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NEET 2015 Jul 25 Question Paper with Solutions
All 180 questions from the NEET 2015 (Jul 25) 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 2015Chemistry PYQs 2015Biology PYQs 2015
- Questions
- 180
- Physics
- 45
- Chemistry
- 45
- Biology
- 90
Physics45 questions
Q136Single correctKinematics
The position vector of a particle as a function of time is given by:
Where R is in meters, t is seconds and and denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?
Where R is in meters, t is seconds and and denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Magnitude of the velocity of particle is 8 meter/second
Approach:
The position components describe uniform circular motion; the radius, speed and acceleration follow directly from the trigonometric form.
Step 1:Squaring and adding the components removes the time dependence.
Step 2:Differentiating the position gives the speed, whose magnitude stays constant.
Step 3:The acceleration points toward the centre and equals the centripetal value.
Final answer: Magnitude of the velocity of particle is 8 meter/second
Q137Single correctElectromagnetic Waves
The energy of the em waves is of the oder of 15 keV. To which part of the spectrum does it belong?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4X-rays
Approach:
The spectral region is identified from the photon energy range characteristic of each type of electromagnetic radiation.
Step 1:X-ray photons span about 100 eV to 100 keV in energy.
Step 2:Ultraviolet photons are of the order of a few eV, far below 15 keV, while -rays exceed about 100 keV.
Final answer: X-rays
Q138Single correctRay Optics
A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively.
The prism will:
The prism will:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3separate the red colour part from the green and blue colours
Approach:
Each colour is tested for total internal reflection at the 45 degree face of the right-angled prism using its critical angle.
Step 1:Total internal reflection occurs at the 45 degree face when the refractive index exceeds the value needed for a 45 degree critical angle.
Step 2:Comparing each index with 1.414 shows green (1.44) and blue (1.47) exceed it and undergo total internal reflection, while red (1.39) does not.
Step 3:Since green and blue follow the same reflected path and red exits separately, the red component is split from the other two.
Final answer: separate the red colour part from the green and blue colours
Q139Single correctKinematics
Two particles A and B, move with constant velocities and . At the initial moment their position vectors are and respectively. The condition for particles A and B for their collision is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Collision requires the relative position vector to point along the relative velocity, so that B approaches A directly along the line joining them.
Step 1:For collision, the relative velocity of B with respect to A must be directed along the line from B to A.
Step 2:Equating the unit vector of the separation with the unit vector of the relative velocity expresses this alignment.
Final answer:
Q140Single correctWave Optics
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 radian
Approach:
The path difference between the edge and midpoint wavelets at the first minimum is found and converted to a phase difference.
Step 1:At the first minimum the full-slit path difference between the two edges equals one wavelength, so between the edge and the midpoint it is half a wavelength.
Step 2:Converting this path difference to phase gives the required value.
Final answer: radian
Q141Single correctMoving Charges and Magnetism
A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV the energy acquired by the alpha particle will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 31 MeV
Approach:
The radius of circular motion is expressed in terms of kinetic energy, charge and mass, and equated for the two particles.
Step 1:For equal radii, the quantity must be the same, so kinetic energy scales as .
Step 2:Substituting the alpha particle charge and mass gives the ratio.
Step 3:Since the proton energy is 1 MeV, the alpha particle energy is the same.
Final answer: 1 MeV
Q142Single correctCurrent Electricity
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 40.5 A
Approach:
The ammeter resistance is the parallel combination of coil and shunt, added in series with the external resistance, then Ohm's law gives the current.
Step 1:The coil and shunt in parallel give the ammeter resistance.
Step 2:Adding the series resistance and applying Ohm's law gives the total current.
Final answer: 0.5 A
Q143Single correctThermal Properties of Matter
The value of coefficient of volume expansion of glycerin is . The fractional chage in the density of glycerin for a rise of 4C in its temperature, is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 10.020
Approach:
The fractional change in density equals the coefficient of volume expansion times the temperature rise.
Step 1:Density varies inversely with volume, so its fractional decrease equals the product of the volume expansion coefficient and the temperature rise.
Step 2:Substituting the given values yields the fractional change.
Final answer: 0.020
Q144Single correctThermodynamics
An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Adiabatic
Approach:
The work done on the gas during compression is the area under the pressure-volume curve, which is compared across the processes.
Step 1:For compression to the same final volume, the process with the steepest P-V curve encloses the largest area.
Step 2:Since the area under the adiabatic curve exceeds that under the isothermal and isobaric curves, the work done on the gas is greatest for the adiabatic process.
Final answer: Adiabatic
Q145Single correctAlternating Current
A series R-C circuit is connected to an alternating voltage source. Consider two situation:
(a) When capacitor is air filled.
(b) When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then :
(a) When capacitor is air filled.
(b) When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The voltage across the capacitor is computed from the impedance, and the effect of inserting the mica dielectric on capacitance and reactance is examined.
Step 1:The current and the capacitor voltage depend on the capacitance through the impedance expression.
Step 2:Filling the capacitor with mica raises the dielectric constant and hence the capacitance, lowering the capacitor voltage.
Final answer:
Q146Single correctDual Nature of Radiation and Matter
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2 m
Approach:
The maximum kinetic energy of the emitted electron is found from the photoelectric equation, then converted to the de Broglie wavelength.
Step 1:The incident photon energy at 500 nm minus the work function gives the maximum kinetic energy.
Step 2:Substituting this energy into the de Broglie relation gives the electron wavelength.
Step 3:Because the maximum kinetic energy is an upper limit, slower electrons give longer wavelengths, so the de Broglie wavelength is at least this value.
Final answer: m
Q147Single correctCurrent Electricity
Two metal wires of identical dimension are connected in series. If and are the conductivities of the metal wires respectively, the effective conductivity of the combination is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The two wires form series resistors; the total resistance of the equivalent wire of double length gives the effective conductivity.
Step 1:The series resistance of the two identical-area wires equals the resistance of an equivalent wire of length .
Step 2:Cancelling common factors and solving for the effective conductivity gives the harmonic-type combination.
Final answer:
Q148Single correctSystem of Particles and Rotational Motion
An automobile moves on a road with a speed of 54 km . The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg . If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 46.66 kg
Approach:
The initial angular velocity of the wheel is found from the linear speed, then angular deceleration gives the average torque.
Step 1:Converting the speed to metres per second and dividing by the radius gives the initial angular velocity.
Step 2:The angular deceleration follows from the wheel coming to rest in 15 s.
Step 3:Multiplying the moment of inertia by the deceleration magnitude gives the average torque.
Final answer: 6.66 kg
Q149Single correctWaves
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m at an angle of 6 with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m) is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1103 Hz
Approach:
Only the component of the source velocity along the source-observer line contributes to the Doppler shift for a stationary observer.
Step 1:The effective approaching speed is the component of the source velocity along the line to the observer.
Step 2:Substituting into the Doppler formula for an approaching source gives the apparent frequency.
Final answer: 103 Hz
Q150Single correctSystem of Particles and Rotational Motion
On a frictionless surface a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle to its initial direction and has a speed . The second block's speed after the collision is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
For equal-mass elastic collisions the two outgoing velocities are perpendicular; energy conservation then fixes the second block's speed.
Step 1:In an elastic collision between equal masses with one initially at rest, the velocity vectors after collision are mutually perpendicular.
Step 2:Substituting the first block speed and solving for the second speed gives the result.
Final answer:
Q151Single correctSystem of Particles and Rotational Motion
Point masses and are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity is minimum, is given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The rotational work is minimised when the moment of inertia is least, which occurs about the centre of mass.
Step 1:The work to set the rod spinning equals the rotational kinetic energy, which is least when the moment of inertia is minimum, i.e. about the centre of mass.
Step 2:Locating the centre of mass with at distance x gives the position of P.
Final answer:
Q152Single correctMechanical Properties and Collisions
A ball is thrown vertically downwards from a height of 20 m with an initial velocity . It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity is : (Take g = 10 m)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 120 m
Approach:
The speed just before impact follows from energy at the top; halving the kinetic energy gives the rebound speed, which must just return the ball to the original height.
Step 1:Losing half the energy halves the kinetic energy, so the rebound speed is the impact speed divided by root two.
Step 2:The rebound speed must equal the speed needed to rise 20 m, namely , which fixes the impact speed and then .
Final answer: 20 m
Q153Single correctNuclei
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The helium nucleus has more kinetic energy than the thorium nucleus.
Approach:
Conservation of momentum for decay from rest fixes equal and opposite momenta, and kinetic energy is then compared through the masses.
Step 1:Since the uranium nucleus is at rest, the thorium and helium nuclei carry equal and opposite momenta.
Step 2:With equal momenta, kinetic energy varies inversely with mass, and the lighter helium nucleus carries the greater kinetic energy.
Final answer: The helium nucleus has more kinetic energy than the thorium nucleus.
Q154Single correctElectromagnetic Induction
An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current if any, induced in the coil.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2The current will reverse its direction as the electron goes past the coil
Approach:
Lenz's law is applied as the electron approaches and then recedes from the coil, with the flux first increasing and then decreasing.
Step 1:As the electron approaches, the magnetic flux linked with the coil grows and the induced current opposes the increase, flowing in one sense.
Step 2:As the electron moves away, the flux decreases and the induced current reverses to oppose the decrease.
Step 3:The induced current therefore reverses its direction as the electron passes the coil.
Final answer: The current will reverse its direction as the electron goes past the coil
Q155Single correctOscillations
A particle is executing a simple harmonic motion. Its maximum acceleration is and maximum velocity is . Then its time period of vibration will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The maximum acceleration and maximum velocity in SHM give the angular frequency, from which the period follows.
Step 1:Dividing the maximum acceleration by the maximum velocity gives the angular frequency.
Step 2:Substituting into the period formula gives the result.
Final answer:
Q156Single correctWave Optics
Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maximum and minima in the interference pattern, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The slit-width ratio gives the intensity ratio, whose square roots give the amplitude ratio, leading to the maximum-to-minimum intensity ratio.
Step 1:Intensity is proportional to slit width, so the intensity ratio matches the width ratio and the amplitude ratio is its square root.
Step 2:Combining amplitudes for maximum and minimum gives the intensity ratio.
Final answer:
Q157Single correctElectrostatic Potential and Capacitance
If potential (in volts) in a region is expressed as V(x, y, z) = 6 xy y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The electric field is the negative gradient of the potential, evaluated at the given point.
Step 1:Taking partial derivatives of the potential gives the field components.
Step 2:Substituting the coordinates (1, 1, 0) evaluates the field.
Final answer:
Q158Single correctElectrostatic Potential and Capacitance
A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The attractive force between the plates is the charge times the field due to one plate, expressed in terms of C, V and d.
Step 1:The force on one plate equals its charge times the field produced by the other plate.
Step 2:Substituting and gives the force in the required variables.
Final answer:
Q159Single correctLaws of Motion
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches the box starts to slip and slides 4.0 m down the plank in 4.0s. The coefficients of static and kinetic friction between the box and the plank will be, respectively :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
Slipping begins at the angle of repose, fixing the static coefficient; the measured slide distance and time fix the acceleration and hence the kinetic coefficient.
Step 1:At the verge of slipping the incline equals the angle of repose, so the static coefficient is the tangent of that angle.
Step 2:Applying the displacement relation for 4.0 m in 4.0 s starting from rest gives the sliding acceleration.
Step 3:Substituting this acceleration into the incline relation isolates the kinetic coefficient.
Final answer: and
Q160Single correctAtoms
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
The longest wavelength of each series corresponds to its smallest energy transition; applying the Rydberg formula to both and taking the ratio yields the answer.
Step 1:The longest Lyman wavelength arises from the transition between the second and first levels.
Step 2:The longest Balmer wavelength arises from the transition between the third and second levels.
Step 3:Dividing the two expressions gives the required ratio.
Final answer:
Q161Single correctSemiconductor Electronics
In the given figure, a diode D is connected to an external resistance and an e.m.f. of . If the barrier potential developed across the diode is , the current in the circuit will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The diode drops its barrier potential; the remaining voltage across the resistor sets the current by Ohm's law.
Step 1:Subtracting the barrier potential from the source emf gives the voltage across the resistor.
Step 2:Dividing the resistor voltage by the resistance gives the loop current.
Final answer:
Q162Single correctGravitation
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3the acceleration of S is always directed towards the centre of the earth.
Approach:
The gravitational force on the satellite is central, which fixes the direction of its acceleration and conserves angular momentum.
Step 1:Earth's gravity is the only force and points from the satellite toward earth's centre, so the acceleration shares that direction at every point of the orbit.
Step 2:Because the force is central the torque about the centre vanishes, so angular momentum is constant in magnitude and direction; the total mechanical energy of a closed orbit is also constant, while the linear momentum changes in magnitude between perigee and apogee.
Final answer: the acceleration of S is always directed towards the centre of the earth.
Q163Single correctSystem of Particles and Rotational Motion
A force is acting at a point . The value of for which angular momentum about origin is conserved is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Angular momentum is conserved when the torque is zero, which requires the force to be parallel to the position vector; matching the component ratios fixes the unknown.
Step 1:Conservation of angular momentum about the origin requires zero torque, so the force vector must be parallel to the position vector.
Step 2:Equating the corresponding component ratios determines the unknown coefficient.
Step 3:Solving the ratio gives the value of the coefficient.
Final answer:
Q164Single correctCurrent Electricity
A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. and a resistance . An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The potential gradient along the wire times the balancing length equals the unknown emf; the gradient follows from the driving current through the wire.
Step 1:The driving battery sends a current through the series combination, and the voltage drop per unit length of the wire defines the potential gradient.
Step 2:At balance the unknown emf equals the gradient multiplied by the balancing length.
Final answer:
Q165Single correctKinetic Theory
of a gas occupies litres at NTP. The specific heat capacity of the gas at constant volume is . If the speed of sound in this gas at NTP is , then the heat capacity at constant pressure is :
(Take gas constant )
(Take gas constant )
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The molar mass follows from the given mass at NTP; the speed-of-sound expression gives the adiabatic ratio, and Mayer-type scaling then yields the molar heat at constant pressure.
Step 1:Since 4.0 g occupies one molar volume at NTP, the molar mass is 4 g per mole.
Step 2:Inserting the speed of sound, temperature, and molar mass into the sound relation gives the adiabatic ratio.
Step 3:Multiplying the constant-volume heat capacity by this ratio gives the constant-pressure value.
Final answer:
Q166Single correctSystem of Particles and Rotational Motion
Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in radius and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Equating the centripetal forces on the two stones relates their speeds through their masses and radii.
Step 1:Setting the centripetal force on the lighter stone equal to that on the heavier stone uses the given masses and radii.
Step 2:Cancelling common factors relates the two speeds directly.
Final answer:
Q167Single correctGravitation
A remote - sensing satellite of earth revolves in a circular orbit at a height of m above the surface of earth. If earth's radius is m and , then the orbital speed of the satellite is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The orbital speed depends on the orbital radius, which is the earth's radius plus the satellite height; expressing GM through surface gravity avoids needing the mass.
Step 1:The orbital radius is the sum of earth's radius and the satellite altitude.
Step 2:Substituting surface gravity, earth's radius, and the orbital radius gives the orbital speed.
Final answer:
Q168Single correctWaves
A string is stretched between fixed points separated by cm. It is observed to have resonant frequencies of and . There are no other resonant frequencies between these two. The lowest resonant frequency for this string is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Resonant frequencies of a fixed string are integer multiples of the fundamental, so consecutive resonances differ by the fundamental, which is also the largest common factor.
Step 1:Adjacent resonances differ by one harmonic, so their difference equals the fundamental frequency.
Step 2:The two given frequencies are the fourth and third harmonics of this fundamental, confirming the value.
Final answer:
Q169Single correctThermodynamics
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is , is temperature of the surroundings to which it rejects heat is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The coefficient of performance for an ideal refrigerator relates the cold and hot reservoir temperatures; solving gives the surrounding temperature.
Step 1:Converting the freezer temperature to kelvin gives the cold reservoir temperature.
Step 2:Rearranging the performance relation expresses the hot reservoir temperature in terms of the cold one.
Step 3:Evaluating and converting back to Celsius gives the surrounding temperature.
Final answer:
Q170Single correctMechanical Properties of Fluids
Water rises to a height 'h' in capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1water rises upto the top of capillary tube and stays there without overflowing
Approach:
The capillary product of rise and radius of curvature is fixed; when the available tube is shorter than the natural rise, the meniscus flattens to a larger radius and the water stops at the top without overflowing.
Step 1:The capillary rise is set by the balance of surface tension against the weight of the raised column, giving a fixed product of height and radius of curvature.
Step 2:When the tube above the surface is shorter than the natural rise, the meniscus adjusts to a larger radius of curvature so the column reaches only the tube top.
Final answer: water rises upto the top of capillary tube and stays there without overflowing
Q171Single correctKinetic Theory
Two vessels separately contain two ideal gases A and B at the same temperature the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Writing the ideal gas law in terms of density and molar mass relates each gas's pressure to its density divided by molar mass; the ratio of the two equations gives the molar mass ratio.
Step 1:Expressing each pressure through its density and molar mass at the common temperature gives two relations.
Step 2:Dividing the relations and inserting the pressure and density ratios isolates the molar mass ratio.
Final answer:
Q172Single correctMechanical Properties of Solids
The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Equal elongation of two identical-geometry wires requires the applied load to scale with Young's modulus.
Step 1:For equal length and cross section, the elongation depends on the load divided by Young's modulus.
Step 2:Setting the two elongations equal makes the load ratio equal to the Young's modulus ratio.
Final answer:
Q173Single correctSemiconductor Electronics
The input signal given to a CE amplifier having a voltage gain of 150 is . The corresponding output signal will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
A common-emitter amplifier multiplies the input amplitude by the gain and shifts the phase by radians.
Step 1:The amplifier multiplies the input amplitude by the voltage gain.
Step 2:A common-emitter stage inverts the signal, adding a phase of to the input phase.
Step 3:Combining the amplitude and phase gives the output signal.
Final answer:
Q174Single correctRay Optics and Optical Instruments
In an astronomical telescope in normal adjustment a straight black line of lenght L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
In normal adjustment the separation of the lenses equals the sum of focal lengths; the eyepiece images the line drawn on the objective, and the ratio of object to image size equals the ratio of focal lengths, which is the telescope magnification.
Step 1:The line on the objective is at distance equal to the sum of focal lengths from the eyepiece, and imaging it gives a linear magnification equal to the focal-length ratio.
Step 2:The telescope angular magnification in normal adjustment equals the same focal-length ratio.
Final answer:
Q175Single correctMechanical Properties of Fluids
The heart of man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury is and then the power of heart in watt is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The pumping power equals the pressure times the volume flow rate, with pressure expressed through the mercury column.
Step 1:Converting the mercury column to pressure uses its density, gravity, and height.
Step 2:Expressing the volume flow rate converts five litres per minute to SI units.
Step 3:Multiplying pressure by the flow rate gives the pumping power.
Final answer:
Q176Single correctPhysical World and Measurement
If dimensions of critical velocity of a liquid flowing through a tube are expressed as , where and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Critical velocity is set by the Reynolds-number combination of viscosity, density, and radius; matching this known form gives the exponents directly.
Step 1:The critical velocity in terms of the Reynolds number is proportional to viscosity divided by the product of density and radius.
Step 2:Reading off the exponents of viscosity, density, and radius gives the required values.
Final answer:
Q177Single correctDual Nature of Radiation and Matter
A photoelectric surface is illuminated successively by monochromatic light of wavelength and . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is :
(h = Planck's constant, c = speed of light)
(h = Planck's constant, c = speed of light)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Writing Einstein's photoelectric equation for both wavelengths and applying the factor-of-three relation between the kinetic energies eliminates the unknowns and yields the work function.
Step 1:The kinetic energy for each illumination follows from the photoelectric equation.
Step 2:Imposing that the second energy is three times the first relates the two expressions.
Step 3:Solving for the work function gives the result.
Final answer:
Q178Single correctMechanical Properties of Fluids
The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The volume inflow through the tube equals the total volume outflow through all holes; equating the two products of area and speed gives the ejection speed.
Step 1:The inflow rate through the tube cross section equals the combined outflow rate through the n holes.
Step 2:Solving for the ejection speed gives the required expression.
Final answer:
Q179Single correctPhysical World and Measurement
If vectors and are functions of time, then the value of t at which they are orthogonal to each other is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Two vectors are orthogonal when their dot product vanishes; evaluating the dot product reduces to a single cosine that must equal zero.
Step 1:Forming the dot product and applying the cosine difference identity collapses it to a single cosine.
Step 2:Setting this cosine to zero for orthogonality fixes the argument at a right angle.
Final answer:
Q180Single correctMoving Charges and Magnetism
A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength . The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of with the direction of the field, the torque required to keep the coil in stable equilibrium will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
The torque on a current coil is the cross product of magnetic moment and field; with the plane inclined at the angle between the normal and the field is , so the torque uses the sine of that angle.
Step 1:Because the plane is inclined at to the field, the coil normal makes with the field, fixing the sine factor.
Step 2:Substituting the turns, current, area, field, and sine factor gives the torque.
Final answer:
Chemistry45 questions
Q91Single correctAlcohols, Phenols and Ethers
Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Identify the named reaction of phenol with chloroform and alkali and the group it installs.
Step 1:Phenol reacts with chloroform in the presence of aqueous sodium hydroxide in the Reimer-Tiemann reaction.
Step 2:Dichlorocarbene generated from chloroform and base attacks the ring ortho to the hydroxyl group, and hydrolysis of the resulting benzal chloride yields an aldehyde.
Final answer:
Q92Single correctEquilibrium
If the equilibrium constant for is K, the equilibrium constant for will be :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Apply the rule that halving the stoichiometric coefficients raises the equilibrium constant to the power one-half.
Step 1:The first reaction has equilibrium constant equal to the ratio of squared NO concentration to the product of nitrogen and oxygen concentrations.
Step 2:The second reaction is the first one with all coefficients divided by two, so its constant is the original constant raised to the power one-half.
Final answer:
Q93Single correctSome Basic Concepts of Chemistry
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample ? (At. Wt.: Mg = 24)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Relate the mass of magnesium oxide produced to the mass of pure magnesium carbonate that must have decomposed, then compute purity.
Step 1:Magnesium carbonate decomposes to magnesium oxide and carbon dioxide. The molar masses are 84 g for the carbonate and 40 g for the oxide.
Step 2:The 8.0 g of magnesium oxide corresponds to a mass of pure carbonate scaled by the ratio of molar masses.
Step 3:Percentage purity is the ratio of pure carbonate mass to total sample mass times one hundred.
Final answer:
Q94Single correctSome Basic Concepts of Chemistry
The number of water molecules is maximum in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 moles of water
Approach:
Convert each quantity to a number of molecules using the mole concept and compare.
Step 1:Eighteen grams of water equals one mole, which contains the Avogadro number of molecules.
Step 2:Eighteen moles contains eighteen times the Avogadro number, far exceeding 18 molecules, 1.8 g (0.1 mol), or 18 g (1 mol).
Final answer: moles of water
Q95Single correctThe p-Block Elements
The formation of the oxide ion from oxygen atom requires first an exothermic and then an endothermic step as shown below :
Thus process of formation of in gas phase is unfavourable even though is isoelectronic with neon. It is due to the fact that
Thus process of formation of in gas phase is unfavourable even though is isoelectronic with neon. It is due to the fact that
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Electron repulsion outweighs the stability gained by achieving noble gas configuration
Approach:
Explain why adding a second electron to a negatively charged species is endothermic.
Step 1:The first electron addition is exothermic, but adding a second electron to the already negative oxide ion forces it into a region of high electron density.
Step 2:The energy needed to overcome this inter-electronic repulsion exceeds the stabilisation from attaining the neon configuration, making the overall step endothermic.
Final answer: Electron repulsion outweighs the stability gained by achieving noble gas configuration
Q96Single correctSolutions
What is the mole fraction of the solute in a 1.00 molal aqueous solution ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Convert molality to mole fraction using one kilogram of solvent water.
Step 1:A 1.00 molal solution contains 1 mole of solute in 1000 g of water, and 1000 g of water corresponds to about 55.5 moles.
Step 2:Mole fraction of solute equals moles of solute divided by total moles.
Final answer:
Q97Single correctChemical Kinetics
The rate constant of the reaction is mole per second. If the concentration of A is 5 M then concentration of B after 20 minutes is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
The units of the rate constant indicate a zero order reaction, so concentration of product equals rate constant times time.
Step 1:The rate constant has units of concentration per time, identifying the reaction as zero order.
Step 2:Converting twenty minutes to seconds and multiplying by the rate constant gives the amount of product formed.
Final answer:
Q98Single correctChemical Bonding and Molecular Structure
Decreasing order of stability of , , and is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Compute bond order for each species from molecular orbital theory and rank by stability.
Step 1:Bond orders are 2.5 for the dioxygenyl cation, 2 for neutral dioxygen, 1.5 for the superoxide ion, and 1 for the peroxide ion.
Step 2:Stability rises with bond order, so the decreasing order of stability follows the decreasing bond orders.
Final answer:
Q99Single correctAldehydes, Ketones and Carboxylic Acids
Which one of the following esters gets hydrolysed most easily under alkaline conditions ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Alkaline ester hydrolysis is accelerated by electron-withdrawing substituents on the aromatic ring.
Step 1:Base-catalysed hydrolysis proceeds through nucleophilic attack of hydroxide at the carbonyl carbon, favoured when electron density there is reduced.
Step 2:Among the substituents, the nitro group is the strongest electron-withdrawing group through resonance and induction, making p-nitrophenyl acetate hydrolyse fastest.
Final answer:
Q100Single correctThe s-Block Elements
On heating which of the following releases most easily ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Thermal stability of carbonates decreases as the cation becomes smaller and more polarising.
Step 1:Alkali metal carbonates of potassium and sodium are highly stable and do not decompose easily, while alkaline earth carbonates decompose more readily.
Step 2:Between the alkaline earth carbonates, the smaller magnesium ion polarises the carbonate ion more strongly than the larger calcium ion, so magnesium carbonate loses carbon dioxide most easily.
Final answer:
Q101Single correctEquilibrium
Which of the following pairs of solution is not an acidic buffer ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 and
Approach:
An acidic buffer requires a weak acid together with the salt of its conjugate base.
Step 1:Acetic acid, carbonic acid, and phosphoric acid are weak acids, so their mixtures with their salts function as acidic buffers.
Step 2:Perchloric acid is a strong acid, and a strong acid with its salt cannot resist pH change, so this pair does not form a buffer.
Final answer: and
Q102Single correctCoordination Compounds
The sum of coordination number and oxidation number of the metal M in the complex (where en is ethylenediamine) is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Determine the coordination number from the bidentate ligands and the oxidation number from charge balance, then add them.
Step 1:Both ethylenediamine ligands and the oxalate ligand are bidentate, giving four plus two donor atoms.
Step 2:The complex ion bears a single positive charge balanced by one chloride; oxalate contributes minus two and the en ligands are neutral, fixing the metal oxidation state at plus three.
Step 3:Adding the coordination number and the oxidation number gives the required total.
Final answer:
Q103Single correctThe p-Block Elements
Which of the statements given below is incorrect ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 is an oxide of fluorine
Approach:
Evaluate each statement and identify the incorrect one based on electronegativity and structure.
Step 1:Dichlorine heptoxide is the anhydride of perchloric acid, ozone is bent, and ONF is isoelectronic with the nitrite ion, so these statements are correct.
Step 2:Fluorine is more electronegative than oxygen, so in the compound formed from oxygen and fluorine the oxygen carries the positive oxidation state; it is therefore oxygen difluoride, a fluoride of oxygen rather than an oxide of fluorine.
Final answer: is an oxide of fluorine
Q104Single correctHydrocarbons
In the reaction with HCl, an alkene reacts in accordance with the Markovnikov's rule, to give a product 1-chloro-1-methylcyclohexane. The possible alkene is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3(A) and (B)
Approach:
Markovnikov addition places the chlorine on the more substituted carbon, so any alkene giving the same tertiary carbocation yields the same product.
Step 1:Both methylenecyclohexane and 1-methylcyclohexene protonate to form the same tertiary carbocation at the ring carbon bearing the methyl group.
Step 2:Chloride then adds to that tertiary carbon, giving 1-chloro-1-methylcyclohexane from each alkene, so both alkenes A and B are possible.
Final answer: (A) and (B)
Q105Single correctHydrocarbons
2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2
Approach:
Protonation of the alkene, a 1,2-methyl shift to the more stable carbocation and loss of a proton give the tetrasubstituted target alkene.
Step 1:Protonation of 3,3-dimethyl-1-butene at the terminal carbon gives a secondary carbocation next to the quaternary carbon.
Step 2:A 1,2-methyl shift converts it into the more stable tertiary carbocation, and loss of a proton then gives the tetrasubstituted alkene 2,3-dimethyl-2-butene.
Final answer:
Q106Single correctAmines
The following reaction
is known by the name
is known by the name
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Schotten-Baumen reaction
Approach:
Identify the named reaction in which an amine is benzoylated by benzoyl chloride in the presence of base.
Step 1:Aniline reacts with benzoyl chloride in the presence of sodium hydroxide to form an anilide.
Step 2:Benzoylation of an amine or phenol with an acid chloride in the presence of aqueous base is known as the Schotten-Baumann reaction.
Final answer: Schotten-Baumen reaction
Q107Single correctGeneral Principles and Processes of Isolation of Elements
In the extraction of copper from its sulphide ore, the metal finally obtained by the reduction of cuprous oxide with :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3copper (I) sulphide
Approach:
Recall the self-reduction step in the pyrometallurgical extraction of copper.
Step 1:During roasting, part of the copper(I) sulphide is converted to copper(I) oxide.
Step 2:The remaining cuprous sulphide then reduces the cuprous oxide to metallic copper in a self-reduction reaction.
Final answer: copper (I) sulphide
Q108Single correctSome Basic Concepts of Chemistry
If Avogadro number is changed from to this would change :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2the mass of one mole of carbon
Approach:
Determine which quantity depends directly on the numerical value of the Avogadro number.
Step 1:The molar mass of an element equals the mass of one Avogadro number of atoms, so it is defined in terms of the Avogadro number.
Step 2:Stoichiometric ratios in equations and the elemental ratios in a compound are fixed by atomic counts and do not depend on the chosen value of the constant, so only the mass of one mole of carbon changes.
Final answer: the mass of one mole of carbon
Q109Single correctThe p-Block Elements
The variation of the boiling point of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the higher boiling point of hydrogen fluoride ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2There is strong hydrogen bonding between HF molecules.
Approach:
Explain the anomalously high boiling point of hydrogen fluoride relative to the other hydrogen halides.
Step 1:The boiling points of the heavier hydrogen halides increase with molar mass due to stronger dispersion forces, giving HI greater than HBr greater than HCl.
Step 2:Hydrogen fluoride lies highest because the small, highly electronegative fluorine atom enables strong intermolecular hydrogen bonding that must be overcome on boiling.
Final answer: There is strong hydrogen bonding between HF molecules.
Q110Single correctHaloalkanes and Haloarenes
Which of the following reaction (s) can be used for the preparation of alkyl halides ?
(I)
(II)
(III)
(IV)
(I)
(II)
(III)
(IV)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(I), (III) and (IV) only
Approach:
Apply the Lucas reagent requirements for converting different classes of alcohols to chlorides.
Step 1:Primary and secondary alcohols react with hydrochloric acid only in the presence of anhydrous zinc chloride, the Lucas reagent.
Step 2:The tertiary alcohol in reaction three reacts readily with hydrochloric acid alone, while the primary alcohol in reaction two without the catalyst does not, so reactions one, three, and four succeed.
Final answer: (I), (III) and (IV) only
Q111Single correctCoordination Compounds
The name of complex ion, is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Hexacyanidoferrate (III) ion
Approach:
Apply IUPAC nomenclature rules to name the complex anion.
Step 1:Six cyanide ligands give the prefix hexacyanido under current IUPAC conventions, and the anionic complex takes the -ate suffix on the Latin stem for iron.
Step 2:The overall charge of minus three with neutral counting fixes iron in the plus three oxidation state, written in Roman numerals.
Final answer: Hexacyanidoferrate (III) ion
Q112Single correctRedox Reactions
Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified for complete oxidation ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Count the electrons each compound donates per mole; the one with the fewest needs the least permanganate.
Step 1:Ferrous sulphate supplies only one electron per formula unit through oxidation of iron(II) to iron(III).
Step 2:The sulphite, oxalate, and nitrite compounds release additional electrons from the oxidisable anion in addition to the iron centre, so they consume more permanganate.
Final answer:
Q113Single correctChemical Bonding and Molecular Structure
In which of the following pairs, both the species are not isostructural ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4
Approach:
Compare the molecular geometries within each pair to find the non-isostructural one.
Step 1:Silicon tetrachloride and the phosphorus tetrachloride cation are both tetrahedral, diamond and silicon carbide share a covalent network, and ammonia and phosphine are both pyramidal, so these pairs are isostructural.
Step 2:Xenon tetrafluoride is square planar while xenon tetroxide is tetrahedral, so this pair is not isostructural.
Final answer:
Q114Single correctPolymers
Caprolactum is used for the manufacture of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Nylon - 6
Approach:
Identify the polymer obtained from the monomer caprolactam.
Step 1:Caprolactam is the cyclic amide (lactam) of 6-aminohexanoic acid.
Step 2:Ring-opening polymerisation of caprolactam yields a single repeating amide unit polymer.
Final answer: Nylon - 6
Q115Single correctCoordination Compounds
The hybridization involved in complex is (At. No. Ni = 28)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1
Approach:
Determine the oxidation state of nickel, the d-electron count, and the effect of the strong-field cyanide ligand.
Step 1:Nickel exists in the +2 state, giving a configuration.
Step 2:Cyanide is a strong-field ligand that pairs the d-electrons, freeing one 3d orbital.
Step 3:The vacant 3d orbital combines with one 4s and two 4p orbitals to give square planar geometry.
Final answer:
Q116Single correctSome Basic Concepts of Chemistry
What is the mass of precipitate formed when 50 mL of 16.9% solution of is mixed with 50 mL of 5.8% NaCl solution ?
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 37 g
Approach:
Find the moles of each reactant, identify the limiting reagent, and compute the mass of AgCl precipitate.
Step 1:A 16.9% solution provides 8.45 g of silver nitrate in 50 mL.
Step 2:A 5.8% solution provides 2.9 g of sodium chloride in 50 mL.
Step 3:The reactants combine in a 1:1 ratio, so 0.05 mol of AgCl forms.
Step 4:Multiplying moles by the molar mass of AgCl gives the precipitate mass.
Final answer: 7 g
Q117Single correctClassification of Elements and Periodicity
Gadolinium belongs to 4f series. It's atomic number is 64. Which of the following is the correct electronic configuration of gadolinium ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
Build the configuration of element 64 accounting for the stability of a half-filled 4f subshell.
Step 1:Xenon accounts for 54 electrons, leaving ten electrons to place.
Step 2:A half-filled 4f subshell with one electron in 5d is energetically favoured.
Final answer:
Q118Single correctHydrocarbons
Which of the following is not the product of dehydration of the alcohol shown below ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 21-(butan-2-yl)cyclohex-1-ene — the cyclohexene with a fully saturated side chain
Approach:
Work out which alkenes an acid-catalysed dehydration of the alcohol can actually give, and identify the structure that no elimination pathway reaches.
Step 1:Protonation and loss of water from the tertiary alcohol place the positive charge on the carbon that carried the OH group; that carbon bears a methyl, an ethyl and the ring.
Step 2:Losing a beta-hydrogen from the methyl gives the exocyclic 1-butene, from the ethyl CH2 gives the 2-butene, and from the ring carbon gives the cyclohexylidene alkene. A double bond lying inside the ring with the side chain left fully saturated is not reachable from this carbocation.
Final answer: 1-(butan-2-yl)cyclohex-1-ene — the cyclohexene with a fully saturated side chain
Q119Single correctStates of Matter
A gas such as carbon monoxide would be most likely to obey the ideal gas law at :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1high temperatures and low pressures.
Approach:
Recall the conditions under which intermolecular forces and molecular volume become negligible.
Step 1:At low pressure the molecules are far apart, making the molecular volume negligible compared with the container volume.
Step 2:At high temperature the kinetic energy dominates over intermolecular attractions.
Final answer: high temperatures and low pressures.
Q120Single correctThe p-Block Elements
The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Al < Ga < In < Tl
Approach:
Apply the inert pair effect down group 13.
Step 1:Descending the group, the inert pair effect grows stronger as the ns2 electrons become reluctant to participate in bonding.
Step 2:Ordering from least to most stable +1 state follows the group order from aluminium to thallium.
Final answer: Al < Ga < In < Tl
Q121Single correctEquilibrium
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 112.65
Approach:
Compute the excess hydroxide after neutralisation and convert to pH.
Step 1:Mixing equal volumes halves each concentration, giving 0.05 M base and 0.005 M acid.
Step 2:Neutralisation leaves an excess hydroxide concentration of 0.045 M.
Step 3:Taking the negative logarithm gives the pOH, and subtracting from 14 gives the pH.
Final answer: 12.65
Q122Single correctThe p-Block Elements
Strong reducing behaviour of is due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Presence of one OH group and two PH bonds
Approach:
Relate the reducing power of hypophosphorous acid to its structure.
Step 1:Hypophosphorous acid contains one hydroxyl group and two phosphorus-hydrogen bonds.
Step 2:The phosphorus-hydrogen bonds are responsible for the reducing character.
Final answer: Presence of one OH group and two PH bonds
Q123Single correctOrganic Chemistry - Basic Principles
The number of structural isomers possible from the molecular formula is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 14
Approach:
Enumerate the primary, secondary, and tertiary amines having the formula C3H9N.
Step 1:Two primary amines arise: propan-1-amine and propan-2-amine.
Step 2:One secondary amine arises: N-methylethanamine.
Step 3:One tertiary amine arises: trimethylamine.
Final answer: 4
Q124Single correctOrganic Chemistry - Reaction Mechanisms
Which of the following statements is not correct for a nucleophile ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Nucleophile is a Lewis acid
Approach:
Recall the defining electronic character of a nucleophile.
Step 1:A nucleophile donates an electron pair, which makes it a Lewis base rather than a Lewis acid.
Step 2:Nucleophiles attack electron-deficient sites and are not electron seeking, so the remaining statements are correct.
Final answer: Nucleophile is a Lewis acid
Q125Single correctCoordination Compounds
Number of possible isomers for the complex will be (en = ethylenediamine)
(1)
(2)
(3)
(4)
SolutionAnswer: Option 33
Approach:
Count the geometrical and optical isomers of the octahedral cation.
Step 1:The cation can be either cis or trans with respect to the two chloride ligands.
Step 2:The cis form is chiral and exists as a pair of enantiomers, while the trans form is achiral.
Final answer: 3
Q126Single correctStructure of Atom
Which is the correct order of increasing energy of the listed orbitals in the atom of titanium ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 33s 3p 3d 4s
Approach:
Apply the (n+l) rule for the orbital energies in titanium.
Step 1:Computing (n+l) gives 3 for 3s, 4 for 3p, 4 for 4s, and 5 for 3d.
Step 2:In titanium the 3d orbitals lie below 4s once filled, so the increasing order places 3d before 4s.
Final answer: 3s 3p 3d 4s
Q127Single correctHaloalkanes and Haloarenes
In an reaction on chiral centres there is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2inversion more than retention leading to partial racemization
Approach:
Recall the stereochemical outcome of nucleophilic substitution through a carbocation intermediate.
Step 1:The reaction proceeds through a planar carbocation, allowing attack from both faces.
Step 2:Tight ion-pair formation makes inverted product slightly more abundant than retained product, giving partial racemisation.
Final answer: inversion more than retention leading to partial racemization
Q128Single correctThe Solid State
The vacant space in bcc lattice cell is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 432 %
Approach:
Use the packing efficiency of a body-centred cubic structure to find the empty space.
Step 1:A body-centred cubic lattice has a packing efficiency of about 68 percent.
Step 2:Subtracting from the total volume gives the empty space.
Final answer: 32 %
Q129Single correctThermodynamics
the heat of combustion of carbon to is kJ/mol. The heat released upon formation of 35.2 g of from carbon and oxygen gas is
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1 kJ
Approach:
Scale the molar heat of combustion by the number of moles in 35.2 g of carbon dioxide.
Step 1:Dividing the mass by the molar mass of carbon dioxide gives the number of moles.
Step 2:Multiplying the moles by the molar heat gives the heat released.
Final answer: kJ
Q130Single correctElectrochemistry
Aqueous solution of which of the following compounds is the best conductor of electric current ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Hydrochloric acid, HCl
Approach:
Compare the degree of ionisation of each solute in water.
Step 1:Hydrochloric acid is a strong acid that ionises completely, releasing the maximum number of ions.
Step 2:Acetic acid and ammonia ionise only partly, while fructose does not ionise at all.
Final answer: Hydrochloric acid, HCl
Q131Single correctHydrocarbons
The oxidation of benzene by in presence of air produces :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2maleic anhydride
Approach:
Recall the industrial vapour-phase oxidation of benzene.
Step 1:Benzene undergoes catalytic oxidation over vanadium pentoxide in air at high temperature.
Step 2:The ring is oxidatively cleaved to give maleic anhydride.
Final answer: maleic anhydride
Q132Single correctAldehydes, Ketones and Carboxylic Acids
Reaction of a carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2hydrazine in presence of feebly acidic solution
Approach:
Identify the reagent that adds to a carbonyl and then eliminates water.
Step 1:Ammonia derivatives such as hydrazine add to the carbonyl carbon as nucleophiles.
Step 2:The intermediate loses water under slightly acidic conditions to form a hydrazone.
Final answer: hydrazine in presence of feebly acidic solution
Q133Single correctAmines
Method by which Aniline cannot be prepared is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
Approach:
Evaluate whether each route gives aniline, focusing on the Gabriel synthesis with an aryl halide.
Step 1:Hydrolysis of phenylisocyanide, Hofmann degradation of benzamide, and reduction of nitrobenzene all give aniline.
Step 2:The Gabriel synthesis fails with chlorobenzene because the aryl halide is inert to nucleophilic substitution by phthalimide.
Final answer: potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution
Q134Single correctOrganic Chemistry - Basic Principles
Two possible stereo-structures of , which are optically active, are called :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Enantiomers
Approach:
Classify the two optically active forms of lactic acid having one chiral centre.
Step 1:Lactic acid has a single chiral carbon, giving two non-superimposable mirror images.
Step 2:Non-superimposable mirror images that are optically active are termed enantiomers.
Final answer: Enantiomers
Q135Single correctThe Solid State
The correct statement regarding defects in crystalline solid is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Frenkel defect is a dislocation defect
Approach:
Assess each statement about Schottky and Frenkel point defects.
Step 1:A Frenkel defect arises when an ion is dislocated from its lattice site to an interstitial position, so it is a dislocation defect.
Step 2:Schottky defects lower density, Frenkel defects keep density unchanged, and Frenkel defects occur in solids with large size differences, not alkaline metal halides.
Final answer: Frenkel defect is a dislocation defect
Biology90 questions
Q1Single correctPrinciples of Inheritance and Variation
In his classic experiments on pea plants, Mendel did not use
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pod length
Approach:
Identify which of the listed traits was not among Mendel's seven contrasting characters of the garden pea.
Step 1:Mendel selected seven pairs of contrasting traits in Pisum sativum: seed shape, seed colour, flower colour, flower position, pod shape, pod colour, and stem height.
Step 2:Pod shape and pod colour were studied, but pod length was not one of the chosen characters.
Final answer: Pod length
Q2Single correctMolecular Basis of Inheritance
Which one of the following is not applicable to RNA?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Chargaff's rule
Approach:
Determine which property is characteristic of double-stranded DNA rather than single-stranded RNA.
Step 1:Chargaff's rule states that in double-stranded DNA the amount of adenine equals thymine and guanine equals cytosine.
Step 2:RNA is typically single-stranded, so equimolar base ratios required by Chargaff's rule do not hold for it.
Final answer: Chargaff's rule
Q3Single correctSexual Reproduction in Flowering Plants
Male gametophyte in angiosperms produces:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Two sperms and a vegetative cell
Approach:
Recall the cellular composition of the mature male gametophyte in angiosperms.
Step 1:The mature pollen grain is the male gametophyte and contains a vegetative cell and a generative cell.
Step 2:The generative cell divides to form two male gametes (sperms), giving a three-celled structure with two sperms and one vegetative cell.
Final answer: Two sperms and a vegetative cell
Q4Single correctCell: The Unit of Life
Which of the following are not membrane-bound?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Ribosomes
Approach:
Identify the organelle that lacks a surrounding membrane.
Step 1:Ribosomes are composed of ribosomal RNA and proteins and have no bounding membrane.
Step 2:Lysosomes, mesosomes, and vacuoles are all bounded by membranes.
Final answer: Ribosomes
Q5Single correctBiomolecules
The chitinous exoskeleton of arthropods is formed by the polymerisation of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2N-acetyl glucosamine
Approach:
Recall the monomer that polymerises to form chitin.
Step 1:Chitin is a structural polysaccharide forming the arthropod exoskeleton.
Step 2:Chitin is a homopolymer of N-acetyl glucosamine units linked by beta-1,4 glycosidic bonds.
Final answer: N-acetyl glucosamine
Q6Single correctMorphology of Flowering Plants
Among china rose, mustard, brinjal, potato, guava, cucumber, onion and tulip, how many plants have superior ovary?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Six
Approach:
Classify each plant by ovary position (superior versus inferior).
Step 1:China rose, mustard, brinjal, potato, onion, and tulip bear hypogynous flowers with a superior ovary.
Step 2:Guava and cucumber bear epigynous flowers with an inferior ovary.
Final answer: Six
Q7Single correctCell: The Unit of Life
The function of the gap junction is to
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1facilitate communication between adjoining cells by connecting the cytoplasm for rapid transfer of ions, small molecules and some large molecules.
Approach:
Recall the role of gap junctions among the types of cell junctions.
Step 1:Gap junctions form channels that directly connect the cytoplasm of adjacent cells.
Step 2:Through these channels, ions, small molecules, and some larger molecules pass rapidly between cells, enabling communication.
Final answer: facilitate communication between adjoining cells by connecting the cytoplasm for rapid transfer of ions, small molecules and some large molecules.
Q8Single correctBiomolecules
Which of the following immunoglobulins does constitute the largest percentage in human milk?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Ig A
Approach:
Identify the predominant antibody class in human milk and colostrum.
Step 1:Secretory IgA is the principal immunoglobulin secreted in mammary glands.
Step 2:Colostrum and milk are rich in IgA, providing passive immunity to the infant.
Final answer: Ig A
Q9Single correctNeural Control and Coordination
In mammalian eye, the 'fovea' is the center of the visual field, where:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4high density of cones occur, but has no rods.
Approach:
Recall the photoreceptor composition of the fovea centralis.
Step 1:The fovea lies at the centre of the macula lutea and is the region of sharpest vision.
Step 2:It contains a high concentration of cones and lacks rods.
Final answer: high density of cones occur, but has no rods.
Q10Single correctBody Fluids and Circulation
Doctors use stethoscope to hear the sounds produced during each cardiac cycle. The second sound is heard when:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Semilunar valves close down after the blood flows into vessels from ventricles
Approach:
Associate each heart sound with the valve event that produces it.
Step 1:The first heart sound (lubb) arises from closure of the atrioventricular valves at the start of ventricular systole.
Step 2:The second heart sound (dupp) arises from closure of the semilunar valves after ventricular ejection of blood into the great vessels.
Final answer: Semilunar valves close down after the blood flows into vessels from ventricles
Q11Single correctSexual Reproduction in Flowering Plants
Coconut water from a tender coconut is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Free nuclear endosperm
Approach:
Identify the botanical nature of coconut water.
Step 1:Coconut endosperm develops first as free nuclear endosperm, where nuclei divide without cell wall formation.
Step 2:The white kernel that forms later is the cellular endosperm.
Final answer: Free nuclear endosperm
Q12Single correctBiotechnology: Principles and Processes
The cutting of DNA at specific locations became possible with the discovery of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Restriction enzymes
Approach:
Recall the enzyme class that cleaves DNA at specific recognition sequences.
Step 1:Restriction endonucleases recognise specific palindromic sequences and cut DNA at or near those sites.
Step 2:Ligases join DNA fragments and probes and markers serve detection or selection roles.
Final answer: Restriction enzymes
Q13Single correctCell: The Unit of Life
Which of the following structures is not found in a prokaryotic cell?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Nuclear envelope
Approach:
Identify the structure absent from prokaryotic cells.
Step 1:Prokaryotic cells lack membrane-bound organelles and an organised nucleus.
Step 2:Ribosomes, mesosomes, and plasma membrane are all present in prokaryotes.
Final answer: Nuclear envelope
Q14Single correctCell Cycle and Cell Division
Arrange the following events of meiosis in correct sequence:
(a) Crossing over
(b) Synapsis
(c) Terminalisation of chiasmata
(d) Disappearance of nucleolus
(a) Crossing over
(b) Synapsis
(c) Terminalisation of chiasmata
(d) Disappearance of nucleolus
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(b), (a), (c), (d)
Approach:
Order the events according to the sub-stages of prophase I.
Step 1:Synapsis between homologous chromosomes occurs in zygotene.
Step 2:Crossing over takes place in pachytene, chiasmata are terminalised through diplotene into diakinesis, and the nucleolus disappears in diakinesis.
Final answer: (b), (a), (c), (d)
Q15Single correctTransport in Plants
A column of water within xylem vessels of tall trees does not break under its weight because of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Tensile strength of water
Approach:
Identify the property of water that keeps the xylem column intact.
Step 1:Cohesive forces among water molecules give the water column high tensile strength.
Step 2:This tensile strength allows the transpiration pull to draw water upward in tall trees without breaking the column.
Final answer: Tensile strength of water
Q16Single correctBiological Classification
The imperfect fungi which are decomposer of litter and help in mineral cycling belong to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Deuteromycetes
Approach:
Match the description of imperfect fungi acting as decomposers to the correct class.
Step 1:Deuteromycetes are called imperfect fungi because only their asexual or vegetative phases are known.
Step 2:Many members act as decomposers of litter and aid in mineral cycling.
Final answer: Deuteromycetes
Q17Single correctBiological Classification
The structures that help some bacteria to attach to rocks and / or host tissues are:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Fimbriae
Approach:
Recall the bacterial surface appendage responsible for attachment.
Step 1:Fimbriae are small bristle-like fibrous outgrowths on the bacterial surface.
Step 2:They enable bacteria to adhere to rocks and to host tissues.
Final answer: Fimbriae
Q18Single correctBiotechnology: Principles and Processes
The DNA molecule to which the gene of interest is integrated for cloning is called:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Vector
Approach:
Recall the term for the DNA carrier used to clone a gene of interest.
Step 1:A cloning vector is a DNA molecule, such as a plasmid, into which foreign DNA is inserted.
Step 2:The recombinant vector replicates within a host, multiplying the gene of interest.
Final answer: Vector
Q19Single correctBiological Classification
Pick up the wrong statement:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Nuclear membrane is present in Monera
Approach:
Test each statement against the defining features of the kingdoms.
Step 1:Monera comprises prokaryotes, which lack a nuclear membrane.
Step 2:Protista show mixed nutrition, some fungi are edible, and animal cells lack cell walls, so the other statements are correct.
Final answer: Nuclear membrane is present in Monera
Q20Single correctAnimal Kingdom
Metagenesis refers to:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Alternation of generation between asexual and sexual phases of an organisms
Approach:
Define metagenesis in the context of cnidarian life cycles.
Step 1:Metagenesis is the alternation of generations between an asexual polyp phase and a sexual medusa phase.
Step 2:This pattern is seen in cnidarians such as Obelia.
Final answer: Alternation of generation between asexual and sexual phases of an organisms
Q21Single correctHuman Reproduction
Which of the following events is not associated with ovulation in human female?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Decrease in estradiol
Approach:
Identify the hormonal change that does not accompany ovulation.
Step 1:At ovulation, the Graffian follicle matures fully and releases the secondary oocyte.
Step 2:Rising estradiol triggers a positive feedback LH surge; estradiol increases rather than decreases at this stage.
Final answer: Decrease in estradiol
Q22Single correctLocomotion and Movement
Which of the following joints would allow no movement?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Fibrous joint
Approach:
Classify the joints by the degree of movement they permit.
Step 1:Fibrous joints, such as the sutures of the skull, hold bones tightly together and allow no movement.
Step 2:Cartilaginous joints allow limited movement, while synovial and ball-and-socket joints are freely movable.
Final answer: Fibrous joint
Q23Single correctMicrobes in Human Welfare
Match List I with List II
| List I | List II |
|---|---|
| (a). | (i). Production of immunosuppressive agents |
| (b). | (ii). Ripening of Swiss cheese |
| (c). | (iii). Commerical production of ethanol |
| (d). | (iv). Production of blood cholesterol lowering agents |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Pair each microbe with the product or role it is industrially associated with.
Step 1:Saccharomyces cerevisiae is used in commercial production of ethanol, and Monascus purpureus yields statins that lower blood cholesterol.
Step 2:Trichoderma polysporum produces the immunosuppressive agent cyclosporin A, and Propionibacterium is involved in ripening of Swiss cheese.
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q24Single correctEcology and Environment
The UN conference of Parties on climate change in the year 2012 was held at:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Doha
Approach:
Recall which city hosted the 2012 session of the UNFCCC Conference of Parties.
Step 1:The Conference of Parties (COP) is the supreme decision-making body of the UN Framework Convention on Climate Change.
Step 2:COP 18 was convened at Doha, Qatar, in November-December 2012.
Final answer: Doha
Q25Single correctHuman Physiology
If you suspect major deficiency of antibodies in a person, to which of the following would you look for confirmatory evidences?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Serum globulins
Approach:
This question links antibodies to the plasma protein fraction in which they occur.
Step 1:Antibodies are immunoglobulins, which belong to the gamma-globulin fraction of plasma proteins.
Step 2:A deficiency of antibodies therefore appears as a reduced serum globulin level rather than altered albumin or fibrinogen.
Final answer: Serum globulins
Q26Single correctPlant Physiology
Chromatophores take part in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Photosynthesis
Approach:
This question tests the function of chromatophores in photosynthetic prokaryotes and algae.
Step 1:Chromatophores are pigment-bearing membranous structures found in blue-green algae and photosynthetic bacteria.
Step 2:Because they house the pigment systems, chromatophores carry out the light-trapping reactions of photosynthesis.
Final answer: Photosynthesis
Q27Single correctEcology and Environment
Acid rain is caused by increase in the atmospheric concentration of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4 and
Approach:
This question identifies the gaseous pollutants responsible for acid precipitation.
Step 1:Sulphur dioxide and oxides of nitrogen released from combustion react with atmospheric water vapour.
Step 2:These acids return to the surface as acid rain, with sulphuric acid contributing the major share.
Final answer: and
Q28Single correctEcology and Environment
During ecological succession:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4the gradual and predictable change in species composition occurs in a given area
Approach:
This question requires the correct description of the process of ecological succession.
Step 1:Ecological succession is defined as the gradual and orderly change in the species composition of a community over time in a given area.
Step 2:Primary succession is slow, species numbers change continuously, and the final stable stage is the climax community rather than the pioneer community.
Final answer: the gradual and predictable change in species composition occurs in a given area
Q29Single correctPlant Physiology
The oxygen evolved during photosynthesis comes from water molecules. Which one of the following pairs of elements is involved in this reaction?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Manganese and Chlorine
Approach:
This question identifies the mineral elements that activate the water-splitting (photolysis) reaction of photosynthesis.
Step 1:Photolysis of water in Photosystem II is catalysed by the oxygen-evolving complex, whose activity depends on manganese ions.
Step 2:Chloride ions are also required for the splitting of water and release of oxygen in this reaction.
Final answer: Manganese and Chlorine
Q30Single correctReproduction
Which of the following pairs is not correctly matched?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Binary fission — Sargassum
Approach:
This question requires matching each vegetative or asexual mode of reproduction with its correct example.
Step 1:Binary fission is a characteristic mode of reproduction in unicellular organisms such as Amoeba, Paramoecium and Euglena, not in the multicellular alga Sargassum.
Step 2:The rhizome of banana, conidia of Penicillium and offsets of water hyacinth are correctly paired with their reproductive modes.
Final answer: Binary fission — Sargassum
Q31Single correctGenetics and Evolution
In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Autosomal recessive
Approach:
This question requires reading a four-generation pedigree to deduce the mode of inheritance.
Step 1:Affected offspring arise from unaffected parents, indicating that the trait is recessive and can be carried silently.
Step 2:Both sexes are affected in nearly equal proportion and transmission is not confined to males, ruling out X-linkage and dominance.
Final answer: Autosomal recessive
Q32Single correctAnimal Kingdom
Which one of the following animals has two separate circulatory pathways?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Whale
Approach:
This question identifies the animal with complete double circulation through a four-chambered heart.
Step 1:A whale is a mammal possessing a four-chambered heart with two atria and two ventricles.
Step 2:This separation keeps oxygenated and deoxygenated blood in two distinct pathways, whereas lizard, shark and frog have incomplete separation.
Final answer: Whale
Q33Single correctMorphology of Flowering Plants
Flowers are unisexual in:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cucumber
Approach:
This question distinguishes a plant bearing unisexual flowers from those with bisexual flowers.
Step 1:Cucumber is monoecious and produces separate staminate and pistillate flowers.
Step 2:China rose, onion and pea bear bisexual flowers containing both stamens and carpels in the same flower.
Final answer: Cucumber
Q34Single correctSexual Reproduction in Flowering Plants
Which one of the followng fruits is parthenocarpic?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Banana
Approach:
This question identifies a seedless fruit formed without fertilisation.
Step 1:Parthenocarpy is the development of fruit without fertilisation, yielding seedless fruits.
Step 2:The cultivated banana develops its edible pulp without fertilisation and is seedless.
Final answer: Banana
Q35Single correctGenetics and Evolution
A pleiotropic gene:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3controls multiple traits in an individual.
Approach:
This question tests the definition of pleiotropy.
Step 1:Pleiotropy is the phenomenon in which a single gene influences several distinct phenotypic traits.
Step 2:The other statements describe unrelated or incorrect ideas and do not define pleiotropy.
Final answer: controls multiple traits in an individual.
Q36Single correctHuman Physiology
Which of the following is not a function of the skeletal system?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Production of body heat
Approach:
This question separates a non-skeletal function from genuine roles of the skeletal system.
Step 1:The skeletal system stores minerals such as calcium and phosphorus, provides a framework for locomotion, and houses red bone marrow that produces erythrocytes.
Step 2:Generation of body heat is primarily a function of muscular and metabolic activity, not of the skeleton.
Final answer: Production of body heat
Q37Single correctAnimal Kingdom
A jawless fish, which lays eggs in fresh water and whose ammocoetes larvae after metamorphosis return to the ocean is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3
Approach:
This question identifies the jawless vertebrate with an ammocoete larval stage.
Step 1:Petromyzon, the lamprey, is an anadromous cyclostome that breeds in fresh water and produces the ammocoete larva.
Step 2:After metamorphosis the lamprey migrates back to the sea, while Myxine and related hagfishes lack this freshwater larval phase.
Final answer:
Q38Single correctSexual Reproduction in Flowering Plants
Filiform apparatus is characteristic feature of:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Synergids
Approach:
This question locates the filiform apparatus within the embryo sac.
Step 1:The filiform apparatus is a set of finger-like cell-wall thickenings at the micropylar end of the synergids.
Step 2:This structure guides the pollen tube into the embryo sac, a role specific to the synergid cells.
Final answer: Synergids
Q39Single correctAnatomy of Flowering Plants
Read the different components from (a) to (d) in the list given below and tell the correct order of the components with reference to their arrangement from outer side to inner side in a woody dicot stem.
(a) Secondary cortex
(b) Wood
(c) Secondary phloem
(d) Phellem
The correct order is:
(a) Secondary cortex
(b) Wood
(c) Secondary phloem
(d) Phellem
The correct order is:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2(d), (a), (c), (b)
Approach:
This question orders the tissues of a woody dicot stem from the outermost to the innermost layer.
Step 1:In a woody dicot stem the protective phellem (cork) lies outermost, followed inward by the secondary cortex.
Step 2:Inside the secondary cortex lie the secondary phloem and then the wood (secondary xylem) at the centre.
Final answer: (d), (a), (c), (b)
Q40Single correctHuman Physiology
Which one of the following hormones is not involved in sugar metabolism ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Aldosterone
Approach:
This question separates a mineral-regulating hormone from those that act on blood sugar.
Step 1:Insulin, glucagon and cortisone all regulate blood glucose by promoting its uptake, release or formation.
Step 2:Aldosterone is secreted by the adrenal cortex and regulates sodium and potassium levels in the body rather than glucose.
Final answer: Aldosterone
Q41Single correctBiotechnology and its Applications
Golden rice is a genetically modified crop plant where the incorporated gene is meant for biosynthesis of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Vitamin A
Approach:
This question recalls the nutritional purpose of golden rice.
Step 1:Golden rice is engineered to synthesise beta-carotene, the precursor of vitamin A, in the rice grain.
Step 2:The aim is to combat vitamin A deficiency in populations dependent on rice.
Final answer: Vitamin A
Q42Single correctStrategies for Enhancement in Food Production
Outbreeding is an important strategy of animal husbandry because it :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2is useful in overcoming inbreeding depression.
Approach:
This question identifies the principal benefit of outbreeding in animal husbandry.
Step 1:Outbreeding introduces new alleles and increases heterozygosity, restoring vigour lost through continued inbreeding.
Step 2:Production of purelines and exposure of recessive alleles for elimination are outcomes of inbreeding rather than outbreeding.
Final answer: is useful in overcoming inbreeding depression.
Q43Single correctGenetics and Evolution
A gene showing codominance has :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3both alleles independently expressed in the heterozygote
Approach:
This question tests the genetic definition of codominance.
Step 1:In codominance neither allele is dominant or recessive; both express their effects fully in the heterozygote.
Step 2:The AB blood group, where both A and B antigens are produced, illustrates this independent expression.
Final answer: both alleles independently expressed in the heterozygote
Q44Single correctHuman Physiology
Which one of the following hormones through synthesised elsewhere is stored and released by the master gland?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Antidiuretic hormone
Approach:
This question identifies a hormone made in the hypothalamus but released from the pituitary.
Step 1:Antidiuretic hormone and oxytocin are synthesised by the hypothalamus and transported to the posterior pituitary.
Step 2:The posterior pituitary, the master gland here, then releases the antidiuretic hormone into circulation.
Final answer: Antidiuretic hormone
Q45Single correctStrategies for Enhancement in Food Production
Increase in concentration of the toxicant at successive trophic levels is known as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Biomagnification
Approach:
This question names the process by which toxicants concentrate up a food chain.
Step 1:Non-degradable toxicants such as DDT accumulate in organisms and pass to higher trophic levels.
Step 2:This progressive increase of a toxicant along a food chain is termed biomagnification.
Final answer: Biomagnification
Q46Single correctEvolution
Industrial melanism is an example of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Natural selection
Approach:
Industrial melanism describes the increase in frequency of dark-coloured peppered moths in industrialised regions where soot-darkened surfaces favour their survival.
Step 1:On soot-covered tree trunks the dark moths are camouflaged while the light moths are conspicuous to predators.
Step 2:Differential survival of the better-camouflaged dark form increases its frequency over generations.
Final answer: Natural selection
Q47Single correctDigestion and Absorption
The primary dentition in human differs from permanent dentition in not having one of the following type of teeth :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Premolars
Approach:
The dental formula of the deciduous (milk) set is compared with that of the permanent set to identify the absent tooth type.
Step 1:The deciduous dental formula is given by the arrangement of incisors, canine and molars per half-jaw.
Step 2:Premolars are absent from this formula, so milk dentition lacks premolars that appear only in the permanent set.
Final answer: Premolars
Q48Single correctPrinciples of Inheritance and Variation
The wheat grain has an embryo with one, large, shield-shaped cotyledon known as:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Scutellum
Approach:
The single, large, shield-shaped cotyledon of the grass (monocot) embryo is identified by name.
Step 1:In the wheat embryo the lone cotyledon lies against the endosperm and is shield-shaped.
Step 2:Coleorrhiza and coleoptile are protective sheaths of the radicle and plumule, not the cotyledon.
Final answer: Scutellum
Q49Single correctExcretory Products and their Elimination
The body cells in cockroach discharge their nitrogenous waste in the haemolymph mainly in the form of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Potassium urate
Approach:
The chemical form in which cockroach tissues release nitrogenous waste into the haemolymph is identified.
Step 1:Cockroaches are uricotelic and excrete waste largely as uric acid and its salts.
Step 2:Body cells release the waste into the haemolymph chiefly as potassium urate, which the Malpighian tubules then process.
Final answer: Potassium urate
Q50Single correctBiomolecules
Which of the following biomolecules does have phosphodiester bond ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Nucleic acids in a nucleotide
Approach:
The biomolecule whose monomers are joined by phosphodiester linkages is identified.
Step 1:Nucleotides in a nucleic acid are connected through a phosphate bridging the sugars of adjacent units.
Step 2:Polysaccharides use glycosidic bonds, polypeptides use peptide bonds and glycerides use ester bonds.
Final answer: Nucleic acids in a nucleotide
Q51Single correctPrinciples of Inheritance and Variation
The term "linkage" was coined by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4T.H. Morgan
Approach:
The scientist who introduced the term linkage for the tendency of genes on the same chromosome to be inherited together is identified.
Step 1:T.H. Morgan, working on Drosophila, described genes that fail to assort independently.
Step 2:Boveri and Sutton proposed the chromosomal theory and Mendel formulated the laws of inheritance.
Final answer: T.H. Morgan
Q52Single correctBiological Classification
Which one is wrong statement ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Mucor has biflagellate zoospores
Approach:
Each statement is assessed and the incorrect one is selected.
Step 1:Mucor is a zygomycete that reproduces asexually by non-motile aplanospores and lacks flagellated zoospores.
Step 2:Gymnosperm endosperm is haploid, brown algae contain chlorophyll a and c with fucoxanthin, and archegonia occur in the listed groups.
Final answer: Mucor has biflagellate zoospores
Q53Single correctHuman Reproduction
Ectopic pregnancies are referred to as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Implantation of embryo at site other than uterus.
Approach:
The defining feature of an ectopic pregnancy is identified.
Step 1:An ectopic pregnancy occurs when the embryo implants outside the uterine cavity, most often in a fallopian tube.
Step 2:The condition is defined by location of implantation rather than embryo defect, hormones or genetics.
Final answer: Implantation of embryo at site other than uterus.
Q54Single correctEcosystem
Most animals that live in deep oceanic waters are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3detritivores
Approach:
The dominant feeding mode of deep-sea fauna is determined from the available food source at great depth.
Step 1:Light does not penetrate to the deep ocean, so primary production by photosynthesis is absent there.
Step 2:Deep-sea animals rely on organic detritus sinking from upper layers, so most are detritivores.
Final answer: detritivores
Q55Single correctHuman Health and Disease
Which of the following diseases is caused by a protozoan ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Babesiosis
Approach:
Identify which of the listed diseases is caused by a protozoan parasite.
Step 1:Babesiosis is caused by the sporozoan protozoan Babesia, transmitted by ticks.
Step 2:Influenza is viral, blastomycosis is fungal and syphilis is bacterial.
Final answer: Babesiosis
Q56Single correctOrganisms and Populations
In which of the following interactions both partners are adversely affected ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Competition
Approach:
The interaction in which both participants experience a negative effect is identified.
Step 1:Competition for shared limiting resources reduces the fitness of both interacting species.
Step 2:Predation and parasitism benefit one partner, while mutualism benefits both.
Final answer: Competition
Q57Single correctMolecular Basis of Inheritance
Identify the correct order of organisation of genetic material from largest to smallest :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Genome, chromosome, gene, nucleotide
Approach:
The levels of genetic organisation are arranged from the most inclusive to the smallest unit.
Step 1:The genome is the complete genetic complement, which is packaged into chromosomes.
Step 2:Each chromosome carries many genes, and a gene is built from a sequence of nucleotides.
Final answer: Genome, chromosome, gene, nucleotide
Q58Single correctPrinciples of Inheritance and Variation
A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 30.25
Approach:
Colour blindness is X-linked recessive; the genotypes of the couple and the carrier daughters are tracked to a grandson generation.
Step 1:The colour-blind father has genotype Y and the normal mother is X X, so all daughters are carriers X and all sons are unaffected.
Step 2:If a carrier daughter marries a normal man X Y, half of her sons inherit and are colour blind.
Step 3:Sons make up half the grandchildren and half of those sons are colour blind, giving an overall fraction of one quarter.
Final answer: 0.25
Q59Single correctPhotosynthesis in Higher Plants
Photosynthesis the light-independent reactions take place at :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Stromal matrix
Approach:
The site within the chloroplast where the dark (Calvin) reactions occur is identified.
Step 1:The light-independent reactions fix carbon using the enzyme RuBisCO, which is located in the chloroplast stroma.
Step 2:The photosystems and the thylakoid lumen are sites of the light-dependent reactions.
Final answer: Stromal matrix
Q60Single correctBiogeochemical Cycles
In which of the following both pairs have correct combination ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Gaseous nutrient cycle — Carbon and Nitrogen ; Sedimentary nutrient cycle — Sulphur and Phosphorus
Approach:
Each nutrient is classified by whether its reservoir is the atmosphere (gaseous cycle) or the Earth's crust (sedimentary cycle).
Step 1:Carbon and nitrogen have their main reservoirs in the atmosphere, making their cycles gaseous.
Step 2:Sulphur and phosphorus are stored mainly in rocks and sediments, making their cycles sedimentary.
Final answer: Gaseous nutrient cycle — Carbon and Nitrogen ; Sedimentary nutrient cycle — Sulphur and Phosphorus
Q61Single correctBiotechnology: Principles and Processes
The introduction of t-DNA into plants involves :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Infection of the plant by Agrobacterium tumefaciens
Approach:
The natural mechanism by which T-DNA is delivered into plant cells is identified.
Step 1:Agrobacterium tumefaciens carries a Ti plasmid whose T-DNA segment transfers into the host plant genome during infection.
Step 2:Soil pH change, cold exposure and standing in water do not mediate T-DNA delivery.
Final answer: Infection of the plant by Agrobacterium tumefaciens
Q62Single correctEvolution
The wings of a bird and the wings of an insect are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1analogous structures and represent convergent evolution
Approach:
The relationship between bird wings and insect wings is classified by their origin and the evolutionary pattern they illustrate.
Step 1:Bird and insect wings differ in structural origin but serve the same function of flight, making them analogous.
Step 2:Analogous organs arise when unrelated lineages adapt to similar functions, which is convergent evolution.
Final answer: analogous structures and represent convergent evolution
Q63Single correctMineral Nutrition / Transport in Plants
Root pressure develops due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Active absorption
Approach:
The cause of root pressure in the xylem is identified.
Step 1:Active absorption of ions into root xylem lowers the water potential and draws water in osmotically.
Step 2:This active uptake generates the positive hydrostatic force called root pressure, independent of transpiration.
Final answer: Active absorption
Q64Single correctExcretory Products and their Elimination
Human urine is usually acidic because :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3hydrogen ions are actively secreted into the filtrate.
Approach:
The renal process responsible for the normally acidic pH of urine is identified.
Step 1:Tubular cells actively secrete hydrogen ions into the filtrate as part of acid-base regulation.
Step 2:The added hydrogen ions lower the pH of the filtrate, making the excreted urine acidic.
Final answer: hydrogen ions are actively secreted into the filtrate.
Q65Single correctCell: The Unit of Life
A protoplast is a cell :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3without cell wall
Approach:
The defining feature of a protoplast relative to an intact plant cell is identified.
Step 1:A protoplast is the living plant cell content bounded by the plasma membrane after the cell wall has been removed.
Step 2:The nucleus and the plasma membrane are both retained in a protoplast; only the wall is stripped away.
Final answer: without cell wall
Q66Single correctBiodiversity and Conservation
The species confined to a particular region and not found elsewhere is termed as :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Endemic
Approach:
The term for a species restricted to a single geographic region is identified.
Step 1:A species found naturally only within one defined area and nowhere else is described as endemic.
Step 2:Alien refers to introduced species, rare to low abundance and keystone to a disproportionate ecological role.
Final answer: Endemic
Q67Single correctBiological Classification
Select the wrong statements :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4The viroids were discovered by D.J. Ivanowski
Approach:
Assess each historical statement about viruses and identify the one that is factually wrong.
Step 1:W.M. Stanley crystallised the tobacco mosaic virus, M.W. Beijerinck coined the term contagium vivum fluidum, and viruses cause both tobacco mosaic disease and AIDS.
Step 2:D.J. Ivanowski demonstrated the infectious agent of tobacco mosaic disease, that is, a virus; viroids were discovered much later, by T.O. Diener.
Final answer: The viroids were discovered by D.J. Ivanowski
Q68Single correctMorphology of Flowering Plants
Axile placentation is present in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Lemon
Approach:
The plant showing axile placentation, in which ovules attach to a central axis of a multilocular ovary, is identified.
Step 1:In axile placentation the placenta is central and the ovary is divided into chambers, as seen in citrus fruits such as lemon.
Step 2:Pea shows marginal placentation while Argemone and Dianthus show parietal and free-central placentation.
Final answer: Lemon
Q69Single correctHuman Reproduction
A childless couple can be assisted to have a child through a technique called GIFT. The full form of this technique is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Gamete intra fallopian transfer
Approach:
GIFT is an assisted reproductive technology used to help couples who cannot conceive naturally.
Step 1:GIFT is the abbreviation for Gamete Intra Fallopian Transfer.
Step 2:In this technique an ovum collected from a donor is transferred into the fallopian tube of a female who cannot produce ova but can provide a suitable environment for fertilization and further development.
Final answer: Gamete intra fallopian transfer
Q70Single correctNeural Control and Coordination
Destruction of the anterior horn cell of the spinal cord would result in loss of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1voluntary motor impulses
Approach:
The location and function of neurons in the spinal cord grey matter determine the effect of their destruction.
Step 1:The anterior (ventral) horn of the spinal cord contains the cell bodies of lower motor neurons that send axons to skeletal muscles.
Step 2:Destruction of these cells, as occurs in poliomyelitis, interrupts the motor pathway controlling voluntary muscle activity.
Final answer: voluntary motor impulses
Q71Single correctBiological Classification
During biological nitrogen fixation, inactivation of nitrogenase by oxygen poisoning is prevented by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Leghaemoglobin
Approach:
Nitrogenase is oxygen sensitive, so the root nodule must maintain a low free oxygen environment.
Step 1:The enzyme nitrogenase is irreversibly inactivated by molecular oxygen.
Step 2:Leghaemoglobin, a pink oxygen-scavenging pigment present in root nodules, binds free oxygen and keeps its concentration low around the enzyme.
Final answer: Leghaemoglobin
Q72Single correctOrganisms and Populations
An association of individuals of different species living in the same habitat and having functional interactions is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Biotic community
Approach:
Ecological terms are distinguished by the level of biological organisation each describes.
Step 1:A population consists of individuals of a single species, so it does not fit a multi-species grouping.
Step 2:An assemblage of populations of different species occupying the same habitat and interacting with one another constitutes a biotic community.
Final answer: Biotic community
Q73Single correctBreathing and Exchange of Gases
Name the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls.
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Emphysema
Approach:
Each respiratory disease has a characteristic site and mechanism of damage.
Step 1:Damage to the alveolar walls reduces the respiratory surface area available for the exchange of gases.
Step 2:Emphysema is the chronic disorder, often caused by cigarette smoking, in which alveolar walls are destroyed and the alveolar surface area is drastically reduced.
Final answer: Emphysema
Q74Single correctMolecular Basis of Inheritance
Balbiani rings are sites of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3RNA and protein synthesis
Approach:
Balbiani rings are structural features of polytene chromosomes that indicate transcriptional activity.
Step 1:Balbiani rings are large puffs on the polytene chromosomes found in cells such as the salivary glands of dipteran larvae.
Step 2:These puffs are regions of intense gene activity where DNA is transcribed into RNA, which is then used for protein synthesis.
Final answer: RNA and protein synthesis
Q75Single correctCell: The Unit of Life
Match the columns and identify the correct option.
| Column-I | Column-II |
|---|---|
| a. Thylakoids | i. Disc-shaped sacs in Golgi apparatus |
| b. Cristae | ii. Condensed structure of DNA |
| c. Cisternae | iii. Flat membranous sacs in stroma |
| d. Chromatin | iv. Infoldings in mitochondria |
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Approach:
Each cellular structure is matched to its correct description by its location and form.
Step 1:Thylakoids are flat membranous sacs located in the stroma of the chloroplast, matching (iii).
Step 2:Cristae are the infoldings of the inner mitochondrial membrane, matching (iv); cisternae are the disc-shaped sacs of the Golgi apparatus, matching (i); chromatin is the condensed form of DNA, matching (ii).
Final answer: (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Q76Single correctCell: The Unit of Life
Cellular organelles with membranes are :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3lysosomes, Golgi apparatus and mitochondria
Approach:
Membrane-bound organelles are distinguished from non-membranous structures such as ribosomes and chromosomes.
Step 1:Ribosomes and chromosomes are naked ribonucleoprotein and nucleoprotein assemblies with no bounding membrane, so neither can belong to a set of membrane-bound organelles.
Step 2:Lysosomes, the Golgi apparatus and mitochondria are all enclosed by membranes, so this combination is correct.
Final answer: lysosomes, Golgi apparatus and mitochondria
Q77Single correctPlant Growth and Development
Auxin can be bioassayed by :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Avena coleoptile curvature
Approach:
A bioassay measures the effect of a hormone using a living plant response specific to that hormone.
Step 1:Hydroponics is a method of soil-less culture and a potometer measures transpiration, so neither serves as an auxin bioassay.
Step 2:The Avena coleoptile curvature test quantifies auxin by the degree of bending of an oat coleoptile, making it the classical auxin bioassay.
Final answer: Avena coleoptile curvature
Q78Single correctHuman Reproduction
Which of the following layers in an antral follicle is acellular ?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3Zona pellucida
Approach:
The layers of an ovarian follicle differ in whether they are composed of cells or of secreted material.
Step 1:Theca interna, stroma and granulosa are all cellular layers of the follicle.
Step 2:The zona pellucida is a transparent acellular glycoprotein layer that surrounds the oocyte.
Final answer: Zona pellucida
Q79Single correctMolecular Basis of Inheritance
Satellite DNA is important because it :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable form parents to children.
Approach:
The significance of satellite DNA derives from its variability and its role in DNA fingerprinting.
Step 1:Satellite DNA consists of highly repetitive non-coding sequences that vary in copy number among individuals.
Step 2:This produces a high degree of polymorphism across a population while remaining constant within an individual and being inherited from parents, which forms the basis of DNA fingerprinting.
Final answer: shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable form parents to children.
Q80Single correctCell: The Unit of Life
Cell wall is absent in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Mycoplasma
Approach:
Presence or absence of a cell wall distinguishes the listed organisms.
Step 1:Funaria is a moss, Nostoc a cyanobacterium and Aspergillus a fungus, all of which possess cell walls.
Step 2:Mycoplasma are the smallest living cells and completely lack a cell wall, being bounded only by a plasma membrane.
Final answer: Mycoplasma
Q81Single correctSexual Reproduction in Flowering Plants
In angiosperms, microsporogenesis and megasporogenesis :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2involve meiosis
Approach:
The defining common feature of these two spore-forming processes is identified.
Step 1:Microsporogenesis occurs in the anther and megasporogenesis in the ovule, so no single location is common to the two processes.
Step 2:Both processes involve the reductional division of diploid mother cells, that is, meiosis, to form haploid spores.
Final answer: involve meiosis
Q82Single correctTransport in Plants
Roots play insignificant role in absorption of water in :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Pistia
Approach:
The habitat of the plant determines the importance of roots in water uptake.
Step 1:Pea, wheat and sunflower are terrestrial plants whose roots are the main organs of water absorption.
Step 2:Pistia is a free-floating hydrophyte that absorbs water directly over its general body surface, so its roots play an insignificant role in water absorption.
Final answer: Pistia
Q83Single correctEnvironmental Issues
Which of the following are most suitable indicators of pollution in the environment?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Lichens
Approach:
Sensitivity of organisms to atmospheric sulphur dioxide determines their value as bioindicators.
Step 1:Lichens are symbiotic associations that are extremely sensitive to gaseous pollutants, particularly sulphur dioxide.
Step 2:Because lichens fail to grow in areas with high sulphur dioxide levels, their presence or absence serves as a reliable indicator of such pollution.
Final answer: Lichens
Q84Single correctHuman Health and Disease
Grafted kidney may be rejected in a patient due to :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Cell-mediated immune response
Approach:
Graft rejection is attributed to the specific arm of the immune system responsible for recognising foreign tissue.
Step 1:Transplanted tissue carries foreign major histocompatibility antigens that are recognised by the recipient's immune system.
Step 2:T-lymphocytes mediate the cell-mediated immune response that attacks and rejects the grafted kidney.
Final answer: Cell-mediated immune response
Q85Single correctAnimal Kingdom
Body having meshwork of cell, internal cavities lined with food filtering flagellated cells and indirect development are the characteristics of phylum :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1Porifera
Approach:
The diagnostic features described are matched to the correct animal phylum.
Step 1:The presence of flagellated collar cells lining internal canals that filter food from water is unique to sponges.
Step 2:These choanocytes, together with a cellular body organisation and indirect development through a larval stage, characterise the phylum Porifera.
Final answer: Porifera
Q86Single correctBiological Classification
In which group of organisms the cell walls form two thin overlapping shells which fit together?
(1)
(2)
(3)
(4)
SolutionAnswer: Option 4Chrysophytes
Approach:
The structure of the cell covering identifies the correct group of protists.
Step 1:Diatoms, which belong to the chrysophytes, possess siliceous cell walls.
Step 2:These walls form two thin overlapping halves, or shells, that fit together like a soap box, characterising the chrysophytes.
Final answer: Chrysophytes
Q87Single correctBiological Classification
Choose the wrong statements:
(1)
(2)
(3)
(4)
SolutionAnswer: Option 2Morels and truffles are poisonous mushrooms
Approach:
The question asks for the incorrect statement among four assertions about fungi.
Step 1:Neurospora (a fungus) is a classical model in biochemical genetics, so statement (1) is correct.
Step 2:Morels and truffles are edible ascomycete fungi, prized as delicacies; calling them poisonous is incorrect.
Step 3:Yeast is a unicellular fungus widely used in fermentation, and Penicillium is multicellular and yields antibiotics, so statements (3) and (4) are correct.
Final answer: Morels and truffles are poisonous mushrooms
Q88Single correctHuman Reproduction
In human females, meiosis-II is not completed until :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1fertilization
Approach:
The timing of meiotic divisions during human oogenesis determines when the second division is completed.
Step 1:The secondary oocyte is released at ovulation arrested in metaphase of the second meiotic division.
Step 2:Meiosis-II resumes and is completed only when a sperm penetrates the oocyte, that is, at fertilization.
Final answer: fertilization
Q89Single correctEcosystem
Eutrophication of water bodies leading to killing of fishes is mainly due to non-availability of :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 3oxygen
Approach:
The cause of fish mortality during eutrophication is traced to changes in dissolved gas levels.
Step 1:Nutrient enrichment causes algal blooms whose subsequent decay is carried out by aerobic decomposers.
Step 2:These decomposers consume the dissolved oxygen, and the resulting oxygen depletion suffocates and kills the fish.
Final answer: oxygen
Q90Single correctDigestion and Absorption
The enzyme that is not present in succus Entericus is :
(1)
(2)
(3)
(4)
SolutionAnswer: Option 1nucleases
Approach:
The composition of intestinal juice is compared with the listed enzymes to find the one absent.
Step 1:Succus entericus, the intestinal juice, contains enzymes such as nucleosidases, lipase and maltase.
Step 2:Nucleases are secreted in the pancreatic juice rather than in the intestinal juice, so they are absent from succus entericus.
Final answer: nucleases
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