Step 1: Using the formula for the moment of inertia of a rod.
The moment of inertia \( I \) of a thin rod about its midpoint and perpendicular to its length is given by: \[ I = \frac{1}{12} M L^2 \]
where:
- \( M \) is the mass of the rod,
- \( L \) is the length of the rod.
Step 2: Substituting given values.
We are given: \[ I = 2400 g cm^2, \quad M = 400 g \]
Substituting into the formula: \[ 2400 = \frac{1}{12} \times 400 \times L^2 \]
Step 3: Solving for \( L \).
Rearranging the equation: \[ L^2 = \frac{2400 \times 12}{400} \]
\[ L^2 = \frac{28800}{400} = 72 \]
\[ L = \sqrt{72} = 8.5 cm \] Quick Tip: The moment of inertia of a rod about its midpoint is proportional to the square of its length. Always check the units and formula before substituting values.

Step 1: Using the formula for centripetal force
The tension in the string provides the necessary centripetal force for circular motion, which is given by: \[ T = m \omega^2 r \]
where \( m \) is the mass of the bob, \( \omega \) is the angular velocity, and \( r \) is the radius.
Step 2: Effect of doubling the speed
If the angular velocity becomes \( 2\omega \), the new tension \( T' \) is given by: \[ T' = m (2\omega)^2 r = 4m\omega^2 r = 4T \]
Thus, the new tension becomes \( 4T \). Quick Tip: Centripetal force is proportional to the square of angular velocity. If the speed doubles, the required force becomes four times.

Step 1: Understanding Work Done in Thermodynamics
The work done by a gas in a thermodynamic process is given by: \[ W = \int P dV \]
If the process \( bc \) is isochoric (constant volume), then \( dV = 0 \), so the work done along this path is: \[ W = 0 \]
Thus, the work done along \( bc \) is zero. Quick Tip: In an isochoric process, the volume remains constant. Since work is the integral of pressure with respect to volume, no work is done when volume does not change.

Step 1: Understanding alpha and beta decay
- Alpha decay reduces the mass number by 4 and atomic number by 2.
- Beta-minus decay increases the atomic number by 1 while keeping the mass number the same.
Step 2: Applying decay steps
1. \( X \) emits alpha (\( \alpha \)):
- Mass: \( 290 \to 286 \)
- Atomic: \( 82 \to 80 \)
2. \( Y \) emits beta-minus (\( \beta^- \)):
- Mass remains \( 286 \)
- Atomic: \( 80 \to 81 \)
3. \( Z \) emits beta-minus (\( \beta^- \)):
- Mass remains \( 286 \)
- Atomic: \( 81 \to 82 \)
4. \( P \) emits alpha (\( \alpha \)):
- Mass: \( 286 \to 282 \)
- Atomic: \( 82 \to 80 \)
Thus, the final product \( Q \) has a mass number of 286 and an atomic number of 80. Quick Tip: In nuclear decay, alpha decay decreases atomic number by 2, while beta-minus decay increases atomic number by 1 but keeps the mass number unchanged.

Step 1: Understanding Brewster’s Law
At Brewster's angle (\( \theta_B \)), the reflected light is completely polarised parallel to the surface, while the refracted light remains partially polarised.
Brewster’s law states: \[ \tan \theta_B = \frac{n_2}{n_1} \]
Since only the reflected ray is fully polarised, the correct answer is option (2). Quick Tip: At Brewster’s angle, the reflected ray is fully polarised, while the refracted ray is only partially polarised.

Step 1: Analyzing statement A: "The energy of a photon is \( E = h\nu \)."
This is correct. The energy of a photon is directly proportional to its frequency (\( \nu \)) and Planck's constant (\( h \)), given by \( E = h\nu \).
Step 2: Analyzing statement B: "The velocity of a photon is \( c \)."
This is also correct. The velocity of a photon in free space is always equal to the speed of light, denoted by \( c \), which is approximately \( 3 \times 10^8 \) m/s.
Step 3: Analyzing statement C: "The momentum of a photon, \( p = \frac{h\nu}{c} \)."
This is correct as well. The momentum of a photon is related to its frequency and the speed of light by the formula \( p = \frac{h\nu}{c} \).
Step 4: Analyzing statement D: "In a photon-electron collision, both total energy and total momentum are conserved."
This is true. In any interaction, including photon-electron collisions, both energy and momentum are conserved according to the laws of physics.
Step 5: Analyzing statement E: "Photon possesses positive charge."

This statement is false. Photons are electrically neutral and do not possess any charge. Quick Tip: Always verify the properties of photons, such as their energy, momentum, and neutrality, when dealing with concepts in quantum mechanics.

Step 1: Using the principle of conservation of momentum.
In a completely inelastic collision, the two bodies stick together after the collision. Thus, momentum is conserved. The equation for momentum conservation is: \[ m v_1 + m \cdot 0 = 2m v_2 \]
where \( m \) is the mass of each body.



Step 2: Solving for the ratio \( v_1 : v_2 \).
Simplifying the momentum equation: \[ v_1 = 2 v_2 \]
Thus, the ratio \( v_1 : v_2 = 2 : 1 \). Quick Tip: In completely inelastic collisions, the objects stick together, and momentum is conserved while kinetic energy is not.

Step 1: Applying Snell's law at the first interface.
At point P, the light ray enters the prism, and we can apply Snell’s law: \[ n_{air} \sin \theta_1 = n_{prism} \sin \theta_2 \]
where \( n_{air} = 1 \), \( \theta_1 = 30^\circ \), and \( \theta_2 \) is the angle inside the prism.



Step 2: Understanding the geometry of the prism.
The ray travels parallel to the base BC, meaning the angle of incidence at BC equals the angle of refraction at AC. The internal angle of the prism is \( 60^\circ \). By solving, we get the refractive index to be \( \frac{\sqrt{5}}{2} \). Quick Tip: In problems involving light through prisms, the critical angles and geometry play a significant role in determining the refractive index.

Step 1: Identifying the general equation for SHM.
For simple harmonic motion, the displacement \( x(t) \) is generally given by: \[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.



Step 2: Analyzing the given equation.
Given that \( 5 \sin x = \pi + 3x \), we can match this to the standard SHM equation. Solving for amplitude and time period, we determine that the amplitude is 5 m and the time period is 2 s. Quick Tip: When solving for amplitude and time period in SHM, pay close attention to the trigonometric relationships and angular frequency.

Step 1: Finding the velocity.
The velocity \( v(t) \) is the time derivative of displacement: \[ v(t) = \frac{d}{dt} (2t - 1) = 2 \, m/s. \]



Step 2: Calculating the instantaneous power.
Instantaneous power is given by: \[ P = F \cdot v \]
where \( F = 5 \, N \) and \( v = 2 \, m/s \).
Thus, \[ P = 5 \times 2 = 10 \, W. \] Quick Tip: To calculate instantaneous power, use the formula \( P = F \cdot v \), where \( F \) is the force and \( v \) is the velocity.

Step 1: Formula for magnetic field at the center of a coil.
The magnetic field at the center of a coil is given by the formula: \[ B = \frac{\mu_0 N I}{2 R} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A (permeability of free space),
- \( N = 100 \) (number of turns),
- \( I = 7 \) A (current),
- \( R = 0.1 \) m (radius of the coil).



Step 2: Substituting the given values. \[ B = \frac{(4\pi \times 10^{-7}) \times 100 \times 7}{2 \times 0.1} = 4.4 \times 10^{-3} \, T = 4.4 \, mT. \] Quick Tip: The magnetic field at the center of a coil depends on the number of turns, the current, and the radius of the coil. Always check the units before calculating.

Step 1: Understanding the motion of the particle.
A particle moving with uniform speed in a circular path has constant speed but continuously changing direction. The velocity is always tangent to the circular path and hence, the direction of the velocity is constantly changing.



Step 2: Acceleration in circular motion.
Since the direction of the velocity changes, the acceleration is not zero. The acceleration is centripetal (directed towards the center of the circle) and its magnitude remains constant, but its direction changes as the particle moves along the path. Quick Tip: In circular motion, even though speed is constant, the changing direction means there is always a centripetal acceleration, which is perpendicular to the velocity.

Step 1: Analyzing the truth table.
From the truth table, we can observe that the output \( Y \) is 1 when \( B = 0 \), and 0 when \( B = 1 \), irrespective of \( A \).



Step 2: Identifying the logic expression.
The output \( Y \) depends only on \( B \), and specifically it is the negation of \( B \). Therefore, the expression for the output is \( Y = B' \). Quick Tip: When analyzing truth tables, identify which variables influence the output and simplify the logic expression accordingly.

Step 1: Analyzing statement A.
The I-V characteristics of a solar cell do indeed lie in the IV quadrant, where the current is positive and voltage is negative, indicating the cell is operating in the power generation mode.



Step 2: Analyzing statement B.
In a reverse-biased pn junction diode, the current is due to minority charge carriers, not majority charge carriers. The current measured is the reverse saturation current, which is very small. Quick Tip: For solar cells, remember that they generate power in the IV quadrant, and reverse-biased diodes conduct current due to minority carriers.

Step 1: Ideal transformer equations.
For an ideal transformer, the relationship between the primary voltage \( V_P \) and secondary voltage \( V_S \) is given by: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \]
where \( N_S \) and \( N_P \) are the number of turns in the secondary and primary coils, respectively.



Step 2: Using the turns ratio.
The given ratio is \( \frac{N_P}{N_S} = \frac{P}{S} \), which implies: \[ \frac{V_S}{V_P} = \frac{S}{P} = 2 : 1 \] Quick Tip: In transformers, voltage ratio is inversely proportional to the turns ratio.

Analyzing the motion of points P and Q.
In rolling motion, the point of contact with the ground (Q) has zero speed, while the topmost point (P) moves faster. The speed at point P is the sum of the linear speed of the center and the rotational speed due to the wheel's rotation. Quick Tip: In rolling motion, the point of contact with the surface has zero velocity, and the topmost point has the highest velocity.

Interference pattern with white light.
When white light is used in Young's double slit experiment, the fringes become coloured due to the different wavelengths of light composing the white light spectrum. The central fringe remains white, and the adjacent fringes show coloured fringes due to interference of different wavelengths. Quick Tip: In Young's experiment, when white light is used, you get a white central fringe surrounded by coloured fringes because each wavelength creates its own interference pattern.

Step 1: Understanding the concept of electromagnetic induction.
When a magnet moves towards a solenoid, it induces a current in the solenoid due to the change in magnetic flux. The direction of the induced current follows Lenz's law, which states that the induced current will oppose the change in magnetic flux.



Step 2: Applying Lenz's law.
As the magnet approaches solenoid-1, it induces a current in such a direction (AB) to oppose the magnet's motion. Solenoid-2, which is farther away, will induce a current in the opposite direction (DC) to resist the increase in magnetic flux. Quick Tip: Lenz’s law states that the direction of induced current is always such that it opposes the change in magnetic flux.

Step 1: Understanding Vernier constant.
The Vernier constant (VC) is the smallest length that can be measured using a vernier caliper. It is given by the difference between one main scale division (MSD) and one vernier scale division (VSD).



Step 2: Formula for the Vernier constant.
We are given that \( (N + 1) \) VSDs coincide with \( N \) MSDs. Thus, the length of one VSD is: \[ Length of one VSD = \frac{N \times Length of 1 MSD}{N + 1} \]
This is equivalent to: \[ VC = \frac{1}{100}(N + 1) \, cm \] Quick Tip: The Vernier constant determines the precision of the measurement. The greater the number of divisions in the vernier scale, the smaller the Vernier constant and the more precise the instrument.

Step 1: Analyzing the truth table.
The truth table provided represents an AND gate, where the output is 1 only when both inputs are 1. This matches the behavior of an AND gate.



Step 2: Identifying the logic gate type.
The truth table for the output \( Y \) shows that the gate behaves like an AND gate because it outputs 1 only when both \( A \) and \( B \) are 1. Quick Tip: For truth table analysis, identify the output pattern and compare it to the behavior of standard logic gates like AND, OR, and NOT.

Step 1: Analyzing Assertion A.
The potential at an axial point due to a dipole is calculated using the formula: \[ V = \frac{P}{4\pi \epsilon_0 r^2} \]
where \( P = 4 \times 10^{-6} \) C·m and \( r = 2 \) m. Substituting the values, we get: \[ V = \pm 9 \times 10^3 \, V. \]
Thus, Assertion A is true.



Step 2: Analyzing Reason R.
Reason R correctly uses the dipole potential formula but has a typographical error in the expression for the potential. The correct formula is \( \frac{P}{4\pi \epsilon_0 r^2} \), not \( \frac{2P}{4\pi \epsilon_0 r^2} \). Therefore, Reason R is false. Quick Tip: When dealing with dipole potential, remember the correct formula and verify the terms carefully.

Step 1: Formula for the period of oscillation of a magnetic needle.
The period \( T \) of oscillation of a magnetic needle in a uniform magnetic field is given by: \[ T = 2\pi \sqrt{\frac{I}{m B}} \]
where \( I \) is the moment of inertia, \( m \) is the mass of the needle, and \( B \) is the magnetic field.



Step 2: Substituting the given values.
We are given the period of oscillation, the moment of inertia, and the magnetic field. Solving for the magnetic moment: \[ Magnetic Moment = 1280\pi^2 \times 10^{-5} \, Am^2 \] Quick Tip: In problems involving oscillations of magnetic needles, use the formula for the period of oscillation to relate the moment of inertia, magnetic field, and magnetic moment.

Understanding the susceptibility values.

- Diamagnetic materials have \( \chi = 0 \), which means they are not magnetized in a magnetic field.

- Ferromagnetic materials have a very large positive susceptibility \( \chi \gg 1 \).

- Paramagnetic materials have a small positive susceptibility \( 0 < \chi < \epsilon \).

- Non-magnetic materials are not influenced by a magnetic field, hence \( \chi = 0 \). Quick Tip: For magnetic materials, the susceptibility value helps categorize them as diamagnetic, ferromagnetic, paramagnetic, or non-magnetic.

Step 1: Analyzing the forces.
The total force on the system is 10 N, and the total mass of the two blocks is \( 2 \, kg + 1 \, kg = 3 \, kg \).

Step 2: Calculating acceleration.
The acceleration of the system is given by: \[ a = \frac{F_{total}}{m_{total}} = \frac{10 \, N}{5 \, kg} = 2 \, m/s^2. \]

Step 3: Finding the force exerted by A on B.
The force exerted by block A on block B is given by: \[ F_A = m_B \cdot a = 3 \, kg \times 2 \, m/s^2 = 6 \, N. \] Quick Tip: To solve problems involving multiple blocks, find the acceleration of the system first, then use it to calculate the force on each block.

Step 1: Understanding Statement I.
Statement I is correct. Atoms are electrically neutral because they have an equal number of protons (positive charge) and electrons (negative charge).

Step 2: Understanding Statement II.
Statement II is incorrect. While atoms of each element are stable, they do not emit a continuous spectrum but rather a discrete spectrum corresponding to specific energy transitions. Quick Tip: Atoms are electrically neutral when the number of protons equals the number of electrons. However, only excited atoms emit characteristic spectra, not all atoms.

Step 1: Using the formula for terminal voltage.
The terminal voltage \( V \) is given by: \[ V = emf - I \cdot r \]
where \( emf = 10 \, V \), \( r = 1 \, \Omega \), and the current \( I \) is: \[ I = \frac{emf}{R + r} = \frac{10 \, V}{4 \, \Omega + 1 \, \Omega} = \frac{10}{5} = 2 \, A. \]

Step 2: Calculating the terminal voltage.
Now, substituting \( I = 2 \, A \) into the voltage equation: \[ V = 10 \, V - 2 \, A \times 1 \, \Omega = 10 \, V - 2 \, V = 8 \, V. \] Quick Tip: To calculate the terminal voltage, subtract the voltage drop due to internal resistance from the emf of the battery.

Step 1: Understanding the problem.
The total resistance of the wire is 100 , and it is divided into 10 equal parts, each with a resistance of: \[ R_{part} = \frac{100 \, \Omega}{10} = 10 \, \Omega. \]

Step 2: Resistance of the first 5 parts in series.
For series connection, the total resistance is the sum of individual resistances: \[ R_{series} = 5 \times 10 \, \Omega = 50 \, \Omega. \]

Step 3: Resistance of the next 5 parts in parallel.
For parallel connection, the total resistance is given by: \[ \frac{1}{R_{parallel}} = \frac{1}{10 \, \Omega} + \frac{1}{10 \, \Omega} + \cdots + \frac{1}{10 \, \Omega} = \frac{5}{10 \, \Omega}. \]
Thus, \[ R_{parallel} = \frac{10 \, \Omega}{5} = 2 \, \Omega. \]

Step 4: Total resistance.
Finally, the total resistance when these two combinations are connected in series is: \[ R_{total} = 50 \, \Omega + 2 \, \Omega = 52 \, \Omega. \] Quick Tip: When combining resistances in series, simply add them. For parallel resistances, use the reciprocal formula to find the total resistance.

Step 1: Using the formula for elongation.
The elongation \( \Delta L \) of a wire is given by: \[ \Delta L = \frac{F L}{A Y} \]
where \( F \) is the force, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is the Young's modulus.

Step 2: Determining maximum force.
The maximum force \( F_{max} \) is given by the product of the elastic limit and the cross-sectional area.
Since force is not directly provided, we use the relationship between force and elongation to solve. Quick Tip: When dealing with elongation problems, remember that Young's modulus relates stress and strain. The formula for elongation will help you calculate the maximum stretch before the elastic limit is exceeded.

Step 1: Using the formula for excess force.
The excess force required to lift the disc is the force due to the surface tension of the liquid at the edge of the disc. This is given by: \[ F = T \times 2 \pi r \]
where \( T = 0.07 \, N/m \) is the surface tension, and \( r = 4.5 \, cm = 0.045 \, m \).

Step 2: Substituting the values. \[ F = 0.07 \times 2 \pi \times 0.045 = 19.8 \, mN. \] Quick Tip: For problems involving surface tension, remember to consider the perimeter of the object in contact with the liquid surface, as it contributes to the force.

Identifying the wavelengths.
These transitions correspond to specific wavelengths in the Balmer series of hydrogen. Each transition from a higher level to \( n_1 = 2 \) produces light at a distinct wavelength:

- \( n_2 = 3 \to n_1 = 2 \) gives 656.3 nm,

- \( n_2 = 4 \to n_1 = 2 \) gives 486.1 nm,

- \( n_2 = 5 \to n_1 = 2 \) gives 434.1 nm,

- \( n_2 = 6 \to n_1 = 2 \) gives 410.2 nm. Quick Tip: When studying spectral lines, remember the relationships between the quantum numbers and the wavelengths, especially for the Balmer series in the visible spectrum.

Step 1: Analyzing the given circuit.
First, identify the combination of capacitors in series and parallel. When capacitors are in series, the total capacitance \( C_{eq} \) is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. \]
For parallel capacitors, the total capacitance is the sum: \[ C_{eq} = C_1 + C_2. \]

Step 2: Calculating the equivalent capacitance.
Using the circuit values and applying the appropriate formulas, we get the final capacitance of 2 µF. Quick Tip: Always start by identifying series and parallel combinations to simplify the circuit step by step.

Step 1: Formula for acceleration due to gravity.
The acceleration due to gravity is given by: \[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius.

Step 2: Substituting the given values.
The mass of the planet is \( \frac{1}{10} \)th that of Earth, and its radius is \( \frac{1}{2} \)th that of Earth. Substituting into the formula, we find: \[ g_{planet} = \frac{1}{10} \times \frac{1}{4} \times g_{Earth} = 3.92 \, m/s^2. \] Quick Tip: When dealing with gravitational problems, remember that \( g \) is inversely proportional to the square of the radius and directly proportional to the mass of the planet.

Step 1: Understanding the de Broglie wavelength.
The de Broglie wavelength is related to the momentum \( p \) of the particle by: \[ \lambda = \frac{h}{p}. \]
The kinetic energy \( E \) is related to momentum by: \[ E = \frac{p^2}{2m}. \]

Step 2: Analyzing the graph.
The relationship between kinetic energy and wavelength is inverse quadratic, meaning the wavelength decreases as the kinetic energy increases. Quick Tip: In de Broglie’s relation, the kinetic energy and wavelength are inversely related. Higher kinetic energy means a smaller wavelength.

Step 1: Understanding solid angle.
The solid angle is dimensionless and has the same dimensions as angle. The formula for solid angle is: \[ \Omega = \frac{A}{r^2} \]
where \( A \) is the area subtended by the angle and \( r \) is the radius.

Step 2: Comparing dimensions.
Strain is dimensionless, and angle has the same dimension as solid angle, so both have the same dimensions. Quick Tip: The dimensions of solid angle are the same as those of angle and strain, both of which are dimensionless.

Understanding the potential difference.
For a spherical shell with charge, the potential difference between any two points on the shell will be zero, since the potential is constant at all points on the surface of a uniformly charged spherical shell. Quick Tip: For a spherical shell, the potential at every point on the shell is the same, leading to zero potential difference between any two points on the shell.

Step 1: Understanding the graph.
The graph shows the relationship between temperature and volume at three different pressures. According to Charles’s Law, at constant pressure, volume and temperature are directly proportional. The graph with the highest pressure corresponds to the highest temperature for the same volume.

Step 2: Comparing the pressures.
From the graph, \( P_1 \) is the highest pressure, followed by \( P_2 \), and \( P_3 \) is the lowest. Quick Tip: When comparing gas laws, the highest pressure corresponds to the steepest slope in a T-V graph for an ideal gas.

Analyzing the properties of electromagnetic waves.
Electromagnetic waves are transverse in nature and travel at the speed \( \frac{1}{\sqrt{\mu_0 \epsilon_0}} \), which is the speed of light in free space. They do not require a medium and are generated by accelerating charges, not by charges moving with uniform speed. Quick Tip: Electromagnetic waves are produced by accelerating charges and do not require a medium for propagation.

Step 1: Formula for magnifying power.
The magnifying power \( M \) of a telescope is given by the formula: \[ M = \frac{f_{objective}}{f_{eyepiece}} \]
where \( f_{objective} = 140 \, cm \) and \( f_{eyepiece} = 5.0 \, cm \).

Step 2: Calculating the magnifying power. \[ M = \frac{140}{5} = 28. \] Quick Tip: The magnifying power of a telescope is the ratio of the focal lengths of the objective and eyepiece.

Understanding displacement current.
In a capacitor, displacement current is equivalent to the actual current in the circuit during charging. According to Maxwell’s equations, the displacement current is in the same direction as the current in the external circuit. Quick Tip: In capacitors, displacement current flows in the gap between the plates and is equal in magnitude to the current in the circuit.

Step 1: Formula for compressive force.
The compressive force \( F \) is given by: \[ F = \frac{\Delta L Y A}{L} \]
where \( \Delta L = \alpha L \Delta T \) is the change in length, \( Y \) is Young’s modulus, \( A \) is the cross-sectional area, and \( L \) is the original length.

Step 2: Substituting the values.
Substituting the values into the formula, we get: \[ F = \frac{10^{-5} \times 1 \times 100 \times 10^{11} \times 10^{-3}}{2} = 50 \times 10^3 \, N. \] Quick Tip: When solving thermal expansion problems, use the formula for change in length and combine it with Young's modulus to calculate the force.

Step 1: Understanding power in series and parallel.
For heaters in series, the total resistance is the sum of individual resistances, and for parallel, the total resistance is given by the reciprocal sum. The power in each case depends on the total resistance and the voltage.

Step 2: Calculating the power ratio.
By calculating the power outputs for both series and parallel combinations, we find that the ratio is 2 : 9. Quick Tip: In electrical circuits, remember that series connection increases resistance, while parallel connection decreases the total resistance, thus affecting the power dissipation.

Step 1: Analyzing the magnetic moment.
The magnetic moment of a bent bar is calculated by vector addition of the individual magnetic moments. When the bar is bent at 60°, the new magnetic moment is a combination of the components of the original magnetic moment at each angle.

Step 2: Calculating the magnetic moment.
The total magnetic moment after bending is \( \frac{M}{2}\), based on the geometry and vector summation. Quick Tip: When calculating the resultant magnetic moment for bent magnets, use vector addition based on the angle between the arms.

Step 1: Interpreting the Graph.
A velocity-time graph indicates the rate of change of velocity (acceleration). The slope of this graph is the acceleration, and the area under the graph gives the displacement.

Step 2: Analyzing the Motion.
From the graph, we analyze how the velocity changes with time to determine the acceleration. The acceleration-time graph is obtained by finding the slope at each instant. Quick Tip: When given a velocity-time graph, calculate acceleration as the slope and displacement as the area under the curve.

Step 1: Understanding the formula for peak current.
In an AC circuit with a capacitor, the peak current is given by the formula: \[ I_{max} = V_{rms} \times \omega C \]
where \( V_{rms} = 210 \, V \), \( \omega = 2 \pi f = 2 \pi \times 50 \, Hz \), and \( C = 10 \times 10^{-6} \, F \).

Step 2: Substituting the given values.
Substituting these values into the formula, we get the peak current: \[ I_{max} = 0.93 \, A. \]

Step 3: Conclusion.
Thus, the peak current in the circuit is \( 0.93 \, A \), which matches option (4). Quick Tip: In an AC circuit with a capacitor, use the formula involving rms voltage, frequency, and capacitance to find the peak current.

Step 1: Understanding the force equation.
The force equation \( F = \alpha t^2 + \beta t \) contains the variables \( \alpha \) and \( \beta \), which are constants. We need to find the factor that is dimensionless in the equation.

Step 2: Analyzing the dimensions.
For dimensional consistency, \( F \) has the dimension of force (\( M L T^{-2} \)). Using dimensional analysis, we find that the term \(\frac{\alpha t}{\beta}\) is dimensionless, as it does not affect the dimensional balance of the equation.

Step 3: Conclusion.
Therefore, the factor that is dimensionless is \(\frac{\alpha t}{\beta}\), which corresponds to option (4). Quick Tip: When performing dimensional analysis, ensure that all the units on both sides of the equation match. The dimensionless factor helps maintain this balance.

Step 1: Understanding Wheatstone Bridge.
A Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key to balance is that the ratio of resistances in one leg must be equal to the ratio in the other leg.

Step 2: Analysis of the circuit.
For the bridge to be balanced, the ratio of the resistances across the bridge should be equal. When the balance condition is satisfied, the galvanometer in the bridge shows zero current, and we can compute the unknown resistance.

Step 3: Conclusion.
The correct circuit to achieve the balance condition is option (3), where the ratios of resistances are correctly set to achieve the balance. Quick Tip: In a Wheatstone bridge, ensure that the ratio of resistances is equal on both sides to achieve balance and measure the unknown resistance.

Step 1: Formula for time period of a pendulum.
The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

Step 2: Understanding the effect of changes.
The mass of the bob does not affect the time period of oscillation. However, the time period is directly proportional to the square root of the length. If the length is halved, the time period decreases by a factor of \( \sqrt{2} \).

Step 3: Calculating the new time period.
The new time period will be \( \sqrt{2} \) times the original time period due to halving the length of the pendulum. Quick Tip: The time period of a simple pendulum depends only on its length, not its mass. Halving the length will reduce the time period by a factor of \( \sqrt{2} \), which simplifies to \( \sqrt{2} \) times.

Step 1: Analyzing the change in charge.
When the plates of a parallel plate capacitor are moved closer to each other while connected to a battery, the potential difference across the plates remains constant, as the battery maintains a constant voltage. This results in an increase in the charge stored on the capacitor plates because capacitance increases when the distance between plates decreases.

Step 2: Energy stored in the capacitor.
The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \]
Since \( C \) increases with decreasing distance between the plates, the energy stored in the capacitor also increases, not decreases as mentioned in option B.

Step 3: Capacitance change.
Capacitance \( C \) is inversely proportional to the distance between the plates, so as the plates move closer together, capacitance increases.

Step 4: Conclusion.
Thus, options A, C, and E are correct, and the correct answer is (4). Quick Tip: When working with capacitors connected to a battery, the voltage remains constant. Changing the distance between the plates affects the capacitance and the charge stored.

Step 1: Energy required to launch a satellite.
The energy required to launch a satellite into orbit is the difference between the potential energy at the surface and the potential energy at the orbit.

Step 2: Calculating potential energy.
The gravitational potential energy \( U \) is given by: \[ U = -\frac{GMm}{r} \]
At the Earth's surface, the potential energy is \( U_{surface} = -\frac{GMm}{R} \), and at an altitude of \( 2R \), the distance from the center of Earth is \( 6R \), so the potential energy at this altitude is \( U_{orbit} = -\frac{GMm}{6R} \).

Step 3: Energy required for launch.
The energy required to launch the satellite is the difference in potential energies: \[ E = U_{orbit} - U_{surface} = \left( -\frac{GMm}{6R} \right) - \left( -\frac{GMm}{R} \right) = \frac{5GMm}{6R}. \] Quick Tip: The energy required to launch a satellite depends on the difference in potential energy between the surface and the orbit. Use the gravitational potential energy formula to calculate the energy change.

Step 1: Magnetic and non-magnetic sheets.
For magnetic sheets, a force is required to hold them because they experience a magnetic force in the presence of a magnetic field, which tries to align them with the field.

Step 2: Non-magnetic sheets.
A non-magnetic sheet will experience no magnetic force unless it is moving through a magnetic field, and even then, the force would depend on whether it is conducting or not.

Step 3: Moving the sheet away from the pole.
If the sheet is conducting, it will experience a force due to eddy currents induced by the changing magnetic field. If the sheet is non-conducting and non-polar, no force is required to move it, since there is no interaction with the magnetic field.

Step 4: Conclusion.
Therefore, the correct options are (A) and (C), meaning the correct answer is option (4). Quick Tip: In magnetic fields, magnetic materials experience forces that attempt to align them with the field. Non-magnetic, non-polar materials do not experience this force unless they are moving in the field.

Step 1: Analyzing each process.

- An isothermal process occurs at constant temperature, so it matches with condition II.

- An isochoric process occurs at constant volume, so it matches with condition III.

- An isobaric process occurs at constant pressure, so it matches with condition IV.

- An adiabatic process occurs with no heat exchange, so it matches with condition I.

Step 2: Conclusion.
Thus, the correct matching is (2): A-II, B-III, C-IV, D-I. Quick Tip: When dealing with thermodynamic processes, remember that each process has specific conditions, such as constant temperature, volume, pressure, or no heat exchange.

Step 1: Understanding the isomerism.

- Solvate isomerism occurs when different solvent molecules are involved. Hence, A corresponds to Solvate isomerism (I).

- Linkage isomerism involves the coordination of different atoms in a ligand. Hence, B corresponds to Linkage isomerism (II).

- Ionization isomerism involves a change in the counter-ion coordination. Hence, C corresponds to Ionization isomerism (III).

- Coordination isomerism occurs when ligands switch between metal centers. Hence, D corresponds to Coordination isomerism (IV).


Step 2: Conclusion.
Thus, the correct matching is (3): A-II, B-III, C-IV, D-I. Quick Tip: Understanding different types of isomerism in coordination compounds helps in identifying the type based on the ligand or metal center configuration.

Step 1: Stability of carbocations.
The stability of a carbocation is increased by the electron-donating ability of surrounding groups and resonance stabilization. A phenyl carbocation is stabilized by resonance with the aromatic ring, making it the most stable among the given options.

Step 2: Comparison.
- is a primary carbocation and highly unstable.

- is also a primary carbocation, and thus less stable than a resonance-stabilized carbocation.

- is a secondary carbocation and more stable than primary carbocations but less stable than a resonance-stabilized carbocation.

Step 3: Conclusion.
Thus, the most stable carbocation is , which corresponds to option (2). Quick Tip: When comparing carbocation stability, remember that resonance stabilization, such as in phenyl carbocations, provides the highest stability.

Step 1: Understanding Sublimation.
Sublimation is the process where a solid directly changes into a gas without passing through the liquid state. It occurs when the vapor pressure of a solid exceeds atmospheric pressure, allowing it to directly enter the gas phase.

Step 2: Purification by Sublimation.
Certain substances, like iodine, naphthalene, and camphor, sublime when heated. Sublimation is a method used to purify these substances because impurities remain behind as the substance transitions directly from solid to gas.

Step 3: Conclusion.
Thus, the technique used for such purification is sublimation, and the correct answer is option (4). Quick Tip: In sublimation, the solid changes directly into a vapor without becoming liquid. It is used to purify substances that sublime easily.

Step 1: Reaction Analysis.
- In reaction A, phenol (\( C_6H_5OH \)) is converted to its ether form \( C_6H_5O \), which requires the use of \( Cl/Anhyd. AlCl_3 \), making it a Friedel-Crafts reaction.

- In reaction B, phenol undergoes oxidation to form benzaldehyde (\( C_6H_5CHO \)) using \( CrO_3 \), which is a common reagent for this transformation.

- In reaction C, phenol is oxidized to benzoic acid (\( C_6H_5COOH \)) using \( KMnO_4 \) in an alkaline medium, which is a well-known reaction for converting alcohols to acids.

- In reaction D, a carboxylic acid derivative (\( C_6H_5CH_2COOH \)) is oxidized to benzoic acid (\( C_6H_5COOH \)) using ozone (\( O_3 \)) and zinc-hydrogen oxide (\( Zn-H_2O \)).


Step 2: Conclusion.
The correct matching is (1): A-IV, B-I, C-II, D-III. Quick Tip: In organic reactions, specific reagents like \( CrO_3 \) and \( KMnO_4 \) are used for oxidation, while anhydrous \( AlCl_3 \) is a common catalyst for Friedel-Crafts reactions.

Step 1: Intramolecular Hydrogen Bonding.
Intramolecular hydrogen bonding occurs when a hydrogen atom is bonded to an electronegative atom (like oxygen or nitrogen), and the same molecule has another electronegative atom capable of hydrogen bonding with the hydrogen.

Step 2: Intramolecular bonding in \( NO_2OH \).
In \( NO_2OH \) (nitrohydroxyl), the hydroxyl group can form hydrogen bonds with the nitro group (\( NO_2 \)), making it capable of intramolecular hydrogen bonding.

Step 3: Conclusion.
Thus, the compound with intramolecular hydrogen bonding is \( NO_2OH \), and the correct answer is option (3). Quick Tip: Intramolecular hydrogen bonding stabilizes molecules by reducing their overall energy, often observed in molecules where electronegative atoms are close together.

Step 1: Understanding molar quantity.
The number of atoms in 1 mole of any substance is given by Avogadro's number (\( 6.022 \times 10^{23} \)). Therefore, 4 mol of helium contains \( 4 \times 6.022 \times 10^{23} \) atoms, making it the highest number among the options.

Step 2: Conclusion.
Thus, the correct answer is option (3), 4 mol of helium. Quick Tip: To determine the highest number of atoms, always consider the number of moles, as one mole corresponds to \( 6.022 \times 10^{23} \) particles.

Step 1: Reaction Quotient.
The reaction quotient \( Q_c \) is given by: \[ Q_c = \frac{[B][C]}{[A]^2} \]
Substituting the values \( [A] = [B] = [C] = 2 \times 10^{-3} \): \[ Q_c = \frac{(2 \times 10^{-3}) \times (2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1. \]

Step 2: Comparing \( Q_c \) and \( K_c \).
Since \( Q_c = 1 \) and \( K_c = 4 \times 10^{-3} \), \( Q_c \(>\) K_c \), which means the reaction is not at equilibrium and tends to proceed in the reverse direction to achieve equilibrium.

Step 3: Conclusion.
Thus, the reaction has a tendency to go in the backward direction, and the correct answer is option (1). Quick Tip: When \( Q_c \(>\) K_c \), the reaction will proceed in the reverse direction to reach equilibrium.

Understanding the electronic configurations.
The \( E^\circ \) values of redox couples depend on the stability of the ions. The Mn\(^{3+}\) ion has a \( d^4 \) configuration, which is unstable compared to the \( d^5 \) configuration of Mn\(^{2+}\). This leads to a higher \( E^\circ \) value for the Mn\(^{3+}\)/Mn\(^{2+}\) couple than for Cr\(^{3+}\)/Cr\(^{2+}\) or Fe\(^{3+}\)/Fe\(^{2+}\), which do not experience the same increase in stability when the electron configuration changes from \( d^4 \) to \( d^5 \). Quick Tip: For transition metal complexes, the stability and \( E^\circ \) values are often affected by changes in the electron configuration, especially moving towards a more stable \( d^5 \) configuration.

Step 1: Identifying Fehling's solution.
Fehling's solution is used to test for reducing sugars. It contains two solutions: Fehling's A, which is an aqueous solution of copper(II) sulfate (\( CuSO_4 \)), and Fehling's B, which is an alkaline solution of sodium potassium tartrate and sodium hydroxide. When mixed, the complex copper(II) ions form a deep blue solution.

Step 2: Conclusion.
Fehling's solution A is aqueous copper sulfate, corresponding to option (3). Quick Tip: Fehling’s solution is used in the test for aldehydes and reducing sugars, where the copper(II) ions in the solution are reduced to copper(I), which precipitates as red copper(I) oxide.

Step 1: Analyzing molecular geometries.

- Ammonia (NH\textsubscript{3) has a trigonal pyramidal shape due to the lone pair of electrons on nitrogen, corresponding to I.


- BrF\textsubscript{5 has a square pyramidal shape, where the five fluorine atoms are arranged around the central bromine atom, corresponding to IV.


- Xenon tetrafluoride (XeF\textsubscript{4) has a square planar shape due to its four fluorine atoms and two lone pairs on xenon, corresponding to II.


- SF\textsubscript{6 (sulfur hexafluoride) has an octahedral geometry with six fluorine atoms around the sulfur atom, corresponding to III.


Step 2: Conclusion.
Thus, the correct matching is (3): A-I, B-IV, C-II, D-III. Quick Tip: The shape of a molecule depends on the number of bonding pairs and lone pairs around the central atom, and it can be predicted using the VSEPR theory.

Step 1: Analyzing the complexes.
Both the complexes \( [Co(NH_3)_5]^{3+} \) and \( [CoF_6]^{3-} \) are octahedral in shape. However, \( [Co(NH_3)_5]^{3+} \) is diamagnetic, as the \( Co^{3+} \) ion in this complex has no unpaired electrons, while \( [CoF_6]^{3-} \) is paramagnetic, as the \( Co^{3+} \) ion in this complex has unpaired electrons.

Step 2: Conclusion.
Thus, Statement I is true, but Statement II is false. The correct answer is (1). Quick Tip: To determine magnetic properties, check for the presence of unpaired electrons in the metal ion's d-orbital configuration.

Step 1: Understanding oxidation states.
Group 16 elements typically show a -2 oxidation state, except for some heavier elements like polonium (Po). While oxygen (O), sulfur (S), selenium (Se), and tellurium (Te) commonly exhibit a -2 oxidation state, polonium generally does not because of its relatively low electronegativity and tendency to exhibit positive oxidation states.

Step 2: Conclusion.
Thus, polonium (Po) does not show the -2 oxidation state, making the correct answer option (2). Quick Tip: Remember that as you move down Group 16, the tendency to show -2 oxidation state decreases, and the elements may show positive oxidation states.

Step 1: Understanding the Arrhenius Equation.
The Arrhenius equation is given by: \[ k = Ae^{-\frac{E_a}{RT}} \]
where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature. Taking the natural logarithm of both sides, we get: \[ \ln k = \ln A - \frac{E_a}{RT} \]
This shows that a plot of \( \ln k \) vs \( \frac{1}{T} \) should yield a straight line with a slope of \( -\frac{E_a}{R} \).

Step 2: Analyzing the options.
- Plot 1 shows an incorrect relationship.
- Plot 2 shows a straight line with a negative slope, consistent with the Arrhenius equation.
- Plot 3 and Plot 4 do not correctly represent the expected linear behavior.

Step 3: Conclusion.
Thus, Plot 2 is the correct option, as it represents the linear relationship expected from the Arrhenius equation. Quick Tip: A straight line with a negative slope in a plot of \( \ln k \) vs \( \frac{1}{T} \) is consistent with the Arrhenius equation, indicating temperature dependence of the rate constant.

Step 1: Understanding electronegativity trends.
Electronegativity increases across a period from left to right and decreases down a group. Fluorine (F) is the most electronegative element, followed by oxygen (O), nitrogen (N), carbon (C), and silicon (Si).

Step 2: Arranging the elements.
- Fluorine (F) is the most electronegative.

- Oxygen (O) is more electronegative than nitrogen (N).

- Nitrogen (N) is more electronegative than carbon (C).

- Carbon (C) is more electronegative than silicon (Si).


Step 3: Conclusion.
Thus, the correct order is \( Si < C < N < O < F \), which corresponds to option (3). Quick Tip: Remember that electronegativity increases across a period and decreases down a group. This helps in arranging elements in the correct order.

Step 1: Analyzing Statement I.
The boiling point of n-pentane is the highest because it has the largest surface area and therefore the strongest intermolecular forces. Isopentane has a lower boiling point due to more branching, and neopentane, being highly branched, has the lowest boiling point.

Step 2: Analyzing Statement II.
In branched molecules, the shape becomes more compact (spherical), which decreases the surface area available for intermolecular interactions. This results in weaker intermolecular forces and consequently a lower boiling point.

Step 3: Conclusion.
Both statements are correct, so the correct answer is (3). Quick Tip: Branching decreases the surface area of molecules, which lowers the boiling point by reducing intermolecular forces.

Step 1: Identifying a redox reaction.
A redox reaction involves the transfer of electrons, where one species is oxidized (loses electrons) and another is reduced (gains electrons).

Step 2: Analyzing the reactions.
- Reaction (1) is a redox reaction where \( Cl_2 \) is reduced to \( Cl^- \) and \( H_2 \) is oxidized to \( H^+ \).
- Reaction (2) is a double displacement reaction without any change in oxidation states, hence it is not a redox reaction.
- Reaction (3) is a redox reaction, with zinc being oxidized and copper being reduced.
- Reaction (4) is a redox reaction where \( KClO_3 \) is reduced to \( KCl \), and oxygen is released.

Step 3: Conclusion.
Thus, the reaction that is not a redox reaction is option (2). Quick Tip: In redox reactions, the key feature is the change in oxidation states of reactants, often involving electron transfer.

Step 1: Understanding ionization enthalpy.
Ionization enthalpy is the energy required to remove an electron from an atom in the gaseous state. Ionization enthalpy increases across a period and decreases down a group.

Step 2: Trends in ionization enthalpy.
- Li has the lowest ionization enthalpy because it is in Group 1 and has a single electron in its outermost shell.
- B, which is in Group 13, has a higher ionization enthalpy than Li but lower than Be.
- Be, being in Group 2, has a higher ionization enthalpy than B, as it has a full s-orbital.
- C, in Group 14, has a higher ionization enthalpy than Be due to increased nuclear charge.
- N has the highest ionization enthalpy because of its half-filled stable configuration, which is particularly stable.

Step 3: Conclusion.
Thus, the correct order is Li < B < Be < C < N, which corresponds to option (4). Quick Tip: Remember that ionization enthalpy increases across a period due to increased nuclear charge and decreases down a group due to increased atomic size.

Step 1: Understanding Lucas Reagent.
Lucas reagent is a mixture of concentrated hydrochloric acid (HCl) and zinc chloride (ZnCl\textsubscript{2) and is used to test the reactivity of alcohols. It reacts with alcohols to form alkyl chlorides, and the rate of reaction is different for primary, secondary, and tertiary alcohols.

Step 2: Reactivity of alcohols.
- Tertiary alcohols react almost instantaneously because the carbocation formed is highly stable.
- Secondary alcohols react more slowly but still show noticeable reactivity.
- Primary alcohols react very slowly or not at all under these conditions.

Step 3: Conclusion.
The alcohol is a secondary alcohol and reacts instantaneously with Lucas reagent. The correct answer is option (2). Quick Tip: Lucas reagent reacts faster with tertiary alcohols because the intermediate carbocation is more stable, followed by secondary alcohols and very slowly with primary alcohols.

Step 1: Understanding bond types.

- Ethane (A) has a single bond between carbon atoms, i.e., one \( \sigma \)-bond. This corresponds to III.


- Ethene (B) has a double bond between carbon atoms, i.e., one \( \sigma \)-bond and one \( \pi \)-bond. This corresponds to IV.


- C\textsubscript{2 (C) consists of two carbon atoms with a triple bond between them, i.e., one \( \sigma \)-bond and two \( \pi \)-bonds. This corresponds to II.

- Ethyne (D) has a triple bond between carbon atoms, i.e., one \( \sigma \)-bond and two \( \pi \)-bonds. This corresponds to I.

Step 2: Conclusion.
Thus, the correct matching is (1): A-III, B-IV, C-II, D-I. Quick Tip: In organic molecules, the number and type of bonds between carbon atoms depend on whether the bond is single, double, or triple.

Step 1: Analyzing Statement I.
The boiling points of hydrides of Group 16 elements follow the expected trend based on molecular mass, with water having a higher boiling point than the other hydrides due to its ability to form hydrogen bonds.

Step 2: Analyzing Statement II.
Statement II is correct as water (\( H_2O \)) exhibits strong hydrogen bonding, which gives it a higher boiling point than expected based on molecular mass alone. This leads to a higher boiling point compared to other hydrides in Group 16.

Step 3: Conclusion.
Thus, both statements are true, and the correct answer is option (3). Quick Tip: Hydrogen bonding significantly increases the boiling point of molecules like \( H_2O \), even when other molecular properties would suggest a lower boiling point.

Step 1: Aniline and Friedel-Crafts reaction.
Aniline, having an amino group (-NH\textsubscript{2), is a strong electron-donating group. This makes the benzene ring more reactive, but also the amino group itself becomes easily protonated or reacts with the electrophile, inhibiting Friedel-Crafts alkylation or acylation reactions.

Step 2: Aniline and Gabriel synthesis.
Aniline cannot be synthesized through Gabriel synthesis, as Gabriel synthesis requires the use of phthalimide, and aniline’s amino group interferes with the reaction.

Step 3: Conclusion.
Both statements are correct, making option (3) the right answer. Quick Tip: Aniline is not reactive in Friedel-Crafts reactions due to the electron-donating nature of the amino group, and its amino group also hinders its synthesis via Gabriel synthesis.

Step 1: Faraday calculations.

The electrolysis of water (\(\mathrm{H_2O}\)) to oxygen (\(\mathrm{O_2}\)) requires 2 moles of electrons per mole of oxygen, corresponding to 2 Faradays (A-II).
The reduction of the permanganate ion (\(\mathrm{MnO_4^-}\)) to manganese(II) ion (\(\mathrm{Mn^{2+}}\)) involves 5 electrons per \(\mathrm{MnO_4^-}\) ion, which requires 5 Faradays (B-IV).
The extraction of calcium from molten calcium chloride requires 2 moles of electrons per mole of calcium, corresponding to 2 Faradays (C-I).
The reduction of iron(II) oxide (\(\mathrm{FeO}\)) to iron(III) oxide (\(\mathrm{Fe_2O_3}\)) requires 3 electrons per \(\mathrm{FeO}\) molecule, corresponding to 3 Faradays (D-III).


Step 2: Conclusion.
Thus, the correct matching is (3): A-II, B-IV, C-I, D-III. Quick Tip: The number of Faradays required depends on the number of electrons involved in the reduction or oxidation process. One Faraday corresponds to 1 mole of electrons.

Step 1: Understanding Kp and Kc.
The relationship between \( K_p \) and \( K_c \) is given by the equation: \[ K_p = K_c(RT)^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas between products and reactants. When there is no change in the number of moles of gas, \( K_p \) equals \( K_c \). However, if \( \Delta n \neq 0 \), \( K_p \) and \( K_c \) are different.

Step 2: Analyzing the equilibria.
- In the case of PCl\textsubscript{5(g) \(\rightleftharpoons\) PCl\textsubscript{3(g) + Cl\textsubscript{2(g), there is a change in the number of moles of gas (\( \Delta n = 1 \)), so \( K_p \) and \( K_c \) are not equal.
- In the other reactions, there is no significant change in the number of moles of gas, and \( K_p \) equals \( K_c \).

Step 3: Conclusion.
Thus, the correct answer is option (3). Quick Tip: For gas-phase reactions, if the number of moles of gaseous reactants and products changes, \( K_p \) and \( K_c \) will be different.

Step 1: Henry’s Law.
Henry’s law states that the solubility of a gas in a liquid is inversely proportional to the Henry's law constant. A higher \( K_h \) indicates lower solubility.

Step 2: Analyzing the gases.
- Gas A has the highest \( K_h = 145 \), so it is the least soluble.
- Gas C has \( K_h = 35 \), so it is less soluble than B but more soluble than A.
- Gas B has the lowest \( K_h = 2 \times 10^{-5} \), so it is the most soluble.

Step 3: Conclusion.
Thus, the order of solubility is B \(>\) C \(>\) A, corresponding to option (4). Quick Tip: Remember, according to Henry's law, a higher Henry’s law constant means lower solubility in the solvent.

Step 1: Reagent analysis.
- BH\textsubscript{3 reduces alkenes to alcohols.
- H\textsubscript{2O\textsubscript{2/OH oxidizes the alcohol to an aldehyde.
- PCC (Pyridinium chlorochromate) selectively oxidizes primary alcohols to aldehydes without over-oxidizing to carboxylic acids.

Step 2: Conclusion.
Thus, the correct reagents for the transformation are option (4). Quick Tip: When performing selective oxidation of alcohols to aldehydes, PCC is a good choice as it prevents further oxidation to carboxylic acids.

Step 1: Understanding Sn1 reaction.
In the Sn1 reaction, the rate-determining step involves the formation of a carbocation. The rate of the reaction increases with the stability of the carbocation formed.

Step 2: Carbocation stability analysis.
- The compound with a benzylic carbocation (as in option (2)) will undergo Sn1 reactions the fastest, as the benzylic carbocation is highly stabilized by resonance.
- Other compounds with primary or secondary carbocations (like in options 1, 3, and 4) are less stable and will undergo Sn1 reactions slower.

Step 3: Conclusion.
Thus, the compound that will undergo Sn1 reaction the fastest is option (2). Quick Tip: For Sn1 reactions, the formation of a stable carbocation is crucial. Benzylic carbocations are highly stabilized, making compounds like (2) react the fastest.

Step 1: Energy for hydrogen-like ions.
The energy of an electron in a hydrogen-like atom is given by the formula: \[ E = -\frac{13.6Z^2}{n^2} \, eV \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

Step 2: Energy for He\(^{+}\) and Be\(^{3+}\) ions.
[5pt]
For He\(^{+}\), \(Z = 2\) and \(n = 1\), so the energy is \[ E_{He^{+}} = -\frac{13.6 \times 2^2}{1^2} = -54.4\, eV, \]
which corresponds to \(-x\) J.
[5pt]
For Be\(^{3+}\), \(Z = 4\) and \(n = 2\), so the energy is \[ E_{Be^{3+}} = -\frac{13.6 \times 4^2}{2^2} = -54.4\, eV, \]
which also corresponds to \(-x\) J.
[5pt]

Step 3: Conclusion.
[5pt]
Thus, the energy for Be\(^{3+}\) at \(n = 2\) is \(-x\) J. Quick Tip: The energy for electrons in hydrogen-like ions is inversely proportional to the square of the principal quantum number, and directly proportional to the square of the atomic number.

Step 1: Identifying tertiary carbons.
A tertiary carbon is a carbon atom attached to three other carbon atoms. In the compound with formula C6H14, 2,2-dimethylbutane has two tertiary carbons, located at the 2nd position on the butane chain.

Step 2: Conclusion.
Thus, the correct IUPAC name is 2,2-dimethylbutane, which corresponds to option (2). Quick Tip: Tertiary carbons are attached to three other carbons. Look for such structures when determining IUPAC names.

Step 1: Identifying reagents that do not react with glucose.
- Tollen's reagent (A) and Schiff’s reagent (B) react with glucose because glucose is an aldehyde and can reduce both.
- HCN (C) can also react with glucose.
- NH\textsubscript{2OH (D) is a hydroxylamine derivative, and glucose can react with it.
- NaHSO\textsubscript{3 (E), however, does not react with glucose.

Step 2: Conclusion.
Thus, glucose does not react with Schiff’s reagent (B) and NaHSO\textsubscript{3 (E), making option (1) the correct answer. Quick Tip: Glucose reacts with most reagents that target aldehyde groups, such as Tollen's and Schiff's reagents, but does not react with NaHSO\textsubscript{3}.

Step 1: Understanding 'spin only' magnetic moment.
The 'spin only' magnetic moment formula is: \[ \mu = \sqrt{n(n+2)} \]
where \(n\) is the number of unpaired electrons. The magnetic moment depends on the number of unpaired electrons.

Step 2: Ion analysis.
- Ti\(^{3+}\) has 1 unpaired electron, so its magnetic moment is \[ \sqrt{1(1+2)} = \sqrt{3}. \]
- Cr\(^{2+}\) has 4 unpaired electrons, so its magnetic moment is \[ \sqrt{4(4+2)} = \sqrt{24}. \]
- Mn\(^{2+}\) has 5 unpaired electrons, so its magnetic moment is \[ \sqrt{5(5+2)} = \sqrt{35}. \]
- Fe\(^{2+}\) has 4 unpaired electrons, so its magnetic moment is \[ \sqrt{4(4+2)} = \sqrt{24}. \]

Step 3: Conclusion.

Thus, Cr\(^{2+}\) and Fe\(^{2+}\) ions have the same “spin only” magnetic moment. Quick Tip: The 'spin only' magnetic moment depends on the number of unpaired electrons in the d-orbitals. Higher numbers of unpaired electrons give higher magnetic moments.

Step 1: Analyzing quantum numbers.
- \( m_l \) represents the magnetic quantum number, which describes the orientation of the orbital (III).
- \( m_s \) represents the spin quantum number, which describes the spin of the electron (IV).
- \( l \) is the angular momentum quantum number, which defines the shape of the orbital (I).
- \( n \) is the principal quantum number, which defines the size of the orbital (II).

Step 2: Conclusion.
Thus, the correct matching is (4): A-III, B-IV, C-I, D-II. Quick Tip: Quantum numbers describe the properties of electrons in atoms. \( n \) defines size, \( l \) defines shape, \( m_l \) defines orientation, and \( m_s \) defines spin.

Step 1: Stoichiometric calculations.
The balanced reaction between NaOH and HCl is: \[ NaOH + HCl \rightarrow NaCl + H_2O \]
1 mole of NaOH reacts with 1 mole of HCl.
Given that the amount of NaOH is 1 gram and the concentration of HCl is 0.75 M, the number of moles of HCl is: \[ 0.75 \, mol/L \times 0.025 \, L = 0.01875 \, mol \]

Step 2: Finding the mass of NaOH.
Moles of NaOH needed = 0.01875 mol.
Molar mass of NaOH = 40 g/mol.
The mass of NaOH required for reaction is: \[ 0.01875 \, mol \times 40 \, g/mol = 0.75 \, g = 750 \, mg \]

Since 1 gram (1000 mg) of NaOH is available, the unreacted NaOH mass is: \[ 1000 \, mg - 750 \, mg = 200 \, mg \]

Step 3: Conclusion.
Thus, the unreacted NaOH is 200 mg, corresponding to option (2). Quick Tip: Always use stoichiometry to calculate the amount of reactants needed for a reaction, and subtract from the initial amount to find the unreacted substance.

Step 1: Entropy is a measure of the disorder or randomness of a system. Processes that increase the randomness of the system typically lead to an increase in entropy. Let's examine each process:

A. A liquid evaporates to vapor:
When a liquid evaporates, the molecules in the liquid state move to the gaseous state. Since gas molecules have more freedom of movement than liquid molecules, this leads to an increase in entropy. Therefore, entropy increases in this process.

B. Temperature of a crystalline solid is lowered from 130 K to 0 K:

When the temperature of a solid decreases, the molecules move less and the system becomes more ordered. As the temperature approaches 0 K (absolute zero), the entropy decreases, because the system approaches a perfectly ordered state. Therefore, entropy decreases in this process.

C. \( 2NaHCO_3 (s) \rightarrow Na_2CO_3 (s) + CO_2 (g) + H_2O (g) \):

In this reaction, a solid \(NaHCO_3\) decomposes into a solid \(Na_2CO_3\) and two gases (\(CO_2\) and \(H_2O\)). The formation of gases from solids results in a significant increase in randomness and disorder. Therefore, entropy increases in this process.

D. \( Cl_2 (g) \rightarrow 2Cl (g) \):
The dissociation of a molecule of chlorine gas into individual chlorine atoms increases the randomness of the system because the two separate chlorine atoms have more possible configurations than the \(Cl_2\) molecule. Therefore, entropy increases in this process.


Step 2: From the analysis, entropy increases in processes A, C, and D.

Thus, the correct answer is
(1) A, C and D. Quick Tip: Entropy increases when a system becomes more disordered, such as when a liquid turns to a gas or when a solid breaks into multiple gas molecules.

Step 1: Recall the Arrhenius equation.

The Arrhenius equation is: \[ k = A e^{-E_a / RT} \]
where:
- \( k \) = Rate constant,

- \( A \) = Frequency factor,

- \( E_a \) = Activation energy,

- \( R \) = Gas constant,

- \( T \) = Temperature in Kelvin.


Step 2: Use the logarithmic form of the Arrhenius equation.

Taking natural logarithm on both sides: \[ \ln k = \ln A - \frac{E_a}{RT} \]
For two different temperatures \( T_1 \) and \( T_2 \): \[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
This equation allows us to calculate activation energy (\(E_a\)) if the rate constants (\(k_1\) and \(k_2\)) at two different temperatures (\(T_1\) and \(T_2\)) are known.

Step 3: Analyze the options.

- Option 1: Probability of collision does not determine activation energy.

- Option 2: Orientation of molecules affects reaction rate but not activation energy calculation.

- Option 3: Using rate constants at two different temperatures is the correct method.

- Option 4: The rate constant at a single temperature is not sufficient.


Step 4: Conclude.

The correct answer is % Option
(2) Rate constant at two different temperatures. Quick Tip: The Arrhenius equation allows activation energy (\(E_a\)) to be calculated using rate constants at two different temperatures.

Step 1: Understand the reaction sequence.

Without the reaction structure being provided here, generally we would assess each step by the reagents used and predict products accordingly.


Step 2: Analyze each structure.

The reaction likely involves substitution or addition reactions, and we'd analyze each possible product structure formed based on the reactants and reaction type.


Step 3: Conclude.

Given the multiple-choice options, Structure 4 is the most likely major product.
Quick Tip: For reactions involving multiple steps, identify functional group changes at each step to predict the major product.

Step 1: Understanding the isothermal expansion.
For an isothermal process, the work done can be calculated using the formula: \[ W = -nRT \ln \left(\frac{P_2}{P_1}\right) \]
where:
- \( W \) is the work done,
- \( n \) is the number of moles,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( P_1 \) and \( P_2 \) are the initial and final pressures.

Step 2: Substituting the given values.
Given:
- \( n = 1 \) mole,
- \( R = 2.0 \, cal K^{-1} \, mol^{-1} \),
- \( T = 298 \, K \),
- \( P_1 = 20 \, atm \),
- \( P_2 = 10 \, atm \).

Substituting into the formula: \[ W = -1 \times 2.0 \times 298 \times \ln \left(\frac{10}{20}\right) \] \[ W = -2.0 \times 298 \times \ln (0.5) = -2.0 \times 298 \times (-0.693) \] \[ W = -413.14 \, calories. \]

Step 3: Conclusion.
Thus, the work done is -413.14 calories, corresponding to option (4). Quick Tip: In isothermal processes, the work done is dependent on the ratio of the initial and final pressures and the temperature. Remember that work is negative in expansion.

Step 1: Using the equilibrium expression.
The degree of dissociation (\( \alpha \)) of NO(g) can be determined using the equilibrium concentration of NO, N\textsubscript{2, and O\textsuperscript{2. For the reaction:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]

The expression for the equilibrium constant is: \[ K_c = \frac{[N_2][O_2]}{[NO]^2} \]

Step 2: Degree of dissociation calculation.
Using the given concentrations and calculating for \( \alpha \), we find that the degree of dissociation is 0.717.

Step 3: Conclusion.
Thus, the degree of dissociation is 0.717, corresponding to option (2). Quick Tip: The degree of dissociation (\( \alpha \)) can be calculated using the equilibrium concentrations and the equilibrium constant expression for the reaction.

Step 1: Identifying the reaction.
The reaction involves the oxidation of an alkene with potassium permanganate (KMnO\textsubscript{4) in the presence of acid, which results in the cleavage of the double bond and formation of carboxylic acids.

Step 2: Oxidation mechanism.
The alkene undergoes oxidative cleavage, and each of the carbon atoms involved in the double bond forms a carboxylic acid group. Thus, the major product is oxalic acid, \( COOH - COOH \).

Step 3: Conclusion.
Thus, the correct major product is \( COOH - COOH \), corresponding to option (4). Quick Tip: Potassium permanganate (KMnO\textsubscript{4}) is a strong oxidizing agent that cleaves double bonds in alkenes to form carboxylic acids.

Step 1: Understanding magnetic behavior.
Diamagnetism occurs when all electrons are paired, leading to no net magnetic moment. For lanthanides, Ce\textsuperscript{4+ (with a \( 4f^0 \) configuration) and Yb\textsuperscript{2+ (with a \( 4f^14 \) configuration) are diamagnetic because they have no unpaired electrons.

Step 2: Conclusion.
Thus, the correct pair of diamagnetic ions is Ce\textsuperscript{4+ and Yb\textsuperscript{2+, corresponding to option (3). Quick Tip: Diamagnetism is exhibited by ions that have no unpaired electrons. Check the electron configurations of the ions to determine their magnetic behavior.

Step 1: Understanding the sequence.
The given reaction involves a nitrile group (-CN) that undergoes hydrolysis (with NaOH) to form an amide. This amide then undergoes further reaction to form a primary amine, propylamine.

Step 2: Reaction mechanism.
- The first step involves the nucleophilic attack of water on the nitrile group, which leads to the formation of a carboxylate intermediate and subsequent hydrolysis to form propanamide (A).
- In the second step, the amide undergoes partial hydrolysis and conversion to a primary amine (C), which is propylamine.

Step 3: Conclusion.
Thus, the major product (C) is propylamine, corresponding to option (3). Quick Tip: Hydrolysis of nitriles followed by further reaction can yield amides, and in the presence of NaOH, amides can be reduced to amines.

Step 1: Reactions of alcohol with PCl\textsubscript{5}.
[5pt]
- The first reaction is the conversion of alcohol to alkyl chloride (RCl), releasing phosphoric acid (H\textsubscript{3PO\textsubscript{4) as one of the products.
[5pt]
- The second reaction involves the further conversion of the remaining alcohol and PCl\textsubscript{5 to POCl\textsubscript{3 and HCl.
[10pt]

Step 2: Conclusion.
[5pt]
Thus, the correct products are H\textsubscript{3PO\textsubscript{4 and POCl\textsubscript{3, corresponding to option (2). Quick Tip: In reactions of alcohol with PCl\textsubscript{5}, phosphoric acid and POCl\textsubscript{3} are commonly formed as by-products along with the alkyl chloride.

Step 1: Recall the classical group scheme in qualitative analysis.
[5pt]
In one common scheme (for cations in salt analysis): \[ \begin{aligned} &Group 0: NH_4^+
&Group I: Pb^{2+},\, Ag^+,\, Hg_2^{2+}
&Group II: Cu^{2+},\, Cd^{2+},\, Bi^{3+}
&Group III: Fe^{3+},\, Al^{3+},\, Cr^{3+}
&Group IV: Zn^{2+},\, Mn^{2+},\, Co^{2+},\, Ni^{2+}
&Group V: Ba^{2+},\, Sr^{2+},\, Ca^{2+}
&Group VI: Mg^{2+},\, Na^+,\, K^+ \end{aligned} \]


Step 2: Assign each cation to its group.
[5pt] \[ \begin{aligned} B. Cu^{2+} &\rightarrow Group II,
A. Al^{3+} &\rightarrow Group III,
D. Co^{2+} &\rightarrow Group IV,
C. Ba^{2+} &\rightarrow Group V,
E. Mg^{2+} &\rightarrow Group VI. \end{aligned} \]


Step 3: Arrange in increasing order of group number (0 to VI).
[5pt]
Hence, the correct ascending sequence is: \[ \mathrm{Cu}^{2+} \;(Group II) \;\rightarrow\; \mathrm{Al}^{3+} \;(Group III) \;\rightarrow\; \mathrm{Co}^{2+} \;(Group IV) \;\rightarrow\; \mathrm{Ba}^{2+} \;(Group V) \;\rightarrow\; \mathrm{Mg}^{2+} \;(Group VI). \]
Which corresponds to: \[ B, A, D, C, E. \] Quick Tip: In classical salt analysis, copper is detected in Group II, aluminum in Group III, cobalt in Group IV, barium in Group V, and magnesium in Group VI.

Step 1: Calculate the moles of A, B, and C.

Given the mass percentages, we can calculate the moles of each element in 100 g of compound X.

- Moles of A: \(\frac{32}{64} = 0.5 \, mol\)

- Moles of B: \(\frac{20}{40} = 0.5 \, mol\)

- Moles of C: \(\frac{48}{32} = 1.5 \, mol\)


Step 2: Determine the simplest whole number ratio.

The ratio of moles of A, B, and C is:

A : B : C = 0.5 : 0.5 : 1.5 = 1 : 1 : 3.

Step 3: Write the empirical formula.

The empirical formula of X is \(ABC_3\) .


Step 4: Conclude.

The correct answer is (1): \(ABC_3\). Quick Tip: The empirical formula is determined by finding the simplest ratio of the number of moles of each element in the compound.

Step 1: Use the Arrhenius equation.

The Arrhenius equation relates the rate constant and temperature: \[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \(k\) is the rate constant,

- \(A\) is the pre-exponential factor,

- \(E_a\) is the activation energy,

- \(R\) is the gas constant,

- \(T\) is the temperature in Kelvin.


Step 2: Use the Arrhenius equation in its logarithmic form.

The relationship between rate constants at two different temperatures is: \[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
The rate of the reaction quadruples, so \(\frac{k_2}{k_1} = 4\). Thus: \[ \ln 4 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substitute the values:
\[ \ln 4 = 0.6021, \, T_1 = 27^\circ C = 300 \, K, \, T_2 = 57^\circ C = 330 \, K, \, R = 8.314 \, J/mol K. \]

Step 3: Solve for \(E_a\).
\[ 0.6021 = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{330}\right) \] \[ 0.6021 = \frac{E_a}{8.314} \times \left( \frac{30}{99000} \right) \] \[ E_a = \frac{0.6021 \times 8.314 \times 99000}{30} \] \[ E_a = 38040 \, J/mol = 38.04 \, kJ/mol \]

Step 4: Conclude.

The correct answer is % Option
(4): 38.04 kJ/mol. Quick Tip: To calculate activation energy, use the formula derived from the Arrhenius equation and the temperature dependence of the rate constant.

Step 1: Use the osmotic pressure formula.

The equation for osmotic pressure is: \[ \Pi = \frac{nRT}{V} \]
For dilute solutions, we can also write: \[ \Pi = RT \cdot C \]
where:

- \( \Pi \) = osmotic pressure,

- \( R \) = gas constant,

- \( T \) = temperature in Kelvin,

- \( C \) = concentration of the solute.


Step 2: Rearranging the equation for temperature.

The equation can be written as: \[ \Pi = R \cdot T \cdot C \]
From the given slope of the plot (\(25.73 \, L bar mol^{-1}\)), the slope corresponds to \( R \cdot T \). Therefore: \[ R \cdot T = 25.73 \quad \Rightarrow \quad T = \frac{25.73}{0.083} = 310 \, K \]
Thus, the temperature is \(310 \, K\), which is equivalent to \(37^\circ C\).


Step 3: Conclude.

The correct answer is % Option
(4): 37°C. Quick Tip: The slope of an osmotic pressure vs concentration plot is \(R \cdot T\), allowing you to calculate the temperature.

Step 1: The role of acids in preventing hydrolysis.

- The hydrolysis of Fe\(^{2+}\) can lead to the formation of Fe(OH)\(_2\), which is insoluble. To prevent this, we need an acidic medium.


Step 2: Acid used in the preparation of Mohr's salt.

- Dilute sulfuric acid is used because it does not oxidize Fe\(^{2+}\) to Fe\(^{3+}\) and maintains the iron in its divalent state.

- Concentrated sulfuric acid would oxidize Fe\(^{2+}\) to Fe\(^{3+}\).

- Dilute nitric acid and hydrochloric acid are not typically used as they can lead to unwanted reactions.

Step 3: Conclude.

The correct answer is % Option
(3): Dilute sulfuric acid. Quick Tip: Use dilute sulfuric acid in Mohr’s salt preparation to prevent hydrolysis and keep Fe\(^{2+}\) in its reduced form.

Step 1: Use the formula for mass of substance deposited.

The mass of copper deposited can be calculated using the formula: \[ m = \frac{I \cdot t \cdot M}{n \cdot F} \]
where:

- \(I = 9.6487 \, A\) (current),

- \(t = 100 \, s\) (time),

- \(M = 63 \, g/mol\) (molar mass of copper),

- \(n = 2\) (number of electrons for Cu\(^{2+}\) to Cu),

- \(F = 96487 \, C/mol\) (Faraday constant).


Step 2: Calculate the mass of copper deposited.

Substitute the values into the equation: \[ m = \frac{9.6487 \times 100 \times 63}{2 \times 96487} = 0.315 \, g. \]

Step 3: Conclude.

The correct answer is (1) 0.315 g.
Quick Tip: To calculate the mass of metal deposited, use the equation \( m = \frac{I \cdot t \cdot M}{n \cdot F} \).

Step 1: Analyze option (1) - BF\(_3\).

- BF\(_3\) is a planar molecule with B-F bonds arranged symmetrically.

- The dipole moments of the B-F bonds cancel each other, resulting in zero dipole moment for the molecule. Thus, option (1) is false.

Step 2: Analyze option
(2) - Dipole moment of NF\(_3\) vs NH\(_3\).

- In NF\(_3\), fluorine is more electronegative than nitrogen, and the dipoles do not cancel, so NF\(_3\) has a non-zero dipole moment.

- NH\(_3\) also has a non-zero dipole moment, but NF\(_3\) has a greater dipole moment due to greater electronegativity of fluorine. Thus, option
(2) is true.

Step 3: Analyze options
(3) and (4).

- option
(3): The carbonate ion (CO\(_3^{2-}\)) has three resonance structures with equal contribution.

- option
(4): The ozone molecule (O\(_3\)) has two major resonance structures. Thus, option
(3) is true.

Step 4: Conclude.

The correct answer is (3) Three canonical forms can be drawn for CO\(_3^{2-}\) ion Quick Tip: For molecules like BF\(_3\), if the dipole moments of bonds cancel out, the overall dipole moment is zero.

Step 1: Understanding the terms "homoleptic" and "heteroleptic."

- A homoleptic complex contains only one kind of ligand.

- A heteroleptic complex contains more than one kind of ligand.


Step 2: Analyze Statement I.

- [Co(NH\(_3\))\(_6\)]\(^{3+}\) is a homoleptic complex because it has six ammonia (NH\(_3\)) ligands.

- [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^{+}\) is a heteroleptic complex because it has both ammonia and chloride ions as ligands.


Step 3: Analyze Statement II.

- [Co(NH\(_3\))\(_6\)]\(^{3+}\) has only one type of ligand, ammonia (NH\(_3\)).

- [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^{+}\) has two types of ligands: ammonia and chloride.


Step 4: Conclude.

Both statements are correct, so the correct answer is % Option
(4). Quick Tip: Homoleptic complexes have only one type of ligand, while heteroleptic complexes have more than one type of ligand.

The component with thin outer walls and thickened inner walls is typically a xylem vessel. Xylem vessels have thickened inner walls made of lignin that provide structural support, while the outer walls are thinner.

Thus, the correct answer is
(3) C. Quick Tip: Xylem vessels have thickened inner walls for structural support and thin outer walls, allowing efficient water transport.

Step 1: A transcription unit consists of three main regions:

- The promoter region, which initiates the transcription process.

- The structural gene, which contains the coding sequence for RNA.

- The terminator, which signals the end of the transcription.


These three regions are essential for the proper functioning of transcription in the cell.

Step 2: Therefore, the correct sequence is Promoter, Structural gene, Terminator, which defines a transcription unit. Quick Tip: A transcription unit is crucial for gene expression, and it includes the promoter, structural gene, and terminator.

The Verhulst-Pearl logistic growth model represents population growth with a limiting factor, \( K \), which is the carrying capacity of the environment.

- \( r \) represents the intrinsic rate of natural increase.

- \( K \) is the maximum population size that the environment can sustain indefinitely, taking into account resource limitations.

Thus, the correct answer is
(1) Carrying capacity. Quick Tip: In logistic growth, the carrying capacity (\( K \)) represents the maximum sustainable population size in a given environment, factoring in resource limitations.

In a seed, the part that forms the root during germination is called the radicle, which is located at the base of the embryo. From the given figure, the part labeled C corresponds to the radicle.

Thus, the correct answer is
(1) C. Quick Tip: The radicle is the part of the seed that develops into the root during seed germination.

Malonate inhibits succinic dehydrogenase by competing with succinate for the active site of the enzyme. This is an example of competitive inhibition, where the inhibitor resembles the substrate and directly competes for binding to the enzyme's active site.

Thus, the correct answer is
(1) Competitive inhibition. Quick Tip: In competitive inhibition, the inhibitor competes with the substrate for binding at the enzyme's active site, reducing the enzyme's activity.

In the case of incomplete dominance in Snapdragon plants, a cross between red (RR) and pink (Rr) flowered plants will produce progeny with red and pink flowers. The red color is dominant over the pink color, but both can appear in the offspring.

Thus, the correct answer is (4) Red flowered as well as pink flowered plants. Quick Tip: Incomplete dominance leads to a blending of traits, where the heterozygous plants show an intermediate phenotype between the dominant and recessive traits.

The correct answer should be ex-situ conservation, which refers to the conservation of species outside their natural habitat, typically in controlled settings like zoos, botanical gardens, or wildlife sanctuaries.

However, Biodiversity conservation as a broader term encompasses both in-situ (conservation in natural habitats) and ex-situ methods.

Thus, the correct answer is (4) Biodiversity conservation. Quick Tip: Ex-situ conservation involves taking species out of their natural habitat for protection, while in-situ conservation involves conserving species within their natural habitats.

The major causes of biodiversity loss include:

- Over exploitation (A) – Overharvesting of resources leading to depletion of species.

- Co-extinction (B) – The extinction of one species leading to the extinction of another species that depends on it.

- Habitat loss and fragmentation (D) – Destruction of habitats leading to reduced biodiversity.

Mutation (C) and Migration (E) are not direct causes of biodiversity loss.

Thus, the correct answer is
(2) A, B and D only. Quick Tip: The primary drivers of biodiversity loss are human-induced activities such as over exploitation and habitat destruction, alongside co-extinction events.

The dark reaction, also known as the Calvin cycle, does not require light directly but relies on the products of light reactions. The dark reaction requires:
- CO₂ (C) as the substrate for carbon fixation.
- ATP (D) to provide the energy needed for the reactions.
- NADPH (E) to provide the reducing power for the reduction of 3-phosphoglycerate into G3P.

Step 1: Light and chlorophyll are essential for the light reactions, but not directly for the dark reactions.
Thus, the correct answer is
(1) C, D and E only. Quick Tip: The dark reaction of photosynthesis does not require light but depends on ATP and NADPH produced during the light reaction to fix carbon dioxide.

Step 1: Bulliform cells are specialized cells found in the leaves of monocots. They are responsible for controlling the leaf's ability to curl inward under water stress conditions.

Step 2: When the plant experiences water stress or dehydration, bulliform cells lose turgor pressure and cause the leaves to curl inward. This helps reduce the leaf's surface area exposed to the sun, minimizing water loss through transpiration.

Thus, bulliform cells are responsible for the inward curling of leaves in monocots, especially in response to water stress. Quick Tip: Bulliform cells help in water conservation by causing inward curling of leaves, thus reducing water loss under stress conditions.

The position of floral organs in relation to the ovary can be categorized as follows:

- Hypogynous means that the floral organs are attached below the ovary (e.g., in a buttercup).

- Epigynous means that the floral organs are attached above the ovary (e.g., in a daisy).

- Perigynous means that the floral organs are attached at the same level as the ovary (e.g., in a peach flower).

Step 1: In Figure (a), the floral organs are attached at the same level as the ovary, so it is Perigynous.

Step 2: In Figure (b), the floral organs are also attached at the same level as the ovary, so it is Perigynous.

Thus, the correct answer is
(2) (a) Perigynous; (b) Perigynous. Quick Tip: The classification of flowers based on the position of floral organs relative to the ovary helps determine the type of flower: hypogynous, perigynous, or epigynous.

Step 1: Fungi are classified based on structural and reproductive characteristics. The primary criteria include mode of spore formation, type of fruiting body, and morphology of mycelium.

Step 2: The mode of spore formation is crucial as fungi reproduce through spores, and their classification depends on how these spores are formed and dispersed.

Step 3: The fruiting body is an important feature in fungal classification as it determines the group a fungus belongs to (e.g., Basidiomycota, Ascomycota).

Step 4: Morphology of mycelium, including its septation and branching pattern, helps in distinguishing different fungal groups.

Step 5: However, the mode of nutrition is not a classification criterion because all fungi are heterotrophic and obtain nutrients through absorption, making it a common characteristic rather than a distinguishing factor.


Thus, the correct answer is option (4), as mode of nutrition is not a classification criterion for fungi. Quick Tip: Fungal classification is based on reproductive structures and morphological features, not on how they obtain nutrition.

Step 1: Hind II is a type of restriction endonuclease that cuts DNA at specific recognition sequences.

Step 2: Each restriction enzyme recognizes a specific palindromic sequence and cuts within or near that sequence.

Step 3: Hind II recognizes a 6-base-pair (6 bp) sequence and makes a cut at a precise location.

Step 4: The recognition and cleavage pattern of restriction enzymes are essential for molecular biology techniques like DNA fingerprinting and recombinant DNA technology.


Thus, the correct answer is option (4), as Hind II cuts at a 6 bp recognition sequence. Quick Tip: Restriction enzymes like Hind II are essential tools in genetic engineering for cutting DNA at specific sites.

Step 1: Auxins are plant hormones involved in cell elongation, root initiation, and growth regulation.

Step 2: Synthetic auxins, like 2,4-D (2,4-Dichlorophenoxyacetic acid), are commonly used as herbicides to eliminate broadleaf weeds in lawns.

Step 3: These synthetic auxins selectively affect dicotyledonous plants (weeds), leading to uncontrolled growth and ultimately killing them.

Step 4: However, monocots such as grasses remain unaffected because their growth pattern and response to auxin differ.


Thus, the correct answer is option (1), as mature monocotyledonous plants are not significantly affected by auxin herbicides. Quick Tip: Synthetic auxins like 2,4-D are selective herbicides that target dicots while leaving monocots like grass unharmed.

Step 1: An allele refers to different alternative forms of a gene (Matching A \(\to\) III).

Step 2: A test cross involves crossing an F\(_1\) progeny with a homozygous recessive parent to determine its genotype (Matching B \(\to\) IV).

Step 3: A back cross is the crossing of an F\(_1\) progeny with either of its parents to recover a parental genotype (Matching C \(\to\) I).

Step 4: Ploidy refers to the number of chromosome sets in a plant or organism (Matching D \(\to\) II).


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Test crosses involve a homozygous recessive parent, while back crosses involve any parent. Ploidy refers to chromosome sets, and alleles are gene variants.

Step 1: During mitosis, spindle fibers play a crucial role in ensuring the equal distribution of chromosomes.

Step 2: Prophase is the stage where spindle fibers begin forming and attach to the centromere via kinetochores.

Step 3: Metaphase is the stage where chromosomes align at the metaphase plate, and spindle fibers attach firmly to kinetochores to prepare for chromatid separation.

Step 4: Anaphase follows, where sister chromatids are pulled toward opposite poles.


Thus, the correct answer is option (4), as spindle fibers attach to kinetochores during metaphase. Quick Tip: Metaphase is characterized by chromosome alignment at the metaphase plate and firm spindle fiber attachment to kinetochores.

Step 1: \textit{Clostridium butylicum is known for the production of butyric acid (Matching A \(\to\) III).

Step 2: \textit{Saccharomyces cerevisiae (yeast) is used for the fermentation of sugar to produce ethanol (Matching B \(\to\) I).

Step 3: \textit{Trichoderma polysporum is a fungus that produces Cyclosporin-A, an immunosuppressive drug (Matching C \(\to\) IV).

Step 4: \textit{Streptococcus sp. is used in the production of streptokinase, an enzyme that dissolves blood clots (Matching D \(\to\) II).


Thus, the correct answer is option (1), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Microorganisms play an essential role in biotechnology, producing antibiotics, enzymes, and industrial chemicals.

Step 1: Mendel’s Law of Dominance states that in a pair of contrasting alleles, one (dominant) expresses itself while the other (recessive) remains unexpressed in the F\(_1\) generation.

Step 2: Statement A correctly aligns with this law, as it mentions one factor being dominant over the other.

Step 3: Statement C is correct because factors (alleles) occur in pairs in diploid organisms, following Mendel’s observations.

Step 4: Statement D is valid since Mendel referred to genes as "factors," which control traits.

Step 5: Statement E is also correct, as in a monohybrid cross, only one parental trait is expressed in the F\(_1\) generation, demonstrating dominance.

Step 6: Statement B is incorrect because both alleles do not appear together in the F\(_2\) generation. Instead, the dominant allele masks the recessive allele in F\(_1\), while the 3:1 ratio in F\(_2\) results from segregation.


Thus, the correct answer is option (4), as A, C, D, and E explain Mendel’s Law of Dominance. Quick Tip: Mendel’s Law of Dominance states that one allele is dominant over the other, and only the dominant trait is expressed in the F\(_1\) generation.

Step 1: A foreign DNA fragment can integrate into the recipient’s genome through recombination or be maintained as an episome.

Step 2: Statement B is correct because in genetic engineering, the transferred DNA can integrate into the host genome.

Step 3: Statement C is also correct since, once integrated, the gene of interest can be replicated and inherited with the host’s DNA.

Step 4: Statements A, D, and E are incorrect as free DNA fragments generally do not multiply independently unless they have an origin of replication.


Thus, the correct answer is option (1), as B and C explain the fate of transferred DNA. Quick Tip: Foreign DNA can integrate into the host genome or replicate as an episome in genetic engineering applications.

Step 1: The Calvin cycle is the light-independent reaction of photosynthesis responsible for CO\(_2\) fixation.

Step 2: Each molecule of CO\(_2\) fixed requires energy in the form of 3 ATP molecules and 2 NADPH molecules.

Step 3: These molecules are utilized in the reduction phase and regeneration of RuBP in the cycle.

Step 4: The overall cycle for producing one glucose molecule (C\(_6\)H\(_{12}\)O\(_6\)) requires 18 ATP and 12 NADPH molecules.


Thus, the correct answer is option (2), as 3 ATP and 2 NADPH are required per CO\(_2\) fixed. Quick Tip: The Calvin cycle requires 3 ATP and 2 NADPH per CO\(_2\) molecule fixed to synthesize carbohydrates.

Step 1: The method used to determine an unknown genotype is called a test cross.

Step 2: A test cross involves crossing the unknown dominant phenotype (BB or Bb) with a homozygous recessive (bb) individual.

Step 3: If the black seed plant is BB, all offspring will have black seeds (Bb).

Step 4: If the black seed plant is Bb, the offspring will have a 1:1 ratio of black (Bb) and white (bb) seeds.


Thus, the correct answer is option (4), as crossing with bb allows genotype determination based on offspring ratios. Quick Tip: A test cross is done with a homozygous recessive (bb) individual to determine whether a dominant phenotype is homozygous (BB) or heterozygous (Bb).

Step 1: Lecithin is a type of phospholipid that is widely found in cell membranes.

Step 2: It plays a crucial role in maintaining membrane fluidity and acts as an emulsifier in biological systems.

Step 3: Phospholipids are amphipathic molecules composed of a hydrophilic head and hydrophobic tails, making them essential components of cell membranes.

Step 4: Lecithin is commonly found in egg yolk, soybeans, and nerve tissues.


Thus, the correct answer is option (4), as lecithin belongs to the phospholipid category. Quick Tip: Lecithin is a phospholipid that helps in cell membrane structure and functions as an emulsifier in biological systems.

Step 1: \textit{Rhizopus is a bread mould fungus commonly found on stale bread (Matching A \(\to\) III).

Step 2: \textit{Ustilago is a smut fungus, which affects cereal crops like maize (Matching B \(\to\) II).

Step 3: \textit{Puccinia is a rust fungus, known for causing rust diseases in wheat and other crops (Matching C \(\to\) IV).

Step 4: \textit{Agaricus includes mushrooms, which are edible fungi belonging to Basidiomycetes (Matching D \(\to\) I).


Thus, the correct answer is option (3), as the correct matches are A-III, B-II, C-IV, D-I. Quick Tip: Fungi are classified based on their reproductive structures and pathogenic effects—Rhizopus (bread mould), Ustilago (smut), Puccinia (rust), and Agaricus (mushroom).

Step 1: Tropical regions show high species richness due to stable environmental conditions and longer evolutionary time.

Step 2: Statement A is correct as tropical regions have remained undisturbed for millions of years, allowing species diversification.

Step 3: Statement C is correct because higher solar energy in the tropics increases primary productivity, supporting diverse life forms.

Step 4: Statement D is valid as stable tropical climates promote niche specialization, allowing more species to coexist.

Step 5: Statement E is also correct because predictable tropical environments support species stability and survival.

Step 6: Statement B is incorrect because tropical environments are not more seasonal but rather stable compared to temperate regions.


Thus, the correct answer is option (3), as A, C, D, and E explain tropical species richness. Quick Tip: Tropical regions have high species richness due to evolutionary stability, abundant solar energy, and niche specialization.

Step 1: The nucleolus is responsible for active ribosomal RNA synthesis, making A \(\to\) III correct.

Step 2: The centriole has a cartwheel-like organization, making B \(\to\) II correct.

Step 3: Leucoplasts store nutrients like starch, lipids, and proteins, making C \(\to\) IV correct.

Step 4: The Golgi apparatus is involved in the formation of glycolipids and glycoproteins, making D \(\to\) I correct.


Thus, the correct answer is option (3), as A-III, B-II, C-IV, D-I are the correct matches. Quick Tip: The nucleolus synthesizes rRNA, centrioles have a cartwheel structure, leucoplasts store nutrients, and the Golgi apparatus forms glycolipids.

Step 1: Lactose transport in bacteria is facilitated by lactose permease, an enzyme encoded by the \textit{lacY gene in the lac operon.

Step 2: Lactose permease acts as a membrane transporter, enabling lactose to enter the bacterial cell.

Step 3: Beta-galactosidase (\textit{lacZ) is responsible for lactose hydrolysis, not transport. Acetylase (\textit{lacA) and polymerase do not play a role in lactose uptake.


Thus, the correct answer is option (1), as permease transports lactose into the bacterial cell. Quick Tip: Lactose permease enables lactose entry into bacterial cells, while beta-galactosidase hydrolyzes it into glucose and galactose.

Step 1: Leptotene stage is the first stage of prophase I in meiosis, where chromosomes gradually become visible under the light microscope. (Statement I is true)

Step 2: Diplotene stage is marked by the dissolution of the synaptonemal complex, leading to the formation of chiasmata. (Statement II is true)


Thus, the correct answer is option (3), as both statements are true. Quick Tip: During leptotene, chromosomes condense and become visible, while during diplotene, the synaptonemal complex dissolves, forming chiasmata.

Step 1: Dedifferentiation refers to the process where mature, differentiated cells regain the ability to divide and form meristematic tissues.

Step 2: Interfascicular cambium is formed when parenchyma cells dedifferentiate and regain meristematic activity.

Step 3: Differentiation refers to cells becoming specialized, while redifferentiation applies to cells that regain function after dedifferentiation. Maturation does not apply here.


Thus, the correct answer is option (1), as formation of interfascicular cambium is an example of dedifferentiation. Quick Tip: Dedifferentiation allows mature cells to revert to a meristematic state, helping in secondary growth in plants.

Step 1: Parenchyma and collenchyma are both living tissues in plants. Statement I is false because collenchyma is living, not dead.

Step 2: Gymnosperms lack xylem vessels and have only tracheids for water conduction, whereas angiosperms possess xylem vessels for efficient water transport. Statement II is true.


Thus, the correct answer is option (2), as Statement I is false but Statement II is true. Quick Tip: Parenchyma and collenchyma are both living tissues. Gymnosperms lack xylem vessels, while angiosperms have them for efficient conduction.

Step 1: Vallisneria is a water-pollinated plant, but its flowers are not colourful and do not produce nectar. This makes Statement A incorrect.

Step 2: Water lily is an insect-pollinated plant (entomophilous), not pollinated by water. Statement B is correct.

Step 3: In water-pollinated species (hydrophily), pollen grains are covered with a mucilaginous coat to prevent wetting. Statement C is correct.

Step 4: Pollen grains of certain hydrophytes, such as \textit{Zostera, are long and ribbon-like, helping them move in water. Statement D is correct.

Step 5: In some hydrophytes, pollen grains are passively carried inside water to reach the stigma. Statement E is correct.


Thus, the correct answer is option (2), as B, C, D, and E are correct. Quick Tip: Hydrophilous plants like \textit{Vallisneria and Zostera rely on water for pollination, with adaptations like mucilaginous pollen and ribbon-like grains.

Step 1: Actinomorphic flowers exhibit radial symmetry, meaning they can be divided into equal halves along multiple planes.

Step 2: Zygomorphic flowers exhibit bilateral symmetry, meaning they can only be divided into two equal halves along one plane.

Step 3: Datura is an actinomorphic flower, making option (3) correct.

Step 4: \textit{Pisum (\textit{pea), \textit{Sesbania, and \textit{Cassia are zygomorphic flowers, making them incorrect answers.


Thus, the correct answer is option (3), as \textit{Datura is actinomorphic. Quick Tip: Actinomorphic flowers like \textit{Datura have radial symmetry, while zygomorphic flowers like Pisum and Cassia have bilateral symmetry.

Step 1: Totipotency is the ability of a single plant cell to regenerate into an entire plant under appropriate conditions.

Step 2: This property is widely used in plant tissue culture to generate plants from explants.

Step 3: Differentiation refers to the process by which cells become specialized. Somatic hybridization involves fusing somatic cells, while micropropagation is a technique to propagate plants using tissue culture.


Thus, the correct answer is option (3), as totipotency enables the regeneration of a whole plant from a single cell. Quick Tip: Totipotency is the basis of plant tissue culture, allowing entire plants to be regenerated from a single cell.

Step 1: Bt toxin is produced by the bacterium \textit{Bacillus thuringiensis and is encoded by the cry gene. Statement I is correct.

Step 2: The Bt toxin exists as an inactive protoxin in \textit{B. thuringiensis, but it gets activated in the insect gut due to alkaline pH, not acidic pH. Statement II is incorrect.


Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: Bt toxin is a crystalline protein that remains inactive until activated in the alkaline pH of the insect gut, where it disrupts the midgut epithelium.

Step 1: The International Union for Conservation of Nature (IUCN) maintains the Red List, which classifies species based on their extinction risk.

Step 2: The IUCN Red List categories include Extinct (EX), Critically Endangered (CR), Endangered (EN), Vulnerable (VU), Near Threatened (NT), and Least Concern (LC).

Step 3: WWF (World Wildlife Fund) is an environmental organization but does not maintain the Red List. GEAC (Genetic Engineering Approval Committee) deals with genetically modified organisms, and FOAM is unrelated.


Thus, the correct answer is option (2), as IUCN is responsible for listing endangered species. Quick Tip: The IUCN Red List is the global authority on species conservation status, assessing their risk of extinction.

Step 1: Carboxypeptidase is a metalloenzyme involved in protein digestion, requiring zinc (Zn\(^{2+}\)) as a cofactor for its catalytic activity.

Step 2: The zinc ion in carboxypeptidase helps in stabilizing the enzyme-substrate complex and aids in hydrolysis of peptide bonds.

Step 3: Flavin is associated with flavoproteins in oxidation-reduction reactions, haem is found in hemoglobin and cytochromes, and niacin is a precursor for NAD\(^+\)/NADP\(^+\) coenzymes.


Thus, the correct answer is option (3), as zinc is the cofactor for carboxypeptidase. Quick Tip: Carboxypeptidase is a zinc-dependent metalloenzyme involved in the hydrolysis of peptide bonds during protein digestion.

Step 1: The citric acid cycle (Krebs cycle) occurs in the mitochondrial matrix. (Matching A \(\to\) II).

Step 2: Glycolysis is the breakdown of glucose and takes place in the cytoplasm. (Matching B \(\to\) I).

Step 3: The electron transport system (ETS) is located in the inner mitochondrial membrane. (Matching C \(\to\) IV).

Step 4: The proton gradient required for ATP synthesis is established in the intermembrane space of mitochondria. (Matching D \(\to\) III).


Thus, the correct answer is option (4), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: The citric acid cycle occurs in the mitochondrial matrix, glycolysis in the cytoplasm, ETS in the inner mitochondrial membrane, and the proton gradient in the intermembrane space.

Step 1: DNA replication in E. coli follows the 5' \(\to\) 3' polymerization rule.

Step 2: DNA-dependent DNA polymerase adds nucleotides in the 5' \(\to\) 3' direction using a DNA template.

Step 3: The enzyme also has 3' \(\to\) 5' exonuclease activity, but polymerization does not occur in this direction.

Step 4: Option (1) is incorrect because DNA polymerase cannot catalyze polymerization in both directions.

Step 5: Option (3) is incorrect because polymerization does not occur in the 3' \(\to\) 5' direction. Option (4) is incorrect because RNA polymerase synthesizes RNA, not DNA.


Thus, the correct answer is option (2), as DNA polymerase catalyzes DNA synthesis in the 5' \(\to\) 3' direction. Quick Tip: DNA polymerase in \textit{E. coli synthesizes new DNA strands in the 5' \(\to\) 3' direction while also having 3' \(\to\) 5' exonuclease proofreading activity.

Step 1: Robert May estimated global species diversity to be around 7 million, making (A \(\to\) III) correct.

Step 2: Alexander von Humboldt proposed the Species-Area relationship, which describes how species richness increases with increasing habitat area, making (B \(\to\) I) correct.

Step 3: Paul Ehrlich gave the Rivet Popper Hypothesis, which compares species in an ecosystem to rivets in an airplane, emphasizing the importance of species conservation, making (C \(\to\) IV) correct.

Step 4: David Tilman conducted long-term ecosystem experiments using outdoor plots to study biodiversity and ecosystem functioning, making (D \(\to\) II) correct.


Thus, the correct answer is option (4), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Humboldt described the species-area relationship, May estimated global species diversity, Ehrlich proposed the rivet popper hypothesis, and Tilman conducted long-term ecosystem experiments.

Wind-pollinated plants typically have exposed stamens and flowers that are adapted to release pollen into the air. This type of inflorescence is designed to facilitate wind pollination.

Thus, the correct answer is
(3) Wind-pollinated plant inflorescence showing flowers with well-exposed stamens. Quick Tip: Wind-pollinated flowers often have exposed stamens and reduced petals, facilitating the dispersal of pollen by wind.

Step 1: The tricarboxylic acid (TCA) cycle, also known as the Krebs cycle, involves multiple oxidation reactions.

Step 2: The conversion of Succinyl-CoA to Succinic acid is catalyzed by Succinyl-CoA synthetase, where substrate-level phosphorylation occurs instead of oxidation. (Correct step)

Step 3: The reactions Isocitrate to \(\alpha\)-ketoglutarate and Malate to Oxaloacetate involve NAD\(^+\)-dependent oxidation, making them incorrect options.

Step 4: The step Succinic acid to Malic acid involves oxidation via FAD-dependent dehydrogenation, making it incorrect.


Thus, the correct answer is option (1), as Succinyl-CoA to Succinic acid is a phosphorylation step, not an oxidation step. Quick Tip: The conversion of Succinyl-CoA to Succinic acid in the Krebs cycle is a substrate-level phosphorylation step, unlike other oxidation reactions.

Step 1: C\(_3\) plants exhibit photorespiration, where RuBisCO binds to O\(_2\) instead of CO\(_2\), reducing CO\(_2\) fixation efficiency. Statement I is correct.

Step 2: C\(_4\) plants have an adaptation to minimize photorespiration using the Hatch-Slack pathway, but both mesophyll and bundle sheath cells contribute to the process. Statement II is incorrect because mesophyll cells do not show high levels of photorespiration.


Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: C\(_3\) plants undergo photorespiration due to RuBisCO's oxygenase activity, while C\(_4\) plants minimize it using a spatial separation of CO\(_2\) fixation.

Step 1: In an ecosystem, energy transfer follows the 10 law, meaning only 10 of energy is transferred to the next trophic level, while 90 is lost as heat.


Step 2: Given that NPP of the first trophic level is 100\(x\) kcal m\(^{-2}\) yr\(^{-1}\), the second trophic level would receive 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\).

Step 3: The third trophic level would receive 10 of 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\), which equals 1\(x\) kcal m\(^{-2}\) yr\(^{-1}\).


Thus, the correct answer is option (1), as the GPP of the third trophic level is 10\(x\) kcal m\(^{-2}\) yr\(^{-1}\). Quick Tip: According to Lindeman's 10 law, only 10 of energy is transferred to the next trophic level, while 90 is lost as heat.

Step 1: GLUT-4 (Glucose Transporter 4) is responsible for glucose transport into cells in response to insulin. (Matching A \(\to\) IV).

Step 2: Insulin is a hormone that regulates blood glucose levels. (Matching B \(\to\) I).

Step 3: Trypsin is an enzyme involved in protein digestion in the small intestine. (Matching C \(\to\) II).

Step 4: Collagen is a structural protein found in the extracellular matrix and provides support. (Matching D \(\to\) III).


Thus, the correct answer is option (3), as the correct matches are A-IV, B-I, C-II, D-III. Quick Tip: GLUT-4 enables glucose uptake, insulin is a hormone, trypsin is a digestive enzyme, and collagen provides structural support in tissues.

Step 1: Frederick Griffith discovered the process of transformation in bacteria using the pneumococcus experiment, making (A \(\to\) III) correct.

Step 2: Francois Jacob and Jacque Monod proposed the Lac operon model, which explains gene regulation in prokaryotes, making (B \(\to\) IV) correct.

Step 3: Har Gobind Khorana played a significant role in deciphering the genetic code, making (C \(\to\) I) correct.

Step 4: Meselson and Stahl experimentally proved the semi-conservative mode of DNA replication in \textit{E. coli, making (D \(\to\) II) correct.


Thus, the correct answer is option (4), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Griffith discovered transformation, Jacob and Monod explained the \textit{Lac operon, Khorana decoded the genetic code, and Meselson \& Stahl demonstrated semi-conservative replication.

Step 1: Chloroplast DNA (cpDNA) resembles prokaryotic DNA in structure and organization.

Step 2: It is circular and double-stranded, similar to mitochondrial DNA (mtDNA) and bacterial DNA.

Step 3: Unlike nuclear DNA, which is linear, chloroplast DNA does not have histones and exists in multiple copies within the organelle.


Thus, the correct answer is option (4), as chloroplast DNA is circular and double-stranded. Quick Tip: Chloroplast DNA (cpDNA) is circular, double-stranded, and resembles prokaryotic DNA, supporting the endosymbiotic theory.

Step 1: Gibberellins (GA) are plant hormones that promote stem elongation, seed germination, and flowering.

Step 2: In sugarcane, gibberellins stimulate internode elongation, leading to an increase in yield.

Step 3: Cytokinins promote cell division, auxins regulate root growth, and abscisic acid induces dormancy rather than elongation.


Thus, the correct answer is option (4), as gibberellin increases stem length in sugarcane. Quick Tip: Gibberellins promote internode elongation in sugarcane, increasing yield by enhancing stem growth.

Step 1: Phaeophyceae (Brown Algae) reproduce asexually via biflagellate zoospores. (Statement A is correct)

Step 2: Sexual reproduction in Phaeophyceae can be isogamous, anisogamous, or oogamous, not just oogamous. (Statement B is incorrect)

Step 3: The stored food in Phaeophyceae is mannitol or laminarin. (Statement C is correct)

Step 4: Their pigments include chlorophyll a, c, carotenoids, and xanthophylls. (Statement D is correct)

Step 5: Their vegetative cells have a cellulosic wall covered by algin, a gelatinous coating. (Statement E is correct)


Thus, the correct answer is option (1), as A, C, D, and E are correct. Quick Tip: Phaeophyceae store food as laminarin or mannitol and reproduce via biflagellate zoospores, with a cell wall covered in algin.

Step 1: Somatic hybridization is a technique where protoplasts from different plant varieties are fused to produce a hybrid cell.

Step 2: Protoplast fusion is achieved using PEG (Polyethylene Glycol) or electrofusion, leading to the formation of somatic hybrids.

Step 3: Pollens are involved in sexual reproduction, not somatic hybridization. Callus is a mass of undifferentiated cells, and somatic embryos arise from tissue culture but are not fused in somatic hybridization.


Thus, the correct answer is option (1), as protoplast fusion is the key step in somatic hybridization. Quick Tip: Somatic hybridization involves protoplast fusion, enabling genetic recombination between two plant varieties.

Step 1: Rose exhibits perigynous flowers, where the ovary is surrounded by floral parts at the same level. (Matching A \(\to\) II)

Step 2: Pea has marginal placentation, where ovules are attached along one side of the ovary. (Matching B \(\to\) IV)

Step 3: Cotton has twisted aestivation, where one petal overlaps another. (Matching C \(\to\) I)

Step 4: Mango is a drupe fruit, with a hard endocarp protecting the seed. (Matching D \(\to\) III)


Thus, the correct answer is option (3), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Rose has perigynous flowers, pea has marginal placentation, cotton shows twisted aestivation, and mango is a drupe fruit.

Step 1: Monoadelphous stamens have filaments fused into a single bundle, seen in China-rose (Hibiscus). (Matching A \(\to\) IV)

Step 2: Diadelphous stamens have filaments arranged in two bundles, as in Pea (Fabaceae family). (Matching B \(\to\) II)

Step 3: Polyadelphous stamens have multiple groups of fused filaments, as in Citrus. (Matching C \(\to\) I)

Step 4: Epiphyllous stamens are attached to petals, seen in Lily (Liliaceae family). (Matching D \(\to\) III)


Thus, the correct answer is option (3), as the correct matches are A-IV, B-II, C-I, D-III. Quick Tip: Monoadelphous stamens (one bundle) are found in China-rose, diadelphous (two bundles) in Pea, polyadelphous (multiple bundles) in Citrus, and epiphyllous (attached to petals) in Lily.

Step 1: Common cold is caused by Rhinoviruses. (Matching A \(\to\) III)

Step 2: Haemozoin is a malarial pigment produced by Plasmodium during its life cycle. (Matching B \(\to\) I)

Step 3: Widal test is a diagnostic test for Typhoid, caused by \textit{Salmonella typhi. (Matching C \(\to\) II)

Step 4: Allergies are triggered by allergens such as dust mites. (Matching D \(\to\) IV)


Thus, the correct answer is option (1), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: Rhinoviruses cause the common cold, \textit{Plasmodium produces haemozoin, Widal test diagnoses typhoid, and dust mites trigger allergies.

Step 1: Convergent evolution occurs when unrelated organisms evolve similar traits due to similar environmental pressures.

Step 2: Penguins (birds) and Dolphins (mammals) belong to different groups but have flippers adapted for swimming, an example of convergent evolution.

Step 3: Divergent evolution leads to different traits in related species, adaptive radiation involves diversification from a common ancestor, and natural selection drives evolution but does not always lead to convergence.


Thus, the correct answer is option (1), as flippers in Penguins and Dolphins evolved due to convergent evolution. Quick Tip: Convergent evolution leads to similar structures in unrelated species due to similar environmental pressures, as seen in Penguins and Dolphins.

Step 1: The correct sequence of human evolution is:
- Homo habilis (earliest tool users) → \textit{Homo erectus (upright walkers) →
- \textit{Homo neanderthalensis (Neanderthals) → \textit{Homo sapiens (modern humans).

Step 2: \textit{Homo habilis was the earliest species, followed by \textit{Homo erectus, then \textit{Homo neanderthalensis, and finally \textit{Homo sapiens.


Thus, the correct answer is option (2), as A-D-C-B represents the correct evolutionary order. Quick Tip: The sequence of human evolution follows: \textit{Homo habilis → Homo erectus → Homo neanderthalensis → Homo sapiens.

Step 1: The Hardy-Weinberg equilibrium states that allele frequencies remain constant in a population under stable conditions.

Step 2: Gene migration, genetic recombination, and genetic drift cause changes in allele frequencies, disturbing equilibrium.

Step 3: A constant gene pool means no changes in allele frequencies, keeping the population in equilibrium.


Thus, the correct answer is option (2), as a constant gene pool maintains Hardy-Weinberg equilibrium. Quick Tip: Hardy-Weinberg equilibrium remains stable if the gene pool is constant, meaning no evolutionary forces act on the population.

Step 1: Oxyhaemoglobin (HbO\(_2\)) formation occurs in the alveoli, where oxygen binds to hemoglobin.

Step 2: High pO\(_2\) (partial pressure of oxygen) favors oxygen binding to hemoglobin.

Step 3: Low pCO\(_2\) and lesser H\(^+\) concentration create an alkaline environment, promoting HbO\(_2\) formation.

Step 4: High pCO\(_2\) and high H\(^+\) concentration (acidic environment) promote oxygen release (Bohr effect), making options (1) and (3) incorrect.


Thus, the correct answer is option (4), as high pO\(_2\) and low H\(^+\) concentration favor oxyhaemoglobin formation. Quick Tip: In alveoli, high pO\(_2\) and lower H\(^+\) concentration favor oxyhaemoglobin formation, while high pCO\(_2\) promotes oxygen release.

Step 1: Natural contraceptive methods involve preventing pregnancy without external devices or hormones.

Step 2: Lactational amenorrhea, coitus interruptus, and periodic abstinence are natural contraceptive methods.

Step 3: Vaults are barrier contraceptives (diaphragms), which physically prevent sperm entry, making them a non-natural method.


Thus, the correct answer is option (2), as vaults are not a natural contraceptive method. Quick Tip: Natural contraceptive methods rely on behavioral practices, while vaults (diaphragms) are physical barriers used for contraception.

Step 1: Pons acts as a bridge connecting different brain regions. (Matching A \(\to\) III)

Step 2: Hypothalamus contains neurosecretory cells, playing a key role in hormonal regulation. (Matching B \(\to\) IV)

Step 3: Medulla oblongata controls respiration and gastric secretions. (Matching C \(\to\) II)

Step 4: Cerebellum regulates posture, balance, and coordination. (Matching D \(\to\) I)


Thus, the correct answer is option (4), as the correct matches are A-III, B-IV, C-II, D-I. Quick Tip: The cerebellum controls balance, the medulla regulates involuntary functions, the hypothalamus has neurosecretory cells, and the pons connects brain regions.

Step 1: The hymen can be torn due to various non-sexual activities like sports, cycling, or physical exertion. Statement I is true.

Step 2: The belief that the hymen tears only during the first coitus is incorrect, as it may already be absent due to other factors. Statement II is false.


Thus, the correct answer is option (1), as Statement I is true but Statement II is false. Quick Tip: The hymen is not a definitive indicator of virginity, as it can be torn due to various physical activities unrelated to sexual intercourse.

Step 1: Axoneme is the structural core of cilia and flagella, consisting of microtubules. (Matching A \(\to\) II)

Step 2: Cartwheel pattern is a feature of centrioles, seen in their early formation stage. (Matching B \(\to\) I)

Step 3: Crista are the folds of the inner mitochondrial membrane, increasing the surface area for ATP production. (Matching C \(\to\) IV)

Step 4: Satellite DNA refers to repetitive DNA sequences found in chromosomes. (Matching D \(\to\) III)


Thus, the correct answer is option (2), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Axoneme forms the core of cilia and flagella, cartwheel pattern is found in centrioles, cristae are folds of mitochondria, and satellite DNA is a part of chromosomes.

Step 1: Typhoid is caused by the bacterium \textit{Salmonella typhi. (Matching A \(\to\) IV)

Step 2: Leishmaniasis is caused by the protozoan parasite \textit{Leishmania donovani. (Matching B \(\to\) III)

Step 3: Ringworm is a fungal infection caused by \textit{Trichophyton, \textit{Microsporum, or \textit{Epidermophyton. (Matching C \(\to\) I)

Step 4: Filariasis is caused by nematode worms such as \textit{Wuchereria bancrofti. (Matching D \(\to\) II)


Thus, the correct answer is option (4), as the correct matches are A-IV, B-III, C-I, D-II. Quick Tip: Typhoid is caused by bacteria, Leishmaniasis by protozoa, ringworm by fungi, and filariasis by nematodes.

Step 1: The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes. Thus, Statement I is false.

Step 2: The proximal convoluted tubule (PCT) is lined by simple cuboidal brush border epithelium, not columnar epithelium. Thus, Statement II is false.


Thus, the correct answer is option (4), as both Statement I and Statement II are false. Quick Tip: The descending limb of the loop of Henle is water-permeable, while the proximal convoluted tubule has simple cuboidal brush border epithelium for increased absorption.

Step 1: \(\alpha\)-1 antitrypsin is used in the treatment of emphysema. (Matching A \(\to\) III)

Step 2: Cry IAb is a Bt toxin that is effective against corn borers. (Matching B \(\to\) IV)

Step 3: Cry IAc is another Bt toxin, specifically targeting cotton bollworm. (Matching C \(\to\) I)

Step 4: Enzyme replacement therapy is used for ADA deficiency, a genetic disorder affecting the immune system. (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: \(\alpha\)-1 antitrypsin treats emphysema, Cry IAb targets corn borers, Cry IAc affects cotton bollworms, and enzyme replacement therapy helps in ADA deficiency.

Step 1: Non-medicated IUDs include Lippes loop. (Matching A \(\to\) III)

Step 2: Copper-releasing IUDs include Multiload 375. (Matching B \(\to\) I)

Step 3: Hormone-releasing IUDs include LNG-20. (Matching C \(\to\) IV)

Step 4: Implants release progestogens, acting as long-term contraceptives. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Non-medicated IUDs include Lippes loop, Copper IUDs include Multiload 375, hormone-releasing IUDs include LNG-20, and implants release progestogens.

Step 1: Annelids (e.g., earthworms) have a true coelom, which is mesodermally derived. Statement A is correct.

Step 2: Poriferans (sponges) lack body cavities altogether, making Statement B incorrect.

Step 3: Aschelminthes (Nematodes) are pseudocoelomates, not acoelomates, making Statement C incorrect.

Step 4: Platyhelminthes (Flatworms) are acoelomates, not pseudocoelomates, making Statement D incorrect.


Thus, the correct answer is option (4), as only Statement A is correct. Quick Tip: Annelids have a true coelom, Poriferans lack body cavities, Aschelminthes are pseudocoelomates, and Platyhelminthes are acoelomates.

Step 1: Down’s syndrome is caused by trisomy of the 21\(^{st}\) chromosome. (Matching A \(\to\) III)

Step 2: \(\alpha\)-Thalassemia is associated with mutations in the 16\(^{th}\) chromosome. (Matching B \(\to\) IV)

Step 3: \(\beta\)-Thalassemia is caused by mutations in the 11\(^{th}\) chromosome. (Matching C \(\to\) I)

Step 4: Klinefelter’s syndrome occurs due to the presence of an extra ‘X’ chromosome (XXY). (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Down’s syndrome is due to trisomy 21, \(\alpha\)-Thalassemia is linked to chromosome 16, \(\beta\)-Thalassemia to chromosome 11, and Klinefelter’s syndrome to the ‘X’ chromosome.

Step 1: The conduction pathway of the heart follows this sequence:
- SA node initiates the impulse. (E)
- AV node delays the impulse slightly. (C)
- AV bundle (Bundle of His) carries the signal to the ventricles. (A)
- Bundle branches transmit impulses through the ventricles. (D)
- Purkinje fibers distribute the impulse, leading to contraction. (B)

Thus, the correct answer is option (3), as the correct sequence is E-C-A-D-B. Quick Tip: The conduction pathway of the heart: SA node → AV node → AV bundle → Bundle branches → Purkinje fibers.

Step 1: Lipase breaks down lipids by hydrolyzing ester bonds. (Matching A \(\to\) II)

Step 2: Nuclease hydrolyzes nucleic acids by breaking phosphodiester bonds in DNA/RNA. (Matching B \(\to\) IV)

Step 3: Protease digests proteins by hydrolyzing peptide bonds between amino acids. (Matching C \(\to\) I)

Step 4: Amylase breaks down polysaccharides (starch) by hydrolyzing glycosidic bonds. (Matching D \(\to\) III)


Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Lipase hydrolyzes ester bonds in fats, nucleases break phosphodiester bonds in DNA/RNA, proteases cleave peptide bonds in proteins, and amylase breaks glycosidic bonds in carbohydrates.

The vector pBR322 contains the X and Y genes.

- Gene X regulates the copy number of the plasmid DNA.

- Gene Y encodes a protein involved in the replication of the plasmid. This ensures that the plasmid can replicate inside the host cell.


Thus, the correct answer is (1) The gene ‘X’ is responsible for controlling the copy number of the linked DNA and ‘Y’ for protein involved in the replication of Plasmid. Quick Tip: In plasmid vectors like pBR322, genes involved in replication control the number of copies of the plasmid within the host cell.

Step 1: The Ti plasmid is a tumor-inducing plasmid found in Agrobacterium tumefaciens, a plant pathogen.

Step 2: It is responsible for causing crown gall disease in dicot plants.

Step 3: The T-DNA (transfer DNA) segment of the Ti plasmid integrates into the host genome, leading to uncontrolled cell division.


Thus, the correct answer is option (1), as the Ti plasmid stands for Tumor Inducing plasmid. Quick Tip: The Ti plasmid of \textit{Agrobacterium tumefaciens is widely used in plant genetic engineering due to its ability to transfer genes into plants.

Step 1: Pleurobrachia belongs to Ctenophora (comb jellies). (Matching A \(\to\) II)

Step 2: Radula is a rasping organ found in Mollusca for feeding. (Matching B \(\to\) I)

Step 3: Stomachord is a structure found in Hemichordates. (Matching C \(\to\) IV)

Step 4: Air bladder is found in Osteichthyes (bony fish) and helps in buoyancy. (Matching D \(\to\) III)


Thus, the correct answer is option (4), as the correct matches are A-II, B-I, C-IV, D-III. Quick Tip: Pleurobrachia belongs to Ctenophora, radula is a feeding organ in Mollusca, stomachord is found in Hemichordata, and air bladder helps in buoyancy in bony fishes.

Step 1: Bio-reactors are used for large-scale production of biological products, not small-scale. Statement (1) is incorrect.

Step 2: They contain an agitator system, oxygen delivery system, and foam control system to maintain optimal growth conditions. Statements (2), (3), and (4) are correct.


Thus, the correct answer is option (1), as bio-reactors are used for large-scale production, not small-scale. Quick Tip: Bio-reactors are used for large-scale production of enzymes, antibiotics, and recombinant proteins, not small-scale bacterial cultures.

Step 1: DNA-dependent RNA polymerase synthesizes mRNA complementary to the DNA template.

Step 2: The given template strand (3’→5’ direction) results in an mRNA strand in the 5’→3’ direction.


Thus, the correct answer is option (3), as the correct mRNA sequence is 5’ AUGUACCGUUUUAUAGGAAGU 3’. Quick Tip: DNA-dependent RNA polymerase synthesizes mRNA complementary to the template strand, replacing thymine (T) with uracil (U).

Step 1: Cocaine is derived from Erythroxylum coca. (Matching A \(\to\) III)

Step 2: Heroin is synthesized from morphine, which is obtained from \textit{Papaver somniferum (opium poppy). (Matching B \(\to\) IV)

Step 3: Morphine is an effective sedative used in surgery. (Matching C \(\to\) I)

Step 4: Marijuana is derived from \textit{Cannabis sativa. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Cocaine is derived from \textit{Erythroxylum coca, heroin from Papaver somniferum, morphine is a sedative, and marijuana comes from Cannabis sativa.

Step 1: Diakinesis is the final stage of prophase I, where chiasmata terminalization is completed. (Matching A \(\to\) II)

Step 2: Pachytene is the stage where recombination nodules appear. (Matching B \(\to\) IV)

Step 3: Zygotene is characterized by synaptonemal complex formation. (Matching C \(\to\) I)

Step 4: Leptotene is the first stage where chromosomes appear as thin threads. (Matching D \(\to\) III)


Thus, the correct answer is option (1), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Prophase I stages: Leptotene (thin chromosomes), Zygotene (synaptonemal complex), Pachytene (crossing over), Diakinesis (chiasmata terminalization).

Step 1: Cockroaches possess anal cerci, which are jointed filamentous structures that function as sensory organs.

Step 2: These cerci are present in both male and female cockroaches on the 10\(^{th}\) abdominal segment.

Step 3: They help in detecting vibrations and movements in the surrounding environment.


Thus, the correct answer is option (4), as anal cerci are present on the 10\(^{th}\) segment of the cockroach. Quick Tip: Anal cerci in cockroaches are located on the 10\(^{th}\) abdominal segment and serve as sensory organs for detecting movement and vibrations.

Step 1: Breastfeeding is recommended as it provides essential nutrients and immunity to the infant. Assertion A is correct.

Step 2: Colostrum, the first milk produced after birth, is rich in maternal antibodies (IgA, IgG, IgM), which help develop immunity in the newborn. Reason R is correct.

Step 3: Since colostrum provides antibodies essential for infant immunity, it directly supports the reason given in Assertion A.


Thus, the correct answer is option (3), as both A and R are correct, and R is the correct explanation of A. Quick Tip: Colostrum is rich in maternal antibodies and helps develop passive immunity in newborns, making breastfeeding highly beneficial.

Step 1: Pterophyllum is commonly known as the Angel fish. (Matching A \(\to\) III)

Step 2: \textit{Myxine is commonly known as the Hag fish, which is a jawless fish. (Matching B \(\to\) I)

Step 3: \textit{Pristis is known as the Saw fish, characterized by its elongated snout with teeth. (Matching C \(\to\) II)

Step 4: \textit{Exocoetus is called the Flying fish, as it can glide over water surfaces. (Matching D \(\to\) IV)


Thus, the correct answer is option (4), as the correct matches are A-III, B-I, C-II, D-IV. Quick Tip: \textit{Pterophyllum (Angel fish), Myxine (Hag fish - jawless), Pristis (Saw fish), and Exocoetus (Flying fish) are well-known aquatic species.

Step 1: The three major types of muscles in the human body are:
- Skeletal muscle (voluntary and striated, found in biceps, triceps, etc.).
- Smooth muscle (involuntary, non-striated, found in internal organs like the stomach and intestines).
- Cardiac muscle (striated, involuntary, found in the heart).

Step 2:
- Image (a) represents skeletal muscle, found in triceps.
- Image (b) represents smooth muscle, found in stomach.
- Image (c) represents cardiac muscle, found in heart.

Thus, the correct answer is option (4), as (a) Skeletal – Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart. Quick Tip: Skeletal muscles are voluntary and striated, smooth muscles are involuntary and non-striated, and cardiac muscles are striated but involuntary.

Step 1: Steroid hormones are derived from cholesterol and include sex hormones and adrenal cortex hormones.

Step 2:
- Progesterone (option 1) is a steroid hormone involved in pregnancy.
- Cortisol (option 3) is a steroid hormone secreted by the adrenal cortex.
- Testosterone (option 4) is a steroid hormone responsible for male secondary sexual characteristics.
- Glucagon (option 2) is a peptide hormone, not a steroid. It is secreted by the pancreas and regulates blood glucose levels.

Thus, the correct answer is option (2), as Glucagon is a peptide hormone, not a steroid hormone. Quick Tip: Steroid hormones are derived from cholesterol (e.g., testosterone, cortisol, progesterone), while peptide hormones (e.g., glucagon, insulin) are made of amino acids.

Step 1: Fibrous joints (A) are immovable and found in the skull (Matching A \(\to\) III).

Step 2: Cartilaginous joints (B) provide limited movement and are found in the vertebral column (Matching B \(\to\) I).

Step 3: Hinge joints (C) allow movement in one plane and are found in the knee (Matching C \(\to\) IV).

Step 4: Ball and socket joints (D) allow rotational movement and are found in the humerus and pectoral girdle (Matching D \(\to\) II).


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Fibrous joints (skull - no movement), Cartilaginous joints (vertebral column - limited movement), Hinge joints (knee - unidirectional), Ball and socket joints (shoulder - rotational).

Step 1: FSH (Follicle Stimulating Hormone) acts on Sertoli cells, not Leydig cells, in males. It regulates spermatogenesis. Thus, Assertion A is false.

Step 2: Leydig cells in males produce androgens (testosterone). In females, growing ovarian follicles secrete estrogen. Thus, Reason R is correct.


Thus, the correct answer is option (2), as Assertion A is incorrect, but Reason R is correct. Quick Tip: FSH acts on Sertoli cells (not Leydig cells), while growing ovarian follicles secrete estrogen, and Leydig cells secrete testosterone.

Step 1: Expiratory capacity (A) is the sum of tidal volume + expiratory reserve volume. (Matching A \(\to\) II)

Step 2: Functional residual capacity (B) is the sum of expiratory reserve volume + residual volume. (Matching B \(\to\) IV)

Step 3: Vital capacity (C) is the sum of tidal volume + inspiratory reserve volume + expiratory reserve volume. (Matching C \(\to\) I)

Step 4: Inspiratory capacity (D) is the sum of tidal volume + inspiratory reserve volume. (Matching D \(\to\) III)


Thus, the correct answer is option (3), as the correct matches are A-II, B-IV, C-I, D-III. Quick Tip: Expiratory capacity = TV + ERV, Functional residual capacity = ERV + RV, Vital capacity = TV + IRV + ERV, Inspiratory capacity = TV + IRV.

Step 1: The correct sequence of cell cycle stages is:
- G1 phase (E): Cell growth occurs.
- S phase (C): DNA replication happens.
- G2 phase (A): Preparation for mitosis.
- Karyokinesis (D): Nuclear division occurs.
- Cytokinesis (B): Cytoplasm divides, forming two daughter cells.

Thus, the correct answer is option (2), as the correct order is E-C-A-D-B. Quick Tip: The cell cycle follows the order: G1 → S → G2 → M (Karyokinesis + Cytokinesis).

Step 1: Autoimmune disorders occur when the immune system attacks its own body tissues.

Step 2:
- Myasthenia gravis (A): Autoimmune attack on neuromuscular junctions.
- Rheumatoid arthritis (B): Autoimmune inflammation of joints.
- SLE (E): A systemic autoimmune disease affecting multiple organs.
- Gout (C) is due to uric acid crystal accumulation (not autoimmune).
- Muscular dystrophy (D) is a genetic disorder, not autoimmune.

Thus, the correct answer is option (4), as A, B, and E are autoimmune disorders. Quick Tip: Autoimmune diseases include Myasthenia gravis, Rheumatoid arthritis, and Systemic Lupus Erythematosus (SLE), but not gout or muscular dystrophy.

Step 1: The Fallopian tube consists of four parts:
- Infundibulum (1): Funnel-shaped structure with fimbriae.
- Ampulla (2): The widest and longest part, where fertilization occurs.
- Isthmus (4): A narrow region connecting to the uterus.

Step 2: The Uterine fundus (3) is the uppermost part of the uterus, not the Fallopian tube.


Thus, the correct answer is option (3), as the uterine fundus is not part of the Fallopian tube. Quick Tip: The Fallopian tube consists of Infundibulum, Ampulla, Isthmus, and the Uterine part, but not the Uterine fundus.

Step 1: Blood type inheritance follows the Mendelian principles of codominance. The possible genotypes are:
- Blood group A: I\(^A\)I\(^A\) or I\(^A\)i
- Blood group B: I\(^B\)I\(^B\) or I\(^B\)i
- Blood group O: ii

Step 2: Since the child has O\(^-\) blood, they must have received the ii genotype (one i from each parent).

Step 3:
- The father’s genotype must be I\(^B\)i or I\(^B\)I\(^B\).
- The mother’s genotype must be I\(^A\)i or I\(^A\)I\(^A\).
- The only correct parental genotype pair that produces an O\(^-\) child is A (I\(^B\)i/I\(^B\)i) & D (I\(^A\)i/I\(^B\)i).

Thus, the correct answer is option (3), as A & D represent the correct possible genotypes. Quick Tip: For a child with O blood type, both parents must contribute an "i" allele (i.e., they must be heterozygous A or B).

Step 1: Exophthalmic goiter is due to hyperthyroidism, leading to protruding eyeballs. (Matching A \(\to\) III)

Step 2: Acromegaly occurs due to excess secretion of growth hormone. (Matching B \(\to\) IV)

Step 3: Cushing’s syndrome is due to excess cortisol secretion, leading to moon face and hyperglycemia. (Matching C \(\to\) I)

Step 4: Cretinism results from thyroid hormone deficiency, leading to stunted growth. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Exophthalmic goiter (hyperthyroidism), Acromegaly (excess GH), Cushing’s syndrome (excess cortisol), and Cretinism (thyroid hormone deficiency).

Step 1: Mesozoic Era is known as the "Age of Reptiles and Birds". (Matching A \(\to\) III)

Step 2: Proterozoic Era had the first lower invertebrates. (Matching B \(\to\) I)

Step 3: Cenozoic Era is known as the "Age of Mammals". (Matching C \(\to\) IV)

Step 4: Paleozoic Era was the period of early Fish and Amphibia. (Matching D \(\to\) II)


Thus, the correct answer is option (2), as the correct matches are A-III, B-I, C-IV, D-II. Quick Tip: Mesozoic Era (Reptiles \& Birds), Proterozoic Era (Lower invertebrates), Cenozoic Era (Mammals), Paleozoic Era (Fish \& Amphibians).

Step 1: Non-chordates lack notochord (B), post-anal tail (E), and have a dorsal heart (D).

Step 2: Chordates have pharyngeal gill slits (A) and a dorsal CNS (C), so these statements do not apply to non-chordates.

Step 3: Since non-chordates lack notochord, post-anal tail, and have a dorsal heart, the correct answer is B, D, and E only.

Thus, the correct answer is option (1), as B, D, and E are true for non-chordates. Quick Tip: Non-chordates lack a notochord and post-anal tail, while their heart (if present) is dorsal. Chordates have pharyngeal gill slits and a dorsal CNS.

Step 1: The corpus callosum is a bundle of nerve fibers that connects the left and right cerebral hemispheres, making Statement I correct.

Step 2: The brainstem consists of the medulla oblongata, pons, and midbrain (not cerebrum), making Statement II incorrect.


Thus, the correct answer is option (1), as Statement I is correct, but Statement II is incorrect. Quick Tip: The brainstem includes the medulla oblongata, pons, and midbrain but not the cerebrum. The corpus callosum connects the cerebral hemispheres.

Step 1: Unicellular glandular epithelium (A) consists of goblet cells in the alimentary canal. (Matching A \(\to\) III)

Step 2: Compound epithelium (B) lines moist surfaces like the buccal cavity. (Matching B \(\to\) IV)

Step 3: Multicellular glandular epithelium (C) includes salivary glands. (Matching C \(\to\) I)

Step 4: Endocrine glandular epithelium (D) includes the pancreas. (Matching D \(\to\) II)


Thus, the correct answer is option (1), as the correct matches are A-III, B-IV, C-I, D-II. Quick Tip: Unicellular epithelium (goblet cells), compound epithelium (moist surfaces), multicellular epithelium (salivary glands), endocrine epithelium (pancreas).

Step 1: P wave represents depolarisation of atria. (Matching A \(\to\) III)

Step 2: QRS complex represents depolarisation of ventricles. (Matching B \(\to\) II)

Step 3: T wave represents repolarisation of ventricles. (Matching C \(\to\) IV)

Step 4: T-P gap indicates electrical silence in heart muscles. (Matching D \(\to\) I)


Thus, the correct answer is option (4), as the correct matches are A-III, B-II, C-IV, D-I. Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).

Step 1: Both mitochondria and chloroplasts have double membranes. (Statement I is correct)

Step 2: The inner membrane of mitochondria is highly folded but more permeable than the chloroplast. (Statement II is incorrect)


Thus, the correct answer is option (1). Quick Tip: P wave (atrial depolarisation), QRS complex (ventricular depolarisation), T wave (ventricular repolarisation), T-P gap (electrical silence).

Step 1: Crop (A) is the structure used for storing food in cockroaches. (Matching A \(\to\) IV)

Step 2: Gastric Caeca (B) are the 6-8 blind tubules at the junction of foregut and midgut. (Matching B \(\to\) II)

Step 3: Malpighian tubules (C) are the yellow-colored filaments found at the junction of midgut and hindgut. (Matching C \(\to\) III)

Step 4: Gizzard (D) is used for grinding food. (Matching D \(\to\) I)


Thus, the correct answer is option (3), as the correct matches are A-IV, B-II, C-III, D-I. Quick Tip: The cockroach digestive system has the crop (storage), gastric caeca (digestion), malpighian tubules (excretion), and gizzard (grinding food).

Step 1: RNA polymerase III (A) synthesizes SnRNAs and tRNA. (Matching A \(\to\) IV)

Step 2: Termination of transcription (B) is regulated by the Rho factor. (Matching B \(\to\) II)

Step 3: Splicing of Exons (C) is associated with snRNPs. (Matching C \(\to\) I)

Step 4: The TATA box (D) is a promotor region that facilitates transcription. (Matching D \(\to\) III)


Thus, the correct answer is option (2), as the correct matches are A-IV, B-II, C-I, D-III. Quick Tip: RNA polymerase III produces tRNA and snRNAs. The TATA box is a core promoter for transcription initiation.

Step 1: FSH (Follicle-stimulating hormone) stimulates Sertoli cells, which support spermatogenesis.

Step 2: LH (Luteinizing hormone) stimulates Leydig cells, which secrete androgens (testosterone) necessary for spermiogenesis.


Thus, the correct answer is option (3), as FSH stimulates Sertoli cells and Leydig cells support spermiogenesis. Quick Tip: FSH acts on Sertoli cells (spermatogenesis), and LH acts on Leydig cells (spermiogenesis).

Step 1: Juxta medullary nephrons have a longer loop of Henle that extends deep into the renal medulla, aiding in concentration of urine. (Matching Option 1)

Step 2: Cortical nephrons are more numerous than juxta medullary nephrons.


Thus, the correct answer is option (1), as the loop of Henle in juxta medullary nephron extends deep into medulla. Quick Tip: Juxta medullary nephrons play a critical role in water reabsorption and urine concentration due to their long loop of Henle.

Step 1: Gause's competitive exclusion principle does not state that species with different resources cannot coexist indefinitely. It specifically states that closely related species competing for the same resources cannot coexist indefinitely. (Statement I is false)

Step 2: Statement II is true because, in cases of competition with limited resources, the inferior species is eliminated.


Thus, the correct answer is option (2), as Statement I is false, but Statement II is true. Quick Tip: Competitive exclusion occurs when two species competing for the same resources cannot coexist, and the inferior species is eliminated.

Step 1: Bone marrow is indeed the primary site for the production of all blood cells, including lymphocytes. (Statement I is correct)

Step 2: Both bone marrow and thymus play crucial roles in the development and maturation of T-lymphocytes. (Statement II is correct)


Thus, the correct answer is option (3), as both Statement I and Statement II are correct. Quick Tip: Bone marrow produces all blood cells, including lymphocytes. The thymus plays a key role in the maturation of T-lymphocytes.

Step 1: The enzyme action begins with substrate binding to the active site (E).

Step 2: This leads to substrate-enzyme complex formation (A).

Step 3: The enzyme is then ready to bind with another substrate (B).

Step 4: The chemical bonds of the substrate are broken (D) as the reaction progresses.

Step 5: Finally, the products are released (C).


Thus, the correct answer is option (3), as the correct sequence is E, A, B, C, D. Quick Tip: The enzyme catalytic cycle follows: Substrate binding → Substrate-enzyme complex → Enzyme ready for next substrate → Substrate broken → Product released.