Step 1: Understanding Transformer Voltage Ratio

In an ideal transformer, the voltage ratio is given by:
\[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \]
Given that \( \frac{N_P}{N_S} = \frac{1}{2} \), we can rewrite this as:
\[ N_S = 2N_P \]


Step 2: Applying the Formula

Substituting in the transformer equation:
\[ \frac{V_S}{V_P} = \frac{N_S}{N_P} = 2 \]
This implies that \( V_S : V_P = 2:1 \).


Conclusion:

Since the voltage ratio is \( \mathbf{2:1} \), the correct answer is \( \mathbf{(1)} \). Quick Tip: For an ideal transformer, the voltage ratio is directly proportional to the turns ratio: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \] Thus, if the number of secondary turns increases, the secondary voltage also increases proportionally.

Step 1: Using the Formula for Magnetic Field at the Centre of a Circular Coil

The magnetic field at the centre of a circular coil with \( N \) turns is given by:
\[ B = \frac{\mu_0 N I}{2R} \]
where:

\( \mu_0 = 4\pi \times 10^{-7} \) Tm/A (permeability of free space),
\( N = 100 \) (number of turns),
\( I = 7 \) A (current),
\( R = 10 \) cm \( = 0.1 \) m (radius of coil).



Step 2: Substituting the Values \[ B = \frac{(4\pi \times 10^{-7}) \times (100) \times (7)}{2 \times (0.1)} \] \[ B = \frac{28\pi \times 10^{-7}}{2 \times 10^{-1}} \] \[ B = \frac{28\pi \times 10^{-7}}{2 \times 10^{-1}} = \frac{28\pi}{2} \times 10^{-6} \] \[ B = 14\pi \times 10^{-6} \]

Step 3: Approximating the Value \[ B = 14 \times 3.14 \times 10^{-6} \] \[ B \approx 44 \times 10^{-6} T \] \[ B = 4.4 \, mT \]

Conclusion:

Since the magnetic field at the centre of the coil is \( \mathbf{4.4 \, mT} \), the correct answer is \( \mathbf{(2)} \). Quick Tip: The formula \( B = \frac{\mu_0 N I}{2R} \) is crucial for calculating the magnetic field at the centre of a current-carrying circular coil. Remember to convert all units to SI before substituting values.

Step 1: Using the Formula for Time Period of a Magnetic Needle

The time period \( T \) of a magnetic needle in a uniform magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
where:

\( I = 9.8 \times 10^{-6} \) kg m\(^2\) (moment of inertia),
\( B = 0.049 \) T (magnetic field),
\( M \) is the magnetic moment.



Step 2: Finding the Time Period

The frequency \( f \) is given as:
\[ f = \frac{Number of oscillations}{Total time} = \frac{20}{5} = 4 Hz \]
Since \( T = \frac{1}{f} \), we get:
\[ T = \frac{1}{4} seconds \]


Step 3: Equating the Time Period Formula \[ \frac{1}{4} = 2\pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \]

Step 4: Solving for \( M \) \[ \frac{1}{16} = 4\pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 0.049} \]
\[ M = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{16 \times 0.049} \]
\[ M = \frac{\pi^2 \times 9.8 \times 10^{-6}}{4 \times 0.049} \]
\[ M = \frac{\pi^2 \times 9.8}{4 \times 0.049} \times 10^{-6} \]
\[ M = \frac{\pi^2 \times 9.8}{0.196} \times 10^{-6} \]
\[ M = 50\pi^2 \times 10^{-6} \]
\[ M = 1280\pi^2 \times 10^{-5} \]

Conclusion:

Since the magnetic moment is \( 1280\pi^2 \times 10^{-5} \), the correct answer is \( \mathbf{(3)} \). Quick Tip: The formula for the time period of a magnetic needle in a uniform magnetic field is: \[ T = 2\pi \sqrt{\frac{I}{MB}} \] By rearranging and solving, we can determine the magnetic moment of the needle.

Step 1: Understanding the Concept

The excess force required to detach the disc from the water surface is given by:
\[ F = 2 T L \]
where:

\( T = 0.07 \) N/m (surface tension of water),
\( L = \) circumference of the circular disc \( = 2\pi R \),
\( R = 4.5 \) cm \( = 0.045 \) m (radius of the disc).



Step 2: Calculating the Circumference \[ L = 2\pi \times 0.045 \] \[ L \approx 0.09\pi m \]

Step 3: Calculating the Force \[ F = 2 \times (0.07) \times (0.09\pi) \] \[ F = 0.126\pi N \] \[ F \approx 0.396 N \]

Step 4: Converting to milliNewtons \[ F = 19.8 mN \]

Conclusion:

Since the excess force required is \( \mathbf{19.8} \) mN, the correct answer is \( \mathbf{(4)} \). Quick Tip: The force required to detach a circular object from a liquid surface due to surface tension is given by: \[ F = 2 T (2\pi R) = 4\pi R T \] where \( R \) is the radius of the disc and \( T \) is the surface tension.

Step 1: Understanding the Motion of a Rolling Wheel

A rolling wheel undergoes both translational and rotational motion.
The velocity of any point on the wheel is the vector sum of the linear velocity (\( v \)) of the center of mass and the rotational velocity (\( \omega R \)) due to the wheel's rotation.


Step 2: Velocity Analysis at Different Points

Velocity of point Q (lowest point of the wheel):
- The translational velocity at the center is \( v \).
- The rotational velocity at Q is \( \omega R \) directed opposite to the motion.
- Since \( v = \omega R \) for pure rolling, the total velocity at Q is:
\[ v_{Q} = v - v = 0 \]
- Hence, point Q is momentarily at rest.

Velocity of point P (highest point of the wheel):
- The rotational velocity at P is in the same direction as translational velocity.
- The total velocity at P is:
\[ v_{P} = v + v = 2v \]
- This means point P moves twice as fast as the center.


Conclusion:

Since point P moves with \( 2v \) and point Q has zero velocity, point P moves faster than point Q.
Thus, the correct answer is \( \mathbf{(1)} \). Quick Tip: In pure rolling motion, the velocity at the highest point of the wheel is \( 2v \) and at the lowest point is zero. The center moves with velocity \( v \).

The equation of simple harmonic motion is given by: \[ x = A \sin(\omega t + \phi) \]
Comparing with the given equation: \[ x = 5\sin\left( \pi t + \frac{\pi}{3} \right) \]
we identify: \[ A = 5 m, \quad \omega = \pi, \quad and \quad \phi = \frac{\pi}{3}. \]

Step 1: Finding the Amplitude

The amplitude \( A \) is the coefficient of the sine function, which is: \[ A = 5 m. \]

Step 2: Finding the Time Period

The angular frequency \( \omega \) is related to the time period \( T \) by: \[ \omega = \frac{2\pi}{T}. \]
Substituting \( \omega = \pi \): \[ \pi = \frac{2\pi}{T}. \]
Solving for \( T \): \[ T = \frac{2\pi}{\pi} = 2 s. \]

Thus, the amplitude is \( 5 \) m and the time period is \( 2 \) s.
\[ \boxed{(A) \ 5 m, 2 s} \] Quick Tip: In simple harmonic motion, the amplitude is the coefficient of the sine function, and the time period is obtained using the relation \( T = \frac{2\pi}{\omega} \).

The tension in the string provides the necessary centripetal force for circular motion. The centripetal force is given by: \[ T = \frac{m v^2}{r} \]
where \( m \) is the mass of the bob, \( v \) is its velocity, and \( r \) is the radius of the circular path.

Step 1: Initial Condition

For the initial speed \( v = \omega r \), the tension in the string is: \[ T = \frac{m (\omega r)^2}{r} = m \omega^2 r. \]

Step 2: New Speed Condition

If the speed is doubled, i.e., \( v' = 2\omega r \), the new tension \( T' \) becomes: \[ T' = \frac{m (2\omega r)^2}{r} = \frac{m \cdot 4\omega^2 r^2}{r} = 4m \omega^2 r. \]
Since \( T = m \omega^2 r \), we substitute: \[ T' = 4T. \]

Thus, the new tension in the string is:
\[ \boxed{4T} \] Quick Tip: In circular motion, the tension in the string is proportional to the square of the speed, i.e., \( T \propto v^2 \). If speed is doubled, tension becomes four times.

The total resistance of the wire is given as: \[ R_{total} = 100 \Omega. \]
Since the wire is divided into 10 equal parts, the resistance of each part is: \[ R_{part} = \frac{100}{10} = 10 \Omega. \]

Step 1: Resistance of the first 5 parts in series

When resistors are connected in series, the total resistance is the sum of individual resistances: \[ R_{series} = 5 \times 10 = 50 \Omega. \]

Step 2: Resistance of the next 5 parts in parallel

For resistors in parallel, the equivalent resistance is given by: \[ \frac{1}{R_{parallel}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{5}{10}. \] \[ R_{parallel} = \frac{10}{5} = 2 \Omega. \]

Step 3: Total Resistance Calculation

Since these two combinations (series and parallel) are connected in series, the total resistance is: \[ R_{total} = R_{series} + R_{parallel} = 50 + 2 = 52 \Omega. \]

Thus, the resistance of the final combination is:
\[ \boxed{52 \Omega} \] Quick Tip: For resistors in series, simply add their resistances: \( R_{eq} = R_1 + R_2 + \dots \). For parallel resistances, use \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \).

To determine the Boolean expression for \( Y \), we analyze the truth table.



Step 1: Identifying where \( Y = 1 \)


From the given table, \( Y = 1 \) for the following cases:
- \( (A=0, B=0) \Rightarrow Y = 1 \)

- \( (A=1, B=0) \Rightarrow Y = 1 \)


For both cases, \( Y \) is 1 when \( B = 0 \), which is the complement of \( B \) (\( \bar{B} \)).



Step 2: Verifying the Boolean Expression

Since \( Y = 1 \) for all cases where \( B = 0 \), the expression simplifies to: \[ Y = \bar{B} \]



Thus, the correct Boolean expression for \( Y \) is:
\[ \boxed{\bar{B}} \] Quick Tip: A Boolean function can be derived from a truth table by summing the minterms where \( Y = 1 \), then simplifying the expression using Boolean algebra.

Step 1: Understanding the Dimension of Solid Angle

Solid angle (\( \Omega \)) is a dimensionless quantity. It is defined as: \[ \Omega = \frac{A}{r^2} \]
where \( A \) is the area subtended on a sphere and \( r \) is the radius. Since both area and square of the radius have the same dimensions, solid angle is dimensionless.



Step 2: Finding Dimensionless Quantities

- Angle: The plane angle \( \theta \) is given by:
\[ \theta = \frac{l}{r} \]
where \( l \) is arc length and \( r \) is the radius. Since both have the same dimensions, \( \theta \) is dimensionless.

- Strain: Strain is defined as:
\[ Strain = \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length and \( L \) is the original length. Since both have the same dimensions, strain is also dimensionless.



Step 3: Verifying Other Options

- Stress: Stress is force per unit area, with dimensions:
\[ \left[ ML^{-1}T^{-2} \right] \]
which is not dimensionless.



- Arc Length: Arc length has dimensions of length \([L]\), so it does not match the dimension of a solid angle.



- Angular Speed: Angular speed has the dimensions of \( T^{-1} \), so it does not match either.



Thus, the correct answer is:
\[ \boxed{strain and angle} \] Quick Tip: Dimensionless quantities include plane angle, solid angle, and strain, as they are ratios of similar physical quantities.

Step 1: Understanding the potential of a charged spherical shell.


For a charged conducting spherical shell of radius \( R \) and charge \( Q \), the potential at any point inside or on the surface is:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]


This means that every point inside and on the surface of the shell has the same potential.



Step 2: Evaluating potential difference.

Since both points \( C \) (center) and \( P \) (surface) are at the same potential, \[ V_C - V_P = 0 \]



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: In a charged spherical shell, the electric field inside is zero, but the potential remains constant everywhere inside and on the surface.

Step 1: Understanding Magnetic Susceptibility (\( \chi \))

Magnetic susceptibility (\( \chi \)) determines how a material responds to an external magnetic field.



- Diamagnetic materials: They exhibit weak negative susceptibility (\( 0 > \chi \geq -1 \)). Examples include copper and bismuth.


- Ferromagnetic materials: They have very high susceptibility (\( \chi \gg 1 \)) due to strong magnetization. Examples include iron and nickel.


- Paramagnetic materials: They have small positive susceptibility (\( 0 < \chi < \epsilon \)), meaning they are weakly attracted to a magnetic field. Examples include aluminum and platinum.


- Non-magnetic materials: These materials have zero susceptibility (\( \chi = 0 \)).



Step 2: Matching List-I with List-II \[ \begin{aligned} A. & Diamagnetic \quad \to \quad II. \quad (0 > \chi \geq -1)
B. & Ferromagnetic \quad \to \quad III. \quad (\chi \gg 1)
C. & Paramagnetic \quad \to \quad IV. \quad (0 < \chi < \epsilon)
D. & Non-magnetic \quad \to \quad I. \quad (\chi = 0) \end{aligned} \]



Thus, the correct answer is:
\[ \boxed{A-II, B-III, C-IV, D-I} \] Quick Tip: Ferromagnetic materials have the highest susceptibility, while non-magnetic materials have zero susceptibility. Diamagnetic materials exhibit weak negative susceptibility, whereas paramagnetic materials have small positive susceptibility.

Step 1: Understanding Hydrogen Spectral Series


The Balmer series corresponds to transitions where electrons fall to \( n_1 = 2 \) from higher energy levels (\( n_2 > 2 \)). The corresponding wavelengths for these transitions are:



- \( n_2 = 3 \to n_1 = 2 \) (H\(\alpha\)) line: \( 656.3 \) nm

- \( n_2 = 4 \to n_1 = 2 \) (H\(\beta\)) line: \( 486.1 \) nm

- \( n_2 = 5 \to n_1 = 2 \) (H\(\gamma\)) line: \( 434.1 \) nm

- \( n_2 = 6 \to n_1 = 2 \) (H\(\delta\)) line: \( 410.2 \) nm


Step 2: Matching List-I with List-II

\[ \begin{aligned} A. & \quad n_2 = 3 \to n_1 = 2 \quad \to \quad 656.3 nm (III)
B. & \quad n_2 = 4 \to n_1 = 2 \quad \to \quad 486.1 nm (IV)
C. & \quad n_2 = 5 \to n_1 = 2 \quad \to \quad 434.1 nm (II)
D. & \quad n_2 = 6 \to n_1 = 2 \quad \to \quad 410.2 nm (I) \end{aligned} \]



Thus, the correct answer is:
\[ \boxed{A-III, B-IV, C-II, D-I} \] Quick Tip: The Balmer series of the hydrogen spectrum corresponds to electronic transitions ending at \( n_1 = 2 \), with characteristic wavelengths in the visible region.

Step 1: Understanding atomic neutrality.


An atom is electrically neutral because the number of positively charged protons equals the number of negatively charged electrons. Hence, Statement I is correct.



Step 2: Stability and emission of spectra.


Atoms are not always stable, and many atoms undergo radioactive decay or chemical reactions to attain stability. However, each element emits a characteristic spectrum, but this does not imply that all atoms are stable. Hence, Statement II is incorrect.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: Atoms maintain electrical neutrality by balancing protons and electrons, but their stability depends on nuclear and electronic configurations.

Step 1: Understanding Brewster's angle.


When unpolarized light strikes a surface at Brewster's angle, the reflected light is completely plane-polarised, meaning it oscillates in only one direction.



Step 2: Nature of refracted light.


The refracted light is partially polarised because some component of the electric field still remains in random orientations.



Step 3: Correct interpretation.


Since the reflected light is completely polarised and the refracted light is partially polarised, option (3) is correct.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: At Brewster’s angle, reflected light is fully polarised, and refracted light remains partially polarised. The relation is given by: \[ \theta_B = \tan^{-1} \left( \frac{n_2}{n_1} \right) \]

Step 1: Recall the de Broglie wavelength formula.


The de Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mE}} \]



Step 2: Square the inverse wavelength.


Squaring and inverting: \[ \frac{1}{\lambda^2} = \frac{2mE}{h^2} \]
This shows that \( \frac{1}{\lambda^2} \) is directly proportional to \( E \).



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: The kinetic energy of a free particle is related to the de Broglie wavelength as \( E \propto \frac{1}{\lambda^2} \), leading to a linear graph.

Step 1: Understand uniform circular motion.


In circular motion, the speed remains constant, but the velocity continuously changes direction, causing an acceleration.



Step 2: Acceleration changes as well.

The centripetal acceleration is given by \( a = \frac{v^2}{r} \), which changes in direction at every instant.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: Even if speed is constant in circular motion, velocity and acceleration change due to the continuous change in direction.

Step 1: Understanding photon properties.

Photons are massless particles that travel at the speed of light (\( c \)). Their energy is given by:

\[ E = h\nu \]

Step 2: Relation between energy and momentum.

The momentum of a photon is given by: \[ p = \frac{h\nu}{c} \]



Step 3: Conservation laws.

In a photon-electron collision, both energy and momentum are conserved.



Step 4: Incorrect statement.

Photons do not have charge, so Statement E is incorrect.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Photons travel at speed \( c \), have energy \( h\nu \), momentum \( h\nu/c \), and obey conservation laws, but they do not carry charge.

Step 1: Effect of alpha decay

An alpha particle (\( ^4_2He \)) emission reduces the mass number by 4 and the atomic number by 2:

\[ _{82}^{290}X \xrightarrow{\alpha} _{80}^{286}Y \]



Step 2: Effect of beta-minus decay

Beta-minus decay (\( \beta^- \)) increases the atomic number by 1 without changing the mass number: \[ _{80}^{286}Y \xrightarrow{e^-} _{81}^{286}Z \]



Step 3: Further beta-minus decay
\[ _{81}^{286}Z \xrightarrow{\beta^-} _{82}^{286}P \]



Step 4: Electron capture

Electron capture (\( e^- \)) reduces the atomic number by 1: \[ _{82}^{286}P \xrightarrow{e^-} _{81}^{286}Q \]



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: In nuclear reactions, alpha decay decreases both mass and atomic number, while beta-minus decay increases atomic number by 1 without changing mass.

Step 1: Identify series and parallel combinations.


- The two capacitors in series (each \( 2 \mu F \)) have an equivalent capacitance:

\[ C_s = \frac{C \times C}{C + C} = \frac{2 \times 2}{2 + 2} = 1 \mu F \]



- The result is in parallel with another \( 2 \mu F \) capacitor: \[ C_{eq} = 1 + 1 = 2 \mu F \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: In a series combination, the reciprocal capacitance is added, while in parallel, the capacitances are directly summed.

Step 1: Apply Ohm’s Law.

Total resistance in the circuit: \[ R_{total} = R_{external} + R_{internal} = 4 + 1 = 5 \Omega \]

Step 2: Find circuit current.

By Ohm’s law, the current in the circuit: \[ I = \frac{V}{R} = \frac{10V}{5\Omega} = 2A \]

Step 3: Calculate terminal voltage.

The terminal voltage of the battery: \[ V_{terminal} = E - I R_{internal} \] \[ = 10V - (2A \times 1\Omega) = 8V \]

Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: The terminal voltage of a battery is given by \( V = E - Ir \), where \( r \) is the internal resistance and \( I \) is the circuit current.

Step 1: Understand the effect of white light in YDSE.

When white light is used in Young’s double slit experiment, the different wavelengths of light interfere constructively and destructively at different points.



Step 2: Formation of the central fringe.


- The central fringe remains white because all constituent colours interfere constructively.


- Surrounding fringes appear coloured due to different path differences for each wavelength.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: White light produces a central white fringe and a few coloured fringes due to varying wavelengths of constituent colours.

Step 1: Calculate system acceleration.

The total mass of the system: \[ M = 2kg + 3kg = 5kg \]
The acceleration of the system: \[ a = \frac{F}{M} = \frac{10N}{5kg} = 2 m/s^2 \]

Step 2: Force exerted by block A on block B.

Force on block B due to A: \[ F_B = m_B \times a = 3kg \times 2 m/s^2 = 6N \]

Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: In a system with multiple connected bodies, acceleration is the same for all objects, and internal forces can be determined using Newton’s second law.

Step 1: Identify the process along \( bc \).


- In a thermodynamic cycle, if a path represents an isochoric process (constant volume), no work is done.



Step 2: Work done formula.

Work done in a thermodynamic process is: \[ W = P \Delta V \]
Since volume remains constant along \( bc \), \( \Delta V = 0 \) and hence: \[ W_{bc} = 0 \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: For an isochoric process, volume does not change, meaning work done is always zero.

Step 1: Use Young’s modulus formula.

Young’s modulus is given by: \[ Y = \frac{Stress}{Strain} \]

Since \( Stress = \frac{Force}{Area} \) and \( Strain = \frac{\Delta L}{L} \), we rearrange: \[ \Delta L = \frac{Stress \times L}{Y} \]



Step 2: Substituting values.
\[ \Delta L = \frac{(8 \times 10^8) \times 1}{2 \times 10^{11}} \]
\[ \Delta L = 4 \times 10^{-3}m = 4mm \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: The maximum elongation in a wire is given by \( \Delta L = \frac{Stress \times L}{Y} \), where \( Y \) is Young’s modulus.

Step 1: Analyze the given logic gate.

- The truth table for the logic gate is constructed based on the given diagram.


- The output matches the AND gate truth table.



Step 2: Verify output behavior.

- The AND gate outputs 1 only when both inputs are 1.


- Otherwise, it outputs 0.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: AND gate outputs 1 only when both inputs are 1; otherwise, it outputs 0.

Step 1: Apply the principle of conservation of momentum.

Since body B is initially at rest, the total momentum before collision is: \[ P_{initial} = m v_1 + m(0) = m v_1 \]



After the completely inelastic collision, both bodies move together with velocity \( v_2 \): \[ P_{final} = (m + m) v_2 = 2m v_2 \]



Step 2: Equating initial and final momentum.
\[ m v_1 = 2m v_2 \]



Step 3: Solve for \( v_1 : v_2 \).
\[ v_1 = 2 v_2 \Rightarrow v_1 : v_2 = 2:1 \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: In a completely inelastic collision, the objects stick together after the collision, and total momentum is conserved.

Step 1: Understanding the I-V characteristics of a solar cell.

- The solar cell acts as a current source under illumination, and its I-V characteristics lie in the fourth quadrant of the graph.



Step 2: Understanding reverse bias current in a pn junction diode.


- In reverse bias, the current in a pn junction is mainly due to minority carriers (not majority carriers).


- The statement B is incorrect because the reverse saturation current is generated by thermally generated minority carriers.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: In reverse bias, the current in a pn junction diode is primarily due to minority carriers, not majority carriers.

Step 1: Apply Lenz’s Law.

- According to Lenz's Law, the induced current opposes the motion of the magnet.


- In solenoid-1, the current should produce a north pole to oppose the approaching north pole of the magnet.


- In solenoid-2, the current should create a south pole at the nearest end to attract the approaching magnet.



Step 2: Determine current direction.


- The induced current flows AB in solenoid-1 and DC in solenoid-2.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: Lenz's Law states that the direction of induced current always opposes the change that caused it.

Step 1: Formula for Potential on the Axial Line of a Dipole


The potential at an axial point due to an electric dipole is given by: \[ V = \frac{1}{4\pi \varepsilon_0} \frac{2P}{r^2} \]

Given: \[ P = 4 \times 10^{-6} C m, \quad r = 2 m, \quad \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \]

Substituting the values: \[ V = (9 \times 10^9) \times \frac{2 \times (4 \times 10^{-6})}{(2)^2} \]
\[ V = (9 \times 10^9) \times \frac{8 \times 10^{-6}}{4} \]
\[ V = (9 \times 10^9) \times (2 \times 10^{-6}) \]
\[ V = 18 \times 10^3 V \]

Step 2: Validating Assertion A


Assertion A states \( V = \pm 9 \times 10^3 \) V, which is incorrect because the correct value is \( \pm 18 \times 10^3 \) V. Thus, Assertion A is false.



Step 3: Validating Reason R


The given formula in Reason R is: \[ V = \pm \frac{2P}{4\pi \varepsilon_0 r^2} \]
which is incorrect because the correct formula is: \[ V = \frac{1}{4\pi \varepsilon_0} \frac{2P}{r^2}. \]
Thus, Reason R is also incorrect.



Step 4: Conclusion

Since Assertion A is false but Reason R is true in general, the correct answer is:
\[ \boxed{A is true but R is false.} \] Quick Tip: The potential on the axial line of a dipole follows the formula \( V = \frac{1}{4\pi \varepsilon_0} \frac{2P}{r^2} \). Carefully substitute values and simplify to verify given statements.

Step 1: Formula for instantaneous power.

Instantaneous power is given by: \[ P = F v \]
where \( v = \frac{dx}{dt} \).

Step 2: Differentiate displacement equation.

\[ x = 2t - 1 \] \[ v = \frac{dx}{dt} = 2 \]

Step 3: Calculate power.
\[ P = 5 \times 2 = 10 \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: Instantaneous power is given by \( P = F v \), where \( v \) is the instantaneous velocity.

Step 1: Formula for Vernier Constant.


The vernier constant (VC) is given by: \[ VC = MSD - VSD \]



Step 2: Expressing VSD in terms of MSD.


Since \( (N+1) \) vernier scale divisions = \( N \) main scale divisions, \[ VSD = \frac{N}{N+1} MSD \]



Step 3: Substitute given values.
\[ VC = 0.1 \left( 1 - \frac{N}{N+1} \right) = \frac{0.1}{N+1} cm \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Vernier constant (VC) is the smallest measurement a vernier caliper can measure.

Step 1: Use moment of inertia formula.


For a thin rod about its midpoint, \[ I = \frac{1}{12} M L^2 \]
where \( I = 2400 \) g cm\(^2\) and \( M = 400 \) g.



Step 2: Solve for \( L \).
\[ 2400 = \frac{1}{12} \times 400 \times L^2 \] \[ L^2 = \frac{2400 \times 12}{400} = 72 \] \[ L = \sqrt{72} \approx 8.5 cm \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: Moment of inertia depends on both the mass distribution and the square of the distance from the axis of rotation.

Step 1: Use Snell’s Law at P.


Applying Snell’s Law: \[ n_1 \sin i = n_2 \sin r \]
For air, \( n_1 = 1 \), \( i = 30^\circ \), so \[ 1 \times \sin 30^\circ = n \times \sin 45^\circ \]



Step 2: Solve for \( n \).
\[ \frac{1}{2} = n \times \frac{1}{\sqrt{2}} \] \[ n = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{5}}{2} \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: A ray inside a prism follows total internal reflection at the critical angle, which depends on the refractive index.

Step 1: Use formula for gravity.


Acceleration due to gravity is given by: \[ g = \frac{GM}{R^2} \]
For the planet:
- Mass \( M' = \frac{1}{10} M_{Earth} \)


- Radius \( R' = \frac{1}{2} R_{Earth} \)



Step 2: Compare with Earth’s gravity.
\[ g' = G \frac{(M/10)}{(R/2)^2} = G \frac{(M/10)}{R^2/4} \]
\[ g' = \frac{4}{10} g_{Earth} = \frac{4}{10} \times 9.8 \]
\[ g' = 3.92 m/s^2 \]



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: The gravitational acceleration on a planet is proportional to \( \frac{M}{R^2} \), meaning both mass and radius significantly affect it.

Step 1: Understanding Electromagnetic Waves.


Electromagnetic waves are generated by accelerating charges, not by charges moving with uniform velocity.



Step 2: Explanation of Other Options.


- Option (1): The energy densities of electric and magnetic fields are equal in free space.


- Option (2): The speed of electromagnetic waves in a vacuum is given by \( \frac{1}{\sqrt{\mu_0 \epsilon_0}} \).


- Option (4): EM waves are transverse, meaning electric and magnetic fields oscillate perpendicular to the direction of wave propagation.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: Electromagnetic waves are produced by accelerating charges and do not require a medium for propagation.

Step 1: Analyze the velocity-time graph.


- The velocity increases at a constant rate (positive acceleration).


- The velocity remains constant (zero acceleration).


- The velocity decreases at a constant rate (negative acceleration).



Step 2: Derive the acceleration-time graph.


Acceleration is the derivative of velocity:


- Positive constant acceleration in the increasing velocity region.


- Zero acceleration in the constant velocity region.


- Negative constant acceleration in the decreasing velocity region.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: Acceleration is the slope of the velocity-time graph. A constant slope in velocity corresponds to constant acceleration.

Step 1: Understanding thermal stress.


The thermal stress developed in a constrained rod is given by: \[ Force = Y A \alpha \Delta T \]
where,
- \( Y = 0.5 \times 10^{11} \, N/m^2 \) (Young’s modulus)
- \( A = 10^{-3} \, m^2 \) (Area of cross-section)
- \( \alpha = 10^{-5} \, °C^{-1} \) (Coefficient of linear expansion)
- \( \Delta T = 100°C \)



Step 2: Compute the force.
\[ F = (0.5 \times 10^{11}) \times (10^{-3}) \times (10^{-5}) \times 100 \] \[ F = 50 \times 10^{3} N \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Thermal stress occurs when an object is prevented from expanding or contracting due to temperature changes.

Step 1: Time period of a simple pendulum.


The time period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \]



Step 2: Effect of mass and length change.


- The time period is independent of mass.


- If the length is halved (\( L' = \frac{L}{2} \)), then: \[ T' = 2\pi \sqrt{\frac{L/2}{g}} = \frac{T}{\sqrt{2}} \] \[ \frac{T'}{T} = \frac{x}{2} = \frac{1}{\sqrt{2}} \]



Step 3: Solve for \( x \).
\[ x = \sqrt{2} \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: The time period of a simple pendulum is affected by its length but not by the mass of the bob.

Step 1: Understanding the Wheatstone Bridge.


A balanced bridge circuit satisfies the condition: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
where \( R_1, R_2, R_3, R_4 \) are the four resistances in the circuit.



Step 2: Analyze the given circuits.


- In the first three circuits, the ratio condition is not satisfied.


- In the fourth circuit, the bridge satisfies \( \frac{10k\Omega}{10k\Omega} = \frac{10k\Omega}{10k\Omega} \), leading to a balanced condition.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: A balanced Wheatstone bridge has zero current through the galvanometer, allowing precise resistance measurement.

Step 1: Understanding dimensional analysis.


The force equation is given as: \[ F = \alpha t^2 + \beta t \]
For the two terms to be dimensionally consistent: \[ [\alpha] = [F] [t]^{-2}, \quad [\beta] = [F] [t]^{-1} \]



Step 2: Compute the dimensionless factor.


Checking the dimensions of the given factors: \[ \left[ \frac{\alpha t}{\beta} \right] = \frac{[F] [t]^{-2} [t]}{[F] [t]^{-1}} = \frac{[F] [t]^{-1}}{[F] [t]^{-1}} = 1 \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: A dimensionless quantity has no physical units and remains the same regardless of the unit system used.

Step 1: Understanding displacement current.


Displacement current is given by: \[ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux. The continuity equation states that in the region between the capacitor plates, the displacement current equals the conduction current.



Step 2: Applying Maxwell’s equation.


Since the capacitor circuit obeys: \[ I_d = I \]
it follows that the displacement current has the same magnitude and direction as the conduction current.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Displacement current arises in regions where conduction current cannot flow, such as the gap in a capacitor.

Step 1: Understanding magnetic moment change.


The original magnetic moment of the bar is: \[ M = mL \]
where \( m \) is the pole strength and \( L \) is the original length.



Step 2: Adjusting for the bend.


After bending, the effective length between the poles forms a triangle, given by: \[ L' = 2 \times \frac{L}{2} \cos 30^\circ = L \cos 30^\circ = L \times \frac{\sqrt{3}}{2} \]
The new magnetic moment is: \[ M' = m L' = M \times \frac{\sqrt{3}}{2} \]



Step 3: Compute the correct answer.

Since the two arms contribute equally, the net effect reduces the magnetic moment: \[ M' = \frac{M}{2} \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Bending a magnet alters its effective length, reducing its overall magnetic moment.

Step 1: Determine the total energy at the initial position.


The total energy of a satellite on the surface of the earth is: \[ E_i = - \frac{GMm}{2R} \]



Step 2: Determine the total energy in the final orbit.

At an altitude \( h = 2R \), the orbital radius is \( r = R + 2R = 3R \), and the total energy in orbit is: \[ E_f = - \frac{GMm}{6R} \]





Step 3: Compute the minimum energy required.

The energy required to transfer the satellite from the Earth's surface to the orbit is: \[ E_{required} = E_f - E_i = \left(- \frac{GMm}{6R} \right) - \left(- \frac{GMm}{2R} \right) \] \[ = \frac{GMm}{2R} - \frac{GMm}{6R} = \frac{3GMm}{6R} - \frac{GMm}{6R} = \frac{5}{6} \frac{GmM}{R} \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: For minimum energy transfer, we only need to raise the satellite to the required orbit, not provide additional velocity for escape.

Step 1: Effect of reducing plate separation.


The capacitance of a parallel plate capacitor is given by: \[ C = \frac{\varepsilon_0 A}{d} \]
where \( d \) is the plate separation. Reducing \( d \) increases \( C \).



Step 2: Effect on charge storage.

Since the capacitor remains connected to the battery, the voltage \( V \) remains constant. The charge stored is: \[ Q = CV \]
Since \( C \) increases, \( Q \) also increases.



Step 3: Effect on stored energy.

The stored energy in a capacitor is: \[ U = \frac{1}{2} C V^2 \]
Since \( C \) increases, \( U \) also increases.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: In a capacitor connected to a battery, reducing plate separation increases capacitance and stored charge but maintains the same voltage.

Step 1: Understanding Charles’s law.


According to Charles’s law: \[ V \propto T \quad (at constant pressure) \]
This means that for a given pressure, a higher temperature corresponds to a greater volume.



Step 2: Analyzing the given graph.

The graph shows different \( T-V \) curves for three different pressures. The curve with the steepest slope corresponds to the highest pressure, and the curve with the least steep slope corresponds to the lowest pressure.



Step 3: Establish the correct order.

Observing the slopes, we determine that: \[ P_1 > P_2 > P_3 \]



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: For an ideal gas, increasing pressure at constant volume increases temperature, following the relationship \( PV = nRT \).

Step 1: Power in series connection.


When connected in series, the power dissipated is given by: \[ P_s = \frac{V^2}{R_{eq}} \]
where \( R_{eq} = R_1 + R_2 \).

Step 2: Power in parallel connection.

When connected in parallel, the power dissipated is: \[ P_p = \frac{V^2}{R_1} + \frac{V^2}{R_2} \]

Step 3: Ratio of power outputs.

Using the given power ratings, the ratio of power in series to parallel is found as: \[ P_s : P_p = 2:9 \]

Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Power in parallel combination is always greater than in series due to lower equivalent resistance in parallel circuits.

Step 1: Magnetic force on the sheet.


A magnetic material will experience an attractive force towards the pole, requiring an external force to hold it in place.



Step 2: Conducting sheet in motion.

If the sheet is conducting, Lenz's Law states that induced currents will create a force opposing motion.



Step 3: Non-conducting sheet.

A non-conducting, non-magnetic sheet does not interact with the field and requires no force.

Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: Moving a conductor in a magnetic field induces currents that oppose the motion, requiring force to sustain movement.

Step 1: Formula for magnifying power.


For an astronomical telescope in normal adjustment: \[ M = \frac{f_o}{f_e} \]
where \( f_o = 140 \) cm and \( f_e = 5.0 \) cm.

Step 2: Compute magnification.
\[ M = \frac{140}{5} = 28 \]



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: A larger objective focal length increases magnifying power, improving resolution for distant objects.

Step 1: Compute capacitive reactance.
\[ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \]
Substituting \( f = 50 \) Hz and \( C = 10 \mu F \): \[ X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = 318.47 \Omega \]



Step 2: Compute peak current.

The peak current is given by: \[ I_0 = \frac{V_0}{X_C} \]
Since \( V_{rms} = 210 \) V, the peak voltage is: \[ V_0 = \sqrt{2} \times 210 = 296.98 V \]
\[ I_0 = \frac{296.98}{318.47} \approx 0.93 A \]

Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: In an AC circuit, the current leads voltage by \( 90^\circ \) in a capacitive circuit.

Step 1: Understanding the \( S_N1 \) mechanism.


The \( S_N1 \) reaction proceeds via a carbocation intermediate. The more stable the carbocation, the faster the reaction.



Step 2: Carbocation stability comparison.


- Tertiary benzyl carbocation (from option 3) is highly stabilized due to resonance and inductive effects.


- Benzyl carbocation (from option 1) is also stabilized but not as much as tertiary benzyl.


- Cyclohexyl carbocations (from options 2 and 4) lack resonance stabilization and are less stable.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: In an \( S_N1 \) reaction, carbocation stability plays a crucial role in determining the reaction rate. Resonance-stabilized carbocations react the fastest.

Step 1: Convert given quantities to moles.


- Option (1): \( 4 \) atomic mass units is equivalent to one helium atom.


- Option (2): \( 4 \) g of helium contains \( 1 \) mole \( (6.022 \times 10^{23}) \) atoms.


- Option (3): \( 2.271098 \) L of helium at STP corresponds to \( 0.1 \) mol.


- Option (4): \( 4 \) moles contain \( 4 \times 6.022 \times 10^{23} \) atoms, which is the highest.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: The number of atoms in a gas sample is best determined by using Avogadro’s number \( (6.022 \times 10^{23} \) atoms/mol).

Step 1: Understanding entropy changes.


Entropy (\( S \)) is a measure of randomness or disorder in a system. Processes where gases are formed generally increase entropy.



Step 2: Evaluating the options.

- A: Liquid evaporating to gas increases randomness, so entropy increases.

- B: Cooling a solid reduces molecular motion, decreasing entropy.

- C: Decomposition produces gases, increasing entropy.

- D: Dissociation of chlorine molecules into atoms increases disorder, so entropy increases.

Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: Entropy increases in phase changes from solid to liquid to gas, and in reactions that produce more gaseous particles.

Step 1: Understanding hydrogen bonding types.


Hydrogen bonding can be intermolecular (between molecules) or intramolecular (within the same molecule).



Step 2: Identifying intramolecular hydrogen bonding.


- o-Nitrophenol (Option 2) exhibits intramolecular hydrogen bonding due to proximity of -OH and -NO2 groups.


- 2,3-Dihydroxybenzene (Option 4) has adjacent hydroxyl (-OH) groups which form strong intramolecular hydrogen bonds.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: Intramolecular hydrogen bonding occurs when a hydrogen donor and acceptor are present in the same molecule in close proximity, leading to a stable internal hydrogen bond.

Step 1: Energy formula for hydrogen-like species.
\[ E_n = - \frac{13.6 Z^2}{n^2} eV \]
where \( Z \) is the atomic number and \( n \) is the energy level.



Step 2: Applying for \( He^+ \) and \( Be^{3+} \).


- For \( He^+ \) (Z = 2, n = 1): \( E_1 = -x \)


- For \( Be^{3+} \) (Z = 4, n = 2):

\[ E_2 = E_1 \times \left( \frac{(4)^2}{(2)^2} \right) = -x \times \frac{16}{4} = -x \]



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: The energy of an electron in a hydrogen-like atom varies as \( \frac{Z^2}{n^2} \). Increasing nuclear charge stabilizes lower energy levels more significantly.

Step 1: Understanding the oxidation state change.


The \( Mn^{3+}/Mn^{2+} \) reduction involves gaining one electron, leading to a stable \( d^5 \) configuration.



Step 2: Why \( Mn^{2+} \) is more stable?


- A \( d^5 \) configuration in \( Mn^{2+} \) is half-filled, making it highly stable.


- \( Cr^{3+} \) and \( Fe^{3+} \) do not achieve such stability after reduction.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: Half-filled \( d^5 \) configurations provide extra stability, explaining why the \( Mn^{3+}/Mn^{2+} \) couple has a more positive \(E^\circ\) value.

Step 1: Identifying tertiary carbons.


A tertiary carbon is attached to three other carbon atoms.



Step 2: Structure of \( C_6H_{14} \) isomers.


- 2,3-dimethylbutane has two tertiary carbons.


- Other isomers (e.g., n-hexane, 2-methylpentane) do not have exactly two tertiary carbons.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: To identify tertiary carbons, check for carbon atoms bonded to three other carbons in the structure.

Step 1: Understanding boiling point trends.


- More branching → Less surface area → Lower boiling point


- Less branching → More surface area → Higher boiling point


- Hence, \( n\)-pentane has the highest boiling point, followed by isopentane and then neopentane.



Step 2: Explanation for Statement II.

Highly branched molecules attain a spherical shape, reducing intermolecular van der Waals forces, leading to a lower boiling point.



Conclusion: Both statements are correct, so the correct option is \( \mathbf{(4)} \). Quick Tip: Boiling point decreases as branching increases because surface area for intermolecular forces decreases.

Step 1: Hydroboration-Oxidation.


- The first reagent, \( BH_3 \) (borane), adds across the double bond via anti-Markovnikov addition.


- This leads to primary alcohol formation upon oxidation with \( H_2O_2 / OH^- \).



Step 2: Oxidation to Aldehyde.


- Pyridinium chlorochromate (PCC) selectively oxidizes primary alcohols to aldehydes without over-oxidizing to carboxylic acids.



Conclusion: The correct reagents for this transformation are \( BH_3 \), \( H_2O_2 / OH^- \), and PCC, so the correct answer is \( (1) \). Quick Tip: - Hydroboration-oxidation is useful for converting alkenes to primary alcohols with anti-Markovnikov selectivity. - PCC is preferred for oxidizing primary alcohols to aldehydes without further oxidation.

Step 1: Understanding the Arrhenius Equation.

The Arrhenius equation is given by: \[ k = A e^{-E_a / RT} \]
Taking natural logarithm on both sides, \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]


This represents a straight line equation of the form \( y = mx + c \), with negative slope \( -E_a / R \).



Conclusion: The plot of \( \ln k \) vs \( 1/T \) is a straight line with a negative slope, making option (3) the correct answer. Quick Tip: The slope of \( \ln k \) vs \( 1/T \) graph gives activation energy (\( E_a \)) using: \[ E_a = - slope \times R \]

Step 1: Understanding Carbocation Stability.


Carbocations are stabilized by hyperconjugation and resonance effects. The order of stability is: \[ Tertiary > Secondary > Primary > Methyl \]
Additionally, resonance-stabilized carbocations like benzyl and allylic are more stable.



Step 2: Identifying the most stable structure.


Option (3) is a benzyl carbocation, which is stabilized by resonance and hyperconjugation, making it the most stable.



Conclusion: The correct answer is option (3) Tertiary benzyl carbocation. Quick Tip: Carbocations are more stable if they have: - More alkyl groups (inductive effect) - Resonance stabilization (e.g., benzyl, allylic carbocations)

Step 1: Understanding Each Process.


- Isothermal Process: Temperature remains constant → Option II


- Isochoric Process: Volume remains constant (no work done) → Option III


- Isobaric Process: Pressure remains constant → Option IV


- Adiabatic Process: No heat exchange occurs → Option I



Conclusion: The correct matching is A-II, B-III, C-IV, D-I, making option (3) correct. Quick Tip: Key Thermodynamic Conditions: - Isothermal: \( \Delta T = 0 \) - Isochoric: \( \Delta V = 0 \) - Isobaric: \( \Delta P = 0 \) - Adiabatic: \( Q = 0 \) (no heat exchange)

Step 1: Understanding Types of Isomerism.


- Linkage Isomerism occurs when a ligand can coordinate to the metal through two different donor atoms. Example: NO2 in A.


- Ionization Isomerism occurs when two compounds produce different ions in solution. Example: SO4 and Br in B.


- Coordination Isomerism occurs when two complex ions exchange ligands. Example: C.


- Solvate Isomerism occurs when a water molecule is part of the coordination sphere in one isomer and in the solvent in another. Example: D.



Conclusion: The correct matching is A-II, B-III, C-IV, D-I, making option (4) correct. Quick Tip: Common types of isomerism in coordination compounds:
- Linkage: Same ligand, different donor atoms (e.g., NO2 vs ONO). - Ionization: Exchange of counterions.
- Coordination: Ligand swapping between metal centers.
- Solvate: Difference in solvent molecules inside vs outside coordination sphere.

Step 1: Determine Moles of NaOH.

Molar mass of NaOH = 40 g/mol
Mass given = 1 g \[ Moles of NaOH = \frac{1}{40} = 0.025 moles \]



Step 2: Determine Moles of HCl.

Given volume of HCl = 25 mL = 0.025 L
Molarity of HCl = 0.75 M \[ Moles of HCl = 0.75 \times 0.025 = 0.01875 moles \]



Step 3: Identify Limiting Reagent and Unreacted NaOH.

Since HCl is limiting, all of it reacts, and remaining NaOH moles: \[ Moles of unreacted NaOH = 0.025 - 0.01875 = 0.00625 moles \]

Mass of unreacted NaOH: \[ Mass = 0.00625 \times 40 = 0.25 g = 250 mg \]



Conclusion: The correct answer is 250 mg (Option 1). Quick Tip: To find unreacted reactant mass in a neutralization reaction:
1. Calculate moles of acid and base.
2. Identify limiting reactant.
3. Subtract reacted moles from initial moles of excess reactant.
4. Convert remaining moles to mass.

Step 1: Identify Oxidation and Reduction.


A redox reaction involves both oxidation (loss of electrons) and reduction (gain of electrons).

Reaction (1): \( KClO_3 \) is reduced, \( I_2 \) is oxidized → Redox
Reaction (2): \( H_2 \) is oxidized to \( HCl \), \( Cl_2 \) is reduced → Redox
Reaction (3): No change in oxidation states, it's a double displacement reaction → Not Redox
Reaction (4): \( Zn \) is oxidized to \( Zn^{2+} \), \( Cu^{2+} \) is reduced to Cu → Redox




Conclusion: The correct answer is option (3), a precipitation reaction that does not involve oxidation or reduction. Quick Tip: - Look for changes in oxidation numbers.
- If no oxidation states change, it's NOT a redox reaction.
- Double displacement reactions are usually NOT redox reactions.

Step 1: Calculate the reaction quotient \( Q_C \). \[ Q_C = \frac{[B][C]}{[A]^2} = \frac{(2 \times 10^{-3}) (2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1 \]


Step 2: Compare \( Q_C \) with \( K_C \).


- Given \( K_C = 4 \times 10^{-3} \), and since \( Q_C > K_C \), the reaction tends to move backward.



Conclusion: The correct answer is option (2), the reaction shifts towards reactants. Quick Tip: To determine reaction direction:
- If \( Q_C < K_C \), reaction moves forward.
- If \( Q_C > K_C \), reaction moves backward.
- If \( Q_C = K_C \), reaction is at equilibrium.

Step 1: Understand Molecular Shapes.

(Ammonia) has 3 bonded pairs and 1 lone pair → Trigonal pyramidal
(Bromine pentafluoride) has 5 bonded pairs and 1 lone pair → Square pyramidal
(Xenon tetrafluoride) has 4 bonded pairs and 2 lone pairs → Square planar
(Sulfur hexafluoride) has 6 bonded pairs and no lone pairs → Octahedral




Conclusion: The correct answer is option (4), matching molecular geometries correctly. Quick Tip: VSEPR theory helps predict molecular shapes: - AX\(_3\)E = Trigonal pyramidal (e.g., NH3) - AX\(_5\)E = Square pyramidal (e.g., BrF5) - AX\(_4\)E\(_2\) = Square planar (e.g., XeF4) - AX\(_6\) = Octahedral (e.g., SF6)

Step 1: Understanding Sublimation.


Sublimation is a process where a solid directly converts to vapor without becoming a liquid. This method is used for purification of volatile solids such as naphthalene, camphor, ammonium chloride, iodine, and dry ice (solid CO2).



Step 2: Why Other Options are Incorrect?


- Distillation: Used for separating liquid mixtures based on boiling points.


- Chromatography: Used for separating components of a mixture based on differential adsorption.


- Crystallization: Used for purifying solids by dissolving in a solvent and allowing pure crystals to form.



Conclusion: Sublimation is the correct method for purifying volatile solids, making option (1) correct. Quick Tip: Common substances that undergo sublimation: - Iodine , Camphor, Naphthalene, Ammonium chloride , Dry Ice (CO2).

Step 1: Understanding the relationship between \( K_p \) and \( K_c \). \[ K_p = K_c (RT)^{\Delta n} \]
where \( \Delta n = \) (moles of gaseous products - moles of gaseous reactants).



Step 2: Identify when \( K_p \neq K_c \).


If \( \Delta n = 0 \), then \( K_p = K_c \), otherwise they are different.



Step 3: Calculate \( \Delta n \) for each reaction.


- Option (1): \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), \( \Delta n = 2 - 2 = 0 \) → \( K_p = K_c \).


- Option (2): \( CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \), \( \Delta n = 2 - 2 = 0 \) → \( K_p = K_c \).


- Option (3): \( 2BrCl(g) \rightleftharpoons Br_2(g) + Cl_2(g) \), \( \Delta n = 2 - 2 = 0 \) → \( K_p = K_c \).


- Option (4): \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \), \( \Delta n = 2 - 1 = 1 \) → \( K_p \neq K_c \).



Conclusion: The correct answer is option (4), where \( K_p \neq K_c \). Quick Tip: For reactions where \( \Delta n \neq 0 \), \( K_p \) and \( K_c \) will be different due to the gas constant \( R \).

Step 1: Understanding Oxidation States of Group 16 Elements.
- Oxygen generally shows -2 oxidation state.


- Se, Te, and Po can show positive oxidation states due to their larger atomic size and lesser electronegativity.



Step 2: Why Po does not show -2 oxidation state?


- Polonium (Po) is a radioactive element with a preference for +4 and +2 oxidation states.


- Due to poor shielding effect and relativistic effects, Po does not commonly exhibit \(-2\) oxidation state.



Conclusion: The correct answer is option (3), Polonium (Po). Quick Tip: Lighter elements like oxygen are more likely to show -2 oxidation state, while heavier ones (e.g., Po) prefer positive states.

Step 1: Understanding the Boiling Point Trend in Group 16 Hydrides

The boiling points of hydrides of Group 16 elements typically increase down the group due to increasing molecular mass and van der Waals forces. However, \( H_2O \) has an anomalously high boiling point due to extensive hydrogen bonding.



The correct trend is: \[ H_2O > H_2Te > H_2Se > H_2S. \]



This confirms that Statement I is true.

Step 2: Explanation of the Boiling Point Anomaly of \( H_2O \)

- Based on molecular mass alone, \( H_2O \) should have the lowest boiling point.


- However, the presence of strong intermolecular hydrogen bonding in \( H_2O \) results in significantly higher boiling point than expected.


- This confirms that Statement II is also true.



Step 3: Conclusion

Since both statements correctly describe the trend and the reasoning, the correct answer is:
\[ \boxed{Both Statement I and Statement II are true.} \] Quick Tip: Hydrogen bonding significantly elevates the boiling points of molecules like \( H_2O \), despite their lower molecular mass compared to heavier hydrides in the same group.

Step 1: Understanding the Friedel-Crafts Alkylation Reaction with Aniline


- The Friedel-Crafts alkylation reaction requires a Lewis acid catalyst such as \( AlCl_3 \).


- Aniline (\( C_6H_5NH_2 \)) has a lone pair on nitrogen, which interacts with \( AlCl_3 \), forming a complex.


- This deactivates the benzene ring, preventing it from undergoing the Friedel-Crafts alkylation reaction.


- Hence, Statement I is true.



Step 2: Understanding Gabriel Synthesis


- Gabriel synthesis is used for the preparation of primary amines.


- It involves the reaction of phthalimide with an alkyl halide followed by hydrolysis.


- However, this method is suitable for preparing only aliphatic amines.


- Aniline (\( C_6H_5NH_2 \)) is an aromatic amine and cannot be prepared using Gabriel synthesis.


- Hence, Statement II is true.

Step 3: Conclusion

Since both statements correctly describe the reactivity of aniline, the correct answer is:
\[ \boxed{Both Statement I and Statement II are true.} \] Quick Tip: Aniline does not undergo Friedel-Crafts reactions due to complex formation with the Lewis acid catalyst. Gabriel synthesis is only suitable for aliphatic amines and cannot be used for preparing aromatic amines.

Step 1: Electronic Configuration of \( [Co(NH_3)_6]^{3+} \) and \( [CoF_6]^{3-} \)



- The oxidation state of cobalt in both complexes is +3.


- \( Co^{3+} \) has an electronic configuration of \( 3d^6 \).



Step 2: Nature of Ligands and Magnetic Properties



- \( [Co(NH_3)_6]^{3+} \) (Diamagnetic):


- \( NH_3 \) is a strong field ligand → induces low spin configuration.


- Electrons pair up in the \( t_{2g} \) orbitals, making the complex diamagnetic.



- \( [CoF_6]^{3-} \) (Paramagnetic):


- \( F^- \) is a weak field ligand → induces high spin configuration.


- Unpaired electrons remain, making the complex paramagnetic.



Conclusion:
Both Statement I and Statement II are correct, hence option (4) is correct. Quick Tip: In octahedral complexes, strong field ligands (like NH\(_3\), CN\(^-\)) cause low spin, while weak field ligands (like F\(^-\), Cl\(^-\)) cause high spin.

Step 1: Understanding Fehling’s Solution



Fehling’s solution is used to test for reducing sugars. It consists of two separate solutions:



- Fehling’s Solution A → Aqueous copper(II) sulphate (CuSO4).


- Fehling’s Solution B → Alkaline sodium potassium tartrate (Rochelle’s salt).



Step 2: Working Principle
- When a reducing sugar (like glucose) is present, Cu\(^{2+}\) ions from Fehling’s solution A are reduced to Cu\(_2\)O, forming a brick-red precipitate.



Conclusion:
Since Fehling’s solution ‘A’ contains aqueous copper(II) sulphate, the correct answer is option (4). Quick Tip: Fehling’s solution is widely used for testing glucose and fructose, as they are reducing sugars.

Step 1: Understanding Henry’s Law



Henry’s law states that the solubility of a gas (\( S \)) in a liquid is inversely proportional to the Henry’s law constant (\( K_H \)):
\[ S \propto \frac{1}{K_H} \]



Step 2: Applying the given values



- \( K_H \) for gas B = \( 2 \times 10^{-5} \) kbar (smallest) → Highest solubility


- \( K_H \) for gas C = 35 kbar → Moderate solubility


- \( K_H \) for gas A = 145 kbar (largest) → Lowest solubility



Conclusion:
Since lower \( K_H \) means higher solubility, the correct order is B > C > A. Quick Tip: A gas with a lower Henry’s law constant (\( K_H \)) is more soluble in water.

Step 1: Understanding Faraday’s Laws of Electrolysis



- The number of Faradays (F) required is determined by the electrons exchanged per reaction.



Step 2: Matching the Conversions with Faradays



- \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) → Requires 2F


- \( MnO_4^- + 5e^- \rightarrow Mn^{2+} \) → Requires 5F


- \( 1.5 \) moles of \( Ca^{2+} + 2e^- \rightarrow Ca \) → Requires 1.5 × 2F = 3F


- \( FeO \rightarrow Fe_2O_3 \) → Involves 1 electron transfer per Fe, total = 1F



Conclusion:
Using the above calculations, the correct match is option (4). Quick Tip: Use Faraday’s law: 1 Faraday deposits 1 mole of a substance per unit charge equivalent.

Step 1: Understanding Quantum Numbers



- Principal Quantum Number (\( n \)) → Size of orbital.


- Azimuthal Quantum Number (\( l \)) → Shape of orbital.


- Magnetic Quantum Number (\( m_l \)) → Orientation of orbital.


- Spin Quantum Number (\( m_s \)) → Spin of electron.



Step 2: Matching with the given information



- \( m_l \) (Magnetic Quantum Number) → III (Orientation of orbital).


- \( m_s \) (Spin Quantum Number) → IV (Spin of electron).


- \( l \) (Azimuthal Quantum Number) → I (Shape of orbital).


- \( n \) (Principal Quantum Number) → II (Size of orbital).



Conclusion:
Thus, the correct match is option (1). Quick Tip: The principal quantum number (\( n \)) determines size, while the azimuthal quantum number (\( l \)) determines shape.

Step 1: Understanding the spin-only magnetic moment



The spin-only magnetic moment is given by:
\[ \mu_s = \sqrt{n(n+2)} \ BM \]

where \( n \) is the number of unpaired electrons.



Step 2: Calculate unpaired electrons for each ion



- \( Ti^{3+} \) (d\(^1\)) → \( n = 1 \) → \( \mu_s = \sqrt{3} \) BM


- \( Cr^{2+} \) (d\(^4\)) → \( n = 4 \) → \( \mu_s = \sqrt{24} \) BM


- \( Mn^{2+} \) (d\(^5\)) → \( n = 5 \) → \( \mu_s = \sqrt{35} \) BM


- \( Fe^{2+} \) (d\(^6\)) → \( n = 4 \) → \( \mu_s = \sqrt{24} \) BM


- \( Sc^{3+} \) (d\(^0\)) → No unpaired electrons → \( \mu_s = 0 \) BM



Step 3: Identify matching values
\( Cr^{2+} \) and \( Fe^{2+} \) both have four unpaired electrons, so they have the same spin-only magnetic moment.



Conclusion:
Thus, the correct pair is B and D (Cr\(^{2+}\) and Fe\(^{2+}\)). Quick Tip: Magnetic moment depends on unpaired electrons. Ions with the same number of unpaired electrons have the same spin-only magnetic moment.

Step 1: Understanding glucose reactions



- Tollen’s reagent (A) → Glucose gives a positive test (reducing sugar).


- Schiff’s reagent (B) → Used for aldehydes, but glucose does not react.


- HCN (C) → Reacts with glucose at carbonyl (-CHO) position to form cyanohydrins.


- Hydroxylamine (\( NH_2OH \)) (D) → Reacts with carbonyl groups to form oximes.


- Sodium bisulfite (\( NaHSO_3 \)) (E) → Reacts with aldehydes, but glucose does not react.



Step 2: Identify non-reacting reagents



Glucose does not react with Schiff’s reagent (B) and sodium bisulfite (E).



Conclusion:
Thus, the correct answer is B and E. Quick Tip: Glucose gives positive tests with Tollen’s reagent, HCN, and hydroxylamine, but not with Schiff’s reagent or sodium bisulfite.

Step 1: Understanding the given reagents



- Ozonolysis (O\(_3\) followed by Zn-H\(_2\)O) → Breaks double bonds to form aldehydes or ketones.


- Chromium trioxide (CrO\(_3\)) → Oxidizes alcohols to carboxylic acids or ketones.


- Alkaline potassium permanganate → Strong oxidation of alkenes or alcohols.



Step 2: Matching with reactions



- A → IV (Ozonolysis reaction)


- B → I (A specific reaction type)


- C → II (Oxidation using CrO\(_3\))


- D → III (Oxidation with KMnO\(_4\) and heat)



Conclusion:
The correct match is A-IV, B-I, C-II, D-III. Quick Tip: Ozonolysis breaks double bonds, CrO\(_3\) performs oxidation, and KMnO\(_4\) is a strong oxidizer in basic medium.

Step 1: Understanding electronegativity trends



Electronegativity increases across a period (left to right) and decreases down a group.



Step 2: Compare electronegativities of given elements



- Silicon (Si): Least electronegative in the group.


- Carbon (C): More electronegative than Si.


- Nitrogen (N): More electronegative than C.


- Oxygen (O): More electronegative than N.


- Fluorine (F): The most electronegative element in the periodic table.



Thus, the correct increasing order is:
\[ Si < C < N < O < F \]

Conclusion:
The correct answer is Si < C < N < O < F. Quick Tip: Electronegativity increases across periods and decreases down groups in the periodic table. Fluorine is the most electronegative element.

Step 1: Understanding the Bonding in Each Molecule



- Ethane (\(C_2H_6\)): Contains a single bond (\(\sigma\)-bond) between two carbon atoms. \(\Rightarrow\) Matches with III.


- Ethene (\(C_2H_4\)): Contains one \(\sigma\)-bond and one \(\pi\)-bond between two carbon atoms. \(\Rightarrow\) Matches with IV.


- Carbon molecule (\(C_2\)): In the excited state, it has one \(\sigma\)-bond and two \(\pi\)-bonds. \(\Rightarrow\) Matches with I.


- Ethyne (\(C_2H_2\)): Contains one \(\sigma\)-bond and two \(\pi\)-bonds. \(\Rightarrow\) Matches with II.



Step 2: Matching List-I with List-II

\[ \begin{aligned} A. & \quad Ethane \quad \to \quad III. One \( \sigma \)-bond
B. & \quad Ethene \quad \to \quad IV. One \( \sigma \)-bond and one \( \pi \)-bond
C. & \quad Carbon molecule, \( C_2 \) \quad \to \quad I. One \( \sigma \)-bond and two \( \pi \)-bonds
D. & \quad Ethyne \quad \to \quad II. Two \( \pi \)-bonds \end{aligned} \]

Step 3: Conclusion

Since the correct matching follows the order \( A-III, B-IV, C-I, D-II \), the correct answer is:
\[ \boxed{A-III, B-IV, C-I, D-II} \] Quick Tip: In hydrocarbons, single bonds are \( \sigma \)-bonds, double bonds contain one \( \sigma \) and one \( \pi \)-bond, and triple bonds contain one \( \sigma \) and two \( \pi \)-bonds. The carbon molecule (\( C_2 \)) in its excited state has one \( \sigma \)-bond and two \( \pi \)-bonds.

Step 1: Understanding ionization enthalpy trend



- Ionization enthalpy increases across a period (left to right) and decreases down a group.


- Be has higher ionization enthalpy than B due to its fully filled s-orbital stability.


- N has the highest ionization energy due to half-filled p-orbital stability.



Step 2: Compare the given elements
\[ Li < B < Be < C < N \]



Conclusion:
The correct answer is Li < B < Be < C < N. Quick Tip: Ionization enthalpy increases left to right across a period and decreases down a group. Exceptions exist due to filled and half-filled orbitals.

Understanding Lucas reagent test



- Lucas reagent (conc. HCl + ZnCl\(_2\)) distinguishes primary, secondary, and tertiary alcohols.


- Tertiary alcohols react instantaneously, producing a cloudy solution immediately.


- Secondary alcohols take some time to react.


- Primary alcohols do not react at room temperature.



Conclusion:
The correct answer is tertiary alcohol. Quick Tip: Tertiary alcohols give instantaneous turbidity in the Lucas test, while secondary alcohols react slowly, and primary alcohols do not react.

Step 1: Arrhenius Equation
\[ k = A e^{-E_a/RT} \]

Taking the natural logarithm on both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]



Step 2: Using values at two different temperatures

The activation energy \( E_a \) can be calculated using:
\[ \ln \left(\frac{k_1}{k_2}\right) = \frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1} \right) \]



Thus, knowing \( k \) at two different temperatures allows us to calculate activation energy.

Conclusion:
The correct answer is Rate constant at two different temperatures. Quick Tip: The Arrhenius equation helps calculate activation energy using rate constants measured at two different temperatures.

Step 1: Understanding Resonance in Carbonate Ion



- The carbonate ion (CO\(_3^{2-}\)) exhibits resonance where the double bond is delocalized.


- It has three equivalent resonance structures, making it the correct answer.



Step 2: Evaluating Other Options



- BF\(_3\) is non-polar due to trigonal planar symmetry, so (1) is incorrect.


- Dipole moment of NH\(_3\) is greater than NF\(_3\), so (2) is incorrect.


- Ozone (O\(_3\)) has only two resonance structures, so (4) is incorrect.



Conclusion:
The correct answer is (3) Three canonical forms for CO\(_3^{2-}\). Quick Tip: Resonance occurs when multiple valid Lewis structures exist for a molecule, delocalizing electrons and stabilizing the species.

Step 1: Understanding Diamagnetism in Lanthanoids



- Diamagnetism occurs when an ion has no unpaired electrons in its electronic configuration.


- Ce\(^{4+}\) (Xe configuration) and Yb\(^{2+}\) (completely filled f-orbitals) are diamagnetic.



Step 2: Evaluating Other Options



- Other lanthanoid ions have unpaired electrons and hence exhibit paramagnetism.



Conclusion:
The correct answer is Ce\(^{4+}\) and Yb\(^{2+}\). Quick Tip: A lanthanoid ion is diamagnetic if it has completely filled or empty f-orbitals.

Step 1: Using Arrhenius Equation


\[ \log \left(\frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2} \right) \]



Step 2: Substituting Values
\[ \log 4 = \frac{E_a}{2.303 \times 8.314} \times \left(\frac{30}{(300 \times 330)}\right) \]

Solving, \[ E_a = 38.04 kJ/mol \]



Conclusion:
The correct answer is 38.04 kJ/mol. Quick Tip: Activation energy determines the sensitivity of reaction rates to temperature changes.

Step 1: Using Arrhenius Equation
\[ \log \left(\frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2} \right) \]

Step 2: Substituting Values
\[ \log 4 = \frac{E_a}{2.303 \times 8.314} \times \left(\frac{30}{(300 \times 330)}\right) \]

Solving, \[ E_a = 38.04 kJ/mol \]

Conclusion:
The correct answer is 38.04 kJ/mol. Quick Tip: Activation energy determines the sensitivity of reaction rates to temperature changes.

Cyclohexene undergoes oxidation with acidic KMnO4 to give benzoic acid.

The reaction proceeds as:
\[ C_6H_{10} + KMnO_4/H^+ \longrightarrow C_6H_5COOH \]

Thus, option (1) is correct. Quick Tip: Oxidation of alkenes with KMnO\(_4\)/H\(^+\) results in cleavage of the double bond, forming carboxylic acids if the alkene is terminal.

Step 1: Formula for Work Done in Isothermal Expansion
\[ W = - nRT \ln \left(\frac{P_1}{P_2}\right) \]



Step 2: Substituting Values
\[ W = - (1) \times (2.0) \times (298) \ln \left(\frac{20}{10}\right) \]
\[ = - 2 \times 298 \times 0.693 \]
\[ = - 413.14 cal \]

Conclusion:
The correct answer is -413.14 cal. Quick Tip: For isothermal expansion, work done is given by \[ W = - nRT \ln \left(\frac{P_1}{P_2}\right) \]

Step 1: Reaction with PCl\(_3\)



- PCl\(_3\) reacts with alcohol (ROH) forming alkyl chloride (RCl) and phosphorous acid (H\(_3\)PO\(_3\)).



Step 2: Reaction with PCl\(_5\)



- PCl\(_5\) reacts with alcohol (ROH) forming alkyl chloride (RCl), hydrochloric acid (HCl), and phosphoryl chloride (POCl\(_3\)).



Conclusion:
The correct answer is H\(_3\)PO\(_3\) and POCl\(_3\). Quick Tip: Phosphorus halides react with alcohols forming alkyl chlorides and phosphorus oxoacids.

Step 1: Understanding Qualitative Analysis Groups


- Group 0: Alkali and alkaline earth metals (e.g., Mg\(^{2+}\))

- Group I: Pb\(^{2+}\), Ag\(^{+}\)

- Group II: Cu\(^{2+}\)

- Group III: Al\(^{3+}\)

- Group IV: Co\(^{2+}\)

- Group V: Ba\(^{2+}\)


Step 2: Arranging the Ions



The increasing group order is B, A, D, C, E. Quick Tip: Qualitative analysis separates cations based on solubility differences in different reagents.

Step 1: Using the Osmotic Pressure Equation
\[ \Pi = CRT \]

where slope \(S = RT\).



Step 2: Calculating Temperature
\[ T = \frac{S}{R} = \frac{25.73}{0.083} \]
\[ T = 310 K = 37^\circ C \]



Conclusion:
The correct answer is 37°C. Quick Tip: Osmotic pressure is directly proportional to temperature and solute concentration.

We are given the following information:

- Percentage of A = 32%

- Percentage of B = 20%

- Percentage of C = \( 100% - (32% + 20%) = 48% \)


We are also given the atomic masses:


- Atomic mass of A = 64 u

- Atomic mass of B = 40 u

- Atomic mass of C = 32 u


Step 1: Calculate the moles of each element



The number of moles of each element is given by: \[ Moles of A = \frac{Mass of A}{Molar mass of A} = \frac{32}{64} = 0.5 moles \] \[ Moles of B = \frac{Mass of B}{Molar mass of B} = \frac{20}{40} = 0.5 moles \] \[ Moles of C = \frac{Mass of C}{Molar mass of C} = \frac{48}{32} = 1.5 moles \]



Step 2: Find the simplest whole number ratio

Now, we divide the moles of each element by the smallest number of moles (which is 0.5): \[ Ratio of A = \frac{0.5}{0.5} = 1 \] \[ Ratio of B = \frac{0.5}{0.5} = 1 \] \[ Ratio of C = \frac{1.5}{0.5} = 3 \]



Step 3: Determine the empirical formula

The simplest whole number ratio of A, B, and C is 1:1:3. Therefore, the empirical formula is \( ABC_3 \).



Thus, the correct answer is:
\[ \boxed{ABC_3} \] Quick Tip: The empirical formula can be determined by calculating the moles of each element based on their percentages and atomic masses, then finding the simplest whole number ratio.

Step 1: Formation of A using PBr\(_3\)



- PBr\(_3\) replaces the hydroxyl (-OH) group with a bromine (Br) atom through nucleophilic substitution.


- The major product A is 1-Bromo-1-methylcyclohexane.



Step 2: Elimination to Form B



- Treating A with alcoholic KOH (alc. KOH, ) leads to -elimination (E2 reaction).


- This removes HBr, forming a double bond at the most stable position.


- The major product B is 1-Methylcyclohexene.



Conclusion:
The correct products are A = 1-Bromo-1-methylcyclohexane, B = 1-Methylcyclohexene. Quick Tip: \textbf{PBr\(_3\) is a mild reagent} for converting alcohols to alkyl bromides, while alc. KOH favors elimination (E2 mechanism).

Step 1: Understanding the role of acid in Mohr’s salt preparation


- Mohr’s salt, FeSO\(_4\).(NH\(_4\))\(_2\)SO\(_4\).6H\(_2\)O, is prone to hydrolysis, forming Fe(OH)\(_2\), which is insoluble.


- Dilute sulphuric acid prevents this hydrolysis by providing excess H\(^+\) ions, shifting equilibrium towards Fe\(^{2+}\) ions in solution.



Conclusion:
The correct answer is dilute sulphuric acid because it prevents the oxidation and hydrolysis of Fe\(^{2+}\) ions. Quick Tip: Always add dilute H\(_2\)SO\(_4\) when preparing Fe\(^{2+}\) solutions to prevent oxidation to Fe\(^{3+}\) and hydrolysis.

Step 1: Define degree of dissociation (\(\alpha\))
\[ \alpha = \frac{moles dissociated at equilibrium}{initial moles} \]



Step 2: Use given concentrations and formula
\[ \alpha = \frac{0.1 - 2.8 \times 10^{-3}}{0.1} \]
\[ = \frac{0.0972}{0.1} = 0.717 \]



Conclusion:
The correct degree of dissociation is 0.717. Quick Tip: For equilibrium problems, always define initial concentration, change, and equilibrium concentration (ICE method) to calculate \(\alpha\).

Step 1: Hydrolysis of Nitrile (-CN) to Carboxylate (-COO\(^-\))



- The first step involves alkaline hydrolysis of butanenitrile (-CN) to form butanoate ion (A).



Step 2: Hoffmann Bromamide Reaction



- Treating butanoate ion with NaOH + Br\(_2\) (Hoffmann reaction) leads to decarboxylation and formation of Propylamine (C).


\[ CH_3 - CH_2 - CH_2 - CN \xrightarrow{OH^-} CH_3 - CH_2 - CH_2 - COO^- \xrightarrow{NaOH, Br_2} CH_3 - CH_2 - CH_2 - NH_2 \]

Conclusion:
The final major product C is Propylamine. Quick Tip: \textbf{Hoffmann Bromamide Reaction} converts an amide (-CONH\(_2\)) to a primary amine (-NH\(_2\)) with one carbon less.

Step 1: Use Faraday’s First Law of Electrolysis
\[ m = \frac{ZIt}{F} \]

where,
- \( m \) = mass of deposited copper


- \( Z \) = electrochemical equivalent = \( \frac{Molar Mass}{nF} = \frac{63}{2 \times 96487} \)


- \( I = 9.6487 \) A


- \( t = 100 \) s


- \( F = 96487 \) C



Step 2: Substituting values
\[ m = \frac{63 \times 9.6487 \times 100}{2 \times 96487} \]
\[ m = 0.315 g \]



Conclusion:
The mass of copper deposited is 0.315 g. Quick Tip: \textbf{Faraday’s Laws of Electrolysis:} The amount of substance deposited is directly proportional to charge passed through the solution.

Step 1: Understanding Homoleptic and Heteroleptic Complexes



- A homoleptic complex contains only one type of ligand.


- Example: [Co(NH\(_3\))\(_6\)]\(^{3+}\) contains only NH\(_3\) ligands.



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


- Example: [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) contains NH\(_3\) and Cl ligands.



Step 2: Evaluating the statements



- Statement I is true: [Co(NH\(_3\))\(_6\)]\(^{3+}\) is homoleptic and [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) is heteroleptic.


- Statement II is true: [Co(NH\(_3\))\(_6\)]\(^{3+}\) has only NH\(_3\) ligands, while [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) has two different ligands (NH\(_3\) and Cl).



Conclusion:
Both statements are correct. Quick Tip: \textbf{Homoleptic complexes} have only one type of ligand, whereas \textbf{heteroleptic complexes} have more than one type.

Understanding Lecithin


Step 1: Defining Lecithin

Lecithin is a phospholipid found in living tissues, especially in the cell membrane.
It plays a crucial role in maintaining membrane fluidity and cell signaling.


Step 2: Evaluating Other Options

Glycerides: These are esters formed from glycerol and fatty acids, differing from phospholipids.
Carbohydrates: Sugars and starches, which serve as energy sources rather than structural components of membranes.
Amino Acids: The building blocks of proteins, not phospholipids.


Conclusion:


Lecithin belongs to the phospholipid category. Quick Tip: \textbf{Phospholipids} are essential components of cell membranes and play a key role in emulsification of fats.

Understanding Causes of Biodiversity Loss


Step 1: Identifying Major Causes of Biodiversity Loss

Over-exploitation (A): Excessive use of species for commercial purposes significantly reduces biodiversity.
Co-extinction (B): If one species goes extinct, species that depend on it also face a high risk of extinction.
Habitat loss and fragmentation (D): The destruction of natural habitats leads to the loss of biodiversity and ecosystem imbalance.


Step 2: Evaluating Other Options

Mutation (C): Mutations contribute to genetic variation rather than biodiversity loss.
Migration (E): Migration can impact local biodiversity but is not a direct cause of biodiversity loss.


Conclusion:

The primary causes of biodiversity loss are over-exploitation, co-extinction, and habitat loss. Quick Tip: \textbf{Biodiversity loss is mainly caused by habitat destruction, over-exploitation, pollution, and climate change.}

Understanding Fungal Classification


Step 1: Basis of Fungal Classification

Fungi are classified based on their spore formation, fruiting bodies, and mycelial structure.


Step 2: Evaluating the Classification Criteria


Mode of Nutrition (Incorrect Criterion):

Most fungi are heterotrophic, including saprophytic, parasitic, or mutualistic forms.
However, fungal classification is based on structural and reproductive characteristics, not nutrition.


Spore Formation (Correct Criterion):

Fungi reproduce via spores, and their type and formation vary among different fungal groups.


Fruiting Body (Correct Criterion):

The structure of the fruiting body differentiates fungal groups such as Basidiomycetes and Ascomycetes.


Morphology of Mycelium (Correct Criterion):

The hyphal structure and mycelial organization are key features in fungal classification.



Conclusion:

Fungal classification is based on spore formation, fruiting body structure, and mycelial morphology, but not mode of nutrition. Quick Tip: \textbf{Fungi are classified based on reproductive structures, not their mode of nutrition.}

Understanding Snapdragon Flower Color Inheritance


Step 1: Incomplete Dominance in Snapdragon Flowers

Snapdragon flowers exhibit incomplete dominance.
A cross between a red-flowered (RR) and a pink-flowered (Rr) plant follows the inheritance pattern:
\[ RR \times Rr \rightarrow Produces Red and Pink flowers in a 1:1 ratio. \]


Step 2: Punnett Square Analysis
\[ \begin{array}{|c|c|c|} \hline & R & r
\hline R & RR (Red) & Rr (Pink)
\hline \end{array} \]


50% of the offspring = Red (RR).
50% of the offspring = Pink (Rr).


Conclusion:

The correct answer is that the offspring will consist of red and pink flowered plants. Quick Tip: \textbf{Incomplete dominance occurs when neither allele is completely dominant, leading to an intermediate phenotype.}

Step 1: Understanding Mendel’s Law of Dominance


- Mendel’s Law of Dominance states that in a pair of alleles, one allele is dominant over the other.


- The dominant allele is expressed in the F1 generation, while the recessive allele is masked but reappears in F2 generation in a 3:1 ratio.



Step 2: Evaluating the Given Statements



- A (True): One allele is dominant while the other is recessive.


- B (False): Mendel’s experiments showed that recessive traits do reappear, but not as independent characters in F2.


- C (True): Genes exist in pairs in diploid organisms.


- D (True): Mendel called the unit of inheritance a "factor" (now known as a gene).


- E (True): Only one of the parental characters appears in F1 generation due to dominance.


Conclusion: The correct answer is A, C, D, and E only. Quick Tip: \textbf{Mendel’s Law of Dominance explains why dominant traits appear in F1 generation, while recessive traits reappear in F2 generation.}

Step 1: Understanding Species Richness in Tropical Regions


- Tropical ecosystems exhibit high biodiversity due to favorable environmental conditions.


- High solar energy availability supports greater primary productivity.



Step 2: Evaluating the Given Statements



- A (True): Stability over millions of years allows more species diversification.


- B (False): Tropical environments are less seasonal, making them more predictable for species survival.


- C (True): More solar energy leads to higher photosynthesis, supporting a larger biomass.


- D (True): Constant environments allow for niche specialization, leading to more species.


- E (True): Predictable climate enables stable ecosystems with complex food webs.


Conclusion: The correct answer is A, C, D, and E only. Quick Tip: \textbf{Tropical regions are biodiversity hotspots due to stable climate, high solar energy, and evolutionary history.}

Step 1: Identifying Thickened Cell Walls in Plant Tissues



- Xylem vessels and tracheids have thickened secondary walls to support water transport.


- The inner walls are highly thickened, while outer walls are relatively thin.



Step 2: Evaluating the Components in the Given Figure



- D: Could be parenchyma, which has thin walls.


- A: Possibly phloem, which has sieve elements with thin walls.


- B: Could be companion cells, supporting phloem transport.


- C: Likely to be xylem vessels, which have thickened secondary walls for conducting water under tension.



Conclusion: The component C has thin outer walls and highly thickened inner walls, making it the correct answer. Quick Tip: \textbf{Xylem vessels have thickened secondary walls to withstand water transport pressure.}

Step 1: Understanding Enzyme Cofactors



- Carboxypeptidase is a metalloenzyme involved in protein digestion.


- It requires a metallic ion as a cofactor to function efficiently.



Step 2: Identifying the Correct Cofactor



- Niacin (Vitamin B3): Functions as a precursor for NAD+ and NADP+.


- Flavin (Riboflavin derivative): Functions in redox reactions.


- Haem: Present in cytochromes, catalase, and hemoglobin.


- Zinc (Correct Answer): Acts as a structural and catalytic cofactor for carboxypeptidase.



Conclusion: Zinc is the required cofactor for carboxypeptidase. Quick Tip: \textbf{Metalloenzymes, such as carboxypeptidase, require metallic ions like Zinc for proper catalytic activity.}

Step 1: Understanding Meiotic Prophase-I Stages


- Leptotene Stage: Chromosomes appear as thin threads and gradually become visible under a microscope.


- Diplotene Stage: Characterized by chiasmata formation and dissolution of the synaptonemal complex.



Step 2: Evaluating the Given Statements



- Statement I (True): Chromosomes first become visible during the leptotene stage.


- Statement II (True): The synaptonemal complex dissolves at the diplotene stage, allowing recombination events to become visible.



Conclusion: Since both statements correctly describe their respective meiotic stages, the answer is option (4). Quick Tip: \textbf{Meiotic Prophase-I is divided into five stages: Leptotene, Zygotene, Pachytene, Diplotene, and Diakinesis. Each stage has distinct chromosomal changes.}

Step 1: Understanding DNA Integration in Host Cells



- When a foreign gene is introduced into an alien organism, two possible fates occur:


- Integration into the genome (B).


- Replication along with host DNA (C).



Step 2: Evaluating the Given Statements



- A (False): DNA alone cannot multiply without a replication origin (ORI site).


- B (True): The foreign gene may integrate into the genome via recombination.


- C (True): Once integrated, it may be replicated and inherited along with host DNA.


- D (False): The foreign gene may integrate into the genome, so it does not necessarily remain separate.


- E (False): DNA without an ORI cannot replicate independently.



Conclusion: The correct answer is B and C only. Quick Tip: \textbf{For successful gene expression, a foreign gene must integrate into the host genome or replicate with a suitable vector.}

Step 1: Understanding Floral Arrangements



- Hypogynous Flowers: Ovary is superior, other floral parts are below.


- Perigynous Flowers: Ovary is semi-inferior, floral parts are around the ovary.


- Epigynous Flowers: Ovary is inferior, floral parts are above.



Step 2: Evaluating the Given Figures



- Figure (a): Ovary appears semi-inferior, surrounded by floral parts → Perigynous.


- Figure (b): Similar structure is seen, indicating Perigynous condition.



Conclusion: The correct answer is Perigynous for both (a) and (b). Quick Tip: \textbf{Floral structures can be classified based on ovary position: Hypogynous (superior ovary), Perigynous (semi-inferior ovary), and Epigynous (inferior ovary).}

Understanding Genetic Terminology and Matching



Step 1: Understanding the Terminology


Allele (III): Different alternative forms of a gene.
Test Cross (IV): Crossing an F\textsubscript{1} hybrid with a homozygous recessive parent to determine genotype.
Back Cross (I): Crossing an F\textsubscript{1} hybrid with any of the original parents.
Ploidy (II): The number of chromosome sets in an organism.


Conclusion:

The correct matching is: \[ A \to III, \quad B \to IV, \quad C \to I, \quad D \to II. \] Quick Tip: \textbf{A test cross is used to determine the genotype of an organism by crossing it with a homozygous recessive parent.}

Step 1: Understanding the Process



- Dedifferentiation occurs when a fully differentiated cell regains its ability to divide.


- Interfascicular cambium originates from parenchyma cells, which were previously non-dividing.



Step 2: Eliminating Incorrect Options



- Redifferentiation (Incorrect): A dedifferentiated cell can later re-specialize, but the formation of interfascicular cambium is not redifferentiation.


- Maturation (Incorrect): Maturation refers to the final functional form of a cell.


- Differentiation (Incorrect): Differentiation is a one-way process where cells become specialized.



Conclusion: The process is dedifferentiation, making the correct answer option (2). Quick Tip: \textbf{Dedifferentiation allows plants to regain lost tissues and heal wounds by forming meristematic cells.}

Step 1: Understanding Actinomorphic Flowers



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


- Datura is an example of an actinomorphic flower.



Step 2: Identifying Incorrect Options



- Cassia (Incorrect): It is zygomorphic (bilaterally symmetrical).


- Pisum (Incorrect): It is zygomorphic.


- Sesbania (Incorrect): It is also zygomorphic.



Conclusion: The correct answer is Datura. Quick Tip: \textbf{Actinomorphic flowers have radial symmetry, meaning they can be divided into equal halves from multiple planes.}

Step 1: Understanding Restriction Enzymes


- Hind II was the first restriction enzyme discovered and recognizes a 6 base-pair (bp) sequence.



Step 2: Identifying Incorrect Options



- 4 bp (Incorrect): Some restriction enzymes recognize 4 bp, but Hind II recognizes 6 bp.


- 8 bp (Incorrect): Hind II does not require an 8 bp sequence.


- 10 bp (Incorrect): Some rare restriction enzymes recognize 10 bp, but not Hind II.



Conclusion: The correct answer is 6 bp. Quick Tip: \textbf{Hind II was the first restriction enzyme discovered and recognizes a 6 bp palindromic sequence.}

Step 1: Understand the dark reaction.


The dark reaction (Calvin Cycle) occurs in the stroma of chloroplasts and does not require light directly.



Step 2: Identify required components.


- CO2 (C): It is the carbon source for glucose synthesis.


- ATP (D) and NADPH (E): These provide energy and reducing power.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Dark reaction relies on ATP and NADPH produced in the light reaction for CO2 fixation.}

Step 1: Understanding Bt toxin.


- Bt toxins are encoded by cry genes and are specific to insect groups.


- Bt toxin exists as an inactive protoxin in *Bacillus thuringiensis* but is activated in the insect's gut.



Step 2: Correct the misconception.


- Activation of Bt toxin occurs in an alkaline pH, not an acidic pH.


Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Bt toxin is only activated in an alkaline pH environment, not acidic.}

Step 1: Understanding logistic growth.


- Logistic growth considers environmental resistance, unlike exponential growth.



Step 2: Role of \( K \).


- \( K \) represents carrying capacity, the maximum sustainable population.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Carrying capacity (K) is the maximum population size an environment can sustain indefinitely.}

Step 1: Understanding totipotency.


- Totipotency is the ability of a single cell to regenerate into a whole plant.



Step 2: Compare with other terms.


- Micropropagation: In-vitro mass propagation.


- Differentiation: Specialization of cells.


- Somatic hybridization: Fusion of protoplasts from different species.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Totipotency is fundamental in plant tissue culture and regeneration.}

Step 1: Understanding conservation methods.


- In-situ conservation: Protecting species in their natural habitat (e.g., National Parks).


- Ex-situ conservation: Protecting species outside their natural habitat (e.g., Zoos, Botanical Gardens).



Step 2: Why biodiversity conservation?


- Conservation includes both in-situ and ex-situ strategies.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Ex-situ conservation involves zoos, botanical gardens, and seed banks.}

Step 1: Understanding the lactose uptake mechanism.


- In bacterial cells, lactose is transported across the cell membrane by Lactose Permease, an enzyme coded by the lacY gene of the lac operon.



Step 2: Role of Permease.


- Permease facilitates active transport of lactose into the bacterial cell, allowing further metabolism by Beta-galactosidase.



Conclusion: The correct option is \( \mathbf{(2) \). Quick Tip: \textbf{Lactose Permease helps bacteria absorb lactose, while Beta-galactosidase hydrolyzes it.}

Step 1: Understanding the process of mitosis.


- During Metaphase, chromosomes align at the metaphase plate, and spindle fibers attach to kinetochores (protein complexes at the centromere).



Step 2: Role of Spindle Fibers.


- Spindle fibers ensure chromatid separation by pulling them towards opposite poles during Anaphase.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Spindle fibers attach to kinetochores during Metaphase to align chromosomes for separation.}

Step 1: Understanding seed anatomy.


- The Radicle is the first structure to emerge during germination and later develops into the root system.



Step 2: Identify the correct part in the diagram.


- The labeled "C" in the given figure corresponds to the Radicle.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{The radicle forms the root, while the plumule develops into the shoot.}

Step 1: Understanding transcription units.


A transcription unit consists of:


- Promoter: Binding site for RNA polymerase.


- Structural gene: Encodes the protein or RNA.


- Terminator: Signals the end of transcription.



Step 2: Why the correct answer?


- Transposons and repressors are regulatory elements, not part of a transcription unit.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: \textbf{A transcription unit in DNA has a promoter, structural gene, and terminator.}

Step 1: Role of ATP and NADPH in the Calvin Cycle.


- The Calvin cycle (C3 cycle) occurs in the stroma of chloroplasts.


- Each CO2 molecule fixed requires:


- 3 ATP for energy.


- 2 NADPH for reduction reactions.



Step 2: Why this requirement?


- ATP is used in the Regeneration Phase.


- NADPH provides electrons for Reduction Phase.


Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: \textbf{For every CO2 fixed in the Calvin cycle, 3 ATP and 2 NADPH are needed.}

Step 1: Understanding the nature of parenchyma and collenchyma.


- Parenchyma is a living tissue and functions in storage, photosynthesis, and secretion.


- Collenchyma is also a living tissue, not a dead tissue.



Step 2: Understanding xylem vessels in plants.


- Gymnosperms lack xylem vessels and transport water via tracheids.


- Angiosperms possess xylem vessels, which improve water conduction.



Conclusion: Statement I is incorrect, but Statement II is correct. Quick Tip: \textbf{Parenchyma and collenchyma are living tissues, while sclerenchyma is dead.}

Step 1: Understanding auxin’s role in plant growth.


- Auxins are growth hormones used as selective herbicides.


- They kill dicot weeds but do not harm monocot grasses like wheat and rice.



Step 2: Why monocots are unaffected?


- Monocots lack broad leaves, so they do not absorb auxins as effectively as dicots.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Auxins are used as selective herbicides to eliminate dicot weeds without harming monocots.}

Step 1: Understanding the concept of a test cross.


- A test cross is performed to determine whether an organism with a dominant phenotype is homozygous (BB) or heterozygous (Bb).


- The organism is crossed with a homozygous recessive (bb) individual.



Step 2: Analyzing the test cross results.


- If all offspring are black-seeded (BB × bb → 100 percent Bb), then the plant is homozygous (BB).


- If a 1:1 ratio of black and white seeds appears (Bb × bb → 50 Bb, 50 bb), the plant is heterozygous (Bb).



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{A test cross always involves crossing with a homozygous recessive (bb) individual.}

Step 1: Understanding enzyme inhibition.


- Competitive inhibition occurs when a molecule similar in structure to the substrate binds to the active site of the enzyme, blocking the actual substrate.



Step 2: Example of Malonate.


- Malonate resembles succinate, the substrate for Succinic dehydrogenase, and binds to its active site.


- This prevents the enzyme from converting succinate to fumarate in the Krebs cycle.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Competitive inhibitors resemble the substrate and compete for the enzyme's active site.}

Step 1: Understanding microbial products.


- Clostridium butylicum produces butyric acid (A-III).


- Saccharomyces cerevisiae (yeast) is used in ethanol fermentation (B-I).


- Trichoderma polysporum is a source of Cyclosporin-A, an immunosuppressive drug (C-IV).


- Streptococcus sp. produces Streptokinase, which helps in dissolving blood clots (D-II).



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Microbes play a crucial role in biotechnology, producing antibiotics, enzymes, and industrial products.}

Step 1: Understanding the role of IUCN.


- IUCN (International Union for Conservation of Nature) is responsible for Red List categorization, which assesses global species extinction risks.


- It classifies species into Endangered (EN), Vulnerable (VU), Critically Endangered (CR), and Extinct (EX) categories.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: \textbf{The IUCN Red List provides data on species conservation status globally.}

Step 1: Understanding the function of cell organelles.


- Nucleolus is responsible for ribosomal RNA (rRNA) synthesis (A-III).


- Centriole has a cartwheel structure, essential for mitotic spindle formation (B-II).


- Leucoplasts store starch, lipids, and proteins (C-IV).


- Golgi apparatus is involved in lipid synthesis, including glycolipids (D-I).



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Cell organelles have specialized functions essential for maintaining cellular life.}

Step 1: Understanding fungal classifications.


- Rhizopus is commonly known as bread mould (A-III).


- Ustilago is a smut fungus, causing diseases in crops (B-II).


- Puccinia is a rust fungus, responsible for rust disease in plants (C-IV).


- Agaricus is a common mushroom, widely consumed (D-I).



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Different fungi cause various plant diseases and also have economic importance.}

Understanding pollination in aquatic plants.


- Statement A is incorrect as Vallisneria flowers are not colourful and do not produce nectar.


- Statement B is correct because water lily is pollinated by insects, not water.


- Statement C is correct as pollen grains in water-pollinated species have protective coatings.


- Statement D is correct since some hydrophytic plants have long ribbon-like pollen grains.


- Statement E is correct as some hydrophytes rely on passive water transport for pollination.



Conclusion: The correct option is \( \mathbf{(3)} \). Quick Tip: \textbf{Water pollination is rare and occurs mainly in submerged hydrophytes like Vallisneria and Hydrilla.}

Understanding bulliform cells.


- Bulliform cells are large, thin-walled epidermal cells found in monocot leaves.


- They help in rolling and folding of leaves to reduce water loss during drought conditions.


- When water availability is low, bulliform cells lose turgidity, causing the leaves to curl inward (Option 4).



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Bulliform cells play a key role in leaf movements and drought resistance in grasses like maize and sugarcane.}

Understanding types of stamens.


- Monoadelphous stamens (A) have filaments fused into a single bundle, found in China-rose (A-IV).


- Diadelphous stamens (B) have filaments fused into two groups, seen in Pea (B-II).


- Polyadelphous stamens (C) have filaments fused into multiple groups, found in Citrus (C-I).


- Epiphyllous stamens (D) are fused with perianth, found in Lily (D-III).



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Different fusion types of stamens help in classifying plants systematically based on their floral morphology.}

Understanding oxidation and non-oxidation steps in the TCA cycle.


- The oxidation steps in the Krebs cycle involve NAD+ or FAD+ as electron acceptors.


- The conversion of Succinyl-CoA to Succinic acid does not involve oxidation. Instead, it is associated with substrate-level phosphorylation, generating GTP or ATP.


- Other steps involve oxidation reactions, either removing hydrogens or electrons.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{TCA cycle produces ATP through substrate-level phosphorylation and energy-rich electron carriers via oxidation steps.}

Understanding chloroplast DNA.


- Chloroplast DNA (cpDNA) is circular and double-stranded, similar to bacterial DNA.


- It is present in multiple copies per chloroplast and encodes essential photosynthetic and metabolic genes.


- Unlike nuclear DNA, cpDNA lacks histones and replicates independently through a semi-autonomous process.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Chloroplasts, like mitochondria, have their own DNA and ribosomes, supporting the endosymbiotic theory.}

Understanding the key discoveries in genetics.


- Frederick Griffith (1928) discovered transformation in bacteria, leading to the idea of a "transforming principle."


- Francois Jacob and Jacque Monod worked on gene regulation and proposed the Lac operon model.


- Har Gobind Khorana helped in deciphering the genetic code.


- Meselson and Stahl experimentally proved the semi-conservative replication of DNA.



Conclusion: The correct option is \( \mathbf{(1)} \). Quick Tip: \textbf{Genetic discoveries have helped decode DNA functions, gene expression, and cellular mechanisms.}

Understanding somatic hybridization.


- Somatic hybridization is a method where protoplasts (cells without cell walls) are fused to form a hybrid cell.


- It is used in plant tissue culture to combine genetic material from two different plant species.


- Fusion of protoplasts is achieved using polyethylene glycol (PEG) or electrofusion.



Conclusion: The correct option is \( \mathbf{(2)} \). Quick Tip: \textbf{Somatic hybridization allows the combination of desirable traits from different plant species.}

Understanding wind pollination characteristics.


- Wind-pollinated plants (anemophilous) have well-exposed stamens to facilitate pollen dispersal.


- Their pollen grains are lightweight and non-sticky, aiding easy transport by wind.


- Examples include grasses, maize, and many cereal crops.



Conclusion: The correct option is \( \mathbf{(4)} \). Quick Tip: \textbf{Wind-pollinated flowers typically have feathery stigmas and produce large amounts of pollen.}

Step 1: Understanding RuBisCO's oxygenation activity.


- In C3 plants, RuBisCO has an affinity for both \( CO_2 \) and \( O_2 \). When \( O_2 \) binds, photorespiration occurs, reducing \( CO_2 \) fixation.



Step 2: Examining photorespiration in C4 plants.


- C4 plants have a special mechanism to minimize photorespiration, but both mesophyll and bundle sheath cells show some level of photorespiration, contrary to the statement.



Conclusion: Since Statement I is true and Statement II is incorrect, the correct answer is (2). Quick Tip: \textbf{C4 plants reduce photorespiration by spatially separating carbon fixation between mesophyll and bundle sheath cells.}

Understanding key contributions in biodiversity.


- Robert May estimated the global species diversity to be around 7 million.


- Alexander von Humboldt formulated the Species-Area relationship, which describes species richness based on area size.


- Paul Ehrlich proposed the Rivet Popper Hypothesis, comparing species in an ecosystem to rivets in an airplane.


- David Tilman conducted long-term ecosystem experiments to study biodiversity and ecosystem stability.



Conclusion: The correct option is (1). Quick Tip: \textbf{Biodiversity is crucial for ecosystem stability; small species losses can have major impacts.}

Step 1: Understanding Energy Transfer in Trophic Levels.


- Energy transfer between trophic levels follows the 10 percent Law of Lindeman, meaning only 10 percent of energy is passed to the next level, while the rest is lost as heat or used in metabolism.



Step 2: Applying the 10 percent Rule.


- The first trophic level (producers) has an NPP of \( 100x \).


- The second trophic level (primary consumers) receives 10 percent of \( 100x \) = \( 10x \).


- The third trophic level (secondary consumers) will receive 10 percent of \( 10x \) = \( x \).


- The Gross Primary Productivity (GPP) at any trophic level includes energy assimilated, which is higher than NPP.


- Since respiration energy is added, the GPP of the third trophic level is estimated at \( 10x \).



Conclusion: The correct answer is (2). Quick Tip: \textbf{In an ecosystem, only about 10 percent of energy is passed from one trophic level to the next. This limits the number of trophic levels in a food chain.}

Step 1: Understanding DNA Replication Directionality.


- DNA replication in E. coli is carried out by DNA-dependent DNA polymerase.


- It adds nucleotides in the 5' to 3' direction using the template strand.



Step 2: Proofreading Activity.


- Although polymerization occurs in \(5' \rightarrow 3'\), DNA polymerase also exhibits \(3' \rightarrow 5'\) exonuclease activity for proofreading.


- However, it does not polymerize in \(3' \rightarrow 5'\) direction.



Conclusion: The correct answer is (3). Quick Tip: \textbf{DNA polymerases can only add nucleotides in the \(5' \rightarrow 3'\) direction. However, some exhibit \(3' \rightarrow 5'\) exonuclease activity for proofreading errors.

Identifying the Cellular Locations of Metabolic Processes.


- Citric Acid Cycle (Krebs Cycle): Takes place in the mitochondrial matrix where oxidation of acetyl-CoA occurs.


- Glycolysis: Occurs in the cytoplasm, breaking glucose into pyruvate.


- Electron Transport System (ETS): Located in the inner mitochondrial membrane, where ATP synthesis happens.


- Proton Gradient Formation: Protons accumulate in the intermembrane space, driving ATP synthesis.


Conclusion: The correct answer is (1). Quick Tip: \textbf{Cellular respiration occurs in multiple compartments of the cell: Glycolysis in cytoplasm, Krebs cycle in mitochondrial matrix, and ETS in the inner membrane.}

Step 1: Understanding the Role of Gibberellins in Plant Growth.


- Gibberellins promote stem elongation by stimulating cell division and elongation in internodes.


- This is particularly useful in crops like sugarcane, where increased internodal length leads to higher yield.



Step 2: Functions of Other Plant Hormones.


- Cytokinins mainly promote cell division and delay senescence.


- Abscisic Acid (ABA) induces dormancy and stress resistance.


- Auxins are involved in phototropism and root development.



Conclusion: The correct answer is (1) Gibberellin. Quick Tip: \textbf{Gibberellins are commonly used in sugarcane and grape cultivation to enhance growth and yield.}

Understanding Phaeophyceae.


- Phaeophyceae (Brown Algae) reproduces asexually via biflagellate zoospores, making Statement A correct.


- Stored food in brown algae is mannitol or laminarin, confirming Statement C is correct.


- Their major pigments include chlorophyll a, c, carotenoids, and xanthophyll, supporting Statement D.


- The vegetative cells possess cellulosic walls covered with algin, validating Statement E.


- However, sexual reproduction can be isogamous, anisogamous, or oogamous, making Statement B incorrect.



Conclusion: The correct answer is (2) A, C, D, and E only. Quick Tip: \textbf{Phaeophyceae (Brown Algae) stores food in the form of mannitol or laminarin, has a gelatinous algin coating, and exhibits multiple types of sexual reproduction.}

Identifying Flower Structures.


- Rose (A) is a perigynous flower, where the ovary is half-inferior, making A-II correct.


- Pea (B) exhibits marginal placentation, where ovules attach along one margin of the ovary, making B-IV correct.


- Cotton (C) has twisted aestivation, a characteristic floral arrangement where petals overlap, making C-I correct.


- Mango (D) produces a drupe, a fleshy fruit with a hard seed, making D-III correct.


Conclusion: The correct answer is (4) A-II, B-IV, C-I, D-III. Quick Tip: \textbf{Flowers have distinct characteristics like aestivation (arrangement of petals), placentation, and fruit type, which help in their classification.}

Understanding Biological Molecules.


- GLUT-4 (A) is a glucose transporter responsible for moving glucose into cells, making A-IV correct.


- Insulin (B) is a hormone that regulates glucose metabolism, making B-I correct.


- Trypsin (C) is an enzyme that digests proteins in the small intestine, making C-II correct.


- Collagen (D) is an intercellular ground substance, providing structural support, making D-III correct.



Conclusion: The correct answer is (4) A-IV, B-I, C-II, D-III. Quick Tip: \textbf{Understanding biological molecules helps in their classification: enzymes catalyze reactions, hormones regulate processes, and transporters facilitate movement across membranes.}

Understanding the sub-phases of Prophase I.


- Diakinesis (A) is the final stage, where chiasmata terminalization completes, confirming A-II.


- Pachytene (B) is marked by the appearance of recombination nodules, ensuring B-IV.


- Zygotene (C) is identified by synaptonemal complex formation, making C-I correct.


- Leptotene (D) is the first stage, where chromosomes appear as thin threads, supporting D-III.


Conclusion: The correct answer is (2) A-II, B-IV, C-I, D-III. Quick Tip: \textbf{Prophase I is the longest and most crucial stage in meiosis, consisting of five sub-phases: Leptotene, Zygotene, Pachytene, Diplotene, and Diakinesis.}

Understanding Coelom in Animal Phyla.


- Annelids (A) are true coelomates, making Statement A correct.


- Poriferans (B) lack a body cavity entirely, so they are not pseudocoelomates, making Statement B incorrect.


- Aschelminthes (C) are pseudocoelomates, meaning Statement C is incorrect.


- Platyhelminthes (D) are acoelomates, not pseudocoelomates, making Statement D incorrect.



Conclusion: The correct answer is (1) A only. Quick Tip: \textbf{Coelomates have a true coelom (e.g., Annelida), pseudocoelomates have a false coelom (e.g., Nematoda), and acoelomates lack a coelom (e.g., Platyhelminthes).}

Understanding Brain Functions.


- Pons (A) functions to connect different regions of the brain, making A-III correct.


- Hypothalamus (B) contains neurosecretory cells, controlling homeostasis, confirming B-IV.


- Medulla (C) regulates respiration and gastric secretions, making C-II correct.


- Cerebellum (D) helps in posture and balance, ensuring D-I is correct.



Conclusion: The correct answer is (1) A-III, B-IV, C-II, D-I. Quick Tip: \textbf{The brainstem (medulla and pons) controls involuntary functions, the hypothalamus maintains homeostasis, and the cerebellum regulates balance and coordination.}

Step 1: Understanding oxyhaemoglobin formation.


- In alveoli, oxygen binds to hemoglobin to form oxyhaemoglobin.


- The binding is favored by high partial pressure of oxygen (\( pO_2 \)) and low hydrogen ion concentration (\( H^+ \)).



Step 2: Effects of \( pCO_2 \) and \( H^+ \) on oxyhaemoglobin.


- Lower \( pCO_2 \) ensures efficient oxygen binding by reducing the Bohr effect.


- Less \( H^+ \) concentration prevents hemoglobin from releasing oxygen prematurely.



Conclusion: The correct answer is (1) High \( pO_2 \) and Lesser \( H^+ \) concentration. Quick Tip: \textbf{Oxyhaemoglobin formation in alveoli is favored by high oxygen levels and lower carbon dioxide levels, as these conditions shift hemoglobin’s affinity towards oxygen.}

Understanding the sequence of cell cycle stages.


- Gap 1 (G1) phase (E): The first growth phase, where the cell prepares for DNA synthesis.


- Synthesis (S) phase (C): DNA replication occurs.


- Gap 2 (G2) phase (A): The cell continues to grow and prepares for mitosis.


- Karyokinesis (D): Nuclear division takes place.


- Cytokinesis (B): The final step where the cytoplasm divides, resulting in two daughter cells.



Conclusion: The correct sequence is E-C-A-D-B. Quick Tip: \textbf{The correct sequence of the cell cycle is: G1 (growth) → S (DNA replication) → G2 (preparation for mitosis) → M (karyokinesis & cytokinesis).}

Step 1: Understanding muscle types.


- Skeletal muscles (voluntary) are attached to bones and responsible for movement (e.g., triceps).


- Smooth muscles (involuntary) are found in internal organs like the stomach.


- Cardiac muscles (involuntary) are found in the heart and control heartbeat.



Step 2: Matching muscle types with locations.


- Triceps are skeletal muscles responsible for arm movement.


- Stomach contains smooth muscles, which contract involuntarily for digestion.


- Heart contains cardiac muscles, which pump blood.


Conclusion: The correct answer is (1) Skeletal – Triceps, Smooth – Stomach, Cardiac – Heart. Quick Tip: \textbf{Skeletal muscles are voluntary, smooth muscles are involuntary, and cardiac muscles are involuntary but have automatic rhythmic contractions.}

Understanding Chromosomal Disorders.


- Down’s syndrome (Trisomy 21) is caused by an extra copy of chromosome 21.


- \(\alpha\)-Thalassemia is linked to mutations in chromosome 16 affecting hemoglobin production.


- \(\beta\)-Thalassemia results from defects in chromosome 11 impacting hemoglobin beta-chain.


- Klinefelter’s syndrome (XXY condition) affects the X chromosome, leading to male infertility.


Conclusion: The correct matching is (2) A-III, B-IV, C-I, D-II. Quick Tip: \textbf{Genetic disorders like Down’s syndrome and Klinefelter’s syndrome arise due to chromosomal abnormalities, while Thalassemia is due to single-gene mutations.}

Understanding Autoimmune Disorders.


- Myasthenia gravis: A neuromuscular autoimmune disorder causing muscle weakness.


- Rheumatoid arthritis: An autoimmune disease where the immune system attacks joints.


- Systemic Lupus Erythematosus (SLE): An autoimmune disease affecting multiple organs.


- Gout and Muscular dystrophy are not autoimmune disorders; gout is a metabolic disorder while muscular dystrophy is genetic.


Conclusion: The correct answer is (1) A, B & E only. Quick Tip: \textbf{Autoimmune diseases occur when the body's immune system mistakenly attacks its own tissues, leading to chronic inflammation and organ damage.}

Understanding the role of FSH and Leydig cells.


- Follicle Stimulating Hormone (FSH) stimulates ovarian follicles in females, but in males, it stimulates Sertoli cells, not Leydig cells.


- Leydig cells are stimulated by Luteinizing Hormone (LH) and are responsible for testosterone production.


- Estrogen is secreted by growing ovarian follicles, and testosterone is secreted by Leydig cells, which makes the Reason (R) true.



Conclusion: The correct answer is (3) A is false but R is true. Quick Tip: \textbf{FSH primarily acts on Sertoli cells in males, whereas LH stimulates Leydig cells to produce androgens.}

Understanding the Loop of Henle and Proximal Convoluted Tubule (PCT).


- Descending limb of Henle's loop is permeable to water and impermeable to electrolytes, so Statement I is false.


- Proximal Convoluted Tubule (PCT) is lined by simple cuboidal epithelium with brush borders, not simple columnar epithelium, so Statement II is also false.


Conclusion: The correct answer is (1) Both Statement I and Statement II are false. Quick Tip: \textbf{The descending limb of Henle’s loop allows water reabsorption, while the PCT is lined by simple cuboidal brush border epithelium for nutrient reabsorption.}

Step 1: Understanding the Importance of Breastfeeding.


- Breastfeeding is highly recommended for newborns as it provides essential nutrients and immunity.


- Colostrum, the first milk, is rich in antibodies such as IgA, which protect against infections.



Step 2: Relationship Between A and R.


- Since colostrum provides passive immunity to the baby, Reason R correctly explains Assertion A.



Conclusion: The correct answer is (4) Both A and R are correct and R is the correct explanation of A. Quick Tip: \textbf{Colostrum is rich in antibodies, particularly IgA, which helps develop the newborn’s immune system and protects against infections.}

Step 1: Understanding the Sources of These Drugs.


- Cocaine is derived from Erythroxylum coca and is a stimulant drug.


- Heroin is derived from Papaver somniferum and is a depressant.


- Morphine is used as a pain-relieving sedative in surgery.


- Marijuana comes from Cannabis sativa and has psychoactive effects.



Conclusion: The correct answer is (3) A-III, B-IV, C-I, D-II. Quick Tip: \textbf{Many drugs are derived from plants, such as Cocaine from Erythroxylum, Heroin from Opium poppy, and Marijuana from Cannabis.}

Understanding the Classification of These Fishes.


- Pterophyllum is commonly known as the Angel fish.


- Myxine belongs to Hagfish, a jawless fish.


- Pristis is a type of Sawfish, which has a long, saw-like snout.


- Exocoetus is a Flying fish, capable of gliding over water surfaces.



Conclusion: The correct answer is (1) A-III, B-I, C-II, D-IV. Quick Tip: \textbf{Angel fish (Pterophyllum) is an aquarium fish, while Hagfish (Myxine) is a jawless marine species. Sawfish (Pristis) has a unique rostrum, and Flying fish (Exocoetus) can glide over water.}

Understanding the Classification of These Organisms.


- Pleurobrachia belongs to Ctenophora, a phylum of marine comb jellies.


- Radula is a toothed feeding structure found in Mollusca.


- Stomochord is found in Hemichordata, a phylum that includes acorn worms.


- Air bladder is a characteristic of Osteichthyes (bony fishes) and helps with buoyancy.



Conclusion: The correct answer is (1) A-II, B-I, C-IV, D-III. Quick Tip: \textbf{Ctenophores like Pleurobrachia are jelly-like marine animals, Molluscs use radula for feeding, Hemichordates have stomochords, and bony fishes use air bladders for buoyancy.}

Understanding Anal Cerci in Cockroaches.


- Cockroaches have anal cerci, a pair of jointed filamentous structures located on the 10th abdominal segment.


- These cerci help in detecting air movements and possible threats.



Conclusion: The correct answer is (1) 10th segment. Quick Tip: \textbf{Cockroaches have segmented abdomens, and anal cerci function as sensory organs to detect vibrations in the air.}

Understanding Convergent Evolution.


- Convergent evolution occurs when unrelated organisms develop similar traits due to adaptation to a similar environment.


- Penguins (birds) and Dolphins (mammals) have evolved streamlined bodies and flippers for swimming, but they belong to different taxonomic groups.



Conclusion: The correct answer is (2) Convergent evolution. Quick Tip: \textbf{Convergent evolution results in functionally similar structures (analogous organs) in unrelated species due to similar environmental pressures.}

Understanding Bio-reactors.


- Bio-reactors are large vessels used in industrial biotechnology for mass production of cells or bio-products.


- Statement (2) is incorrect because bio-reactors are used for large-scale production, not small-scale cultures.



Conclusion: The correct answer is (2) Bio-reactors are used to produce small-scale bacterial cultures. Quick Tip: \textbf{Bio-reactors provide controlled environments (temperature, pH, nutrients) to enhance microbial growth and product yield.}

Understanding Transcription.


- The template strand (3' to 5') is used by RNA polymerase to synthesize an mRNA strand (5' to 3').


- The complementary RNA sequence follows A-U and G-C pairing rules.



Conclusion: The correct answer is (4) 5'-AUGUACCGUUUAUAGGUAAGU-3'. Quick Tip: \textbf{Transcription follows the base-pairing rules: Adenine (A) pairs with Uracil (U), Thymine (T) pairs with Adenine (A), and Cytosine (C) pairs with Guanine (G).}

Understanding Joint Types.


- Fibrous joints (A) are immovable (e.g., skull sutures).


- Cartilaginous joints (B) allow limited movement (e.g., intervertebral discs).


- Hinge joints (C) allow movement in one plane (e.g., knee).


- Ball and socket joints (D) allow rotational movement (e.g., shoulder joint).



Conclusion: The correct answer is (3) A-III, B-I, C-IV, D-II. Quick Tip: \textbf{Joints are classified based on movement: Fibrous (immovable), Cartilaginous (partially movable), and Synovial (freely movable).}

Understanding the Matching.


- \(\alpha\)-I antitrypsin is used in treating emphysema.


- Cry IAb and Cry IAc are Bacillus thuringiensis (Bt) toxins used to control corn borers and cotton bollworms, respectively.


- Enzyme replacement therapy is used in treating Adenosine Deaminase (ADA) deficiency.



Conclusion: The correct answer is (2) A-III, B-IV, C-I, D-II. Quick Tip: \textbf{Bt crops like Bt cotton contain Cry proteins, which provide resistance against insects by forming pores in their gut lining.}

Understanding the Matching.


- Axoneme forms the structural framework of cilia and flagella.


- Cartwheel pattern is observed in centrioles.


- Cristae are the folds of the inner mitochondrial membrane.


- Satellite regions are associated with chromosomes.



Conclusion: The correct answer is (3) A-II, B-I, C-IV, D-III. Quick Tip: \textbf{Mitochondria, the powerhouse of the cell, have cristae that increase the surface area for ATP production.}

Understanding Heart's Electrical Conduction System.


- The Sinoatrial (SA) node (E) initiates the action potential.


- The signal passes to the Atrioventricular (AV) node (C).


- From the AV node, it moves through the AV bundle (A).


- Then it travels down the Bundle branches (D).


- Finally, it reaches the Purkinje fibers (B), causing ventricular contraction.



Conclusion: The correct answer is (4) E-C-A-D-B. Quick Tip: \textbf{The SA node is the natural pacemaker of the heart, maintaining rhythmic contractions.}

Step 1: Understanding the Hymen and Its Significance


The hymen is a thin membrane that partially covers the vaginal opening.
Its presence or absence is not a reliable indicator of virginity, as it can be torn due to various activities such as sports, cycling, physical activity, or medical procedures.
Hence, Statement I is true.


Step 2: Evaluating Statement II


The hymen does not always tear during first coitus.
It can stretch without tearing, and some individuals are born with very little or no hymenal tissue.
Since the hymen can be lost due to non-sexual activities, Statement II is false.


Conclusion:



Since Statement I is true but Statement II is false, the correct answer is \( \mathbf{(2)} \). Quick Tip: The hymen is not a definitive indicator of virginity, and its tearing can result from various non-sexual activities. Medical professionals do not use it as a conclusive test.

Understanding the Matching.


- Common cold is caused by Rhinoviruses.


- Haemozoin is a byproduct of Plasmodium, the malarial parasite.


- Widal test is used for typhoid diagnosis.


- Allergies are triggered by dust mites and other allergens.



Conclusion: The correct answer is (2) A-III, B-I, C-II, D-IV. Quick Tip: \textbf{Malaria causes cyclic fever due to the periodic release of haemozoin from RBCs.}

Understanding Disease Classification.


- Typhoid is caused by Salmonella typhi (Bacteria).


- Leishmaniasis is caused by Leishmania (Protozoa).


- Ringworm is a fungal infection caused by dermatophytes.


- Filariasis is caused by Wuchereria bancrofti (Nematode).



Conclusion: The correct answer is (1) A-IV, B-III, C-I, D-II. Quick Tip: \textbf{Leishmaniasis is transmitted by sandflies and causes skin ulcers or visceral infections.}

Understanding Lung Capacity.


- Expiratory Capacity includes Tidal volume + Expiratory reserve volume.


- Functional Residual Capacity includes Expiratory reserve volume + Residual volume.


- Vital Capacity includes Expiratory reserve volume + Tidal volume + Inspiratory reserve volume.


- Inspiratory Capacity includes Tidal volume + Inspiratory reserve volume.



Conclusion: The correct answer is (4) A-II, B-IV, C-I, D-III. Quick Tip: \textbf{Vital capacity is the maximum amount of air a person can exhale after a maximum inhalation.}

Understanding Hardy-Weinberg Equilibrium.


- Hardy-Weinberg Equilibrium states that allele frequencies remain constant in a population unless acted upon by evolutionary forces.


- Factors affecting it include: Mutation, Natural Selection, Genetic Drift, Gene Flow, and Genetic Recombination.


- Constant gene pool implies no evolution, meaning no change in allele frequencies.



Conclusion: The correct answer is (3) Constant gene pool. Quick Tip: \textbf{If all Hardy-Weinberg conditions are met, evolution does not occur, maintaining genetic stability.}

Understanding Enzyme Specificity.


- Lipase hydrolyzes ester bonds in fats.


- Nuclease hydrolyzes phosphodiester bonds in DNA/RNA.


- Protease hydrolyzes peptide bonds in proteins.


- Amylase hydrolyzes glycosidic bonds in starch.



Conclusion: The correct answer is (2) A-II, B-IV, C-I, D-III. Quick Tip: \textbf{Enzymes lower the activation energy of biochemical reactions, making them highly specific.}

S Understanding the Fallopian Tube.


- The Fallopian tube consists of four parts:


Infundibulum, Ampulla, Isthmus, and Interstitial Part.


- The Uterine Fundus is part of the uterus, not the Fallopian tube.



Conclusion: The correct answer is (4) Uterine fundus. Quick Tip: \textbf{Fertilization usually occurs in the ampulla of the Fallopian tube.}

Understanding Different Types of IUDs.


- Non-medicated IUDs (e.g., Lippes loop) function as a physical barrier to implantation.


- Copper releasing IUDs (e.g., Multiload 375) release copper ions, which are toxic to sperm.


- Hormone-releasing IUDs (e.g., LNG-20) release hormones like levonorgestrel, preventing ovulation.


- Implants contain progestogens that provide long-term contraception.



Conclusion: The correct answer is (3) A-III, B-I, C-IV, D-II. Quick Tip: \textbf{Copper IUDs increase the uterine environment's spermicidal effect, making fertilization unlikely.}

Understanding Steroid Hormones.


- Steroid hormones are derived from cholesterol and include Testosterone, Progesterone, and Cortisol.


- Glucagon, however, is a peptide hormone that regulates blood glucose levels.



Conclusion: The correct answer is (3) Glucagon. Quick Tip: \textbf{Peptide hormones like glucagon act through second messengers, unlike steroid hormones which enter the cell.}

Understanding pBR322 Plasmid.


- pBR322 is a commonly used cloning vector.


- Gene ‘X’ (ori region) controls the copy number of the plasmid.


- Gene ‘Y’ (replication-related gene) helps in plasmid replication.



Conclusion: The correct answer is (1). Quick Tip: \textbf{Plasmids are self-replicating circular DNA molecules used in genetic engineering.}

Understanding Natural and Artificial Contraception



Step 1: Identifying Natural Contraceptive Methods

Periodic Abstinence: Avoiding intercourse during the fertile window (ovulation period).
Lactational Amenorrhea: Temporary infertility occurring post-childbirth due to hormonal changes.
Coitus Interruptus: The withdrawal method, where the male withdraws before ejaculation.


Step 2: Identifying Artificial Contraceptive Methods

Vaults (Diaphragms):

Vaults are artificial barrier methods that physically prevent sperm from entering the uterus.
Since they are not natural methods, they do not belong to this category.



Conclusion:


The correct answer is \( \mathbf{(3)} \) Vaults. Quick Tip: \textbf{Natural contraceptive methods are based on physiological cycles and behavioral patterns.}