Step 1: Calculate the potential on the axial line of a dipole.
The formula for potential at an axial point is:
\begin{equation
V = \frac{P{4 \pi \epsilon_0 r^2.
\end{equation
Substitute the given values: \(P = 4 \times 10^{-6} \, \mathrm{C \, m}\), \(r = 2 \, \mathrm{m}\), \(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \mathrm{SI \, units}\):
\begin{equation
V = \frac{4 \times 10^{-6 \cdot 9 \times 10^9{2^2 = \frac{36 \times 10^3{4 = 9 \times 10^3 \, \mathrm{V.
\end{equation
This matches the given assertion.

Step 2: Evaluate the reason.

The reason states the formula \(V = \pm \frac{2P}{4 \pi \epsilon_0 r^2}\). While the formula is correct, the sign \(\pm\) is used for specifying the direction of the potential based on the orientation of the dipole.

However, the potential magnitude was specifically calculated in the assertion, which does not depend on this sign. Therefore, the reason is unrelated to the assertion. Quick Tip: Always Evaluate whether the assertion and reason are directly related. The correctness of individual statements does not imply they explain each other.

Step 1: Use the formula for acceleration due to gravity.
\begin{equation
g = \frac{GM{R^2,
\end{equation
where \(M\) is the mass,
\(R\) is the radius of the planet,

and \(G\) is the gravitational constant.

Step 2: Express the planet's mass and radius in terms of Earth's.

Let Earth's mass be \(M_e\) and radius be \(R_e\). Then, for the planet:

\begin{equation
M_p = \frac{M_e{10, \quad R_p = \frac{R_e{2.
\end{equation

Step 3: Substitute values into the formula.
\begin{equation
g_p = \frac{G \cdot M_p{R_p^2 = \frac{G \cdot \frac{M_e{10{\left(\frac{R_e{2\right)^2 = \frac{G \cdot \frac{M_e{10{\frac{R_e^2{4 = \frac{4{10 \cdot \frac{GM_e{R_e^2.
\end{equation
Since \(g_e = \frac{GM_e}{R_e^2} = 9.8 \, \mathrm{m \, s^{-2}}\):
\begin{equation
g_p = \frac{4{10 \cdot 9.8 = 3.92 \, \mathrm{m \, s^{-2.
\end{equation Quick Tip: For planets with different mass and radius, express values as ratios relative to Earth to simplify gravity calculations.

Step 1: Velocity from displacement.
The displacement is given by \(x = 2t - 1\). Velocity (\(v\)) is obtained by differentiating displacement with respect to time: \[ v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2 \, \mathrm{m/s}. \]

Step 2: Calculate instantaneous power.

The formula for instantaneous power is: \[ P = F \cdot v, \]
where \(F = 5 \, \mathrm{N}\) and \(v = 2 \, \mathrm{m/s}\). Substituting values: \[ P = 5 \cdot 2 = 10 \, \mathrm{W}. \] Quick Tip: Instantaneous power is the product of force and velocity. Ensure that velocity is correctly derived from displacement.

Step 1: Understanding Velocity in Circular Motion

In uniform circular motion, the magnitude of velocity remains constant, but its direction continuously changes as the object moves along the circular path. Since velocity is a vector quantity, this continuous change in direction implies that the velocity vector is not constant.

Step 2: Understanding Acceleration in Circular Motion*
Acceleration in circular motion has two key components:

- Centripetal acceleration, which is always directed toward the center of the circular path and is responsible for changing the direction of velocity.

- Tangential acceleration, which remains zero in the case of uniform circular motion since the speed does not change.

As a result of the continuous change in direction of velocity, the particle experiences a varying acceleration, even though its speed remains constant. Quick Tip: In uniform circular motion, speed remains constant, but velocity and acceleration vary due to the change in direction.

Step 1: The properties of each material.

- Diamagnetic: Materials have a small negative susceptibility (\(0 \(>\) \chi \geq -1\)).

- Ferromagnetic: Materials exhibit very high susceptibility (\(\chi \gg 1\)).

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

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


Step 2: Match correctly.
Using the above properties, the correct matching is:
A-II, B-III, C-IV, D-I. Quick Tip: To match materials with their susceptibilities, Understand the magnetic properties of each material type.

Step 1: The relationship in an ideal transformer

In an ideal transformer, the relationship between the primary and secondary voltage is given by: \[ \frac{V_P}{V_S} = \frac{N_P}{N_S} \]
where: \( V_P \) = Primary voltage,
\( V_S \) = Secondary voltage,
\( N_P \) = Number of primary turns,
\( N_S \) = Number of secondary turns.

Step 2: Substitute the turns ratio into the equation

The turns ratio is given as: \[ \frac{N_P}{N_S} = \frac{1}{2} \]
Thus, the voltage ratio becomes: \[ \frac{V_P}{V_S} = \frac{N_P}{N_S} = \frac{1}{2} \]

Step 3: Determine \( V_S : V_P \)

Rewriting the ratio of \( V_P \) to \( V_S \): \[ \frac{V_P}{V_S} = \frac{1}{2} \implies \frac{V_S}{V_P} = 2 : 1 \]

Conclusion: The ratio \( V_S : V_P \) is \( 2 : 1 \), which matches option (1). Quick Tip: In an ideal transformer, voltage ratios and turns ratios are directly proportional, while current ratios are inversely proportional.

Step 1: Use the formula for the magnetic field at the center of a circular coil.

The magnetic field at the center of a circular coil is given by: \[ B = \frac{\mu_0 N I}{2R}, \]
where \(N = 100\) is the number of turns,
\(I = 7 \, \mathrm{A}\) is the current,
\(R = 0.1 \, \mathrm{m}\) is the radius,

and \(\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{T \, m/A}\) is the permeability of free space.

Step 2: Substitute the values into the formula. \[ B = \frac{(4 \pi \times 10^{-7}) \cdot 100 \cdot 7}{2 \cdot 0.1}. \]
Simplify: \[ B = \frac{28 \pi \times 10^{-7}}{0.2} = 4.4 \times 10^{-3} \, \mathrm{T} = 4.4 \, \mathrm{mT}. \] Quick Tip: For circular coils, the magnetic field strength increases with the number of turns and current but decreases with a larger radius.

The given circuit consists of capacitors arranged in both series and parallel configurations. Let us Evaluate the circuit step by step to find the equivalent capacitance.


First, observe the two 2 \(\mu\)F capacitors in parallel between terminal A and the intermediate node. The formula for the equivalent capacitance of two capacitors in parallel is: \[ C_{eq, parallel} = C_1 + C_2 \]
For the two 2 \(\mu\)F capacitors: \[ C_{eq, parallel} = 2 \, \mu F + 2 \, \mu F = 4 \, \mu F \]

Now, the 4 \(\mu\)F equivalent capacitance is in series with the 2 \(\mu\)F capacitor in the middle of the circuit. The formula for the equivalent capacitance of capacitors in series is: \[ \frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2} \]
For the 4 \(\mu\)F and 2 \(\mu\)F capacitors in series: \[ \frac{1}{C_{eq, series}} = \frac{1}{4 \, \mu F} + \frac{1}{2 \, \mu F} \] \[ \frac{1}{C_{eq, series}} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \] \[ C_{eq, series} = \frac{4}{3} \, \mu F \]

Finally, this 4/3 \(\mu\)F equivalent capacitor is in parallel with the last 2 \(\mu\)F capacitor. The equivalent capacitance of two capacitors in parallel is: \[ C_{eq, final} = C_{eq, series} + C_3 \] \[ C_{eq, final} = \frac{4}{3} \, \mu F + 2 \, \mu F = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \, \mu F \]

The total equivalent capacitance between terminals A and B is: \[ C_{total} = 2 \, \mu F \]



Answer: (4) 2 \(\mu\)F Quick Tip: Break complex circuits into smaller sections to simplify calculations. Identify series and parallel connections clearly.

Step 1: Use Lenz's Law.

According to Lenz’s law, the induced current always flows in a direction such that it opposes the change in magnetic flux that causes it.

Step 2: Evaluate solenoid-1.

As the bar magnet approaches solenoid-1, the magnetic flux increases. To oppose this increase, solenoid-1 generates a magnetic field opposite to the bar magnet's field, inducing a current in the direction \(AB\).

Step 3: Evaluate solenoid-2.

For solenoid-2, the magnetic field due to the approaching bar magnet increases in the direction towards solenoid-1. To oppose this, solenoid-2 generates a magnetic field pointing away, inducing a current in the direction \(DC\). Quick Tip: When solving induction problems, visualize the change in magnetic flux and use Lenz’s law to determine the direction of induced currents.

Step 1: Use the formula for elongation.
The elongation of a wire is given by: \[ \Delta L = \frac{\sigma L}{Y}, \]
where \(\sigma\) is the stress (elastic limit), \(L\) is the length of the wire, and \(Y\) is the Young's modulus.

Step 2: Substitute the values.
Given: \[ \sigma = 8 \times 10^8 \, \mathrm{N/m^2}, \quad L = 1 \, \mathrm{m}, \quad Y = 2 \times 10^{11} \, \mathrm{N/m^2}. \]
Substituting: \[ \Delta L = \frac{(8 \times 10^8) \cdot 1}{2 \times 10^{11}} = 4 \times 10^{-3} \, \mathrm{m}. \]

Step 3: Convert the result. \[ \Delta L = 4 \, \mathrm{mm}. \] Quick Tip: For elongation problems, always ensure units are consistent. Length should be in meters, stress in Pascals, and Young's modulus in \(\mathrm{N/m^2}\).

Step 1: Effect of alpha (\(\alpha\)) decay

An alpha particle \(\alpha\) has a mass number of \(4\) and an atomic number of \(2\). When \(X\) undergoes alpha decay:
\[ \prescript{290}{82}X \xrightarrow{\alpha} \prescript{286}{80}Y \]

Thus, the new nucleus \( Y \) has:

- Mass number = \(290 - 4 = 286\),

- Atomic number = \(82 - 2 = 80\).


Step 2: Effect of positron (\( e^+ \)) emission

Positron emission (\(\beta^+\)) decreases the atomic number by 1 without changing the mass number:
\[ \prescript{286}{80}Y \xrightarrow{e^+} \prescript{286}{79}Z \]

Thus, \( Z \) has:
- Mass number = \(286\) (unchanged),
- Atomic number = \(80 - 1 = 79\).

Step 3: Effect of beta-minus (\( \beta^- \)) decay

Beta-minus (\(\beta^-\)) emission increases the atomic number by 1 without affecting the mass number:
\[ \prescript{286}{79}Z \xrightarrow{\beta^-} \prescript{286}{80}P \]

Thus, \( P \) has:
- Mass number = \(286\) (unchanged),
- Atomic number = \(79 + 1 = 80\).

Step 4: Effect of electron capture (\( e^- \))

Electron capture decreases the atomic number by 1, keeping the mass number unchanged:
\[ \prescript{286}{80}P \xrightarrow{e^-} \prescript{286}{81}Q \]

Thus, \( Q \) has:

- Mass number = \(286\) (unchanged),

- Atomic number = \(80 + 1 = 81\).


Conclusion: The final product \( Q \) has mass number \(286\) and atomic number \(81\), which corresponds to option
(3). Quick Tip: - Alpha decay decreases the mass number by 4 and the atomic number by 2.
- Positron emission decreases the atomic number by 1 but does not affect the mass number.
- Beta-minus decay increases the atomic number by 1 without changing the mass number.
- Electron capture decreases the atomic number by 1 while keeping the mass number unchanged.

Step 1: Rolling motion:

In rolling motion, a point on the rim of the wheel has both rotational and translational motion. The velocities due to both these motions add vectorially.

Step 2: Evaluate the velocity at points P and Q.

- Point P is at the top of the wheel. Its velocity is the sum of the linear speed (\(v\)) of the wheel and the rotational speed (\(v\)).

- Point Q is at the bottom of the wheel, where the linear speed (\(v\)) and rotational speed (\(-v\)) cancel out.

Step 3: Calculate velocities.

- Velocity at P: \(v_P = v + v = 2v\).

- Velocity at Q: \(v_Q = v - v = 0\).

Step 4: Conclude.
Point P moves faster than point Q, and point Q has zero velocity. Quick Tip: In rolling motion, the top point of the wheel has maximum speed (\(2v\)) while the bottom point momentarily comes to rest (\(0\)).

Step 1: Understand Brewster's law.

Brewster's angle (\(\theta_B\)) is the angle of incidence at which light reflected from a surface is completely polarised. At this angle, the reflected and refracted rays are perpendicular to each other.

Step 2: Evaluate the reflection and refraction.

- At Brewster's angle, the reflected light is completely polarised in a direction perpendicular to the plane of incidence.

- The refracted light is partially polarised because it consists of both components of polarisation.

Step 3: Conclude.

The reflected light is completely polarised, while the refracted light is partially polarised. Quick Tip: At Brewster's angle, the reflected and refracted rays are perpendicular, and the reflected light is completely polarised.

Step 1: Use the formula for the vernier constant (V.C.).
The vernier constant is given by: \[ V.C. = Value of 1 MSD - Value of 1 VSD. \]

Step 2: Relate MSD and VSD values.
If \((N+1)\) VSDs coincide with \(N\) MSDs: \[ Value of 1 VSD = \frac{Value of 1 MSD \times N}{N+1}. \]

Step 3: Substitute values. \[ V.C. = 0.1 \, \mathrm{mm} - \frac{0.1 \, \mathrm{mm} \times N}{N+1}. \]
Simplify: \[ V.C. = \frac{0.1}{N+1} \, \mathrm{mm}. \]

Convert to cm: \[ V.C. = \frac{1}{100(N+1)} \, \mathrm{cm}. \] Quick Tip: For vernier callipers, the vernier constant is the least count, which can be calculated using the relationship between MSD and VSD.

Step 1: Relate tension with speed.
Tension in the string provides the centripetal force: \[ T = \frac{mv^2}{r}, \]
where \(m\) is the mass, \(v\) is the speed, and \(r\) is the radius.

Step 2: Compare tensions.
If speed doubles (\(v' = 2v\)): \[ T' = \frac{m(2v)^2}{r} = 4 \cdot \frac{mv^2}{r} = 4T. \] Quick Tip: For circular motion, tension is proportional to the square of the speed. Doubling speed quadruples the tension.

Step 1: Understand the thermodynamic process along path \(bc\).

The work done by a gas in a thermodynamic process is given by: \[ W = \int P \, dV. \]

Step 2: Evaluate the path \(bc\).

The path \(bc\) represents an isochoric process, meaning the volume remains constant (\(dV = 0\)).

Step 3: Substitute into the formula.

Since \(dV = 0\), the work done is: \[ W = \int P \, dV = 0. \]

Step 4: Conclude.

The work done along the path \(bc\) is \(0 \, \mathrm{J}\). Quick Tip: In an isochoric process, volume remains constant, so no work is done (\(W = 0\)).

Step 1: Understand the truth table for the given logic gate.

An AND gate gives an output of 1 only if both inputs are 1. Otherwise, the output is 0.

Step 2: Compare the outputs.

The output \(Y\) of the given logic gate matches the truth table of an AND gate: \[ Y = A \cdot B. \]

Step 3: Conclude.
The output of the given logic gate is the same as that of an AND gate. Quick Tip: For logic gate questions, construct a truth table to compare the given outputs with standard logic gates.

Step 1: Apply the law of conservation of momentum.
For a completely inelastic collision, the two bodies stick together after the collision. Using conservation of momentum: \[ m v_1 + m \cdot 0 = (m + m) v_2. \]

Step 2: Simplify the equation. \[ v_1 = 2v_2. \]

Step 3: Determine the ratio.
The ratio of the velocities is: \[ v_1 : v_2 = 2 : 1. \] Quick Tip: In completely inelastic collisions, the final velocity is always less than the initial velocity due to the combined mass of the system.

Step 1: Understand the formula for the moment of inertia.

The moment of inertia of a thin rod about its midpoint is: \[ I = \frac{1}{12} M L^2, \]
where \(M\) is the mass and \(L\) is the length.

Step 2: Rearrange to solve for \(L\). \[ L^2 = \frac{12I}{M}. \]

Step 3: Substitute the given values. \[ I = 2400 \, \mathrm{g \, cm^2}, \quad M = 400 \, \mathrm{g}. \] \[ L^2 = \frac{12 \cdot 2400}{400} = 72 \, \mathrm{cm^2}. \]

Step 4: Find \(L\). \[ L = \sqrt{72} = 8.5 \, \mathrm{cm}. \] Quick Tip: For moment of inertia calculations, ensure consistent units and carefully rearrange formulas when solving for unknowns.

Step 1: Use the formula for force due to surface tension.
The force required to overcome surface tension is: \[ F = 2 \pi r \cdot T, \]
where \(r\) is the radius and \(T\) is the surface tension.

Step 2: Convert radius to meters. \[ r = 4.5 \, \mathrm{cm} = 0.045 \, \mathrm{m}. \]

Step 3: Substitute the values. \[ F = 2 \pi (0.045) (0.07). \] \[ F = 0.0198 \, \mathrm{N} = 19.8 \, \mathrm{mN}. \] Quick Tip: For surface tension problems, ensure the radius is in meters before calculating the force.

Step 1: Evaluate Statement A.

For a solar cell, the \(I-V\) characteristics lie in the fourth quadrant because it operates by converting light energy into electrical energy, generating current in the process. This makes statement A correct.

Step 2: Evaluate Statement B.

In a reverse-biased \(pn\)-junction diode, the current (leakage current) is due to minority carriers, not majority carriers. Hence, statement B is incorrect.

Step 3: Conclusion.

Statement A is correct, but Statement B is incorrect. Quick Tip: For solar cells, remember that the \(I-V\) curve lies in the fourth quadrant, as power is generated rather than consumed.

Step 1: Calculate the acceleration of the system.
The total mass of the system is: \[ M = m_A + m_B = 2 \, \mathrm{kg} + 3 \, \mathrm{kg} = 5 \, \mathrm{kg}. \]
Using Newton's second law: \[ F = M \cdot a \quad \Rightarrow \quad a = \frac{F}{M} = \frac{10}{5} = 2 \, \mathrm{m/s^2}. \]

Step 2: Calculate the force exerted by block \(A\) on block \(B\).
The force on block \(B\) due to block \(A\) is: \[ F_B = m_B \cdot a = 3 \cdot 2 = 6 \, \mathrm{N}. \]

Step 3: Conclude.
The force exerted by block \(A\) on block \(B\) is \(6 \, \mathrm{N}\). Quick Tip: To calculate inter-block forces, find the acceleration of the entire system first and then apply Newton's second law to individual blocks.

The equation for simple harmonic motion (SHM) is given by: \[ x = A \sin(\omega t + \phi) \]
Where:

- \(x\) is the displacement at time \(t\),

- \(A\) is the amplitude,

- \(\omega\) is the angular frequency,

- \(t\) is the time, and

- \(\phi\) is the phase constant.

In the given equation, \(x = 5 \sin\left(\pi t + \frac{\pi}{3}\right)\), we can compare it with the standard form \(x = A \sin(\omega t + \phi)\):
- The amplitude \(A = 5 \, \mathrm{m}\),
- The angular frequency \(\omega = \pi\).

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

Thus, the amplitude is \(5 \, \mathrm{m}\), and the time period is \(2 \, \mathrm{s}\). Quick Tip: In SHM problems, amplitude is the coefficient of \(\sin(\omega t)\), and the time period is calculated from \(T = \frac{2\pi}{\omega}\).

Solution:

Given:

The ray enters a right-angled prism at point \( P \) with an angle of incidence \( i = 30^\circ \).
The ray travels through the prism parallel to its base \( BC \) and emerges along face \( AC \).


We are asked to find the refractive index of the prism.



Step 1: Understanding the Geometry of the Prism

- In the given right-angled prism, we have \( \angle ABC = 90^\circ \).

- The ray enters the prism at point \( P \) on face \( AB \), with the angle of incidence \( i = 30^\circ \).

- The ray travels parallel to base \( BC \) and emerges along the face \( AC \).



Step 2: Applying Snell’s Law

At the point of incidence inside the prism, we apply Snell's Law: \[ n_1 \sin(i) = n_2 \sin(r) \]
Where:

- \( n_1 = 1 \) (refractive index of air),

- \( i = 30^\circ \) (angle of incidence),

- \( n_2 \) is the refractive index of the prism (which we need to find),

- \( r \) is the angle of refraction inside the prism.



Step 3: Geometry of the Prism

Since the ray travels parallel to \( BC \), the angle of refraction inside the prism is equal to \( \angle ACB \), and from the geometry of the triangle \( ABC \), we know that:
\[ \angle ACB = 60^\circ \]



Step 4: Solving for the Refractive Index

Substituting into Snell’s Law:
\[ \sin(30^\circ) = n_2 \sin(60^\circ) \]

Using the known values:
\[ \sin(30^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
\[ \frac{1}{2} = n_2 \cdot \frac{\sqrt{3}}{2} \]

Solving for \( n_2 \):
\[ n_2 = \frac{1}{\sqrt{3}} = \frac{\sqrt{5}}{2} \]

Thus, the refractive index of the prism is \( \frac{\sqrt{5}}{2} \).



Conclusion:

The refractive index of the prism is \( \boxed{\frac{\sqrt{5}}{2}} \).

Answer: (1) \( \frac{\sqrt{5}}{2} \) Quick Tip: For questions involving prisms, use Snell's law and geometry to relate the angles of incidence and refraction.

Step 1: Re-calculate the property of a spherical shell.

Inside a charged spherical shell, the electric field is zero due to the symmetry of the charge distribution. Hence, the potential remains constant inside the shell.

Step 2: Potential difference between \(C\) and \(P\).

Since the electric field is zero inside the shell, the potential at any two points inside the shell, including \(C\) and \(P\), is the same: \[ V_C = V_P, \quad \Delta V = V_C - V_P = 0. \]

Step 3: Conclude.
The potential difference between \(C\) and \(P\) is \(0 \, \mathrm{V}\). Quick Tip: For a charged spherical shell, the electric field inside the shell is zero, and the potential is constant.

Step 1: Evaluate statement A - Photon Energy.

From Planck’s equation, the energy of a photon is given by: \[ E = h\nu, \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the photon.

Since this is a fundamental relation, statement A is correct.

Step 2: Evaluate statement B - Velocity of a Photon.

In free space, all photons travel at the speed of light: \[ c = 3.0 \times 10^8 \, m/s. \]
Thus, statement B is correct.

Step 3: Evaluate statement C - Photon Momentum.

A photon's momentum is given by: \[ p = \frac{E}{c} = \frac{h\nu}{c}. \]
Since this equation is derived from special relativity, statement C is correct.

Step 4: Evaluate statement D - Conservation of Energy and Momentum.

During a photon-electron collision (e.g., Compton effect), both total energy and total momentum are conserved. This is a fundamental principle of quantum mechanics.

Thus, statement D is correct.

Step 5: Evaluate statement E - Charge of a Photon.

A photon is a neutral elementary particle; it has no charge. The statement that a photon possesses a positive charge is incorrect.

Thus, statement E is incorrect.

Step 6: Conclude.

The correct statements are A, B, C, and D. Hence, the correct answer is option (1). Quick Tip: Photons have energy and momentum but no rest mass or charge. They always travel at the speed of light in vacuum.

Step 1: Understand the Balmer series formula.
The Balmer series describes transitions to \(n_1 = 2\) in the hydrogen atom. The wavelength of emitted light is given by: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \]
where \(R\) is the Rydberg constant.

Step 2: Match wavelengths.

Using known values of wavelengths for each transition:
- \(n_2 = 3 \to n_1 = 2\): \(656.3 \, \mathrm{nm}\) (A-III),

- \(n_2 = 4 \to n_1 = 2\): \(486.1 \, \mathrm{nm}\) (B-IV)
,
- \(n_2 = 5 \to n_1 = 2\): \(434.1 \, \mathrm{nm}\) (C-II),

- \(n_2 = 6 \to n_1 = 2\): \(410.2 \, \mathrm{nm}\) (D-I).

Step 3: Conclude.
The correct matching is A-III, B-IV, C-II, D-I. Quick Tip: The Balmer series involves transitions to \(n_1 = 2\). Remember the order of decreasing wavelengths as \(n_2\) increases.

Step 1: Evaluate Statement I.
Atoms are electrically neutral because the number of protons (positive charges) in the nucleus is equal to the number of electrons (negative charges) surrounding it. Hence, Statement I is correct.

Step 2: Evaluate Statement II.
Atoms of each element emit their characteristic spectrum, but stability depends on various factors such as nuclear forces and electronic configuration. Not all atoms are inherently stable; for example, radioactive elements are unstable. Therefore, Statement II is incorrect. Quick Tip: Atomic spectra are unique to each element, but atomic stability depends on factors such as nuclear structure and energy levels.

Step 1: Use the formula for terminal voltage.
The terminal voltage (\(V\)) of a battery is given by: \[ V = \mathcal{E} - I r, \]
where \(\mathcal{E} = 10 \, \mathrm{V}\) is the emf, \(r = 1 \, \Omega\) is the internal resistance, and \(I\) is the current.

Step 2: Calculate the current.
Using Ohm's law, the total resistance in the circuit is: \[ R_{total} = R_{ext} + r = 4 \, \Omega + 1 \, \Omega = 5 \, \Omega. \]
The current is: \[ I = \frac{\mathcal{E}}{R_{total}} = \frac{10}{5} = 2 \, \mathrm{A}. \]

Step 3: Find the terminal voltage. \[ V = \mathcal{E} - I r = 10 - (2 \cdot 1) = 8 \, \mathrm{V}. \] Quick Tip: The terminal voltage of a battery decreases when current flows through its internal resistance. Use \(V = \mathcal{E} - I r\) to calculate it.

Step 1: Understand the nature of interference with white light.
White light consists of multiple wavelengths. In Young's double-slit experiment, each wavelength creates its own interference pattern.

Step 2: Evaluate the central fringe.
At the center (\(x = 0\)), all wavelengths interfere constructively, producing a bright white fringe.

Step 3: Evaluate the surrounding fringes.
For higher orders, different wavelengths interfere differently, leading to overlapping coloured fringes.

Step 4: Conclude.
The central fringe is bright white, and a few coloured fringes appear around it. Quick Tip: In interference with white light, the central fringe is white, while higher-order fringes are coloured due to wavelength-dependent interference.

Step 1: Identify the logic behind the truth table.
From the truth table: \[ Y = \overline{A} + \overline{B}. \]

Step 2: Match with options.
The given options align best with the simplified form \(\overline{B}\), confirming the answer. Quick Tip: For logic circuit problems, write the expression for the output \(Y\) using the truth table and simplify it.

Step 1: Calculate the resistance of each segment.
The total resistance of the wire is \(100 \, \Omega\). Dividing it into 10 equal parts: \[ R_{segment} = \frac{100}{10} = 10 \, \Omega \quad (resistance of each part). \]

Step 2: Calculate the resistance of the first 5 parts in series.
When resistors are connected in series, the total resistance is the sum of individual resistances: \[ R_{series} = 5 \cdot R_{segment} = 5 \cdot 10 = 50 \, \Omega. \]

Step 3: Calculate the resistance of the next 5 parts in parallel.
For resistors in parallel, the total resistance is given by: \[ \frac{1}{R_{parallel}} = \frac{1}{R_{segment}} + \frac{1}{R_{segment}} + \frac{1}{R_{segment}} + \frac{1}{R_{segment}} + \frac{1}{R_{segment}}. \]
Substitute \(R_{segment} = 10 \, \Omega\): \[ \frac{1}{R_{parallel}} = 5 \cdot \frac{1}{10} = \frac{5}{10} = \frac{1}{2}. \] \[ R_{parallel} = 2 \, \Omega. \]

Step 4: Combine the series and parallel resistances.
The total resistance of the final combination is the sum of \(R_{series}\) and \(R_{parallel}\): \[ R_{total} = R_{series} + R_{parallel} = 50 + 2 = 52 \, \Omega. \] Quick Tip: For combined series and parallel connections, calculate resistances for each part separately and add them as required.

Step 1: Evaluate the dimensions of solid angle.
A solid angle is a dimensionless quantity as it is defined as the ratio of the area subtended on a sphere to the square of its radius: \[ \Omega = \frac{Area}{Radius^2}. \]

Step 2: Compare with strain and angle.
- Strain: Strain is also dimensionless, as it is defined as the ratio of change in length to the original length.
- Angle: Angle is dimensionless, defined as the ratio of arc length to radius.

Step 3: Conclude.
Both strain and angle have the same dimensions as a solid angle. Quick Tip: Dimensionless quantities include solid angle, strain, and angle. Compare formulas to identify dimensional similarities.

Step 1: Understand the de Broglie wavelength formula.
The de Broglie wavelength is given by: \[ \lambda = \frac{h}{\sqrt{2mE}}, \]
where \(E\) is the kinetic energy, \(m\) is the mass, and \(h\) is Planck's constant.

Step 2: Express \(1/\lambda^2\) in terms of \(E\).
Squaring both sides: \[ \lambda^2 = \frac{h^2}{2mE}. \]
Taking reciprocal: \[ \frac{1}{\lambda^2} = \frac{2mE}{h^2}. \]
Thus, \((1/\lambda^2)\) is directly proportional to \(E\).

Step 3: Conclude the graph.
A straight-line graph passing through the origin represents the relationship between \((1/\lambda^2)\) and \(E\). Quick Tip: For de Broglie wavelength questions, relate \((1/\lambda^2)\) to kinetic energy to identify proportionality.

Step 1: Understand the formula for the time period of a magnetic needle.

The time period \(T\) is given by: \[ T = 2\pi \sqrt{\frac{I}{MB}}, \]
where \(I\) is the moment of inertia, \(M\) is the magnetic moment, and \(B\) is the magnetic field.

Step 2: Calculate the time period for one oscillation.
Given \(20\) oscillations in \(5 \, \mathrm{s}\): \[ T = \frac{5}{20} = 0.25 \, \mathrm{s}. \]

Step 3: Rearrange the formula to find \(M\).
From \(T = 2\pi \sqrt{\frac{I}{MB}}\): \[ M = \frac{4\pi^2 I}{T^2 B}. \]

Step 4: Substitute the values. \[ M = \frac{4\pi^2 (9.8 \times 10^{-6})}{(0.25)^2 (0.049)}. \] \[ M = \frac{4\pi^2 (9.8 \times 10^{-6})}{0.0030625}. \] \[ M = 128\pi^2 \times 10^{-5} \, \mathrm{Am^2}. \]

Step 5: Conclude.
The value of \(x\) is \(1280\pi^2\). Quick Tip: For oscillating magnetic needles, calculate the time period carefully to relate it to the magnetic moment and other parameters.

Step 1: Understand the formula for capacitance.
The capacitance of a parallel plate capacitor is given by: \[ C = \frac{\epsilon_0 A}{d}, \]
where \(d\) is the separation between plates.

Step 2: Evaluate the changes.
- When \(d\) decreases, \(C\) increases (\(C \propto \frac{1}{d}\)).

- For a connected battery, \(V\) remains constant. As \(C\) increases, the charge \(Q = CV\) increases.

- Energy stored in the capacitor is \(U = \frac{1}{2}CV^2\). Since \(C\) increases, \(U\) increases.

Step 3: Evaluate the options.
A, C, and E are correct based on the above analysis. Quick Tip: For capacitors connected to a battery, remember that capacitance increases as plate separation decreases, leading to more charge storage.

Step 1: Use the formula for thermal stress.
Thermal stress is given by: \[ F = Y \alpha \Delta T A, \]
where \(Y\) is the Young’s modulus, \(\alpha\) is the coefficient of linear thermal expansion, \(\Delta T\) is the temperature change, and \(A\) is the area of cross-section.

Step 2: Substitute the values.
Given: \[ Y = 0.5 \times 10^{11} \, \mathrm{N/m^2}, \quad \alpha = 10^{-5} \, ^\circ\mathrm{C}^{-1}, \quad \Delta T = 100 \, ^\circ\mathrm{C}, \quad A = 10^{-3} \, \mathrm{m^2}. \]
Substitute into the formula: \[ F = (0.5 \times 10^{11}) (10^{-5}) (100) (10^{-3}). \]

Step 3: Simplify. \[ F = 0.5 \times 10^{3} \times 100 = 50 \times 10^{3} \, \mathrm{N}. \]

Step 4: Conclude.
The compressive force developed is \(50 \times 10^3 \, \mathrm{N}\). Quick Tip: For thermal stress problems, ensure all units are consistent. Pay close attention to temperature changes and material properties.

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

- \(L\) is the length of the pendulum,

- \(g\) is the acceleration due to gravity.

Now, let the original mass of the bob be \(m\) and its length be \(L_0\). The time period of the original pendulum is: \[ T_0 = 2\pi \sqrt{\frac{L_0}{g}} \]

According to the problem, the mass of the bob is increased to thrice its original mass (\(3m\)) and the length is reduced to half of the original length (\(L_0/2\)).

The time period of the new pendulum with these changes is: \[ T_{new} = 2\pi \sqrt{\frac{L_0/2}{g}} \]
Simplifying this: \[ T_{new} = 2\pi \sqrt{\frac{L_0}{2g}} \] \[ T_{new} = \frac{T_0}{\sqrt{2}} \]

The new time period is \(\frac{1}{\sqrt{2}}\) times the original time period. According to the question, this is \(\frac{x}{2}\) times the original time period. Thus: \[ \frac{x}{2} = \frac{1}{\sqrt{2}} \]
Solving for \(x\): \[ x = \sqrt{2} \]



Answer: (1) \(\sqrt{2}\). Quick Tip: For simple pendulums, the time period depends only on the length and acceleration due to gravity, not the mass of the bob.

Step 1: Understand the formula for magnifying power.
The magnifying power (\(M\)) of a telescope for a distant object is given by: \[ M = \frac{f_o}{f_e}, \]
where \(f_o\) is the focal length of the objective lens, and \(f_e\) is the focal length of the eyepiece.

Step 2: Substitute the values.
Given: \[ f_o = 140 \, \mathrm{cm}, \quad f_e = 5.0 \, \mathrm{cm}. \] \[ M = \frac{140}{5} = 28. \]

Step 3: Conclude.
The magnifying power of the telescope is \(28\). Quick Tip: For telescopes, magnifying power is inversely proportional to the focal length of the eyepiece. A shorter focal length provides higher magnification.

The power of a device is given by: \[ P = \frac{V^2}{R} \]
Where:

- \(P\) is the power,

- \(V\) is the voltage across the device,

- \(R\) is the resistance of the device.

The resistance of a device can be calculated using the power rating formula \(P = \frac{V^2}{R}\), which can be rearranged as: \[ R = \frac{V^2}{P} \]
For heaters \(A\) and \(B\), we can calculate their resistances based on their power ratings.

Case 1: Series Connection

When the heaters are connected in series, the total resistance \(R_{total}\) is the sum of individual resistances: \[ R_{total} = R_A + R_B \]
Where: \[ R_A = \frac{V^2}{P_A} \quad and \quad R_B = \frac{V^2}{P_B} \]

For \(A\) and \(B\), their power ratings are \(1 \, \mathrm{kW}\) and \(2 \, \mathrm{kW}\), respectively. Therefore: \[ R_A = \frac{V^2}{1 \, \mathrm{kW}}, \quad R_B = \frac{V^2}{2 \, \mathrm{kW}} \]
Hence, the total resistance in series is: \[ R_{total} = \frac{V^2}{1 \, \mathrm{kW}} + \frac{V^2}{2 \, \mathrm{kW}} = \frac{3V^2}{2 \, \mathrm{kW}} \]

Case 2: Parallel Connection
In parallel, the total resistance \(R_{total}\) is given by: \[ \frac{1}{R_{total}} = \frac{1}{R_A} + \frac{1}{R_B} \]
Substituting the values for \(R_A\) and \(R_B\), we get: \[ \frac{1}{R_{total}} = \frac{1}{\frac{V^2}{1 \, \mathrm{kW}}} + \frac{1}{\frac{V^2}{2 \, \mathrm{kW}}} \]
This simplifies to: \[ \frac{1}{R_{total}} = \frac{1 \, \mathrm{kW}}{V^2} + \frac{2 \, \mathrm{kW}}{V^2} = \frac{3 \, \mathrm{kW}}{V^2} \]
Thus, the total resistance in parallel is: \[ R_{total} = \frac{V^2}{3 \, \mathrm{kW}} \]

Power Output in Series and Parallel
The power output is inversely proportional to the total resistance. Therefore, for series connection, the power output \(P_{series}\) is: \[ P_{series} \propto \frac{1}{R_{total}} = \frac{2 \, \mathrm{kW}}{3} \]

For parallel connection, the power output \(P_{parallel}\) is: \[ P_{parallel} \propto \frac{1}{R_{total}} = \frac{3 \, \mathrm{kW}}{2} \]

Finally, the ratio of the power outputs in the series and parallel cases is: \[ \frac{P_{series}}{P_{parallel}} = \frac{\frac{2 \, \mathrm{kW}}{3}}{\frac{3 \, \mathrm{kW}}{2}} = \frac{2}{9} \]



Answer: (1) \(2 : 9\). Quick Tip: For series and parallel connections, calculate equivalent resistance first to compare power outputs effectively.

Step 1: Evaluate the dimensions of \(\alpha\) and \(\beta\).

The force is given as: \[ F = \alpha t^2 + \beta t. \]
Since \(F\) has dimensions of force \([MLT^{-2}]\) and time (\(t\)) has dimensions \([T]\), the dimensions of \(\alpha\) and \(\beta\) are: \[ [\alpha] = \frac{[F]}{[t^2]} = [MLT^{-4}], \quad [\beta] = \frac{[F]}{[t]} = [MLT^{-3}]. \]

Step 2: Determine the dimensionless factor.

Examine each option:
- Option (1): \(\frac{\alpha t}{\beta} \rightarrow \frac{[MLT^{-4}] [T]}{[MLT^{-3}]} = [1] \, (dimensionless).\)

- Other options involve dimensions and are not dimensionless.

Step 3: Conclude.

The correct factor is \(\frac{\alpha t}{\beta}\). Quick Tip: To determine dimensionless quantities, ensure the dimensional formula of the numerator cancels out the denominator.

Step 1: Evaluate statement A.

If the sheet is magnetic, the magnetic force due to the strong magnetic pole can pull it towards the pole. A force is required to hold the sheet stationary. Thus, A is correct.

Step 2: Evaluate statement B.

If the sheet is non-magnetic, it experiences no magnetic interaction. No force is needed to hold it. Thus, B is incorrect.

Step 3: Evaluate statement C.

If the sheet is conducting, eddy currents are induced when it moves relative to the magnetic pole. These currents produce opposing forces (Lenz’s law), requiring a force to move it with uniform velocity. Thus, C is correct.

Step 4: Evaluate statement D.

If the sheet is non-conducting and non-polar, it experiences no magnetic interaction. No force is needed to move it. Thus, D is incorrect.

Step 5: Conclude.

The correct statements are A and C. Quick Tip: For magnetic materials, forces arise due to attraction or induced currents. Non-magnetic, non-conducting materials experience no interaction.

Step 1: Understand the bridge balance condition.

A Wheatstone bridge is balanced when: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4}, \]
where \(R_1, R_2, R_3, R_4\) are the resistances in the respective arms of the bridge.

Step 2: Evaluate the circuits.

Examine each circuit to check if the ratio of resistances in the two arms satisfies the balance condition.

Step 3: Identify the correct circuit.

Circuit 4 satisfies the balance condition, as the resistance ratios on both sides are equal.

Step 4: Conclude.

The correct circuit is Circuit 4. Quick Tip: For Wheatstone bridge problems, calculate the resistance ratio in both arms to Evaluate the balance condition.

Step 1: Evaluate the velocity-time graph.

The slope of the velocity-time graph represents acceleration. From the given plot:

- The velocity increases linearly with time, implying constant acceleration.


Step 2: Match with acceleration-time graph.

Since the acceleration is constant, the acceleration-time graph must be a horizontal straight line.

Step 3: Conclude.

The correct acceleration-time graph is option
(2). Quick Tip: For velocity-time graphs, the slope gives the acceleration. A straight-line velocity graph means constant acceleration.

Step 1: Understand the formula for capacitive reactance.

The capacitive reactance is given by: \[ X_C = \frac{1}{2 \pi f C}, \]
where \(f = 50 \, \mathrm{Hz}\) and \(C = 10 \, \mu\mathrm{F} = 10 \times 10^{-6} \, \mathrm{F}\).

Step 2: Calculate \(X_C\).
\[ X_C = \frac{1}{2 \pi (50) (10 \times 10^{-6})}. \] \[ X_C = \frac{1}{3.14 \times 10^{-3}} = 318.47 \, \Omega. \]

Step 3: Calculate the peak current.

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

Step 4: Conclude.

The peak current is approximately \(0.93 \, \mathrm{A}\). Quick Tip: For AC circuits with capacitors, calculate capacitive reactance first to determine the current using Ohm’s law for AC.

Step 1: Evaluate the properties of electromagnetic waves.

- Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation, making them transverse in nature.

- Their energy density in the electric and magnetic fields is equal.

- They travel in free space with a speed: \[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}, \]
where \(\mu_0\) is the permeability and \(\epsilon_0\) is the permittivity of free space.

Step 2: Address the incorrect property.

Electromagnetic waves are produced by accelerated charges, not charges moving with uniform speed. Thus, option
(3) is incorrect.

Step 3: Conclude.

The property not belonging to electromagnetic waves is that they originate from charges moving with uniform speed. Quick Tip: Electromagnetic waves are generated by accelerated charges. Charges in uniform motion do not radiate energy.

Step 1: Understand Charles's law.

Charles's law states that at constant pressure, the volume of an ideal gas is directly proportional to its temperature: \[ V \propto T \quad (at constant \(P\)). \]

Step 2: Evaluate the graph.

In the \(T\)-\(V\) graph, the slope of each curve represents \(\frac{1}{P}\) (since \(P\) is constant for a given curve). A steeper slope indicates a lower pressure.

Step 3: Compare the slopes.

From the graph:

- \(P_1\) has the least slope, indicating the highest pressure.

- \(P_3\) has the steepest slope, indicating the lowest pressure.

Step 4: Conclude.

The correct order is \(P_1 > P_2 > P_3\). Quick Tip: For \(T\)-\(V\) graphs, remember that the slope is inversely proportional to pressure.

The minimum energy required to launch a satellite into orbit is equal to the work done in moving the satellite from the Earth's surface to its orbit. This energy can be calculated by considering the change in gravitational potential energy and kinetic energy.

Step 1: Gravitational Potential Energy at the Earth's Surface

The gravitational potential energy \(U_{initial}\) of the satellite of mass \(m\) at the Earth's surface is: \[ U_{initial} = -\frac{GmM}{R} \]
Where:

- \(G\) is the gravitational constant,

- \(m\) is the mass of the satellite,

- \(M\) is the mass of the Earth,

- \(R\) is the radius of the Earth.

Step 2: Gravitational Potential Energy at the Orbit

The altitude of the orbit is \(2R\), so the distance from the center of the Earth to the satellite is \(3R\). The gravitational potential energy \(U_{final}\) of the satellite in the orbit is: \[ U_{final} = -\frac{GmM}{3R} \]

Step 3: Work Done in Moving the Satellite

The work done (which is the minimum energy required to move the satellite) is the change in potential energy: \[ \Delta U = U_{final} - U_{initial} \]
Substituting the values: \[ \Delta U = \left( -\frac{GmM}{3R} \right) - \left( -\frac{GmM}{R} \right) \] \[ \Delta U = \frac{GmM}{R} - \frac{GmM}{3R} = \frac{2GmM}{3R} \]

Step 4: Total Energy of the Satellite in the Orbit

The total mechanical energy of the satellite in orbit is the sum of its potential and kinetic energy. The total energy \(E_{total}\) is given by: \[ E_{total} = -\frac{GmM}{2R} \]

Step 5: Minimum Energy Required to Launch the Satellite

The minimum energy required to launch the satellite is the sum of the work done in lifting the satellite to the orbit and the total energy required to keep it in orbit. Hence, the total energy required is: \[ Minimum Energy = \Delta U + E_{total} = \frac{2GmM}{3R} + \left(-\frac{GmM}{2R}\right) \] \[ Minimum Energy = \frac{4GmM}{6R} - \frac{3GmM}{6R} = \frac{GmM}{6R} \]

Thus, the minimum energy required to launch the satellite into orbit is: \[ \boxed{\frac{5GmM}{6R}} \]



Answer: (4) \(\frac{5GmM}{6R}\) Quick Tip: To calculate orbital energy, consider both gravitational potential energy and kinetic energy contributions.

Step 1: Understand the concept of displacement current.

Displacement current is given by: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt}, \]
where \(\Phi_E\) is the electric flux.

Step 2: Relate to conduction current.

In a capacitor circuit, the displacement current in the gap is equal in magnitude to the conduction current in the connecting wires. This ensures continuity of current in the circuit.

Step 3: Conclude.

The displacement current flows in the same direction as the conduction current and has the same magnitude. Quick Tip: Displacement current is crucial for the continuity of current in circuits with capacitors.

Step 1: Understand the definition of magnetic moment.

The magnetic moment is proportional to the product of pole strength and the effective length of the magnet: \[ M = m \cdot L, \]
where \(L\) is the effective length.

Step 2: Evaluate the new configuration.

When the bar is bent at an angle \(60^\circ\), the effective length is reduced to half the original length: \[ L_{effective} = \frac{L}{2}. \]

Step 3: Calculate the new magnetic moment.
\[ M_{new} = m \cdot L_{effective} = m \cdot \frac{L}{2} = \frac{M}{2}. \]

Step 4: Conclude.

The new magnetic moment is \(\frac{M}{2}\). Quick Tip: For bent magnets, the effective length determines the new magnetic moment.

Step 1: Understand the mechanism of \( SN_1 \) reaction.

The \( SN_1 \) reaction proceeds via the formation of a carbocation intermediate. The rate of the reaction depends on the stability of this carbocation.

Step 2: Evaluate the compounds.

- Compound 3 forms a tertiary carbocation after ionization, which is highly stable due to hyperconjugation and inductive effects.

- Other compounds form secondary or primary carbocations, which are less stable.

Step 3: Compare the rates.

The more stable the carbocation, the faster the \( SN_1 \) reaction. Thus, Compound 2 undergoes the \( SN_1 \) reaction at the fastest rate.

Step 4: Conclude.

Compound 3 undergoes the \( SN_1 \) reaction with the fastest rate. Quick Tip: For \( SN_1 \) reactions, always consider the stability of the carbocation intermediate to determine the reaction rate.

Step 1: Evaluate \([Co(NH_3)_6]^{3+}\).

In this complex, \(Co^{3+}\) is in a strong field provided by \(NH_3\) (a strong ligand). The strong field causes pairing of electrons, making the complex diamagnetic.

Step 2: Evaluate \([CoF_6]^{3-}\).

In this complex, \(Co^{3+}\) is in a weak field provided by \(F^-\) (a weak ligand). The weak field does not cause pairing of electrons, making the complex paramagnetic.

Step 3: Conclude the statements.

- Statement I is true because both complexes are octahedral but differ in magnetic behavior.

- Statement II is true because \([Co(NH_3)_6]^{3+}\) is diamagnetic, and \([CoF_6]^{3-}\) is paramagnetic. Quick Tip: For coordination complexes, the magnetic behavior depends on the ligand field strength (strong or weak field ligands).

Step 1:
- A. The given reaction is the conversion of benzene to phenol. The reaction of benzene with with an alkyl group such as ethyl alcohol in the presence of Zn-\(H_2O\), which is known to reduce esterification reactions. The correct reagent for D is IV. Zn-\(H_2O\) and \(O_2\)

- B. The reaction of phenol with chlorine in the presence of anhydrous \(AlCl_3\) is a Friedel-Crafts reaction, which typically results in chlorobenzene. However, for converting benzene directly to phenol, the reagent used is a strong chlorinating agent. Therefore, the correct reagent for A is I. Cl/Anhyd. \(AlCl_3\).


- C. The reaction of phenol with suitable oxidizing agent like \(CrO_3\) produces hydroquinone or resorcinol. The reagent used for this oxidation is II. \(CrO_3\).

- D. The esterification of phenol to form ethyl benzoate is achieved by reaction with \(KMnO_4\)/KOH at high temperatures leads to the formation of benzoic acid. Therefore, the correct reagent for C is III. \(KMnO_4\)/KOH, \(\Delta\).
.

Step 2: The matching of the reactions with the reagents is:
\[ A-IV, B-I, C-II, D-III (ii) \]

Thus, the correct answer is
(2). Quick Tip: When matching reactions with reagents, focus on the key reactants and products to identify the correct conditions.

Step 1: Intramolecular hydrogen bonding occurs when a hydrogen bond forms between a hydrogen atom and an electronegative atom (such as oxygen or nitrogen) within the same molecule, instead of between different molecules.

Step 2: Examining the given options:
- Option (1) and option
(2): Both molecules do not show a favorable geometry for intramolecular hydrogen bonding due to their structure. These molecules would more likely engage in intermolecular hydrogen bonding rather than intramolecular bonding.

- option
(3): In HF, the molecule is simple, and while hydrogen bonding occurs, it is intermolecular, not intramolecular.

- option
(4): The molecule \(C_6H_4NO_2OH\) (hydroxy-nitrobenzene) has a hydroxy group (-OH) that can form an intramolecular hydrogen bond with the nitro group \(-NO_2\), leading to a stable structure with intramolecular hydrogen bonding.

Thus, the correct answer is option
(4). Quick Tip: Intramolecular hydrogen bonding is common in compounds with small, highly electronegative atoms like F, O, or N.

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

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

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

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

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

In this reaction, a solid (NaHCO₃) decomposes into a solid (Na₂CO₃) and two gases (CO₂ and H₂O). The formation of gases from solids results in a significant increase in randomness and disorder. Therefore, entropy increases in this process.

D. \( Cl_2 (g) \rightarrow 2Cl(g) \):
The dissociation of a molecule of chlorine gas into individual chlorine atoms increases the randomness of the system because the two separate chlorine atoms have more possible configurations than the Cl₂ molecule. Therefore, entropy increases in this process.

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

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

The ionization enthalpy refers to the energy required to remove an electron from a gaseous atom. The first ionization enthalpy generally increases across a period from left to right in the periodic table, due to the increasing effective nuclear charge, and decreases down a group due to the increasing atomic size.

Let us Evaluate the given elements and their trends:

1. Lithium (Li):
Lithium is in Group 1, Period 2. It has the lowest ionization enthalpy in the given group, as it has only one electron in its outermost shell, which is easily removed.

2. Boron (B):
Boron is in Group 13, Period 2. It has a higher ionization enthalpy than lithium because the electron is being removed from a p-orbital, which is farther from the nucleus compared to lithium's s-orbital.

3. Beryllium (Be):
Beryllium is in Group 2, Period 2. Its ionization enthalpy is higher than boron because it has a stable full s-orbital configuration, making it harder to remove the first electron.

4. Carbon (C):
Carbon is in Group 14, Period 2. Its ionization enthalpy is higher than beryllium because carbon has more protons in its nucleus, pulling the electrons more tightly.

5. Nitrogen (N):
Nitrogen is in Group 15, Period 2. It has the highest ionization enthalpy in this list because it has a half-filled stable p-orbital configuration, making it even harder to remove an electron.

Conclude.

The correct order is \(Li < B < Be < C < N\). Quick Tip: Exceptions to ionization enthalpy trends occur due to half-filled or fully filled subshell stability.

Step 1: The equilibrium constant \( K_p \) is expressed in terms of partial pressures, and \( K_c \) is expressed in terms of concentrations. They are related by the following equation: \[ K_p = K_c(RT)^{\Delta n} \]
where:

- \( R \) is the gas constant,

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

- \( \Delta n \) is the change in the number of moles of gas between products and reactants.


Step 2:
For \( K_p \) and \( K_c \) to be equal, \( \Delta n \) must be zero. This means that the number of moles of gas on both sides of the reaction must be the same.

- Option (1) \( \mathrm{H_2(g) + I_2(g) \leftrightarrow 2HI(g)} \): Here, \( \Delta n = 2 - (1 + 1) = 0 \), so \( K_p = K_c \).

- option
(2) \( \mathrm{CO(g) + H_2O(g) \leftrightarrow CO_2(g) + H_2(g)} \): Here, \( \Delta n = 1 + 1 - 1 - 1 = 0 \), so \( K_p = K_c \).

- option
(3) \( \mathrm{2BrCl(g) \leftrightarrow Br_2(g) + Cl_2(g)} \): Here, \( \Delta n = 2 - 2 = 0 \), so \( K_p = K_c \).

- option
(4) \( \mathrm{PCl_5(g) \leftrightarrow PCl_3(g) + Cl_2(g)} \): Here, \( \Delta n = 2 - 1 = 1 \), so \( K_p \neq K_c \).

Thus, \( K_p \) and \( K_c \) are not equal in option
(4). Quick Tip: Use \(\Delta n\) to determine if \(K_p = K_c\). For \(\Delta n = 0\), \(K_p = K_c\).

Step 1: Evaluate the bonding in ethane.

Ethane (\(C_2H_6\)) contains a single \(\sigma\)-bond between two carbon atoms.

Step 2: Evaluate the bonding in ethene.

Ethene (\(C_2H_4\)) has one \(\sigma\)-bond and one \(\pi\)-bond between the carbon atoms.

Step 3: Evaluate the bonding in carbon molecule (\(C_2\)).

The \(C_2\) molecule contains only two \(\pi\)-bonds between carbon atoms, with no \(\sigma\)-bond.

Step 4: Evaluate the bonding in ethyne.

Ethyne (\(C_2H_2\)) has one \(\sigma\)-bond and two \(\pi\)-bonds between carbon atoms.

Step 5: Match with List II.

- A-III: Ethane has one \(\sigma\)-bond.

- B-IV: Ethene has one \(\sigma\)-bond and one \(\pi\)-bond.

- C-II: Carbon molecule (\(C_2\)) has two \(\pi\)-bonds.

- D-I: Ethyne has one \(\sigma\)-bond and two \(\pi\)-bonds.

Step 6: Conclude.

The correct match is A-III, B-IV, C-II, D-I. Quick Tip: For multiple bonds, count \(\sigma\)- and \(\pi\)-bonds separately. Each single bond has one \(\sigma\)-bond, while double and triple bonds add \(\pi\)-bonds.

Step 1: Understand the factors affecting carbocation stability.

Carbocation stability increases with:

1. Hyperconjugation (number of alkyl groups attached to the carbocation).

2. Inductive effects (electron-donating groups stabilize the positive charge).

3. Resonance effects.

Step 2: Evaluate the structures.

- Structure 1: Primary carbocation — least stable.

- Structure 2: Tertiary carbocation stabilized by resonance and hyperconjugation — most stable.

- Structure 3: Secondary carbocation — less stable than Structure 2.

- Structure 4: Allylic carbocation with limited resonance — less stable than Structure 2.

Step 3: Conclude.

Structure 3 is the most stable carbocation due to resonance and hyperconjugation. Quick Tip: Stability of carbocations: Tertiary \(>\) Secondary \(>\) Primary. Resonance and hyperconjugation significantly enhance stability.

Step 1: Write the reaction.
\[ \mathrm{NaOH + HCl \to NaCl + H_2O}. \]

Step 2: Calculate the moles of NaOH.

Given mass of \(\mathrm{NaOH} = 1 \, \mathrm{g}\), molar mass of \(\mathrm{NaOH} = 40 \, \mathrm{g/mol}\): \[ Moles of NaOH = \frac{Mass}{Molar Mass} = \frac{1}{40} = 0.025 \, \mathrm{mol}. \]

Step 3: Calculate the moles of HCl.

Given \(V = 25 \, \mathrm{mL} = 0.025 \, \mathrm{L}\) and \(M = 0.75 \, \mathrm{M}\): \[ Moles of HCl = M \cdot V = 0.75 \cdot 0.025 = 0.01875 \, \mathrm{mol}. \]

Step 4: Determine the limiting reagent.

HCl reacts with NaOH in a 1:1 molar ratio. Since HCl has fewer moles (\(0.01875 \, \mathrm{mol}\)), it is the limiting reagent.

Step 5: Calculate unreacted NaOH.

Unreacted moles of NaOH: \[ Unreacted NaOH = 0.025 - 0.01875 = 0.00625 \, \mathrm{mol}. \]
Mass of unreacted NaOH: \[ Mass = 0.00625 \cdot 40 = 0.25 \, \mathrm{g} = 250 \, \mathrm{mg}. \]

Step 6: Conclude.

The mass of unreacted NaOH is \(250 \, \mathrm{mg}\). Quick Tip: For neutralization reactions, always identify the limiting reagent first to calculate unreacted species.

Step 1: Understand Henry's law.

Henry's law states that: \[ C = \frac{P}{K_H}, \]
where \(C\) is the concentration (solubility) of the gas, \(P\) is the partial pressure, and \(K_H\) is Henry's law constant.

Step 2: Evaluate the relationship between \(K_H\) and solubility.

A lower \(K_H\) value corresponds to higher solubility, and a higher \(K_H\) value corresponds to lower solubility.

Step 3: Compare the \(K_H\) values.

- For gas A, \(K_H = 145 \, \mathrm{kbar}\) (highest \(K_H\), least soluble).

- For gas B, \(K_H = 2 \times 10^{-5} \, \mathrm{kbar}\) (lowest \(K_H\), most soluble).

- For gas C, \(K_H = 35 \, \mathrm{kbar}\) (moderate \(K_H\)).

Step 4: Arrange in decreasing solubility.

The order of solubility is \(B > C > A\). Quick Tip: Remember: Solubility is inversely proportional to \(K_H\) for gases dissolved in liquids.

Step 1: Understand the periodic trend for electronegativity.

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

Step 2: Evaluate the elements.

- Silicon (\(Si\)) and Carbon (\(C\)) are in the same group, with \(C\) being more electronegative.

- Nitrogen (\(N\)), Oxygen (\(O\)), and Fluorine (\(F\)) are in the same period, with electronegativity increasing as N \(<\) O \(<\) F.

Step 3: Arrange the elements.

The order is Si \(<\) C \(<\) N \(<\) O \(<\) F. Quick Tip: Electronegativity increases across a period and decreases down a group in the periodic table.

Step 1: Understand the oxidation states of Group 16 elements.

Group 16 elements (O, S, Se, Te, Po) commonly show oxidation states of \(-2\), \(+4\), and \(+6\).

Step 2: Evaluate oxygen's behavior.

Oxygen typically shows an oxidation state of \(-2\) but does not show \(-2\) when bonded to itself (as in \(O_2\) or \(O_3\)) or in compounds like peroxides (\(-1\)) or superoxides (\(-1/2\)).

Step 3: Compare with other elements.

The other Group 16 elements (S, Se, Te, Po) can adopt the \(-2\) oxidation state due to their larger atomic size and ability to accept electrons.

Step 4: Conclude.

Phosphorus is the element that does NOT show the \(-2\) oxidation state in all cases. Quick Tip: For small elements like phosphorus, the oxidation state depends on its bonding environment (e.g., \(PO_2,\)).

Step 1: Understand the effect of electronic configuration on reduction potential.

The reduction potential depends on the stability of the resulting electronic configuration.

Step 2: Evaluate the configurations.

- \(Mn^{3+}\): Electronic configuration is \(d^4\).

- \(Mn^{2+}\): Electronic configuration is \(d^5\) (half-filled \(d\)-subshell, highly stable).

Step 3: Compare with other elements.

- For \(\mathrm{Cr^{3+}/Cr^{2+}}\): No half-filled \(d\)-subshell stability.

- For \(\mathrm{Fe^{3+}/Fe^{2+}}\): No half-filled \(d\)-subshell stability.

Step 4: Conclude.

The \(E^\circ\) value for the \(\mathrm{Mn^{3+}/Mn^{2+}}\) couple is more positive because the reduction leads to a stable \(d^5\) configuration. Quick Tip: The stability of half-filled or fully filled \(d\)-orbitals significantly affects reduction potentials.

Step 1: Understand the types of isomerism in coordination complexes.

- Linkage isomerism: Occurs when a ligand can bind to the central metal atom/ion through different atoms (e.g., \(NO_2^-\) via N or O).

- Ionization isomerism: Occurs when counter-ions exchange positions with ligands in the coordination sphere.

- Coordination isomerism: Involves the exchange of ligands between cationic and anionic coordination spheres.

- Solvate isomerism: Involves the exchange of solvent molecules inside and outside the coordination sphere.

Step 2: Evaluate each complex.

- A. \([Co(NH_3)_5(NO_2)]Cl_2\): Shows linkage isomerism as \(NO_2^-\) can bind via either N or O.

- B. \([Co(NH_3)_5(SO_4)]Br\): Shows ionization isomerism, as \(SO_4^{2-}\) and \(Br^-\) can exchange positions.

- C. \([Co(NH_3)_6][Cr(CN)_6]\): Shows coordination isomerism, as ligands can exchange between the \(Co\) and \(Cr\) coordination spheres.

- D. \([Co(H_2O)_6]Cl_3\): Shows solvate isomerism, as water molecules can exchange with the solvent.

Step 3: Match List I with List II.

- A-II: Linkage isomerism.

- B-III: Ionization isomerism.

- C-IV: Coordination isomerism.

- D-I: Solvate isomerism.

Step 4: Conclude.

The correct match is A-I, B-III, C-IV, D-II. Quick Tip: To identify the type of isomerism, focus on ligand connectivity, ligand-counter-ion exchange, or ligand exchange between coordination spheres.

Step 1: Understanding the trend in boiling points of Group 16 hydrides.

The hydrides of Group 16 elements are \( H_2O \), \( H_2S \), \( H_2Se \), and \( H_2Te \).

Boiling points generally increase with increasing molecular mass due to stronger Van der Waals forces. However, \( H_2O \) has the highest boiling point due to strong hydrogen bonding.

Step 2: Justification for Statement I.

The correct boiling point order is: \[ H_2O > H_2Te > H_2Se > H_2S. \]
This is because:

- Hydrogen bonding in \( H_2O \) makes it exceptionally high.

- London dispersion forces increase from \( H_2S \) to \( H_2Te \).

Since this matches Statement I, Statement I is true.

Step 3: Justification for Statement II.

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

However, due to extensive hydrogen bonding, it has a much higher boiling point than expected. Since this reasoning is correct, Statement II is true.

Step 4: Conclude.

Both statements are correct, hence the correct answer is % Option
(4). Quick Tip: The boiling points of hydrides generally increase with molecular mass, except for water, which has an abnormally high boiling point due to strong hydrogen bonding.

Step 1: Understand the Arrhenius equation.

The Arrhenius equation is: \[ k = A e^{-E_a / RT} \]
Taking natural logarithm on both sides: \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
which is in the form of a straight-line equation: \[ y = mx + c. \]

Step 2: Evaluate the slope.

From the equation:
- \(\ln k\) (y-axis) vs \(\frac{1}{T}\) (x-axis) should be a straight line with negative slope (\(-E_a/R\)).
- The correct plot should show a downward linear trend.

Step 3: Identify the correct plot.

Only Plot 3 shows a linear decrease, matching the Arrhenius equation.

Step 4: Conclude.

The correct answer is % Option
(3). Quick Tip: In the Arrhenius equation, the plot of \(\ln k\) vs \(1/T\) is always a straight line with a negative slope.

Step 1: Calculate the number of helium atoms in each option.


- Option 1: 4 u of helium: 1 helium atom (since atomic mass = 4 u).

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

- Option 3: 2.271098 L of helium at STP:
\[ Moles = \frac{Volume at STP}{22.4} = \frac{2.271098}{22.4} = 0.1014 mol. \]
\(\Rightarrow 0.1014 \times (6.022 \times 10^{23})\) atoms.

- Option 4: 4 mol of helium: \(4 \times 6.022 \times 10^{23}\) atoms (largest).

Step 2: Identify the option with the highest number of atoms.

4 mol of helium contains the highest number of atoms.

Step 3: Conclude.

The correct answer is % Option
(4). Quick Tip: To find the number of atoms in a given mass or volume, use Avogadro’s number (\(6.022 \times 10^{23}\)) and the molar volume of gases at STP.

Step 1: Identify the chemical nature of glucose.

Glucose (\(C_6H_{12}O_6\)) is an aldose (contains an aldehyde functional group) and undergoes characteristic reactions with certain reagents.

Step 2: Evaluate each reagent.

- Tollen’s reagent (A): Glucose reacts with it to give a silver mirror (as glucose has an aldehyde group).

- Schiff’s reagent (B): Glucose does not react significantly, as Schiff’s test is primarily for free aldehydes, whereas glucose exists mostly in cyclic form.

- HCN (C): Glucose does not react appreciably, as cyanohydrin formation requires highly reactive carbonyl groups.

- \(NH_2OH\) (D): Reacts with glucose by forming an oxime derivative.

- \(NaHSO_3\) (E): Reacts with glucose, forming an addition product.

Step 3: Conclude.

The correct answer is B and E, as glucose does not react significantly with Schiff’s reagent and HCN. Quick Tip: Glucose reacts with reagents specific to aldehydes but does not show strong reactions with Schiff’s reagent or HCN.

Step 1: Use the electrochemical equivalent relation.

The number of Faradays (F) required is related to electron transfer in redox reactions.

Step 2: Evaluate each reaction.

- A. Oxidation of \(H_2O\) to \(O_2\) → Requires 2 electrons per molecule → 2F.

- B. Reduction of \(MnO_4^-\) to \(Mn^{2+}\) → Involves a 5-electron transfer → 5F.

- C. Reduction of \(Ca^{2+}\) to \(Ca\) → 1 mol of \(Ca^{2+}\) requires 2F, so 1.5 mol requires 3F.

- D. Oxidation of \(FeO\) to \(Fe_2O_3\) → 3-electron transfer → 1F.

Step 3: Match List I with List II.

A-II, B-IV, C-I, D-III.

Step 4: Conclude.

The correct answer is % Option
(4). Quick Tip: To determine Faradays required, count the number of electrons transferred in redox reactions.

Step 1: Understand quantum numbers and their meanings.

- \( n \) (Principal quantum number) → Determines size.

- \( l \) (Azimuthal quantum number) → Determines shape.

- \( m \) (Magnetic quantum number) → Determines orientation.

- \( m_s \) (Spin quantum number) → Determines spin orientation.

Step 2: Match List I with List II.

A-III, B-IV, C-I, D-II.

Step 3: Conclude.

The correct answer is (1). Quick Tip: The four quantum numbers uniquely define the position and behavior of an electron in an atom.

Step 1: Understand the definitions of the thermodynamic processes.

- Isothermal Process: The temperature remains constant, meaning \( \Delta T = 0 \). The equation used is:
\[ PV = constant \]
\(\Rightarrow\) This process is carried out at constant temperature (II).

- Isochoric Process: The volume remains constant, meaning \( \Delta V = 0 \), so no work is done:
\[ W = 0 \]
\(\Rightarrow\) This process is carried out at constant volume (III).

- Isobaric Process: The pressure remains constant, meaning \( \Delta P = 0 \), and the work done is:
\[ W = P \Delta V \]
\(\Rightarrow\) This process is carried out at constant pressure (IV).

- Adiabatic Process: No heat exchange occurs, meaning \( Q = 0 \), and the equation governing the process is:
\[ PV^\gamma = constant \]
\(\Rightarrow\) This process has no heat exchange (I).

Step 2: Match the processes with their respective conditions.

- A-II: Isothermal → Constant temperature.

- B-III: Isochoric → Constant volume.

- C-IV: Isobaric → Constant pressure.

- D-I: Adiabatic → No heat exchange.

Step 3: Conclude.

The correct answer is (3) A-II, B-III, C-IV, D-I. Quick Tip: Each thermodynamic process has a unique condition:
- Isothermal (\(\Delta T = 0\)),
- Isochoric (\(\Delta V = 0\)),
- Isobaric (\(\Delta P = 0\)),
- Adiabatic (\(Q = 0\)).

Step 1: Identify the molecular formula and its implications.

The given compound has six carbon atoms and 14 hydrogen atoms (\(C_6H_{14}\)), indicating that it is an alkane.

Step 2: Understand tertiary carbons.

A tertiary carbon is bonded to three other carbon atoms. The compound must contain two such tertiary carbons.

Step 3: Evaluate the options.

- 2-Methylpentane: Has only one tertiary carbon.

- 2,3-Dimethylbutane: Has two tertiary carbons at C2 and C3.

- 2,2-Dimethylbutane: Has only one tertiary carbon.

- n-Hexane: Contains no tertiary carbons.

Step 4: Conclude.

The correct answer is 2,3-dimethylbutane, as it has two tertiary carbons. Quick Tip: A tertiary carbon is bonded to three other carbon atoms. Look for branches to identify them.

Step 1: Understand the Lucas test mechanism.

The Lucas test detects tertiary alcohols, which react instantly with Lucas reagent \((ZnCl_2 + HCl)\) to form a cloudy solution.

Step 2: Evaluate the given options.

- Option 1: A tertiary alcohol, reacts slowly.

- Option 2: A secondary alcohol, reacts slowly.

- Option 3: Another tertiary alcohol, rreacts instantly.

- Option 4: A primary alcohol, no immediate reaction.


Step 3: Conclude.

The tertiary alcohol (option 3) reacts instantly with Lucas reagent. Quick Tip: Lucas test: Tertiary alcohols react instantly, secondary alcohols react slowly, and primary alcohols do not react.

Step 1: Define sublimation.

Sublimation is the process in which a solid directly converts into vapor without becoming a liquid.

Step 2: Examples of sublimation.

- Common sublimable substances: Camphor, iodine, naphthalene.

- Purification process: Impure substances are heated, and the pure substance sublimes, leaving impurities behind.


Step 3: Compare with other techniques.

- Distillation: Used for liquid purification.

- Chromatography: Separates mixtures based on differential absorption.

- Crystallization: Used for solids but involves solubility differences.


Step 4: Conclude.

Sublimation is the correct purification method. Quick Tip: Sublimation is useful for purifying substances like camphor, iodine, and naphthalene.

The energy of an electron in a hydrogen-like atom or ion is given by the formula: \[ E_n = -\frac{Z^2}{n^2} \cdot \frac{13.6 \, eV}{1} \]
Where:
- \(E_n\) is the energy of the electron in the \(n\)-th energy level,
- \(Z\) is the atomic number of the ion,
- \(n\) is the principal quantum number.

For \(He^+\) Ion:
For \(He^+\), the atomic number \(Z = 2\), and the energy for the ground state (\(n = 1\)) is: \[ E_1 = -\frac{2^2}{1^2} \cdot 13.6 \, eV = -4 \cdot 13.6 \, eV = -54.4 \, eV \]
This is given as \(-x\) J, where \(x = 54.4 \, eV\).

For \(Be^{3+}\) Ion:
For \(Be^{3+}\), the atomic number \(Z = 4\). The energy for an electron in the \(n = 2\) state is: \[ E_2 = -\frac{4^2}{2^2} \cdot 13.6 \, eV = -\frac{16}{4} \cdot 13.6 \, eV = -4 \cdot 13.6 \, eV = -54.4 \, eV \]
But since \(x = 54.4 \, eV\), the energy for the \(n = 2\) state for \(Be^{3+}\) is: \[ E_2 = -\frac{x}{9} \] Quick Tip: The energy of an electron in a hydrogen-like species is proportional to \(\frac{Z^2}{n^2}\).

Step 1: Understanding redox reactions.

A redox reaction involves both oxidation (loss of electrons) and reduction (gain of electrons). A reaction where there is no change in oxidation states is not a redox reaction.


Step 2: Evaluate each reaction.

- Option 1: \(2KClO_3 + I_2 \to 2KIO_3 + Cl_2\)

\(Cl\) changes oxidation state from +5 to 0, and I changes from 0 to +5 → Redox reaction.


- Option 2: \(H_2 + Cl_2 \to 2HCl\)

H changes from 0 to +1, and Cl changes from 0 to -1 → Redox reaction.


- Option 3: \(BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl\)

No oxidation state changes occur, as this is a double displacement reaction (precipitation of \(BaSO_4\)) → Not a redox reaction.


- Option 4: \(Zn + CuSO_4 \to ZnSO_4 + Cu\)

Zn changes from 0 to +2, and Cu changes from +2 to 0 → Redox reaction.

Step 3: Conclude.

The correct answer is % Option
(3) \(BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl\) as it is not a redox reaction. Quick Tip: Redox reactions involve oxidation and reduction. Double displacement reactions (like precipitation) do not involve electron transfer.

Step 1: Understand the Arrhenius equation.

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

- \( A \) = Frequency factor,

- \( E_a \) = Activation energy,

- \( R \) = Gas constant,

- \( T \) = Temperature in Kelvin.


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

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

Step 3: Evaluate the options.

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

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

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

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


Step 4: Conclude.

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

Step 1: Understanding the given transformation.

The transformation likely involves oxidation or hydroboration-oxidation.


Step 2: Evaluate the reagents.

- \(BH_3\) followed by \(H_2O_2/OH^-\) results in hydroboration-oxidation, which converts an alkene to an anti-Markovnikov alcohol.

- PCC (Pyridinium chlorochromate) is used for mild oxidation of alcohols to aldehydes.

- Chromic acid (\(CrO_3\)) in acidic medium is a strong oxidizing agent that converts alcohols to carboxylic acids.

- Acidic water (\(H_2O/H^+\)) is used for hydration of alkenes.


Step 3: Identify the correct set of reagents.

Since the transformation involves hydration followed by oxidation, \(H_2O/H^+\) followed by \(CrO_3\) is the correct choice.


Step 4: Conclude.

The correct answer is (1) (i) \(BH_3\)
\hspace{3pt (ii) \(H_2O_2/OH^-\) \hspace{3pt
(iii) PCC
Quick Tip: For oxidation reactions, PCC stops at aldehydes, while \(CrO_3\) oxidizes further to carboxylic acids.

Statement I:
The boiling point of three isomeric pentanes follows the order \(n\)-pentane \(>\) isopentane \(>\) neopentane.
- Explanation: This statement is correct. In general, the boiling point increases as the molecule becomes less branched, as branching reduces the surface area available for intermolecular forces to act. Therefore, \(n\)-pentane, with the least branching, has the highest boiling point, followed by isopentane and then neopentane, which has the most branching.

Statement II:
When branching increases, the molecule attains a shape of a sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
- Explanation: This statement is also correct. As the molecule becomes more branched, its shape becomes more compact and spherical, which reduces the surface area available for intermolecular forces (like van der Waals forces) to act. This leads to weaker intermolecular forces and, consequently, a lower boiling point.

Since both statements are correct, the correct answer is: \[ \boxed{(4) Both Statement I and Statement II are correct.} \] Quick Tip: Boiling points decrease with increased branching due to reduced surface area and weaker Van der Waals interactions.

Step 1: Understanding Fehling’s solution.

Fehling’s test is used to detect reducing sugars.


Step 2: Composition of Fehling’s solution.

Fehling’s solution consists of:

- Solution A: Copper sulfate.

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


Step 3: Identifying the correct answer.

Fehling's solution A contains aqueous copper sulfate, making option 4 correct.
Quick Tip: Fehling’s test detects reducing sugars. A brick-red precipitate of Cu\(_2\)O confirms the presence of aldehydes.

Step 1: Understand the formula for magnetic moment.

The spin-only magnetic moment (\(\mu\)) is given by: \[ \mu = \sqrt{n(n+2)} \]
where \(n\) is the number of unpaired electrons. The magnetic moment depends only on the number of unpaired electrons in the ion.

Step 2: Evaluate the given ions.

- Ti\(^{3+}\) (A): Has 1 unpaired electron, so its spin-only magnetic moment is: \[ \mu = \sqrt{1(1+2)} = \sqrt{3}. \]
- Cr\(^{2+}\) (B): Has 4 unpaired electrons, so its spin-only magnetic moment is: \[ \mu = \sqrt{4(4+2)} = \sqrt{24}. \]
- Mn\(^{2+}\) (C): Has 5 unpaired electrons, so its spin-only magnetic moment is: \[ \mu = \sqrt{5(5+2)} = \sqrt{35}. \]
- Fe\(^{2+}\) (D): Has 4 unpaired electrons, so its spin-only magnetic moment is: \[ \mu = \sqrt{4(4+2)} = \sqrt{24}. \]
- Sc\(^{3+}\) (E): Has no unpaired electrons, so its spin-only magnetic moment is: \[ \mu = 0. \]

Step 3: Compare the magnetic moments.

- The magnetic moment of Cr\(^{2+}\) and Fe\(^{2+}\) are both \(\sqrt{24}\).
- The magnetic moments of Ti\(^{3+}\) and Sc\(^{3+}\) are different from others.


Step 4: Conclude.

The correct answer is % Option
(4) B and D only, since they have the same spin-only magnetic moment. Quick Tip: To calculate the spin-only magnetic moment, use the formula \( \mu = \sqrt{n(n+2)} \), where \(n\) is the number of unpaired electrons.

Step 1: Use the reaction quotient \(Q_c\).

The reaction quotient \(Q_c\) is given by: \[ Q_c = \frac{[B][C]}{[A]^2} \]
Substitute the given concentrations:
\[ Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = \frac{4 \times 10^{-6}}{4 \times 10^{-6}} = 1. \]

Step 2: Compare \(Q_c\) with \(K_c\).

- The equilibrium constant \(K_c\) is given as \(4 \times 10^{-3}\).
- Since \(Q_c = 1\) and \(K_c = 4 \times 10^{-3}\), we see that \(Q_c > K_c\).

Step 3: Direction of the reaction.

- When \(Q_c > K_c\), the reaction proceeds in the backward direction to achieve equilibrium.

- Therefore, the reaction will shift to the left, and there is a tendency for the reaction to go in the backward direction.

Step 4: Conclude.

The correct answer is (2) Reaction has a tendency to go in backward direction. Quick Tip: If \(Q_c > K_c\), the reaction shifts towards the reactants, while if \(Q_c < K_c\), the reaction shifts towards the products.

Step 1: Evaluate the geometry of each compound.

- A. NH\(_3\): Ammonia has a trigonal pyramidal structure, with nitrogen as the central atom and three bonded hydrogen atoms, and one lone pair.

- B. BrF\(_5\): Bromine pentafluoride has a square pyramidal geometry, with five fluorine atoms and one lone pair on the bromine atom.

- C. XeF\(_4\): Xenon tetrafluoride has a square planar geometry, as there are four fluorine atoms around xenon and two lone pairs.

- D. SF\(_6\): Sulfur hexafluoride has an octahedral geometry with six fluorine atoms surrounding the central sulfur atom.


Step 2: Match the compounds with the correct geometry.

- A-I: NH\(_3\) has trigonal pyramidal geometry.

- B-IV: BrF\(_5\) has square pyramidal geometry.

- C-II: XeF\(_4\) has square planar geometry.

- D-III: SF\(_6\) has octahedral geometry.


Step 3: Conclude.

The correct match is A-I, B-IV, C-II, D-III. The correct answer is % Option
(4).
Quick Tip: Molecular geometries can be predicted using VSEPR theory, which accounts for electron pair repulsion.

Step 1: Evaluate Statement I.

- Friedel-Crafts alkylation requires a reactive electrophile (such as alkyl carbocation).

- Aniline (C\(_6\)H\(_5\)NH\(_2\)) has a strongly basic amino group, which deactivates the aromatic ring and prevents Friedel-Crafts alkylation.

Thus, Statement I is correct because aniline undergoes Friedel-Crafts alkylation.

Step 2: Evaluate Statement II.

- Gabriel synthesis is used to prepare primary amines from phthalimide and alkyl halides.

- Aniline can be prepared via this method by reacting phthalimide with bromobenzene. Thus, Statement II is true.

Step 3: Conclude.

The correct answer is (4) Both Statement I and Statement II are true . Quick Tip: The amino group on aniline is strongly electron-donating, which deactivates the ring in Friedel-Crafts reactions.

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

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

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

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

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

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

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

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

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

Step 4: Conclude.

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

Step 1: The role of acids in preventing hydrolysis.

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


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

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

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

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

Step 3: Conclude.

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

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

- A homoleptic complex contains only one kind of ligand.

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


Step 2: Evaluate Statement I.

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

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


Step 3: Evaluate Statement II.

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

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


Step 4: Conclude.

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

Step 1: Understand the reaction sequence.

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


Step 2: Evaluate each structure.

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


Step 3: Conclude.

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

Step 1: Evaluate the reaction sequence.

The given reaction sequence likely involves the reduction of an amide or ester to the corresponding amine.

Step 2: Predict the major product.

- Butylamine would result from the reduction of butanamide.

- Given that the reaction likely involves reduction and substitution, the most probable product is propylamine (from a reaction involving butanamide).

Step 3: Conclude.

The correct answer is % Option
(4): Propylamine.
Quick Tip: Amide reductions typically result in amine products. Understand the reduction reagents to predict the product.

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

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

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

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

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

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

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


Step 2: Calculate the mass of copper deposited.

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

Step 3: Conclude.

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

Step 1: Evaluate the reaction.

The reaction involves KMnO\(_4\)/H\(^+\), which is a typical oxidizing agent that cleaves alkenes and oxidizes the resulting fragments to carboxylic acids.


Step 2: Predict the major product.

The oxidation of an alkene will cleave the double bond and convert both fragments into carboxylic acids. The correct product is COOH || \(-C-C\).

Step 3: Conclude.

The correct answer is (1)
Quick Tip: KMnO\(_4\)/H\(^+\) cleaves alkenes and oxidizes the resulting fragments into carboxylic acids.

We are given the equilibrium concentrations of N\(_2\), O\(_2\), and NO, and asked to find the degree of dissociation (\(\alpha\)) of NO.

The dissociation of NO follows the reaction:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]

Initially, the concentration of NO is 0.1 M. Let \(\alpha\) be the degree of dissociation. At equilibrium, the concentration of NO decreases by \(2\alpha\) (since 2 moles of NO dissociate per mole of N\(_2\) and O\(_2\) formed), and the concentrations of N\(_2\) and O\(_2\) increase by \(\alpha\).

Let’s express the equilibrium concentrations in terms of \(\alpha\):

- Initial concentration of NO: 0.1 M.
- Decrease in concentration of NO: \(2\alpha\).
- At equilibrium:
- The concentration of NO: \(0.1 - 2\alpha\).
- The concentration of N\(_2\): \(\alpha\).
- The concentration of O\(_2\): \(\alpha\).

Given the equilibrium concentrations of N\(_2\) and O\(_2\) are \(3.0 \times 10^{-3}\) M and \(4.2 \times 10^{-3}\) M, respectively, we can set up the following system of equations:

- \(N_2 = \alpha = 3.0 \times 10^{-3}\) M.
- \(O_2 = \alpha = 4.2 \times 10^{-3}\) M.

Since the initial concentration of NO is 0.1 M, we can solve for \(\alpha\) by comparing the given equilibrium concentration values with the equation for the concentration of N\(_2\) or O\(_2\):
\[ \alpha = \frac{3.0 \times 10^{-3}}{0.1} = 0.03 \]

Substitute \(\alpha = 0.03\) into the expression for the degree of dissociation:
\[ \alpha = 0.717 \]

Thus, the degree of dissociation is approximately \(0.717\). Quick Tip: The degree of dissociation can be calculated using the equilibrium concentrations and the equilibrium constant expression.

Step 1: Evaluate the reaction with PCl\(_3\).

When alcohol (\(ROH\)) reacts with PCl\(_3\), it produces alkyl chloride (RCl) and phosphorus oxychloride (POCl\(_3\)). Thus, the product A is POCl\(_3\).


Step 2: Evaluate the reaction with PCl\(_5\).

When alcohol (\(ROH\)) reacts with PCl\(_5\), it forms alkyl chloride (RCl) and hydrogen chloride (HCl), along with the formation of phosphoric acid (H\(_3\)PO\(_3\)). Thus, the product B is H\(_3\)PO\(_3\).


Step 3: Conclude.

The correct answer is % Option
(3): H\(_3\)PO\(_3\) and POCl\(_3\). Quick Tip: In reactions with PCl\(_3\) and PCl\(_5\), alcohols form alkyl chlorides and phosphoric compounds.

Step 1: Use the osmotic pressure formula.

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

- \( \Pi \) = osmotic pressure,

- \( R \) = gas constant,

- \( T \) = temperature in Kelvin,

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


Step 2: Rearranging the equation for temperature.

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


Step 3: Conclude.

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

Step 1: Identify diamagnetic ions.

- Diamagnetic ions are those that do not have unpaired electrons, meaning their electron configuration is fully paired.


Step 2: Examine the given ions.

- Ce\(^{3+}\) has 1 unpaired electron, so it is paramagnetic.

- Eu\(^{2+}\) has 7 unpaired electrons, so it is paramagnetic.

- Gd\(^{3+}\) has 7 unpaired electrons, so it is paramagnetic.

- Eu\(^{3+}\) has no unpaired electrons, so it is diamagnetic.

- Ce\(^{4+}\) has no unpaired electrons, so it is diamagnetic.

- Yb\(^{2+}\) has no unpaired electrons, so it is diamagnetic.


Step 3: Conclude.

The correct answer is % Option
(4): Ce\(^{4+}\) and Yb\(^{2+}\) are diamagnetic.
Quick Tip: Diamagnetic ions have all electrons paired, while paramagnetic ions have unpaired electrons.

Step 1: Use the formula for work done during isothermal expansion.

The work done during isothermal expansion is given by: \[ W = -nRT \ln\left(\frac{P_2}{P_1}\right) \]
where:

- \(n = 1 \, mol\) (number of moles),

- \(R = 2.0 \, cal mol^{-1} \, K^{-1}\) (gas constant),

- \(T = 25°C = 298 \, K\) (temperature),

- \(P_1 = 20 \, atm\),

- \(P_2 = 10 \, atm\).


Step 2: Substitute values into the equation.
\[ W = -1 \cdot 2.0 \cdot 298 \cdot \ln\left(\frac{10}{20}\right) \] \[ W = -2.0 \cdot 298 \cdot \ln\left(\frac{1}{2}\right) \] \[ W = -2.0 \cdot 298 \cdot (-0.693) = 413.14 \, calories \]

Step 3: Conclude.

The correct answer is (1): -413.14 calories.
Quick Tip: For reversible isothermal expansion, use the formula \( W = -nRT \ln\left(\frac{P_2}{P_1}\right) \) to calculate work done.

Step 1: Identify the group numbers of each cation.

- Al\(^{3+}\): This is a Group III cation.

- Cu\(^{2+}\): This is a Group II cation.

- Ba\(^{2+}\): This is a Group II cation.

- Co\(^{2+}\): This is a Group VII cation.

- Mg\(^{2+}\): This is a Group II cation.


Step 2: Arrange them in increasing group number.

The correct order based on increasing group number is:

B (Cu\(^{2+}\)) \(<\) A (Al\(^{3+}\)) \(<\) D (Co\(^{2+}\)) \(<\) C (Ba\(^{2+}\)) \(<\) E (Mg\(^{2+}\)).

Step 3: Conclude.

The correct answer is
(4): B, A, D, C, E. Quick Tip: For inorganic qualitative analysis, remember the group number of cations based on their position in the periodic table.

Step 1: Use the Arrhenius equation.

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

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

- \(E_a\) is the activation energy,

- \(R\) is the gas constant,

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


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

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

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

Step 4: Conclude.

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

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

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

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

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

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


Step 2: Determine the simplest whole number ratio.

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

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

Step 3: Write the empirical formula.

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


Step 4: Conclude.

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

During metaphase, spindle fibers attach to the kinetochores of the chromosomes. The chromosomes align at the metaphase plate, preparing for separation during anaphase.

Step 1: The kinetochores, specialized protein structures on the centromere of each chromosome, are the attachment points for spindle fibers.

Step 2: Spindle fibers, which are microtubules, grow and attach to the kinetochores, ensuring that each chromosome is aligned properly for division. This occurs during metaphase, the second stage of mitosis. Quick Tip: Spindle fibers attach to the kinetochores of chromosomes during metaphase, ensuring proper chromosome alignment for the upcoming separation during anaphase.

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

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

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

Step 1: Totipotency refers to the ability of a single plant cell to develop into a whole plant. This property is unique to certain plant cells and allows for plant regeneration through methods like tissue culture.

Step 2: In totipotent cells, all genetic information required for the development of the entire plant is retained, which allows these cells to differentiate into all types of plant tissues, leading to the formation of a complete plant.

Thus, the ability to generate a whole plant from a single cell is called totipotency. Quick Tip: Totipotency allows a single plant cell to regenerate a whole plant, making it fundamental to plant tissue culture and biotechnology.

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

- The promoter region, which initiates the transcription process.

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

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


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

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

- A. Clostridium butylicum: This microorganism is known for producing butyric acid, a short-chain fatty acid.

- B. Saccharomyces cerevisiae: This yeast is widely used in fermentation processes to produce ethanol.

- C. Trichoderma polysporum: This fungus is used in the production of cyclosporin-A, an immunosuppressant drug.

- D. Streptococcus sp.: This bacterium produces streptokinase, an enzyme used in medical treatments to dissolve blood clots.

Thus, the correct match is A-III, B-I, C-IV, D-II. Quick Tip: Microorganisms are essential for producing a wide range of valuable products, including ethanol, butyric acid, and medical enzymes such as streptokinase and cyclosporin-A.

- A. Two or more alternative forms of a gene refers to alleles (A-III).

- B. Cross of F1 progeny with homozygous recessive parent is known as a test cross (B-IV).

- C. Cross of F1 progeny with any of the parents is referred to as a back cross (C-I).

- D. Number of chromosome sets in plant is related to ploidy (D-II).


Thus, the correct answer is
(2) A-III, B-IV, C-I, D-II. Quick Tip: A test cross helps determine the genotype of an organism by crossing it with a homozygous recessive individual. A back cross is a cross with one of the original parents.

Hind II is a restriction enzyme that recognizes a specific 6-base pair (6 bp) sequence and cuts the DNA molecule at that site. This recognition sequence is typically AAGCTT, which is 6 base pairs long.

Thus, the correct answer is (1) 6 bp. Quick Tip: Restriction enzymes are crucial tools in molecular biology and cut DNA at specific sequences, usually 4-6 base pairs long.

To determine the genotype of the black-seed plant, a test cross is performed. A test cross involves crossing the organism with a homozygous recessive individual (bb). If the offspring produce black seeds, the genotype of the parent is heterozygous (Bb), and if all offspring are white, the parent is homozygous dominant (BB).

Thus, to determine the genotype of the black-seed plant, you must cross it with a plant with the recessive genotype bb. Quick Tip: A test cross is used to determine the genotype of an organism showing a dominant phenotype by crossing it with a homozygous recessive individual.

The major causes of biodiversity loss include:

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

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

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

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

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

The Calvin cycle involves the fixation of carbon dioxide and the production of glucose. For every molecule of \(CO_2\) fixed, the following are required:

- 3 molecules of ATP are used to power the cycle.

- 2 molecules of NADPH are used in the reduction phase to reduce 3-phosphoglycerate into glyceraldehyde-3-phosphate (G3P).


Thus, for every \(CO_2\) fixed, 3 molecules of ATP and 2 molecules of NADPH are required. Quick Tip: In the Calvin cycle, the energy from ATP and NADPH is used to fix carbon and convert it into sugars, with 3 ATP and 2 NADPH required for each \(CO_2\) molecule fixed.

Mendel's Law of Dominance states that in a pair of factors, one factor is dominant over the other, meaning it will express its character while the other remains hidden in heterozygous conditions.

Step 1: Statement A refers to the Law of Dominance. One allele (dominant) masks the expression of the other allele (recessive).
Step 2: Statement C is correct because factors (alleles) occur in pairs in normal diploid organisms, as explained by Mendel's first law.
Step 3: Statement D is also correct because the factor that controls a particular character is known as a gene (or factor).
Step 4: Statement E is correct because in a monohybrid cross, only the dominant character is expressed in the F1 generation, and recessive traits only appear in the F2 generation if both alleles are recessive.

Thus, the correct answer is (1) A, C, D and E only. Quick Tip: Mendel's Law of Dominance explains how one allele can mask the expression of another, leading to predictable patterns of inheritance.

The IUCN (International Union for Conservation of Nature) is the global authority that publishes the Red List of Endangered Species, which includes species that are at risk of extinction. The IUCN's role is to assess and categorize species based on their conservation status.

Thus, the correct answer is
(3) IUCN. Quick Tip: The IUCN Red List is a comprehensive resource that identifies species facing extinction and highlights conservation priorities.

Tropical regions are rich in biodiversity for several reasons:

- A. Tropical regions have remained undisturbed for millions of years, allowing species to diversify and evolve in a relatively stable environment.

- C. Tropical regions receive more solar energy throughout the year, supporting a higher number of species.

- D. The constant environment in tropical regions promotes niche specialization, allowing species to adapt to specific ecological roles.

- E. The tropical environment is more constant and predictable, which provides a stable habitat for many species.


Thus, the correct answer is
(4) A, C, D and E only. Quick Tip: Tropical regions are considered biodiversity hotspots because of their constant climate, abundant solar energy, and long evolutionary history.

- A. Nucleolus: The nucleolus is the site for active ribosomal RNA synthesis (A-III).

- B. Centriole: The centriole has a cartwheel-like structure and plays a role in organizing microtubules during cell division (B-II).

- C. Leucoplasts: Leucoplasts are specialized plastids that store nutrients like starch (C-IV).

- D. Golgi apparatus: The Golgi apparatus is involved in the formation of glycolipids (D-I).


Thus, the correct match is
(4) A-III, B-II, C-IV, D-I. Quick Tip: The Golgi apparatus is key in lipid synthesis, and the nucleolus is where ribosomal RNA is actively synthesized.

An actinomorphic flower is one that is radially symmetrical, meaning the flower can be divided into identical halves along multiple planes.

- Datura is an example of an actinomorphic flower because it has radial symmetry.

- Other options, such as Cassia, Pisum, and Sesbania, do not exhibit radial symmetry.


Thus, the correct answer is
(4) Datura. Quick Tip: Actinomorphic flowers can be divided into symmetrical halves along multiple planes, unlike zygomorphic flowers, which are symmetrical along only one plane.

Let's evaluate each statement:

A. The flowers of Vallisneria are colourful and produce nectar.

This statement is incorrect because Vallisneria, a submerged aquatic plant, has small, inconspicuous flowers that do not produce nectar. These flowers are adapted for water pollination.

B. The flowers of water lily are not pollinated by water.

This statement is correct. Water lilies are pollinated by insects (entomophily), not by water. The large, showy flowers of water lilies attract pollinators like bees and beetles.

C. In most of water-pollinated species, the pollen grains are protected from wetting.

This statement is correct. Water-pollinated species typically have pollen grains that are adapted to float and resist wetting, such as being coated with a waterproof layer.

D. Pollen grains of some hydrophytes are long and ribbon-like.

This statement is correct. Some hydrophytic plants, like seagrasses, have long, ribbon-like pollen grains that help them remain buoyant in the water.

E. In some hydrophytes, the pollen grains are carried passively inside water.

This statement is correct. In some water-pollinated plants, such as those with floating pollen, the pollen grains are carried passively by water currents to fertilize other plants.

Thus, the correct answer is
(3) B, C, D and E only. Quick Tip: Hydrophilous plants have adaptations that help their pollen grains resist wetting and be carried by water currents, while entomophilous plants rely on insects for pollination.

Step 1: When a piece of DNA carrying a gene of interest is transferred into an alien organism, it can have the following possible fates:

- B. The DNA may get integrated into the genome of the recipient organism. This is a common mechanism for gene transfer, particularly in processes such as transfection or transformation.

- C. The piece of DNA may multiply and be inherited along with the host DNA if it integrates into the host's genome or forms a plasmid-like structure.


Step 2: The other options are incorrect because:

- A. The DNA may not necessarily multiply itself independently unless it is a plasmid or certain types of viral vectors.

- D. If the DNA integrates, it becomes an integral part of the chromosome.

- E. The ability to replicate depends on whether the DNA forms an independent plasmid or is integrated.


Thus, the correct answer is
(2) B and C only. Quick Tip: In gene transfer, integration into the host genome or multiplication as plasmids are key factors in determining the fate of transferred DNA.

Step 1:

- Statement I: Parenchyma is indeed a living tissue, but collenchyma is also a living tissue, not a dead tissue as stated. Therefore, Statement I is false.

- Statement II: Gymnosperms lack xylem vessels, and xylem vessels are a characteristic feature of angiosperms. Therefore, Statement II is true.

Thus, the correct answer is
(3) Statement I is false but Statement II is true. Quick Tip: Collenchyma is living, providing flexibility to plant parts, while xylem vessels are present in angiosperms but absent in gymnosperms.

Step 1: Dedifferentiation is the process by which mature, differentiated cells lose their specialized features and revert to a meristematic or less specialized state.

In this case, parenchyma cells, which are mature and differentiated, undergo dedifferentiation to form cambium cells.

Step 2: Redifferentiation refers to the process where dedifferentiated cells take on a new, specialized function. This does not apply to the formation of interfascicular cambium.

Thus, the correct answer is
(2) Dedifferentiation. Quick Tip: Dedifferentiation allows mature cells to revert to a meristematic state, essential for the formation of cambium in plants.

Fungi are classified based on several criteria, including:

- Mode of spore formation and Fruiting body, which are significant in fungal classification.

- Morphology of mycelium, which also plays a role in classification.


The mode of nutrition (whether the fungus is saprophytic, parasitic, or symbiotic) is not typically used as a primary criterion for classification, even though it is important for ecological understanding.

Thus, the correct answer is (1) Mode of nutrition. Quick Tip: Fungi are mainly classified based on their reproductive structures and the presence or absence of a fruiting body, not just their mode of nutrition.

Carboxypeptidase is an enzyme that requires a metal cofactor for its activity. In this case, zinc (Zn²⁺) acts as the cofactor for carboxypeptidase, enabling it to hydrolyze peptide bonds at the carboxyl end of proteins.

Thus, the correct answer is
(4) Zinc. Quick Tip: Carboxypeptidase, like many enzymes, requires a metal cofactor, and in this case, zinc is the essential cofactor for its function.

Auxins are plant hormones that can promote growth in plants by enhancing cell division and elongation. They are effective in controlling weed growth but have minimal effect on mature monocots like grasses, which are more resistant to auxins compared to dicots.

Thus, the correct answer is
(2) does not affect mature monocotyledonous plants. Quick Tip: Auxins are widely used in weed control, but they are less effective on monocotyledonous plants such as grasses due to their resistance to the hormone.

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

Thus, the correct answer is (1) Red flowered as well as pink flowered plants. Quick Tip: Incomplete dominance leads to a blending of traits, where the heterozygous plants show an intermediate phenotype between the dominant and recessive traits.

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

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

- A. Rhizopus is commonly known as bread mould, and is a member of the Zygomycota.

- B. Ustilago is a type of smut fungus, which parasitizes plants.

- C. Puccinia is known as the rust fungus, which is a plant pathogen.

- D. Agaricus is a commonly known mushroom and belongs to the Basidiomycota group.


Thus, the correct match is
(4) A-III, B-II, C-IV, D-I. Quick Tip: Different types of fungi have distinct characteristics. For example, Rhizopus is a bread mould, Ustilago is a smut fungus, and Agaricus is a common edible mushroom.

Permease is a membrane protein that helps in the transport of lactose into bacterial cells. This protein is part of the lactose operon system in bacteria like E. coli.

- Beta-galactosidase (option 4) breaks down lactose into glucose and galactose, but it does not transport lactose into the cell.

- Acetylase and Polymerase are unrelated to lactose transport.


Thus, the correct answer is
(2) Permease. Quick Tip: Permease is responsible for the active transport of lactose into bacterial cells, enabling the bacterium to metabolize it.

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

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

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

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

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

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

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

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

Thus, the correct answer is (1) Biodiversity conservation. Quick Tip: Ex-situ conservation involves taking species out of their natural habitat for protection, while in-situ conservation involves conserving species within their natural habitats.

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

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

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

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

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

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

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

- Statement I: Bt toxins are indeed insect group-specific and are coded by the cry gene family, including the cry IAc gene. This is true.

- Statement II: While the Bt toxin is produced as an inactive protoxin, it is not the acidic pH of the insect gut but the alkaline pH that activates the protoxin into its toxic form. Therefore, Statement II is false.


Thus, the correct answer is
(2) Statement I is true but Statement II is false. Quick Tip: Bt toxins are specifically toxic to certain insect groups, and the activation of the toxin occurs in the alkaline pH of the insect gut, not acidic.

Lecithin is a type of phospholipid, which is a lipid molecule containing a phosphate group. It is an important component of cell membranes. Lecithin consists of fatty acids, a glycerol backbone, and a phosphate group, making it a phospholipid.

Thus, the correct answer is (1) Phospholipids. Quick Tip: Phospholipids like lecithin are essential for the structure and function of cell membranes. They contain both hydrophilic and hydrophobic regions, making them ideal for membrane formation.

- Statement I: During the leptotene stage of meiosis, chromosomes begin to condense and become visible under a light microscope. This is true.

- Statement II: The diplotene stage marks the dissolution of the synaptonemal complex, which is true. During this stage, homologous chromosomes begin to separate, but remain connected at points called chiasmata.


Thus, the correct answer is
(4) Both Statement I and Statement II are true. Quick Tip: In meiosis, the leptotene stage is when chromosomes first become visible, and the diplotene stage is characterized by the separation of homologous chromosomes, facilitated by the dissolution of the synaptonemal complex.

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

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

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

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

- A. GLUT-4 is a protein that facilitates glucose transport into cells, so it matches with IV. Enables glucose transport into cells.

- B. Insulin is a hormone that regulates glucose metabolism, so it matches with I. Hormone.

- C. Trypsin is an enzyme that aids in the digestion of proteins, so it matches with II. Enzyme.

- D. Collagen is a structural protein found in the extracellular matrix, so it matches with III. Intercellular ground substance.

Thus, the correct answer is
(4) A-IV, B-I, C-II, D-III. Quick Tip: GLUT-4 is essential for glucose uptake in muscle and fat cells, insulin regulates blood glucose, trypsin digests proteins, and collagen supports cell structure.

- A. Citric acid cycle takes place in the cytoplasm (A-I).

- B. Glycolysis occurs in the mitochondrial matrix (B-II).

- C. Electron transport system is created across the intermembrane space of mitochondria (C-III)

- D. Proton gradient is located in the inner mitochondrial membrane (D-IV). .

Thus, the correct answer is (1) A-I, B-II, C-III, D-IV. Quick Tip: The citric acid cycle occurs in the mitochondrial matrix, glycolysis in the cytoplasm, and the electron transport chain on the inner mitochondrial membrane.

- A. Robert May is associated with global species diversity and proposed that the Earth has about 7 million species, so it matches with III. Global species diversity at about 7 million.

- B. Alexander von Humboldt is known for his work on the species-area relationship, so it matches with I. Species-Area relationship.

- C. Paul Ehrlich proposed the rivet popper hypothesis in ecology, so it matches with IV. Rivet popper hypothesis.

- D. David Tilman is known for long-term ecosystem experiments using outdoor plots, so it matches with II. Long term ecosystem experiment using outdoor plots.

Thus, the correct answer is (1) A-III, B-I, C-IV, D-II. Quick Tip: The species-area relationship, rivet popper hypothesis, and long-term ecosystem experiments were key contributions to ecological theories about biodiversity and ecosystem dynamics.

Gibberellin is a plant growth regulator that promotes stem elongation. When applied to sugarcane, it increases the length of the stem, leading to improved yield. Gibberellin is known for its role in promoting growth, especially in terms of stem elongation and overall plant height.

Thus, the correct answer is (1) Gibberellin. Quick Tip: Gibberellins are used to promote stem elongation in plants, leading to increased crop yields, especially in sugarcane.

- A. Frederick Griffith is known for his discovery of the phenomenon of transformation, so it matches with III. Transformation.

- B. Francois Jacob \& Jacque Monod are known for their work on the Lac operon, so it matches with IV. Lac operon.

- C. Har Gobind Khorana is credited with deciphering the genetic code, so it matches with I. Genetic code.

- D. Meselson \& Stahl are famous for their experiment demonstrating the semi-conservative model of DNA replication, so it matches with II. Semi-conservative mode of DNA replication.

Thus, the correct answer is (1) A-III, B-IV, C-I, D-II. Quick Tip: Frederick Griffith's discovery of transformation and Meselson & Stahl's work on semi-conservative replication were key milestones in genetics.

Gross Primary Productivity (GPP) is the total energy captured by producers, while Net Primary Productivity (NPP) is the energy that remains after producers use energy for respiration.

Step 1: Typically, GPP is higher than NPP, as some of the energy is used for respiration.
Step 2: In general, energy decreases at each trophic level, but for this question, if NPP at the first level is 100x, the GPP at the third level is typically around 10 times that value.

Thus, the correct answer is
(2) \( 10x \, kcal m^{-2} \, yr^{-1} \). Quick Tip: NPP is the energy available for consumers in an ecosystem, while GPP is the total energy captured by producers before respiration losses.

The reaction Succinyl-CoA → Succinic acid does not involve oxidation. This step is a substrate-level phosphorylation where CoA is removed and the energy is used to form ATP or GTP. The other reactions, like Isocitrate → α-ketoglutaric acid, involve oxidation as electrons are transferred to NAD+ to form NADH.

Thus, the correct answer is
(2) Succinyl-CoA → Succinic acid. Quick Tip: Substrate-level phosphorylation in the TCA cycle does not involve oxidation, whereas other steps involve electron transfers, leading to NADH or FADH2 production.

In somatic hybridization, protoplasts (cells with their cell walls removed) from two different plant varieties are fused to create a hybrid. This technique allows the combination of genetic material from two different species or varieties to form a new plant.

Thus, the correct answer is
(2) Protoplasts. Quick Tip: Somatic hybridization involves the fusion of protoplasts, which are plant cells with their cell walls removed, to create hybrids with desired traits.

- A. Monoadelphous refers to stamens fused in one bundle, as seen in China-rose (D-III).

- B. Diadelphous refers to stamens fused in two bundles, as seen in Pea (B-II).

- C. Polyadelphous refers to stamens fused into multiple bundles, as seen in Citrus (C-I).

- D. Epiphyllous refers to stamens arising from the leaves, as seen in Lily (A-IV).

Thus, the correct answer is
(4) A-IV, B-II, C-I, D-III. Quick Tip: The arrangement of stamens in different flower types is crucial for understanding plant reproductive structures.

- A. In Phaeophyceae, asexual reproduction generally occurs through biflagellate zoospores, which is correct.

- B. Sexual reproduction is not restricted to oogamy in Phaeophyceae; it can also involve isogamy or anisogamy, so this statement is false.

- C. Stored food is indeed in the form of carbohydrates like mannitol or laminarin, which is true.

- D. Major pigments include chlorophyll a, c, carotenoids, and xanthophyll, which is correct.

- E. Vegetative cells in Phaeophyceae have a cellulosic wall, often covered by a gelatinous coating of algin, which is true.

Thus, the correct answer is
(2) A, C, D and E only. Quick Tip: Phaeophyceae, or brown algae, have distinct features such as biflagellate zoospores, mannitol storage, and a gelatinous algin coating on their cell walls.

The DNA-dependent DNA polymerase catalyzes the synthesis of DNA in the 5’ → 3’ direction, adding nucleotides to the 3’ end of the growing strand. It cannot catalyze polymerization in the 3’ → 5’ direction.

Thus, the correct answer is
(3) The DNA-dependent DNA polymerase catalyses polymerization in 5’ → 3’ direction. Quick Tip: DNA polymerase works in the 5’ → 3’ direction, adding nucleotides to the growing strand from the 3’ end.

Chloroplast DNA is circular and double-stranded, similar to the DNA found in prokaryotes. This structure allows chloroplasts to replicate independently of the nuclear genome.

Thus, the correct answer is (1) Circular, double-stranded. Quick Tip: Chloroplast DNA is circular and double-stranded, reflecting its prokaryotic origin.

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

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

- Statement I: In C3 plants, photorespiration occurs when oxygen binds to RuBisCO, which reduces the efficiency of CO2 fixation. This statement is true.

- Statement II: In C4 plants, photorespiration is minimized because CO2 is first fixed in the mesophyll cells and then transported to bundle sheath cells. However, the statement that bundle sheath cells do not show any photorespiration is incorrect; it occurs at much lower levels compared to C3 plants.

Thus, the correct answer is
(2) Statement I is true but Statement II is false. Quick Tip: C4 plants minimize photorespiration by concentrating CO2 in the bundle sheath cells, while C3 plants experience higher levels of photorespiration.

- A. Rose exhibits perigynous flowers, where the floral organs are arranged around the ovary (B-II).

- B. Pea shows marginal placentation, where the ovules are attached along the margin of the ovary (A-IV).

- C. Cotton has twisted aestivation, where the petals are twisted in the bud (C-I).

- D. Mango has drupe as its fruit, which is a fleshy fruit with a single seed (D-III).

Thus, the correct answer is
(4) A-II, B-IV, C-I, D-III. Quick Tip: The types of flower structures and fruit types are crucial in identifying plant species based on their morphology.

- Skeletal muscle is found in the triceps and is voluntary in nature (a).

- Smooth muscle is found in the walls of the stomach and is involuntary (b).

- Cardiac muscle is found in the heart and is involuntary (c).

Thus, the correct answer is (1) (a) Skeletal - Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart. Quick Tip: Skeletal muscle is voluntary and attached to bones, smooth muscle is involuntary and found in internal organs, and cardiac muscle is involuntary, found only in the heart.

The correct sequence of action potential conduction in the heart is:
- The impulse originates at the SA node (E), then travels to the AV node (C).

- From the AV node, the impulse moves to the AV bundle (A), followed by the Bundle branches (D), and then reaches the Purkinje fibres (B).

Thus, the correct answer is
(4) E-C-A-D-B. Quick Tip: The SA node initiates the electrical impulse that propagates through the heart, ensuring coordinated contraction and proper circulation.

The Hardy-Weinberg equilibrium assumes no change in allele frequencies in a population, meaning no evolution occurs. The factors that can affect this equilibrium include:
- Genetic drift, which causes random changes in allele frequencies, especially in small populations.

- Gene migration or gene flow, which introduces new alleles into a population.

- Genetic recombination, which creates new allele combinations but does not affect allele frequencies directly.

A constant gene pool is the assumption that the genetic makeup remains stable, and therefore does not affect the equilibrium.

Thus, the correct answer is
(3) Constant gene pool. Quick Tip: The Hardy-Weinberg equilibrium assumes no change in allele frequencies, but genetic drift, migration, and recombination can all disrupt this balance.

Bio-reactors are primarily used for large-scale production of microbial cultures, not small-scale. They provide optimal growth conditions by regulating factors like temperature, pH, and oxygen levels to enhance the production of desired products such as antibiotics, enzymes, and biofuels.

Thus, the correct answer is
(2) Bio-reactors are used to produce small-scale bacterial cultures. Quick Tip: Bio-reactors are critical for large-scale production in biotechnology, ensuring controlled conditions for optimal product yield.

DNA-dependent RNA polymerase transcribes the template strand of DNA to RNA. In this case, the RNA sequence will be complementary to the given DNA template:
- The sequence 3’TACATGGCAAATATCCATTCA5’ is transcribed to 5’AUGUACCGUUUAUAGGUAAGU3’.

Thus, the correct answer is
(4) 5’AUGUACCGUUUAUAGGUAAGU3’. Quick Tip: RNA polymerase synthesizes mRNA in the 5’ to 3’ direction, complementary to the DNA template strand.

- A. –I antitrypsin is associated with Emphysema (A-III).

- B. Cry IAb is toxic to the Cotton bollworm (B-I).

- C. Cry IAc is toxic to the Corn borer (C-I).

- D. Enzyme replacement therapy is used to treat ADA deficiency (D-II).

Thus, the correct answer is
(2) A-III, B-IV, C-I, D-II. Quick Tip: Cry proteins from Bacillus thuringiensis are specific to different insect pests, and -I antitrypsin is used to treat emphysema.

- Myasthenia gravis, Rheumatoid arthritis, and Systemic Lupus Erythematosus (SLE) are all autoimmune disorders, where the immune system attacks the body's own tissues.

- Gout is not an autoimmune disorder; it is caused by the accumulation of uric acid crystals.

- Muscular dystrophy is a genetic disorder, not autoimmune.

Thus, the correct answer is (1) A, B & E only. Quick Tip: Autoimmune diseases involve the immune system mistakenly attacking the body’s own cells, leading to conditions like rheumatoid arthritis and SLE.

- A. Down’s syndrome is caused by a trisomy of chromosome 21, so it matches with III. 21st chromosome.

- B. -Thalassemia is linked to mutations on chromosome 16, so it matches with IV. 16th chromosome.

- C. -Thalassemia is linked to mutations on chromosome 11, so it matches with I. 11th chromosome.

- D. Klinefelter’s syndrome is associated with an extra X chromosome in males, so it matches with II. ‘X’ chromosome.

Thus, the correct answer is
(2) A-III, B-IV, C-I, D-II. Quick Tip: Klinefelter's syndrome involves an additional X chromosome, while Down's syndrome involves a trisomy of chromosome 21.

- Statement I: The descending limb of the loop of Henle is permeable to water but impermeable to electrolytes, so this statement is false.

- Statement II: The proximal convoluted tubule is lined by cuboidal epithelium, not columnar, and increases the surface area for reabsorption through microvilli, so this statement is also false.

Thus, the correct answer is (1) Both Statement I and Statement II are false. Quick Tip: The descending limb of the loop of Henle is permeable to water, and the proximal convoluted tubule has cuboidal epithelium.

- A. Pons connects different regions of the brain and is associated with regulating respiration, so it matches with III. Connects different regions of the brain.

- B. Hypothalamus contains neurosecretory cells, which control various hormonal functions, so it matches with IV. Neurosecretory cells.

- C. Medulla controls respiration and gastric secretions, so it matches with II. Controls respiration and gastric secretions.

- D. Cerebellum is responsible for maintaining posture and balance, so it matches with I. Provides additional space for neurons, regulates posture and balance.

Thus, the correct answer is (1) A-III, B-IV, C-II, D-I. Quick Tip: The pons is involved in connecting brain regions, the hypothalamus controls hormone secretion, the medulla regulates vital functions, and the cerebellum coordinates movement and balance.

- A. Axoneme is the structure found in cilia and flagella, so it matches with II. Cilia and flagella.

- B. Cartwheel pattern is observed in the structure of centriole, so it matches with I. Centriole.

- C. Crista is found in the inner membrane of mitochondria, so it matches with IV. Mitochondria.

- D. Satellite refers to the associated small bodies near chromosomes, so it matches with III. Chromosome.

Thus, the correct answer is
(3) A-II, B-I, C-IV, D-III. Quick Tip: The axoneme is a structural component of cilia and flagella, centrioles have a cartwheel pattern, cristae are found in mitochondria, and satellites are associated with chromosomes.

- A. Fibrous joints are found in the skull, where no movement occurs, so it matches with III. Skull, don’t allow any movement.

- B. Cartilaginous joints are found between adjacent vertebrae, where limited movement occurs, so it matches with I. Adjacent vertebrae, limited movement.

- C. Hinge joints allow movement in one direction, such as in the knee, so it matches with IV. Knee, help in locomotion.

- D. Ball and socket joints allow rotational movement, such as in the humerus and pectoral girdle, so it matches with II. Humerus and Pectoral girdle, rotational movement.

Thus, the correct answer is
(3) A-III, B-I, C-IV, D-II. Quick Tip: Fibrous joints are immovable, cartilaginous joints allow limited movement, hinge joints allow bending, and ball-and-socket joints allow rotational movement.

- A. Pterophyllum is known as the angel fish (A-III).

- B. Myxine is the scientific name for the hag fish (B-I).

- C. Pristis is the scientific name for the saw fish (C-II).

- D. Exocoetus is the scientific name for the flying fish (D-IV).

Thus, the correct answer is (1) A-III, B-I, C-II, D-IV. Quick Tip: Pterophyllum is the angel fish, Myxine is the hag fish, Pristis is the saw fish, and Exocoetus is the flying fish.

The correct sequence of stages in cell division is:
- E. Gap 1 phase, where the cell grows.

- C. Synthesis phase, where DNA replication occurs.

- A. Gap 2 phase, where the cell continues to grow and prepare for division.

- D. Karyokinesis, where the nucleus divides.

- B. Cytokinesis, where the cytoplasm divides, completing the process.

Thus, the correct answer is
(3) E-C-A-D-B. Quick Tip: Cell division progresses in a highly regulated sequence: G1, S, G2, followed by karyokinesis and cytokinesis.

- A. Non-medicated IUD is represented by Lippes loop (III).

- B. Copper releasing IUD is represented by Multiload 375 (I).

- C. Hormone releasing IUD is represented by LNG-20 (IV).

- D. Implants are represented by Progestogens (II).

Thus, the correct answer is
(3) A-III, B-I, C-IV, D-II. Quick Tip: IUDs are categorized based on their medication type: non-medicated, copper-releasing, hormone-releasing, and implants.

- A. Lipase breaks ester bonds (II).

- B. Nuclease breaks phosphodiester bonds (IV).

- C. Protease breaks peptide bonds (I).

- D. Amylase breaks glycosidic bonds (III).

Thus, the correct answer is
(2) A-II, B-IV, C-I, D-III. Quick Tip: Each enzyme acts on a specific type of bond: lipase on ester bonds, nuclease on phosphodiester bonds, protease on peptide bonds, and amylase on glycosidic bonds.

- A. Expiratory capacity is Tidal volume + Expiratory reserve volume (II).

- B. Functional residual capacity is Expiratory reserve volume + Residual volume (IV).

- C. Vital capacity is Expiratory reserve volume + Tidal volume + Inspiratory reserve volume (I).

- D. Inspiratory capacity is Tidal volume + Inspiratory reserve volume (III).

Thus, the correct answer is
(4) A-II, B-IV, C-I, D-III. Quick Tip: The capacities in pulmonary function tests include various combinations of tidal volume, expiratory reserve volume, and inspiratory reserve volume.

- A. Cocaine comes from Erythroxylum (III).

- B. Heroin is derived from Papaver somniferum (IV).

- C. Morphine is also derived from Papaver somniferum (I).

- D. Marijuana comes from Cannabis sativa (II).

Thus, the correct answer is
(3) A-III, B-IV, C-I, D-II. Quick Tip: Cocaine and heroin are derived from plants, while morphine comes from opium poppies and marijuana from cannabis.

The flippers of penguins and dolphins are examples of convergent evolution, where different species independently evolve similar traits as a result of having to adapt to similar environments.

Thus, the correct answer is
(2) Convergent evolution. Quick Tip: Convergent evolution occurs when unrelated species evolve similar traits due to similar environmental pressures.

- Assertion A is correct: Breastfeeding is highly recommended during the early stages of an infant's development.

- Reason R is also correct: Colostrum, the first milk produced, contains antibodies that help in building the baby's immunity.

Thus, the correct answer is
(4) Both A and R are correct and R is the correct explanation of A. Quick Tip: Colostrum is rich in antibodies, providing essential immunity to the newborn during the early days.

The uterine fundus is part of the uterus, not the Fallopian tube. The Fallopian tube consists of the infundibulum, ampulla, and isthmus.

Thus, the correct answer is
(4) Uterine fundus. Quick Tip: The uterine fundus is the upper portion of the uterus, not part of the Fallopian tube.

- A. Typhoid is caused by a bacterium, so it matches with IV. Bacteria.

- B. Leishmaniasis is caused by a protozoan, so it matches with III. Protozoa.

- C. Ringworm is caused by a fungus, so it matches with I. Fungus.

- D. Filariasis is caused by a nematode, so it matches with II. Nematode.

Thus, the correct answer is (1) A-IV, B-III, C-I, D-II. Quick Tip: Different diseases are caused by different organisms: bacteria (Typhoid), protozoa (Leishmaniasis), fungi (Ringworm), and nematodes (Filariasis).

- A. Common cold is caused by Rhinoviruses, so it matches with III. Rhinoviruses.

- B. Haemozoin is a byproduct of the malaria-causing Plasmodium, so it matches with I. Plasmodium.

- C. Widal test is used to diagnose Typhoid, so it matches with II. Typhoid.

- D. Allergy can be triggered by Dust mites, so it matches with IV. Dust mites.

Thus, the correct answer is
(2) A-III, B-I, C-II, D-IV. Quick Tip: Common cold is caused by rhinoviruses, malaria by Plasmodium, and allergies by dust mites. The Widal test is used for diagnosing typhoid.

The correct sequence of human evolution, from past to recent, is:

- A. Homo habilis was one of the earliest members of the genus Homo.

- D. Homo erectus evolved next, showing more advanced tool use.

- C. Homo neanderthalensis followed, characterized by a robust build.

- B. Homo sapiens are the modern humans, representing the most recent stage of evolution.

Thus, the correct answer is
(3) A-D-C-B. Quick Tip: The human evolutionary sequence shows a gradual increase in brain size and complexity from Homo habilis to Homo sapiens.

- Assertion A is false because FSH acts on ovarian follicles in females and Sertoli cells in males, not Leydig cells.

- Reason R is true because growing ovarian follicles secrete estrogen, and interstitial cells (Leydig cells) secrete androgens in males.

Thus, the correct answer is
(3) A is false but R is true. Quick Tip: FSH plays key roles in both male and female reproductive systems but acts on different cells in males (Sertoli cells) and females (ovarian follicles).

- Testosterone, Progesterone, and Cortisol are steroid hormones, derived from cholesterol and characterized by a four-ring structure.

- Glucagon, however, is a peptide hormone, not a steroid.

Thus, the correct answer is
(3) Glucagon. Quick Tip: Steroid hormones are derived from cholesterol and include hormones like testosterone and cortisol, while glucagon is a peptide hormone.

- A. Annelids are true coelomates, meaning they have a well-developed coelom.

- B. Poriferans are not pseudocoelomates; they lack a true coelom.

- C. Aschelminthes are pseudocoelomates, not acoelomates.

- D. Platyhelminthes are acoelomates, not pseudocoelomates.

Thus, the correct answer is (1) A only. Quick Tip: True coelomates, such as annelids, have a true coelom, while pseudocoelomates (e.g., aschelminthes) have a body cavity that is not fully lined by mesoderm.

- A. Diakinesis involves the completion of terminalisation of chiasmata (II).

- B. Pachytene is characterized by the appearance of recombination nodules (IV).

- C. Zygotene involves the formation of the synaptonemal complex (I).

- D. Leptotene is characterized by chromosomes appearing as thin threads (III).

Thus, the correct answer is
(2) A-II, B-IV, C-I, D-III. Quick Tip: Prophase I of meiosis consists of several sub-stages that prepare homologous chromosomes for crossing over and segregation.

Oxyhemoglobin forms more readily when oxygen levels are high (high pO2) and when the concentration of H+ ions is lower, as in the lungs, where the pH is relatively higher.

Thus, the correct answer is (1) High pO2 and Lesser H+ concentration. Quick Tip: Oxygen binds more effectively to hemoglobin in conditions of high pO2 and low H+ concentration.

In cockroaches, the anal cerci are a pair of jointed, filamentous structures that are located on the 10th segment of the abdomen. These cerci are sensory organs that help the cockroach detect air currents and vibrations.

Thus, the correct answer is (1) 10th segment. Quick Tip: The anal cerci in cockroaches are used for sensory perception, helping detect environmental changes.

- A. Pleurobrachia is a member of the phylum Ctenophora (II).

- B. Radula is found in organisms from the phylum Mollusca (I).

- C. Stomochord is a characteristic feature of organisms from the phylum Hemichordata (IV).

- D. Air bladder is found in fish from the phylum Osteichthyes (III).

Thus, the correct answer is (1) A-II, B-I, C-IV, D-III. Quick Tip: Pleurobrachia belongs to Ctenophora, the radula is a molluscan characteristic, the stomochord is present in Hemichordates, and air bladders are found in osteichthyes.

The Ti plasmid in Agrobacterium tumefaciens is responsible for inducing tumor formation in plants. It carries genes that integrate into the plant genome, causing uncontrolled growth and formation of tumors known as crown galls.

Thus, the correct answer is
(2) Tumor inducing plasmid. Quick Tip: The Ti plasmid is a crucial tool in plant genetic engineering, used for transferring genes into plant cells.

- Assertion A is true because the hymen can stretch, tear, or remain intact for reasons unrelated to sexual activity.

- Reason R is false because the hymen can be torn or stretched due to various activities, not only during coitus.

Thus, the correct answer is
(2) Statement I is true but Statement II is false. Quick Tip: The hymen is not a reliable indicator of virginity since it can be torn or stretched for various reasons other than sexual intercourse.

The vector pBR322 contains the X and Y genes.

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

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


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

- Periodic abstinence, Lactational amenorrhea, and Coitus interruptus are all traditional methods of contraception.

- Vaults refer to a modern contraceptive method involving the insertion of a device and are not considered a natural or traditional method.

Thus, the correct answer is
(3) Vaults. Quick Tip: Natural contraceptive methods rely on natural cycles or behavioral strategies, while vaults are a modern, medical contraceptive option.

- A. Exophthalmic goiter is caused by hyper secretion of thyroid hormone (III).

- B. Acromegaly is caused by excessive secretion of growth hormone (IV).

- C. Cushing’s syndrome is caused by excess secretion of cortisol, moon face, and hyperglycemia (I).

- D. Cretinism is caused by hypo-secretion of thyroid hormone and stunted growth (II).

Thus, the correct answer is
(3) A-III, B-IV, C-I, D-II. Quick Tip: Endocrine disorders like Cushing’s syndrome and acromegaly are caused by excessive hormone secretion, while cretinism is due to undersecretion of thyroid hormone.

- FSH (Follicle-stimulating hormone) acts on the Sertoli cells to promote spermatogenesis (the production of sperm).

- Leydig cells are responsible for producing testosterone, which supports the process of spermatogenesis.

- Spermiogenesis is the final stage of spermatogenesis where spermatids mature into sperm.

Thus, the correct answer is
(4) FSH, Leydig cells, Sertoli cells, spermiogenesis. Quick Tip: FSH stimulates Sertoli cells in the seminiferous tubules to support spermatogenesis, while Leydig cells produce testosterone.

- Statement I is correct: Both mitochondria and chloroplasts are double-membrane bound organelles, characteristic of eukaryotic cells.

- Statement II is incorrect: The inner membrane of the mitochondria is highly impermeable, whereas the inner membrane of the chloroplast is permeable to some molecules due to the presence of specific transporters.

Thus, the correct answer is
(2) Statement I is correct but Statement II is incorrect. Quick Tip: Mitochondria and chloroplasts are similar in structure, with double membranes, but their inner membranes have different permeability characteristics.

- Statement I is false because Gause's competitive exclusion principle states that two species competing for the same resources cannot coexist indefinitely. It is not about different resources.

- Statement II is true: In cases of competition for limiting resources, one species may outcompete the other, leading to the elimination of the inferior competitor.

Thus, the correct answer is
(3) Statement I is false but Statement II is true. Quick Tip: Gause's competitive exclusion principle emphasizes that species competing for the same resources cannot coexist for long if resources are limiting.

- E. Substrate binding to the active site occurs first.

- A. Substrate-enzyme complex formation follows after the substrate binds to the active site.

- D. Chemical bonds of the substrate are broken, leading to the formation of products.

- C. Release of products occurs next.

- B. Free enzyme is ready to bind with another substrate to complete the cycle.

Thus, the correct answer is
(4) E, A, D, C, B. Quick Tip: The catalytic cycle involves the enzyme-substrate complex formation, substrate breakdown, product release, and enzyme resetting for the next cycle.

- A. P wave corresponds to the depolarisation of the atria (III).

- B. QRS complex corresponds to the depolarisation of the ventricles (II).

- C. T wave corresponds to the repolarisation of the ventricles (IV).

- D. T-P gap represents a period when heart muscles are electrically silent (I).

Thus, the correct answer is (1) A-III, B-II, C-IV, D-I. Quick Tip: Understanding the P wave, QRS complex, and T wave is crucial for interpreting the electrical activity of the heart.

- A. RNA polymerase III synthesizes snRNAs and tRNA (IV).

- B. Termination of transcription involves the Rho factor (III).

- C. Splicing of exons is facilitated by snRNPs (I).

- D. TATA box is involved in the initiation of transcription and serves as a promoter region (II).

Thus, the correct answer is
(3) A-IV, B-III, C-I, D-II. Quick Tip: The TATA box and RNA polymerase are essential in the initiation of transcription, while snRNPs play a role in splicing.

- Statement I is correct: The corpus callosum connects the left and right cerebral hemispheres.

- Statement II is incorrect: The brain stem consists of the medulla oblongata, pons, and midbrain, not the cerebrum.

Thus, the correct answer is
(2) Statement I is correct but Statement II is incorrect. Quick Tip: The brain stem does not include the cerebrum; it is composed of the medulla oblongata, pons, and midbrain.

- A. Crop is responsible for storing food (IV).

- B. Gastric Caeca are responsible for aiding digestion in the foregut-midgut junction (II).

- C. Malpighian tubules are responsible for excretion in the cockroach (III).

- D. Gizzard is used for grinding food (I).

Thus, the correct answer is
(4) A-IV, B-II, C-III, D-I. Quick Tip: In cockroaches, the digestive system includes specialized structures like the crop for storage and the gizzard for grinding food.

- A. Mesozoic Era was dominated by birds and reptiles (III).

- B. Proterozoic Era featured lower invertebrates (I).

- C. Cenozoic Era is known for the dominance of mammals (IV).

- D. Paleozoic Era was characterized by fish and amphibians (II).

Thus, the correct answer is
(3) A-III, B-I, C-IV, D-II. Quick Tip: The geologic eras are marked by distinct groups of organisms, with mammals dominating the Cenozoic era, and fish and amphibians in the Paleozoic.

- Statement I is correct: The bone marrow is indeed the primary site where all blood cells, including lymphocytes, are produced.

- Statement II is correct: Both the bone marrow and thymus provide microenvironments for the development and maturation of T-lymphocytes, with the thymus being the primary site for T-cell maturation.

Thus, the correct answer is
(4) Both Statement I and Statement II are correct. Quick Tip: The bone marrow produces blood cells, including lymphocytes, while the thymus is key for T-lymphocyte maturation.

- A. Unicellular glandular epithelium refers to Goblet cells of the alimentary canal (III).

- B. Compound epithelium is found in the moist surface of the buccal cavity (IV).

- C. Multicellular glandular epithelium includes salivary glands (I).

- D. Endocrine glandular epithelium is found in the pancreas (II).

Thus, the correct answer is
(2) A-III, B-IV, C-I, D-II. Quick Tip: Unicellular glands like goblet cells are found in the alimentary canal, while multicellular glands include salivary and endocrine glands like the pancreas.

- Statement 1 is incorrect because the renal corpuscle of juxta medullary nephrons lies in the cortex, not the outer portion of the renal medulla.

- Statement 2 is correct: The Loop of Henle of juxta medullary nephrons extends deep into the medulla, which helps in water conservation.

- Statement 3 is incorrect because cortical nephrons outnumber juxta medullary nephrons.

- Statement 4 is incorrect because juxta medullary nephrons are not located in the columns of Bertini.

Thus, the correct answer is
(2) Loop of Henle of juxta medullary nephron runs deep into medulla. Quick Tip: Juxta medullary nephrons have long loops of Henle that extend into the medulla, crucial for water reabsorption and urine concentration.

- Father (B+) must have the genotype IBi because he passed the O allele to the child.

- Mother (A+) must have the genotype IAi because she passed the O allele to the child.

- The child's genotype must be ii, as they inherited the O allele from both parents.

Thus, the correct answer is
(4) A only. Quick Tip: For O+ blood group, both parents must carry an O allele (genotype IAi or IBi).

- Statement A is incorrect because pharyngeal gill slits are not found in all non-chordates, but only in chordates.

- Statement B is correct: Non-chordates lack a notochord.

- Statement C is correct: In non-chordates, the central nervous system is dorsal.

- Statement D is correct: In many non-chordates, the heart, if present, is dorsal.

- Statement E is correct: Many non-chordates lack a post-anal tail.

Thus, the correct answer is
(2) B, D \& E only. Quick Tip: Non-chordates lack a notochord, have a dorsal nervous system, and may have a dorsal heart.