Frequently Asked Questions
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