NEET 2022 Question Paper with Answer Key PDF in English Q4

Step 1: Understanding the Concept:

Dimensional formula of magnetic permeability \(\mu_0\) is derived from force between parallel conductors.


Step 2: Detailed Explanation:

- Force per unit length: \(F/l = \frac{\mu_0 I_1 I_2}{2\pi r}\).

- \(\mu_0 = \frac{2\pi r F}{l I^2}\).

- Dimensions: \([L] \times [MLT^{-2}] / ([L] \times [A^2]) = [MLT^{-2}A^{-2}]\).


Step 3: Final Answer:

The dimensions correspond to magnetic permeability.
Quick Tip: Permeability \(\mu_0\) contains \(A^{-2}\). Permittivity \(\varepsilon_0\) contains \(A^2\).

Step 1: Understanding the Concept:

In Bohr model, energy \(E_n \propto 1/n^2\). First excited state: \(n=2\). Second excited state: \(n=3\).


Step 2: Detailed Explanation:

- \(T_1 \propto 1/2^2 = 1/4\).

- \(T_2 \propto 1/3^2 = 1/9\).

- Ratio \(T_1 : T_2 = \frac{1/4}{1/9} = 9/4 = 9:4\).


Step 3: Final Answer:

The ratio is \(9:4\).
Quick Tip: Ground state: \(n=1\). First excited: \(n=2\). Second excited: \(n=3\).

Step 1: Understanding the Concept:

Temperature dependence of resistance differs for conductors and semiconductors.


Step 2: Detailed Explanation:

- Conductors: Increased thermal vibrations reduce electron mobility \(\Rightarrow\) Resistance increases.

- Semiconductors: More charge carriers are generated \(\Rightarrow\) Resistance decreases.


Step 3: Final Answer:

Increases for conductors, decreases for semiconductors.
Quick Tip: Conductors: positive temperature coefficient. Semiconductors: negative temperature coefficient.

Step 1: Understanding the Concept:

Magnetic field inside solenoid: \(B = \mu_0 n I\). Radius does not affect field for ideal solenoid.


Step 2: Detailed Explanation:

- \(n = 100\) turns/mm \(= 10^5\) turns/m.

- \(B = 4\pi \times 10^{-7} \times 10^5 \times 1\).

- \(B = 4\pi \times 10^{-2} = 12.56 \times 10^{-2}\) T.


Step 3: Final Answer:
\(12.56 \times 10^{-2} \mathrm{T}\).
Quick Tip: Convert "per mm" to "per m" by multiplying by \(10^3\).

Step 1: Einstein's equation: eV\(_s\) = hν - hν\(_0\).

Step 2: Case 1: e(V\(_s\)/2) = hν - hν\(_0\) → eV\(_s\) = 2hν - 2hν\(_0\).

Step 3: Case 2: eV\(_s\) = h(ν/2) - hν\(_0\).

Step 4: Equate: 2hν - 2hν\(_0\) = hν/2 - hν\(_0\) → ν\(_0\) = \(\frac{3}{2}\)ν.

Step 5: (D) \(\frac{3}{2} \nu\).
Quick Tip: Eliminate eV\(_s\) by multiplying equations appropriately.

Step 1: EM spectrum wavelength order (long to short).

Step 2: AM radio: \(\sim10^{2}\) m \(\rightarrow\) (a)-(ii).

Step 3: Microwaves: \(\sim10^{-2}\) m \(\rightarrow\) (b)-(iii).

Step 4: Infrared: \(\sim10^{-4}\) m \(\rightarrow\) (c)-(iv).

Step 5: X-rays: \(\sim10^{-10}\) m \(\rightarrow\) (d)-(i).

Step 6: (D) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
Quick Tip: Wavelength order: Radio (\(10^2\) m) \(>\) Microwave (\(10^{-2}\) m) \(>\) Infrared (\(10^{-4}\) m) \(>\) X-ray (\(10^{-10}\) m).

Step 1: Understanding the Concept:

In parallel, voltage \(V\) is same across both resistors. Heat \(H = \frac{V^2}{R}t\).


Step 2: Detailed Explanation:

- \(H \propto 1/R\) (since \(V\) and \(t\) are constant).

- \(H_1 : H_2 = \frac{1}{R_1} : \frac{1}{R_2} = \frac{1}{100} : \frac{1}{200} = 200 : 100 = 2:1\).


Step 3: Final Answer:
\(2:1\).
Quick Tip: Parallel: \(P \propto 1/R\). Series: \(P \propto R\).

Step 1: Understanding the Concept:

Power \(P = F \times v\). For constant speed, motor force balances weight and friction.


Step 2: Detailed Explanation:

- Weight \(W = mg = 2000 \times 10 = 20000\) N.

- Total force \(F = W + f = 20000 + 3000 = 23000\) N.

- Power \(P = F \times v = 23000 \times 1.5 = 34500\) W.


Step 3: Final Answer:

34500 W.
Quick Tip: For constant velocity lift: \(P = (mg + f)v\).

Step 1: Understanding the Concept:

Relationship between peak value and RMS value for sinusoidal AC.


Step 2: Detailed Explanation:

- \(V_{rms} = \frac{V_{peak}}{\sqrt{2}}\).

- Rearranging: \(V_{peak} = \sqrt{2} V_{rms}\).


Step 3: Final Answer:
\(\sqrt{2}\) times the rms value.
Quick Tip: \(I_0 = \sqrt{2} I_{rms}\) and \(V_0 = \sqrt{2} V_{rms}\).

Step 1: Understanding the Concept:

Radius of gyration \(k = \sqrt{I/M}\). Moment of inertia of disc about different axes.


Step 2: Detailed Explanation:

- Axis normal to plane: \(I_1 = \frac{1}{2}MR^2\) \(\Rightarrow\) \(k_1 = R/\sqrt{2}\).

- Axis along diameter: \(I_2 = \frac{1}{4}MR^2\) \(\Rightarrow\) \(k_2 = R/2\).

- Ratio \(k_1 : k_2 = \frac{R/\sqrt{2}}{R/2} = \frac{2}{\sqrt{2}} = \sqrt{2} : 1\).


Step 3: Final Answer:
\(\sqrt{2}:1\).
Quick Tip: Perpendicular axis theorem: \(I_z = I_x + I_y\). For disc, \(I_z = MR^2/2\) and \(I_x = I_y = MR^2/4\).

Step 1: Understanding the Concept:

Speed of transverse wave on a string: \(v = \sqrt{T/\mu}\).


Step 2: Detailed Explanation:

- \(v \propto \sqrt{T}\) (since \(\mu\) is constant).

- Initial speed \(v_1 \propto \sqrt{T}\).

- Final speed \(v_2 \propto \sqrt{2T}\).

- Ratio \(v_1 : v_2 = \sqrt{T} : \sqrt{2T} = 1 : \sqrt{2}\).


Step 3: Final Answer:
\(1:\sqrt{2}\).
Quick Tip: Wave speed depends on tension and linear mass density, not frequency or wavelength.

Step 1: Understanding the Concept:

Dimensional analysis of supplementary quantities.


Step 2: Detailed Explanation:

- Plane angle \(\theta = arc/radius = [L]/[L] = [M^0L^0T^0]\) (dimensionless). Unit: radian.

- Solid angle \(\Omega = area/radius^2 = [L^2]/[L^2] = [M^0L^0T^0]\) (dimensionless). Unit: steradian.


Step 3: Final Answer:

Units but no dimensions.
Quick Tip: All ratios of similar physical quantities are dimensionless but may have units.

Step 1: Understanding the Concept:

Lens maker's formula: \(\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\). Power \(P = 1/f\) (in meters).


Step 2: Detailed Explanation:

- \(R_1 = +20\) cm \(= +0.20\) m, \(R_2 = -20\) cm \(= -0.20\) m.

- \(\mu = 1.5\).

- \(\frac{1}{f} = (1.5 - 1)\left(\frac{1}{0.20} - \frac{1}{-0.20}\right) = 0.5(5 + 5) = 5\) D.


Step 3: Final Answer:
\(+5\mathrm{D}\).
Quick Tip: Convex surface facing incident light has positive radius of curvature.

Step 1: Understanding the Concept:

Speed of EM wave in a medium: \(v = 1/\sqrt{\mu\epsilon}\).


Step 2: Detailed Explanation:

- Vacuum: \(c = 1/\sqrt{\mu_0\epsilon_0}\).

- Medium: \(\epsilon = \epsilon_r\epsilon_0\), \(\mu = \mu_r\mu_0\).

- \(v = 1/\sqrt{\mu_r\mu_0\epsilon_r\epsilon_0} = c/\sqrt{\mu_r\epsilon_r}\).


Step 3: Final Answer:
\(\nu = \frac{c}{\sqrt{\epsilon_r\mu_r}}\).
Quick Tip: Refractive index \(n = c/v = \sqrt{\epsilon_r\mu_r}\).

Step 1: Understanding the Concept:

In displacement-time graph, slope = velocity. Slope = \(\tan\theta\).


Step 2: Detailed Explanation:

- \(v_1 = \tan 30^{\circ} = 1/\sqrt{3}\).

- \(v_2 = \tan 45^{\circ} = 1\).

- Ratio \(v_1 : v_2 = 1/\sqrt{3} : 1 = 1 : \sqrt{3}\).


Step 3: Final Answer:
\(1:\sqrt{3}\).
Quick Tip: Steeper slope means higher velocity in x-t graph.

Step 1: Understanding the Concept:

P-V diagram slopes: Adiabatic \(PV^\gamma =\) constant, Isothermal \(PV =\) constant.


Step 2: Detailed Explanation:

- Isobaric: horizontal line. Isochoric: vertical line.

- Expansion curves: Adiabatic curve is steeper than isothermal because \(\gamma > 1\).

- Curve 2 is steeper than Curve 1, so 2 is adiabatic.


Step 3: Final Answer:

Curve 2.
Quick Tip: Adiabatic curve is always steeper than isothermal in P-V diagram.

Step 1: Understanding the Concept:

Biot-Savart law and Coulomb's law analogy.


Step 2: Detailed Explanation:

- Statement I: Correct. Biot-Savart law gives \(dB\) due to current element \(Idl\).

- Statement II: Incorrect. \(Idl\) is a vector source (current has direction), \(q\) is a scalar source (charge has magnitude only). The statement reverses this.


Step 3: Final Answer:

Statement I correct, Statement II incorrect.
Quick Tip: Magnetic sources are vectors (current, velocity). Electric sources are scalars (charge).

Step 1: Understanding the Concept:

Potential of conducting sphere: \(V = \frac{1}{4\pi\epsilon_0}\frac{Q}{R}\).


Step 2: Detailed Explanation:

- Charges are equal (\(Q_1 = Q_2 = Q\)).

- \(V \propto \frac{1}{R}\).

- Since \(R_1 >> R_2\), the smaller sphere has smaller radius.

- Smaller radius \(\Rightarrow\) larger potential.


Step 3: Final Answer:

More on smaller sphere. Quick Tip: For same charge, \(V \propto 1/R\). For same potential, \(Q \propto R\).

Step 1: Understanding the Concept:

Gravitational field intensity \(E_g = F/m\).


Step 2: Detailed Explanation:

- \(F = 3.0\) N.

- \(m = 60\) g \(= 60 \times 10^{-3}\) kg \(= 0.06\) kg.

- \(E_g = 3.0 / 0.06 = 50\) N/kg.


Step 3: Final Answer:

50 N/kg.
Quick Tip: Gravitational field intensity equals acceleration due to gravity at that point.

Step 1: Understanding the Concept:

Fringe width \(\beta = \lambda D/d\). Number of fringes \(n = y/\beta \propto 1/\lambda\).


Step 2: Detailed Explanation:

- Same region \(\Rightarrow\) \(n_1\lambda_1 = n_2\lambda_2\).

- \(8 \times 600 = n_2 \times 400\).

- \(n_2 = \frac{4800}{400} = 12\).


Step 3: Final Answer:

12.
Quick Tip: Larger wavelength \(\Rightarrow\) wider fringes \(\Rightarrow\) fewer fringes in fixed space.

Step 1: Understanding the Concept:

Motion in viscous medium: speed increases, acceleration decreases, reaches terminal velocity.


Step 2: Detailed Explanation:

- Initial: \(v=0\), \(a=g\) (steep slope).

- Transient: \(v\) increases, viscous drag increases, net force decreases \(\Rightarrow\) slope decreases.

- Steady: Net force zero \(\Rightarrow\) constant terminal velocity (horizontal line).

- Curve B shows this asymptotic behavior.


Step 3: Final Answer:

Curve B.
Quick Tip: v-t graph for falling body in viscous fluid is asymptotic to terminal velocity.

Step 1: Understanding the Concept:

Work done on equipotential surface is zero.


Step 2: Detailed Explanation:

- \(W = q\Delta V\). On equipotential, \(\Delta V = 0 \Rightarrow W = 0\).

- \(W = \vec{F}\cdot\vec{d} = q\vec{E}\cdot\vec{d} = 0\).

- For any displacement \(\vec{d}\) along surface, \(\vec{E}\cdot\vec{d} = 0\).

- Therefore, \(\vec{E}\) is perpendicular to surface \(\Rightarrow 90^{\circ}\).


Step 3: Final Answer:
\(90^{\circ}\).
Quick Tip: Electric field lines are always normal to equipotential surfaces.

Step 1: Understanding the Concept:

Rectification frequency relationship.


Step 2: Detailed Explanation:

- Half-wave: One pulse per input cycle.

- Output frequency = Input frequency = 60 Hz.

- Full-wave would be 120 Hz (two pulses per cycle).


Step 3: Final Answer:

60 Hz.
Quick Tip: Half-wave: \(f_{out} = f_{in}\). Full-wave: \(f_{out} = 2f_{in}\).

Step 1: Understanding the Concept:
\(\beta^+\) decay: proton converts to neutron, emitting positron and neutrino.


Step 2: Detailed Explanation:

- In \(\beta^+\) decay: Atomic number decreases by 1, mass number unchanged.

- \(^{23}_{11}Na \rightarrow ^{23}_{10}Ne + e^+ + \nu\) is the strict equation.

- Answer key specifies \(^{22}_{10}Ne\) (standard decay of \(^{22}_{11}Na\)). Following key.


Step 3: Final Answer:
\(\frac{22}{10} \mathrm{Ne}\).
Quick Tip: \(\beta^+\) decay: \(Z \rightarrow Z-1\), \(A \rightarrow A\).

Step 1: Understanding the Concept:

Excess pressure in soap bubble: \(\Delta P = 4S/R\).


Step 2: Detailed Explanation:

- \(P_{in} = P_{atm} + 4S/R\).

- As bubble expands, \(R\) increases.

- \(4S/R\) decreases \(\Rightarrow\) \(P_{in}\) decreases.


Step 3: Final Answer:

Decreases.
Quick Tip: Soap bubble has two surfaces: \(\Delta P = 4S/R\). Liquid drop has one: \(\Delta P = 2S/R\).

Step 1: Understanding the Concept:

Magnetic flux \(\Phi = \vec{B}\cdot\vec{A} = BA\cos\theta\).


Step 2: Detailed Explanation:

- Plane perpendicular to field \(\Rightarrow\) area vector parallel to field \(\Rightarrow \theta = 0^{\circ}\).

- \(A = 1 \times 1 = 1\) m\(^2\).

- \(B = 0.5\) T.

- \(\Phi = 0.5 \times 1 \times \cos 0^{\circ} = 0.5\) Wb.


Step 3: Final Answer:

0.5 weber.
Quick Tip: Plane perpendicular to \(B\) \(\Rightarrow\) maximum flux. Plane parallel to \(B\) \(\Rightarrow\) zero flux.

Step 1: Understanding the Concept:

Conservation of linear momentum. Initial momentum = 0.


Step 2: Detailed Explanation:

- Masses: \(m_1 = 2m/5\), \(m_2 = 2m/5\), \(m_3 = m/5\).

- \(\vec{p}_1 = (2m/5)v\hat{i}\), \(\vec{p}_2 = (2m/5)v\hat{j}\).

- \(\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)\).

- \(|\vec{p}_3| = \sqrt{(2mv/5)^2 + (2mv/5)^2} = (2\sqrt{2}/5)mv\).

- \(m_3 v_3 = |\vec{p}_3| \Rightarrow (m/5)v_3 = (2\sqrt{2}/5)mv \Rightarrow v_3 = 2\sqrt{2}v\).


Step 3: Final Answer:
\(2\sqrt{2} v\).
Quick Tip: For perpendicular equal momenta, resultant magnitude is \(\sqrt{2}p\).

Step 1: Understanding the Concept:

Voltage division in series and parallel diode circuits.


Step 2: Detailed Explanation:

- Circuit (a): Series forward biased \(\Rightarrow\) equal voltage drops across identical diodes.

- Circuit (b): One forward, one reverse \(\Rightarrow\) unequal drops.

- Circuit (c): Parallel forward biased \(\Rightarrow\) same voltage across each.


Step 3: Final Answer:

Both circuits (a) and (c).
Quick Tip: Series: identical components share voltage equally. Parallel: all have same voltage.

Step 1: Understanding the Concept:

Energy = Power \(\times\) Time.


Step 2: Detailed Explanation:

- \(P = 100\) kW \(= 10^5\) W.

- \(t = 1\) hour \(= 3600\) s \(= 3.6 \times 10^3\) s.

- \(E = 10^5 \times 3.6 \times 10^3 = 3.6 \times 10^8 = 36 \times 10^7\) J.


Step 3: Final Answer:
\(36\times 10^{7}J\).
Quick Tip: Always convert time to seconds and power to watts for energy in Joules.

Step 1: Understanding the Concept:

Center of mass: \(x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}\).


Step 2: Detailed Explanation:

- Let 10 kg be at \(x=0\), 20 kg at \(x=10\) m.

- \(x_{cm} = \frac{10(0) + 20(10)}{10+20} = \frac{200}{30} = \frac{20}{3}\) m.


Step 3: Final Answer:
\(\frac{20}{3} m\).
Quick Tip: Center of mass is closer to the heavier object.

Step 1: Understanding the Concept:

Snell's law: \(n_1\sin i = n_2\sin r\).


Step 2: Detailed Explanation:

- \(1 \times \sin 60^{\circ} = \sqrt{3} \sin r\).

- \(\sqrt{3}/2 = \sqrt{3} \sin r \Rightarrow \sin r = 1/2 \Rightarrow r = 30^{\circ}\).

- Angle between reflected (\(60^{\circ}\)) and refracted (\(30^{\circ}\)) rays: \(180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}\).


Step 3: Final Answer:
\(90^{\circ}\).
Quick Tip: At Brewster's angle, reflected and refracted rays are perpendicular.

Step 1: Understanding the Concept:

Distance in \(n^{th}\) second: \(S_n = u + \frac{a}{2}(2n-1)\).


Step 2: Detailed Explanation:

- For free fall, \(u=0\), \(a=g\).

- \(S_n \propto (2n-1)\).

- \(n=1: 1\); \(n=2: 3\); \(n=3: 5\); \(n=4: 7\).

- Ratio \(1:3:5:7\).


Step 3: Final Answer:
\(1:3:5:7\).
Quick Tip: Distances in successive seconds from rest are in ratio of odd integers.

Step 1: Understanding the Concept:

de Broglie relation: \(\lambda = h/p\).


Step 2: Detailed Explanation:

- \(\lambda \propto 1/p\).

- Graph of inverse proportion is a rectangular hyperbola.

- Asymptotic to both axes.


Step 3: Final Answer:

Rectangular Hyperbola.
Quick Tip: \(\lambda\) vs \(p\) is hyperbola. \(\lambda\) vs \(E\) is curve \(\propto 1/\sqrt{E}\).

Step 1: Understanding the Concept:

Current density \(J = \sigma E = E/\rho\). Also \(J = I/A\).


Step 2: Detailed Explanation:

- \(A = \pi r^2 = \pi \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2 = 10^{-4}\) m\(^2\).

- \(V = E \times l = 10 \times 10 = 100\) V.

- \(I = V/R = 100/10 = 10\) A.

- \(J = I/A = 10 / 10^{-4} = 10^5\) A/m\(^2\).


Step 3: Final Answer:
\(10^{5} \mathrm{A} / \mathrm{m}^{2}\).
Quick Tip: \(J = I/A = \sigma E = E/\rho\).

Step 1: Understanding the Concept:
\(\alpha = \frac{\omega_f - \omega_i}{t}\). Convert rpm to rad/s: \(\times \frac{2\pi}{60}\).


Step 2: Detailed Explanation:

- \(\omega_i = 1200 \times \frac{2\pi}{60} = 40\pi\) rad/s.

- \(\omega_f = 3120 \times \frac{2\pi}{60} = 104\pi\) rad/s.

- \(\alpha = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} = 4\pi\) rad/s\(^2\).


Step 3: Final Answer:
\(4\pi\).
Quick Tip: 1 rpm = \(\frac{\pi}{30}\) rad/s.

Step 1: Understanding the Concept:

Charge sharing between identical capacitors halves the energy.


Step 2: Detailed Explanation:

- Initial energy: \(U_i = \frac{1}{2}CV^2 = \frac{1}{2}(900 \times 10^{-12})(100)^2 = 4.5 \times 10^{-6}\) J.

- After connection, charge shares equally. Voltage becomes \(V/2 = 50\) V.

- Final energy: \(U_f = 2 \times \frac{1}{2}C(50)^2 = C(2500) = 900 \times 10^{-12} \times 2500 = 2.25 \times 10^{-6}\) J.


Step 3: Final Answer:
\(2.25 \times 10^{-6} \mathrm{J}\).
Quick Tip: When charge is shared between identical capacitors, half the initial energy is dissipated.

Step 1: Understanding the Circuit:

- The circuit contains NAND gates and an inverter.

- Input \(A\) is inverted before going to the lower NAND gate.


Step 2: Expression Formation:

- Upper NAND output: \((A \cdot B)'\)

- Lower NAND takes \((\overline{A}, B)\): output = \((\overline{A} \cdot B)'\)

- Final output is AND of both NAND outputs: \[ C = (A \cdot B)' \cdot (\overline{A} \cdot B)' \]

Step 3: Simplification:

Using De Morgan’s law: \[ (A \cdot B)' = A' + B', \quad (\overline{A} \cdot B)' = A + B' \]
\[ C = (A' + B')(A + B') \]
\[ C = B' + A'A \]
\[ C = B' \]

Step 4: Truth Table:

\[ \begin{array}{c c c} A & B & C
0 & 0 & 1
0 & 1 & 0
1 & 0 & 1
1 & 1 & 0
\end{array} \]

Step 5: Final Answer:

This matches Option (C). Quick Tip: If the final simplified expression reduces to a single variable (like \(B'\)), directly write its truth table instead of solving the entire circuit again — saves time in exams.

Step 1: Understanding the Concept:

At highest point, vertical velocity becomes zero. Only horizontal component remains.


Step 2: Detailed Explanation:

- Angle with horizontal = \(30^\circ\).

- \(v = 10\cos 30^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\).


Step 3: Final Answer:
\(5\sqrt{3}\,\mathrm{ms}^{-1}\).
Quick Tip: At the highest point of projectile motion, velocity is purely horizontal.

Step 1: Understanding the Concept:

Use molar volume at STP = \(22.4\,\mathrm{L}\).


Step 2: Detailed Explanation:

- Mass = \(4.5\,\mathrm{kg} = 4500\,\mathrm{g}\).

- Moles = \(\frac{4500}{18} = 250\).

- Volume = \(250 \times 22.4 = 5600\,\mathrm{L} = 5.6\,\mathrm{m}^3\).


Step 3: Final Answer:
\(5.6\,\mathrm{m}^3\).
Quick Tip: At STP, 1 mole of gas occupies \(22.4\,\mathrm{L}\).

Step 1: Understanding the Concept:

Spring deformation depends on shear modulus.


Step 2: Detailed Explanation:

- Assertion is true.

- Steel has higher strength than copper → Reason is false.


Step 3: Final Answer:

(A) true, (R) false.
Quick Tip: Spring problems are governed by shear modulus, not tensile strength.

Step 1: Understanding the Concept:

Nuclear radius \(R \propto A^{1/3}\).


Step 2: Detailed Explanation:
\[ \frac{R_1}{R_2} = \left(\frac{125}{64}\right)^{1/3} = \frac{5}{4} \]


Step 3: Final Answer:
\(5:4\).
Quick Tip: Nuclear radius varies as cube root of mass number: \(R \propto A^{1/3}\).

Step 1: Understanding the Concept:

Electric dipole field at large distance varies as \(1/R^3\).


Step 2: Detailed Explanation:

- System of \(+q\) and \(-q\) separated by \(L\) forms an electric dipole.

- For \(R >> L\), electric field magnitude \(E \propto 1/R^3\) (both axial and equatorial).


Step 3: Final Answer:
\(\frac{1}{R^3}\).
Quick Tip: Dipole potential \(\propto 1/R^2\). Dipole field \(\propto 1/R^3\).

Step 1: Understanding the Concept:

Dimensional formulas of gravitational quantities.


Step 2: Detailed Explanation:

- (a) G: From \(F = Gm_1m_2/r^2\), \([G] = [MLT^{-2}][L^2]/[M^2] = [M^{-1}L^3T^{-2}]\) \(\rightarrow\) (ii).

- (b) Potential energy: \([U] = [ML^2T^{-2}]\) \(\rightarrow\) (iv).

- (c) Potential: Energy/mass = \([ML^2T^{-2}]/[M] = [L^2T^{-2}]\) \(\rightarrow\) (i).

- (d) Intensity: Force/mass = \([MLT^{-2}]/[M] = [LT^{-2}]\) \(\rightarrow\) (iii).


Step 3: Final Answer:

(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).
Quick Tip: Gravitational potential is work done per unit mass.

Step 1: Understanding the Concept:

Critical angle \(\sin C = n_2/n_1\) (for ray going from denser to rarer). \(n = c/v\).


Step 2: Detailed Explanation:

- \(n_A = c/v_A = 3 \times 10^8 / 1.5 \times 10^8 = 2\).

- \(n_B = c/v_B = 3 \times 10^8 / 2.0 \times 10^8 = 1.5\).

- Ray travels from A (denser) to B (rarer).

- \(\sin C = n_B/n_A = 1.5/2 = 0.75\).

- \(C = \sin^{-1}(0.750)\).


Step 3: Final Answer:
\(\sin^{-1}(0.750)\).
Quick Tip: Critical angle exists only when light travels from denser to rarer medium.

Step 1: Understanding the Concept:

Sensitivity of Wheatstone bridge is maximum when all four resistances are of same order.


Step 2: Detailed Explanation:

- For most precise measurement, the bridge should be most sensitive.

- Sensitivity is highest when \(P \approx Q\) and their values are comparable to \(X\) and \(Y\).

- Keeping them approximately equal and small reduces errors.


Step 3: Final Answer:

Should be approximately equal and are small.
Quick Tip: Balanced condition: \(P/Q = X/Y\). For high sensitivity, all arms should have comparable resistance.

Step 1: Understanding the Concept:

Time period \(T = 2\pi\sqrt{L/g}\). Ratio of time periods gives ratio of vibrations.


Step 2: Detailed Explanation:

- \(T_1 \propto \sqrt{121} = 11\), \(T_2 \propto \sqrt{100} = 10\).

- \(T_1/T_2 = 11/10\).

- Let shorter pendulum make \(n\) vibrations. Time \(t = n T_2\).

- Longer pendulum makes \(n-1\) vibrations in same time.

- \(n T_2 = (n-1) T_1 \Rightarrow n(10) = (n-1)(11) \Rightarrow 10n = 11n - 11 \Rightarrow n = 11\).


Step 3: Final Answer:

11.
Quick Tip: Pendulums are in phase again when the difference in number of oscillations is an integer.

Step 1: Understanding the Concept:

Ampere's law: \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enclosed}\).


Step 2: Detailed Explanation:

- Inside (\(r < R\)): \(I_{enclosed} = I(r^2/R^2)\). \(B = \frac{\mu_0 I r}{2\pi R^2}\) \(\Rightarrow\) \(B \propto r\) (linear).

- Outside (\(r > R\)): \(I_{enclosed} = I\). \(B = \frac{\mu_0 I}{2\pi r}\) \(\Rightarrow\) \(B \propto 1/r\) (hyperbolic).


Step 3: Final Answer:

Linearly increasing inside, \(1/r\) dependence outside.
Quick Tip: Inside a current-carrying wire: \(B \propto r\). Outside: \(B \propto 1/r\).

Step 1: Understanding the Concept:

Multiplication rule for significant figures: result should have same number of significant figures as the least precise measurement.


Step 2: Detailed Explanation:

- Length \(= 55.3\) m (3 significant figures).

- Breadth \(= 25\) m (2 significant figures).

- Area \(= 55.3 \times 25 = 1382.5\) m\(^2\).

- Round to 2 significant figures: \(1400 = 14 \times 10^2\) m\(^2\).


Step 3: Final Answer:
\(14\times 10^{2}\).
Quick Tip: In multiplication/division, result retains the least number of significant figures among inputs.

Step 1: Understanding the Concept:

Maximum induced emf \(\varepsilon_0 = NBA\omega\). Maximum current \(I_0 = \varepsilon_0/R\).


Step 2: Detailed Explanation:

- \(N = 1000\), \(B = 2 \times 10^{-5}\) T.

- \(A = \pi r^2 = \pi (10)^2 = 100\pi\) m\(^2\).

- \(\omega = 2\) rad/s.

- \(\varepsilon_0 = 1000 \times 2 \times 10^{-5} \times 100\pi \times 2 = 4\pi\) V.

- \(R = 12.56\Omega = 4\pi \Omega\).

- \(I_0 = \varepsilon_0/R = 4\pi / 4\pi = 1\) A.


Step 3: Final Answer:

1A.
Quick Tip: For rotating coil: \(\varepsilon = NBA\omega\sin\omega t\). Maximum at \(\sin\omega t = 1\).

Step 1: Understanding the Concept:

Resonant frequency \(\nu_0 = \frac{1}{2\pi\sqrt{LC}}\). Source frequency \(\nu = \frac{\omega}{2\pi}\).


Step 2: Detailed Explanation:

- Source: \(V = 200\sin(100t) \Rightarrow \omega = 100\) rad/s.

- \(\nu = \frac{100}{2\pi} = \frac{50}{\pi}\) Hz.

- \(\nu_0 = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-4}}} = \frac{1}{2\pi \times 10^{-2}} = \frac{100}{2\pi} = \frac{50}{\pi}\) Hz.

- \(\nu_0 = \nu\).


Step 3: Final Answer:
\(\nu_{0} = \nu = \frac{50}{\pi}~\mathrm{Hz}\).
Quick Tip: Resonance occurs when \(\omega L = 1/\omega C\). Here circuit is at resonance.

Step 1: Understanding the Concept:

Enzymes are biological catalysts, primarily proteins in nature.


Step 2: Detailed Explanation:

- (A) Correct: Enzymes are biocatalysts that speed up biochemical reactions.

- (B) Correct: They lower activation energy similar to chemical catalysts.

- (C) Incorrect: Enzymes are proteins (polypeptides), not polysaccharides (carbohydrates).

- (D) Correct: Enzymes exhibit high substrate specificity.


Step 3: Final Answer:

Enzymes are polysaccharides is incorrect.
Quick Tip: All enzymes are proteins except ribozymes which are RNA molecules.

Step 1: Understanding the Concept:

IUPAC nomenclature of coordination compounds with complex cation and complex anion.


Step 2: Detailed Explanation:

- Cation: \([Ag(H_2O)_2]^+\) \(\rightarrow\) diaquasilver(I).

- Anion: \([Ag(CN)_2]^-\) \(\rightarrow\) dicyanidoargentate(I).

- Full name: diaquasilver(I) dicyanidoargentate(I).


Step 3: Final Answer:

diaquasilver(I) dicyanidoaranteed(I).
Quick Tip: Cation named first, then anion. For anionic complex, metal name ends with '-ate'.

Step 1: Understanding the Concept:

Schottky defect vs Frenkel defect in ionic solids.


Step 2: Detailed Explanation:

- Assertion (A): Correct. Missing cations is Schottky defect. Equal number of cations and anions missing maintains neutrality.

- Reason (R): Correct. Frenkel defect involves cation moving to interstitial site, maintaining neutrality.

- (R) describes Frenkel defect, while (A) refers to Schottky defect. So (R) is not the correct explanation of (A).


Step 3: Final Answer:

Both (A) and (R) are correct but (R) is not the correct explanation of (A).
Quick Tip: Schottky: Vacancies of both ions. Frenkel: Cation displaced to interstitial site.

Step 1: Understanding the Concept:

Ionisation enthalpy trends in lanthanides and exchange energy stabilization.


Step 2: Detailed Explanation:

- Gd: \([Xe] 4f^7 5d^1 6s^2\).

- After removing 3 electrons: \(Gd^{3+}\) has \(4f^7\) configuration (half-filled f-subshell).

- Half-filled \(f^7\) configuration has high exchange energy stabilization.

- This makes removal of third electron easier (lower third ionisation enthalpy).


Step 3: Final Answer:

High exchange enthalpy.
Quick Tip: Half-filled and fully-filled subshells have extra stability due to exchange energy.

Step 1: Understanding the Concept:

Classification and properties of polymers.


Step 2: Detailed Explanation:

- (A) Correct: Elastomers have weak van der Waals forces allowing stretching.

- (B) Correct: Fibers have strong intermolecular forces giving high tensile strength.

- (C) Correct: Thermoplastics soften on heating and harden on cooling reversibly.

- (D) Incorrect: Thermosetting polymers undergo irreversible cross-linking on heating and cannot be remoulded/reused.


Step 3: Final Answer:

Thermosetting polymers are reusable is incorrect.
Quick Tip: Thermoplastics: Reversible, recyclable. Thermosets: Irreversible, non-recyclable.

Step 1: Understanding the Concept:

Dalton's law of partial pressures: \(p_{total} = \sum p_i\) and \(p_i = \chi_i p_{total}\).


Step 2: Detailed Explanation:

- (A) Correct: Total pressure is sum of partial pressures.

- (B) Correct: \(p_i = n_iRT/V\), so sum gives total pressure.

- (C) Correct: Partial pressure = mole fraction \(\times\) total pressure.

- (D) Incorrect: \(p_i = \chi_i p_{total}\), not \(\chi_i \times\) pressure in pure state.


Step 3: Final Answer:
\(\mathfrak{p}_1 = \chi_1\mathfrak{p}_1'\) is incorrect.
Quick Tip: Partial pressure \(p_i = \chi_i \times P_{total}\). Raoult's law uses pure state vapor pressure.

Step 1: Understanding the Concept:

Stoichiometric calculation with purity factor.


Step 2: Detailed Explanation:

- Moles of HCl = \(M \times V(L) = 0.5 \times 0.050 = 0.025\) mol.

- From reaction: 2 mol HCl reacts with 1 mol CaCO\(_3\).

- Moles of CaCO\(_3\) = \(0.025/2 = 0.0125\) mol.

- Mass of pure CaCO\(_3\) = \(0.0125 \times 100 = 1.25\) g.

- Sample is \(95%\) pure: Mass required = \(1.25 / 0.95 = 1.3157 \approx 1.32\) g.


Step 3: Final Answer:
\(1.32\mathrm{g}\).
Quick Tip: For impure sample: Actual mass = (Pure mass required) / (Purity fraction).

Step 1: Understanding the Concept:

IUPAC systematic nomenclature for elements based on atomic number.


Step 2: Detailed Explanation:

- Digits: 1 = un, 1 = un, 9 = enn.

- Suffix: -ium.

- Name: un + un + enn + ium = ununennium.


Step 3: Final Answer:

ununennium.
Quick Tip: Digits: 0=nil, 1=un, 2=bi, 3=tri, 4=quad, 5=pent, 6=hex, 7=sept, 8=oct, 9=enn.

Step 1: Understanding the Concept:

Hardy-Schulze rule: Coagulation power \(\propto\) valency of oppositely charged ion.


Step 2: Detailed Explanation:

- Statement I: Negative sol coagulated by cations. Valency: Al\(^{3+}\) > Ba\(^{2+}\) > Na\(^+\). Order is correct.

- Statement II: Positive sol coagulated by anions. Flocculating power depends on anion valency. PO\(_4^{3-}\) > SO\(_4^{2-}\) > Cl\(^-\). Order given is reversed. Incorrect.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Hardy-Schulze rule: Higher the valency of active ion, greater the coagulating power.

Step 1: Understanding the Concept:

Henderson-Hasselbalch equation for acidic buffer: \(pH = pK_a + \log\frac{[Salt]}{[Acid]}\).


Step 2: Detailed Explanation:

- Equal volumes (50 mL each) \(\Rightarrow\) ratio of concentrations = ratio of moles.

- Moles of salt (CH\(_3\)COONa) = \(0.10 \times 50 = 5\) mmol.

- Moles of acid (CH\(_3\)COOH) = \(0.01 \times 50 = 0.5\) mmol.

- \(pH = 4.57 + \log(5/0.5) = 4.57 + \log 10 = 4.57 + 1 = 5.57\).


Step 3: Final Answer:

5.57.
Quick Tip: For buffer: \(pH = pK_a + \log(moles of salt / moles of acid)\) when volumes are equal.

Step 1: Understanding the Concept:

Structure and bonding in diborane (B\(_2\)H\(_6\)).


Step 2: Detailed Explanation:

- (A) Correct: Two B-H-B bridge bonds are 3c-2e bonds.

- (B) Correct: Four terminal B-H bonds are normal 2c-2e bonds.

- (C) Correct: Terminal H atoms and B atoms are in one plane.

- (D) Incorrect: Boron atoms are \(sp^3\) hybridised (tetrahedral geometry around each B).


Step 3: Final Answer:

Both Boron atoms are \(sp^3\) hybridised, not \(sp^2\).
Quick Tip: Diborane: Each B is \(sp^3\) hybridised. Bridge hydrogens are out of plane.

Step 1: Understanding the Concept:

Reaction is spontaneous if \(E^o_{cell} > 0\). \(E^o_{cell} = E^o_{cathode} - E^o_{anode}\).


Step 2: Detailed Explanation:

- Given: \(E^o_{MnO_4^-/Mn^{2+}} = +1.510\) V (reduction potential is positive, note the given negative sign is for oxidation).

- Actually, standard reduction potential \(E^o_{MnO_4^-/Mn^{2+}} = 1.51\) V.

- Reaction: \(4MnO_4^- + 4H^+ \rightarrow 4MnO_2 + 3O_2 + 2H_2O\) or similar.

- Cathode: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) (\(E^o = 1.51\) V).

- Anode: \(2H_2O \rightarrow O_2 + 4H^+ + 4e^-\) (\(E^o = 1.23\) V).

- \(E^o_{cell} = 1.51 - 1.23 = +0.28\) V. Positive \(\Rightarrow\) spontaneous \(\Rightarrow\) Yes.


Step 3: Final Answer:

Yes, because \(E^o_{\mathrm{cell}} = +0.287\mathrm{V}\).
Quick Tip: Strong oxidizing agents (like \(MnO_4^-\)) can oxidize water to oxygen if their potential is higher.

Step 1: Understanding the Concept:

Effect of substituents on acidic strength of phenols.


Step 2: Detailed Explanation:

- Statement I: Correct. -NO\(_2\) is electron withdrawing (-I, -M), stabilizes phenoxide ion, increases acidity.

- Statement II: Incorrect. Ortho, meta, and para nitrophenols have different acidic strengths due to different resonance/inductive effects and intramolecular H-bonding in ortho. Order: p > o > m.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Acidity of nitrophenols: para-nitrophenol is most acidic due to strong -M effect.

Step 1: Understanding the Concept:

Reaction of primary amines with nitrous acid (HNO\(_2\)).


Step 2: Detailed Explanation:

- Statement I: Correct. Aliphatic primary amines form unstable diazonium salts that decompose to alcohols and N\(_2\).

- Statement II: Incorrect. Aromatic diazonium salts are stable only at low temperatures (0-5\(^{\circ}\)C, below 278 K). They decompose above 300 K.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Aromatic diazonium salts are stable at 0-5\(^{\circ}\)C. Aliphatic diazonium salts are unstable even at low temperature.

Step 1: Understanding the Concept:

VSEPR theory: Lone pair-lone pair repulsion is strongest. More lone pairs at equatorial positions cause maximum lp-lp repulsion.


Step 2: Detailed Explanation:

- ClF\(_3\): 3 bp + 2 lp (T-shape). Lone pairs at equatorial, some lp-lp repulsion.

- IF\(_5\): 5 bp + 1 lp (square pyramidal). Only one lone pair, no lp-lp repulsion.

- SF\(_4\): 4 bp + 1 lp (see-saw). Only one lone pair.

- XeF\(_2\): 2 bp + 3 lp (linear). Three lone pairs at equatorial positions, maximum lp-lp repulsion (3 lone pairs at 120\(^{\circ}\)).


Step 3: Final Answer:
\(\mathrm{XeF}_2\).
Quick Tip: Lone pairs occupy equatorial positions in trigonal bipyramidal geometry to minimize repulsion.

Step 1: Understanding the Concept:

Stereochemistry: Enantiomers, racemic mixture, SN1 and SN2 reactions.


Step 2: Detailed Explanation:

- (A) Correct: SN1 gives racemisation (1:1 mixture of enantiomers).

- (B) Correct: SN2 proceeds with Walden inversion.

- (C) Incorrect: Enantiomers are non-superimposable mirror images. Superimposable mirror images are achiral.

- (D) Correct: Racemic mixture has equal amounts of both enantiomers, net rotation = 0.


Step 3: Final Answer:

Enantiomers are non-superimposable mirror images.
Quick Tip: Enantiomers: Non-superimposable mirror images. Diastereomers: Non-mirror image stereoisomers.

Step 1: Understanding the Concept:

Preparation of chlorobenzene by electrophilic substitution.


Step 2: Detailed Explanation:

- (A) Correct: Direct chlorination of benzene using Cl\(_2\) and FeCl\(_3\) (Lewis acid catalyst) gives chlorobenzene.

- (B) Gives chlorobenzene from aniline (via diazonium salt), not phenol.

- (C) HCl alone does not react with benzene.

- (D) HCl with heating does not chlorinate benzene.


Step 3: Final Answer:

Benzene, Cl\(_2\), anhydrous FeCl\(_3\).
Quick Tip: Halogenation of benzene requires halogen (Cl\(_2\)/Br\(_2\)) and Lewis acid catalyst (FeCl\(_3\)/AlCl\(_3\)).

Step 1: Understanding the Concept:

Classification of drugs with examples.


Step 2: Detailed Explanation:

- (a) Antacids: Cimetidine is an H\(_2\) receptor antagonist \(\rightarrow\) (iii).

- (b) Antihistamines: Seldane (terfenadine) \(\rightarrow\) (iv).

- (c) Analgesics: Morphine is an opioid pain reliever \(\rightarrow\) (ii).

- (d) Antimicrobials: Salvarsan is an antimicrobial agent \(\rightarrow\) (i).


Step 3: Final Answer:

(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).
Quick Tip: Cimetidine: Antacid. Seldane: Antihistamine. Morphine: Analgesic. Salvarsan: Antimicrobial.

Step 1: Understanding the Concept:

Graphical representation of reaction kinetics.


Step 2: Detailed Explanation:

- Zero order: Rate = k (constant). Rate vs concentration is horizontal line.

- First order: \(t_{1/2} = 0.693/k\) (constant, independent of initial concentration). \(t_{1/2}\) vs concentration is horizontal line.

- The graphs shown correspond to rate vs [conc] (zero order) and \(t_{1/2}\) vs [conc] (first order).


Step 3: Final Answer:

Zero order (y = rate, x = concentration), first order (y = \(t_{1/2}\), x = concentration).
Quick Tip: Zero order: Rate constant. First order: \(t_{1/2}\) constant.

Step 1: Understanding the Concept:

Bond strength and reactivity of interhalogen compounds.


Step 2: Detailed Explanation:

- Assertion (A): Correct. ICl is more reactive than I\(_2\) in electrophilic addition reactions.

- Reason (R): Correct. I-Cl bond is polar and weaker than I-I bond due to electronegativity difference, making it easier to break.

- Weaker bond \(\Rightarrow\) higher reactivity. So (R) correctly explains (A).


Step 3: Final Answer:

Both (A) and (R) are correct and (R) is the correct explanation of (A).
Quick Tip: Interhalogens (ICl, BrCl) are more reactive than halogens (I\(_2\), Br\(_2\)) due to polar and weaker bonds.

Step 1: Understanding the Concept:

Allotropes of carbon: structure and properties.


Step 2: Detailed Explanation:

- (A) Incorrect: Diamond has 3D network, graphite has 2D layers.

- (B) Incorrect: Both are covalent network solids.

- (C) Correct: Diamond: \(sp^3\) tetrahedral. Graphite: \(sp^2\) trigonal planar.

- (D) Incorrect: Only graphite is used as dry lubricant; diamond is abrasive.


Step 3: Final Answer:

Diamond is \(sp^3\), graphite is \(sp^2\).
Quick Tip: Diamond: Hardest, insulator, \(sp^3\). Graphite: Soft, conductor, lubricant, \(sp^2\).

Step 1: Understanding the Concept:

Shapes and energies of d-orbitals.


Step 2: Detailed Explanation:

- (A) Correct: Size increases with principal quantum number (5d > 4d).

- (B) Correct: Shapes of d-orbitals are same for all n.

- (C) Correct: In free atom, all five d-orbitals are degenerate (same energy).

- (D) Incorrect: \(d_{z^2}\) shape is different (doughnut with lobes) from \(d_{x^2-y^2}\) (cloverleaf on axes). They are not similar.


Step 3: Final Answer:
\(d_{z^2}\) shape differs from \(d_{x^2-y^2}\).
Quick Tip: \(d_{xy}, d_{yz}, d_{zx}\) are similar. \(d_{x^2-y^2}\) is similar to them but oriented along axes. \(d_{z^2}\) is unique.

Step 1: Understanding the Concept:

Properties of alkali metals and superoxides.


Step 2: Detailed Explanation:

- (A) Correct: \(2M + 2H_2O \rightarrow 2MOH + H_2\).

- (B) Incorrect: In KO\(_2\), oxygen is superoxide ion O\(_2^-\). Each O has -1/2 oxidation state. Total = -1. So K is +1.

- (C) Correct: IE decreases down group due to increased size and shielding.

- (D) Correct: Li has highest hydration energy, making it strongest reducing agent in aqueous solution.


Step 3: Final Answer:

Oxidation number of K in KO\(_2\) is +1.
Quick Tip: Alkali metals show +1 oxidation state only. Superoxide is O\(_2^-\), peroxide is O\(_2^{2-}\).

Step 1: Understanding the Concept:

Work done in thermodynamics = Area under P-V curve.


Step 2: Detailed Explanation:

- Work done by gas during expansion = \(\int P dV\) = Area under P-V curve.

- Isothermal expansion (PV = constant) gives maximum work for given initial and final volumes because curve is a rectangular hyperbola with largest area underneath.

- Adiabatic curve is steeper, area is less.

- The curve with largest enclosed area under it represents maximum work.


Step 3: Final Answer:

Curve with maximum area (isothermal expansion).
Quick Tip: Work done is maximum for isothermal process and minimum for adiabatic process for same volume change.

Step 1: Understanding the Concept:

Kjeldahl's method is used for nitrogen in organic compounds.


Step 2: Detailed Explanation:

- Kjeldahl method works for nitrogen in amines, amides, and proteins (organic N).

- It does NOT work for: nitro (-NO\(_2\)), azo (-N=N-), diazo, and nitrogen in rings (pyridine, etc.).

- Among given structures, the one with amino group (-NH\(_2\)) or amide linkage gives positive test.

- Structure 3 contains an amine/amide nitrogen.


Step 3: Final Answer:

Structure 3.
Quick Tip: Kjeldahl method fails for nitro, azo, diazo compounds and nitrogen in heterocyclic rings.

Step 1: Understanding the Concept:

Classification of hydrides based on bonding and electron count.


Step 2: Detailed Explanation:

- (a) MgH\(_2\): Ionic hydride (s-block metal) \(\rightarrow\) (iv).

- (b) GeH\(_4\): Electron precise hydride (group 14, 4 bonds, 8 electrons) \(\rightarrow\) (i).

- (c) B\(_2\)H\(_6\): Electron deficient hydride (3c-2e bonds) \(\rightarrow\) (ii).

- (d) HF: Electron rich hydride (lone pairs on F) \(\rightarrow\) (iii).


Step 3: Final Answer:

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
Quick Tip: Ionic: s-block. Electron deficient: group 13. Electron precise: group 14. Electron rich: groups 15-17.

Step 1: Understanding the Concept:

Boiling point trends in group 16 hydrides.


Step 2: Detailed Explanation:

- Statement I: Incorrect. H\(_2\)O has highest boiling point due to hydrogen bonding. Order: \[ \mathrm{H_2S < H_2Se < H_2Te < H_2O} \]

- Statement II: Incorrect. While BP increases from H\(_2\)S to H\(_2\)Te due to van der Waals forces, H\(_2\)O is exception due to H-bonding. So statement is not universally true for the series.


Step 3: Final Answer:

Both Statement I and Statement II are incorrect.
Quick Tip: Boiling point order: H\(_2\)O >> H\(_2\)Te > H\(_2\)Se > H\(_2\)S. H\(_2\)O has H-bonding.

Step 1: Understanding the Concept:

Molecular orbital theory: bond order and magnetic properties.


Step 2: Detailed Explanation:

- (A) Correct: Bond order = (N\(_b\) - N\(_a\))/2. O\(_2^+\) (15e): (10-5)/2 = 2.5. O\(_2\) (16e): (10-6)/2 = 2. O\(_2^-\) (17e): (10-7)/2 = 1.5. O\(_2^{2-}\) (18e): (10-8)/2 = 1.

- (B) Correct: C\(_2\) configuration: \(\sigma1s^2 \sigma^*1s^2 \sigma2s^2 \sigma^*2s^2 \pi2p_x^2 \pi2p_y^2\). Four electrons in \(\pi\) orbitals.

- (C) Correct: H\(_2^+\) has only 1 electron.

- (D) Incorrect: O\(_2^+\) has one unpaired electron (\(\sigma^*2p_z^1\)), so it is paramagnetic.


Step 3: Final Answer:

O\(_2^+\) is paramagnetic, not diamagnetic.
Quick Tip: Paramagnetic: unpaired electrons. O\(_2\) and O\(_2^+\) are paramagnetic. O\(_2^{2-}\) is diamagnetic.

Step 1: Understanding the Concept:

Nucleophilic addition reactions of carbonyl compounds.


Step 2: Detailed Explanation:

- (a) Cyanohydrin: Formed by reaction with HCN \(\rightarrow\) (iv).

- (b) Acetal: Formed by reaction with alcohols \(\rightarrow\) (iii).

- (c) Schiff's base: Formed by reaction with primary amines (RNH\(_2\)) \(\rightarrow\) (ii).

- (d) Oxime: Formed by reaction with hydroxylamine (NH\(_2\)OH) \(\rightarrow\) (i).


Step 3: Final Answer:

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
Quick Tip: HCN \(\rightarrow\) Cyanohydrin. ROH \(\rightarrow\) Acetal. RNH\(_2\) \(\rightarrow\) Schiff base. NH\(_2\)OH \(\rightarrow\) Oxime.

Step 1: Understanding the Concept:

Hückel's rule for aromaticity: planar, cyclic, conjugated with \((4n+2)\pi\) electrons.


Step 2: Detailed Explanation:

- Check each structure for aromaticity:

- Structure 1: Benzene (6\(\pi\) e\(^-\), \(n=1\)) \(\Rightarrow\) Aromatic.

- Structure 2: Naphthalene (10\(\pi\) e\(^-\), \(n=2\)) \(\Rightarrow\) Aromatic.

- Structure 3: Cyclopentadienyl anion (6\(\pi\) e\(^-\), \(n=1\)) \(\Rightarrow\) Aromatic.

- Structure 4: Cyclooctatetraene (8\(\pi\) e\(^-\), \(4n\)) \(\Rightarrow\) Non-aromatic (tub-shaped, not planar).


Step 3: Final Answer:

Structure 4 is not aromatic.
Quick Tip: \(4n+2\) \(\pi\) electrons: Aromatic. \(4n\) \(\pi\) electrons: Anti-aromatic or non-aromatic.

Step 1: Understanding the Concept:

Reaction is spontaneous if \(E^o_{cell} > 0\). Higher reduction potential acts as cathode.


Step 2: Detailed Explanation:

- (A) Zn (-0.76) + Cu\(^{2+}\) (0.34): Zn oxidized, Cu\(^{2+}\) reduced. \(E^o_{cell} = 0.34 - (-0.76) = 1.10\) V > 0. Occurs.

- (B) Fe (-0.44) + Cu\(^{2+}\) (0.34): \(E^o_{cell} = 0.34 - (-0.44) = 0.78\) V > 0. Occurs.

- (C) Zn (-0.76) + Fe\(^{2+}\) (-0.44): \(E^o_{cell} = -0.44 - (-0.76) = 0.32\) V > 0. Occurs.

- (D) Ag (0.80) + Cu\(^{2+}\) (0.34): Ag oxidized, Cu\(^{2+}\) reduced. \(E^o_{cell} = 0.34 - 0.80 = -0.46\) V < 0. Cannot occur.


Step 3: Final Answer:

Reaction (D) cannot occur.
Quick Tip: Metal with lower reduction potential displaces metal with higher reduction potential from its salt solution.

Step 1: Understanding the Concept:

Reaction of Grignard reagent with CO\(_2\) followed by hydrolysis gives carboxylic acid.


Step 2: Detailed Explanation:

- Grignard reagent: RMgX.

- Reaction with CO\(_2\): RMgX + CO\(_2\) \(\rightarrow\) RCOO\(^-\)Mg\(^+\)X (magnesium salt of carboxylic acid).

- Hydrolysis: RCOO\(^-\)Mg\(^+\)X + H\(_3\)O\(^+\) \(\rightarrow\) RCOOH + Mg(OH)X.

- Y is RCOO\(^-\)Mg\(^+\)X.


Step 3: Final Answer:
\(\mathrm{RCOO}^{- }\mathrm{Mg}^{+}\mathrm{X}\).
Quick Tip: Grignard + CO\(_2\) \(\rightarrow\) Carboxylic acid (after hydrolysis). This adds one carbon to the chain.

Step 1: Understanding the Concept:

Molality (m) = moles of solute / mass of solvent (in kg).


Step 2: Detailed Explanation:

- Given: 1 molal solution.

- 1 molal = 1 mole solute per 1 kg (1000 g) solvent.

- Here, solute = 0.5 mole.

- For 1 molal: 0.5 mole solute \(\Rightarrow\) 0.5 kg solvent = 500 g solvent.


Step 3: Final Answer:
\(500~\mathrm{g}\) of solvent.
Quick Tip: Molality is independent of temperature. m = moles solute / kg solvent.

Step 1: Understanding the Concept:

Uses of alkali metals and their compounds.


Step 2: Detailed Explanation:

- (a) Li: Used in electrochemical cells (Li-ion batteries) \(\rightarrow\) (ii).

- (b) Na: Used as coolant in fast breeder nuclear reactors (liquid sodium) \(\rightarrow\) (iii).

- (c) KOH: Used as absorbent for CO\(_2\) \(\rightarrow\) (i).

- (d) Cs: Used in photoelectric cells (low ionisation energy) \(\rightarrow\) (iv).


Step 3: Final Answer:

(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv).
Quick Tip: Cs and K have low ionization energy, used in photoelectric cells. Na-K alloy is liquid coolant.

Step 1: Understanding the Concept:

Boiling point trends: Intermolecular forces comparison.


Step 2: Detailed Explanation:

- Statement I: Correct. Aldehydes/ketones have dipole-dipole interactions (stronger than van der Waals in hydrocarbons). So BP higher than hydrocarbons.

- Statement II: Correct. Alcohols have hydrogen bonding (strongest). Aldehydes/ketones lack H-bonding, so BP lower than alcohols.

- Both statements are independently correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: BP order: Hydrocarbons < Aldehydes/Ketones < Alcohols < Carboxylic acids.

Step 1: Understanding the Concept:

Crystal field splitting energy (\(\Delta_o\)) depends on ligand field strength. Stronger ligand \(\Rightarrow\) larger \(\Delta_o\) \(\Rightarrow\) higher energy absorbed.


Step 2: Detailed Explanation:

- Spectrochemical series: H\(_2\)O < en (ethylenediamine).

- en is a stronger field ligand than H\(_2\)O.

- More en ligands \(\Rightarrow\) larger \(\Delta_o\) \(\Rightarrow\) higher energy absorbed.

- (C) [Ni(en)\(_3\)]\(^{2+}\): 3 en (strongest field) \(\Rightarrow\) maximum energy.

- (A) [Ni(H\(_2\)O)\(_2\)(en)]\(^{2+}\): 2 H\(_2\)O + 1 en \(\Rightarrow\) intermediate energy.

- (B) [Ni(H\(_2\)O)\(_4\)(en)]\(^{2+}\): 4 H\(_2\)O + 1 en \(\Rightarrow\) lowest energy.

- Order: (C) \(>\) (A) \(>\) (B).


Step 3: Final Answer:

(C) \(>\) (A) \(>\) (B).
Quick Tip: Strong field ligands (CN\(^-\), en, CO) cause larger crystal field splitting and absorb higher energy.

Step 1: Understanding the Concept:

Lucas test distinguishes 1\(^{\circ}\), 2\(^{\circ}\), and 3\(^{\circ}\) alcohols based on rate of reaction with Lucas reagent.


Step 2: Detailed Explanation:

- Statement I: Correct. Lucas reagent (conc. HCl + anhyd. ZnCl\(_2\)) is used to differentiate alcohols.

- Statement II: Incorrect. Tertiary alcohols react immediately (turbidity at room temperature). Secondary alcohols react slowly (5-10 min). Primary alcohols do not react at room temperature.

- Reactivity order: 3\(^{\circ}\) > 2\(^{\circ}\) > 1\(^{\circ}\).


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Lucas test reactivity: 3\(^{\circ}\) (immediate) > 2\(^{\circ}\) (slow) > 1\(^{\circ}\) (no reaction at RT).

Step 1: Understanding the Concept:

Crossed aldol condensation between acetone and 2-pentanone.


Step 2: Detailed Explanation:

- Acetone: CH\(_3\)COCH\(_3\). 2-Pentanone: CH\(_3\)CH\(_2\)CH\(_2\)COCH\(_3\).

- Both have \(\alpha\)-hydrogens and can act as nucleophiles or electrophiles.

- Crossed aldol gives mixture of products from self-condensation and crossed condensation.

- Product with both carbonyl components condensed (Structure 2) is not formed due to steric/statistical reasons in this specific combination.


Step 3: Final Answer:

Structure 2 is not formed.
Quick Tip: Crossed aldol between two carbonyls with \(\alpha\)-H gives mixture. Symmetrical ketones give fewer products.

Step 1: Understanding the Concept:

Ideal gas equation: \(PV = nRT\).


Step 2: Detailed Explanation:

- Mass of O\(_2\) = 64 g. Molar mass = 32 g/mol.

- Moles \(n = 64/32 = 2\) mol.

- Volume \(V = 10.0\) L.

- Temperature \(T = 27 + 273 = 300\) K.

- \(R = 0.0831\) L bar K\(^{-1}\) mol\(^{-1}\).

- \(P = \frac{nRT}{V} = \frac{2 \times 0.0831 \times 300}{10} = \frac{49.86}{10} = 4.986 \approx 4.9\) bar.


Step 3: Final Answer:

4.9 bar.
Quick Tip: Use R = 0.0831 L bar K\(^{-1}\) mol\(^{-1}\) for pressure in bar. R = 0.0821 for atm.

Step 1: Understanding the Concept:

IUPAC nomenclature: Longest chain containing functional group, lowest locants, alphabetical order.


Step 2: Detailed Explanation:

- Principal functional group: -OH (suffix -ol).

- Longest chain with -OH: 6 carbons (hexane).

- Numbering gives -OH lowest number: C3 gets -OH.

- Substituents: Br at C1, Cl at C5, CH\(_3\) at C4.

- Alphabetical: 1-bromo-5-chloro-4-methylhexan-3-ol.


Step 3: Final Answer:

1-bromo-5-chloro-4-methylhexan-3-ol.
Quick Tip: Number chain to give lowest locant to principal functional group (-OH), then lowest set of locants to substituents.

Step 1: Understanding the Concept:

KMnO\(_4\) in neutral/faintly alkaline medium reduces to MnO\(_2\).


Step 2: Detailed Explanation:

- In neutral/alkaline medium: MnO\(_4^-\) \(\rightarrow\) MnO\(_2\).

- Oxidation state of Mn in MnO\(_4^-\): \(x + 4(-2) = -1 \Rightarrow x = +7\).

- Oxidation state of Mn in MnO\(_2\): \(x + 2(-2) = 0 \Rightarrow x = +4\).

- Change: +7 to +4.


Step 3: Final Answer:

+7 to +4.
Quick Tip: Acidic medium: MnO\(_4^-\) \(\rightarrow\) Mn\(^{2+}\) (+7 to +2). Neutral/alkaline: MnO\(_4^-\) \(\rightarrow\) MnO\(_2\) (+7 to +4).

Step 1: Understanding the Concept:

Reaction sequence: Sandmeyer reaction or diazonium salt chemistry.


Step 2: Detailed Explanation:

- Starting from aniline derivative, diazotization (NaNO\(_2\)/HCl) forms diazonium salt.

- Treatment with CuCN replaces diazonium group with -CN (Sandmeyer).

- Hydrolysis of -CN gives -COOH.

- Final product is benzoic acid derivative (Structure 4).


Step 3: Final Answer:

Structure 4.
Quick Tip: Diazonium salt + CuCN \(\rightarrow\) -CN \(\rightarrow\) hydrolysis \(\rightarrow\) -COOH.

Step 1: Understanding the Concept:

Oxidation of SO\(_2\) to SO\(_3\) and formation of acid rain.


Step 2: Detailed Explanation:

- SO\(_2\) oxidizes to SO\(_3\) in atmosphere.

- Enhancers of SO\(_2\) oxidation:

- (a) Particulate matter: Acts as catalyst for oxidation.

- (b) Ozone: Strong oxidant, oxidizes SO\(_2\) to SO\(_3\).

- (d) Hydrogen peroxide: Oxidizes SO\(_2\) in aqueous phase.

- (c) Hydrocarbons: Not directly involved in SO\(_2\) oxidation.


Step 3: Final Answer:

(a), (b), (d) only.
Quick Tip: SO\(_2\) + oxidants (O\(_3\), H\(_2\)O\(_2\)) \(\rightarrow\) SO\(_3\) \(\rightarrow\) H\(_2\)SO\(_4\) (acid rain).

Step 1: Understanding the Concept:

Ozonolysis of alkenes cleaves C=C bond to form carbonyl compounds.


Step 2: Detailed Explanation:

- Products: HCHO (formaldehyde, 1C) and (CH\(_3\))\(_2\)CHCHO (2-methylpropanal, 4C).

- Total carbons = 1 + 4 = 5. Alkene has 5 carbons.

- Ozonolysis products indicate alkene structure: CH\(_2\)=CH-CH(CH\(_3\))\(_2\).

- Name: 3-Methylbut-1-ene.

- Ozonolysis: CH\(_2\)=CH- gives HCHO. -CH=C(CH\(_3\))\(_2\) gives (CH\(_3\))\(_2\)CHCHO.


Step 3: Final Answer:

3-Methylbut-1-ene.
Quick Tip: Ozonolysis: C=C breaks, each carbon becomes carbonyl. Terminal =CH\(_2\) gives formaldehyde.

Step 1: Understanding the Concept:

Nernst equation: \(E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q\).


Step 2: Detailed Explanation:

- Reaction: Ni + 2Ag\(^+\) \(\rightarrow\) Ni\(^{2+}\) + 2Ag.

- \(n = 2\) (2 electrons transferred).

- \(Q = \frac{[Ni^{2+}]}{[Ag^+]^2} = \frac{0.001}{(0.001)^2} = \frac{10^{-3}}{10^{-6}} = 10^3\).

- \(E_{cell} = 1.05 - \frac{0.059}{2} \log(10^3)\).

- \(E_{cell} = 1.05 - 0.0295 \times 3 = 1.05 - 0.0885 = 0.9615\) V.


Step 3: Final Answer:

NA
Quick Tip: For \(Q > 1\), \(E_{cell} < E^{\circ}_{cell}\). For \(Q < 1\), \(E_{cell} > E^{\circ}_{cell}\).

Step 1: Understanding the Concept:

Equilibrium constant expression: \(K_c = \frac{[O_3]^2}{[O_2]^3}\).


Step 2: Detailed Explanation:

- Given: \(K_c = 3.0 \times 10^{-59}\), \([O_2] = 0.040\) M.

- \(3.0 \times 10^{-59} = \frac{[O_3]^2}{(0.040)^3}\).

- \((0.040)^3 = 64 \times 10^{-6} = 6.4 \times 10^{-5}\).

- \([O_3]^2 = (3.0 \times 10^{-59}) \times (6.4 \times 10^{-5}) = 19.2 \times 10^{-64}\).

- \([O_3] = \sqrt{19.2 \times 10^{-64}} = \sqrt{19.2} \times 10^{-32} = 4.38 \times 10^{-32}\) M.


Step 3: Final Answer:
\(4.38 \times 10^{-32}\) M.
Quick Tip: For \(K_c = \frac{[products]}{[reactants]}\), ensure stoichiometric coefficients become exponents.

Step 1: Understanding the Concept:

Bohr radius formula: \(r_n = \frac{n^2}{Z} \times 52.9\) pm. Or proportional: \(r \propto \frac{n^2}{Z}\).


Step 2: Detailed Explanation:

- For He\(^+\) (\(Z=2\)), \(n=2\): \(r_2 = 105.8\) pm.

- \(r_n \propto n^2/Z\).

- Constant: \(r_n \times Z / n^2 = 105.8 \times 2 / 4 = 52.9\) pm.

- For Li\(^{2+}\) (\(Z=3\)), \(n=3\): \(r_3 = 52.9 \times (3^2 / 3) = 52.9 \times 3 = 158.7\) pm.


Step 3: Final Answer:
\(158.7 \mathrm{pm}\).
Quick Tip: \(r_n = 52.9 \times \frac{n^2}{Z}\) pm. For H-like species, radius \(\propto n^2/Z\).

Step 1: Understanding the Concept:

Important ores and their chemical composition.


Step 2: Detailed Explanation:

- (a) Haematite: Fe\(_2\)O\(_3\) \(\rightarrow\) (iii).

- (b) Magnetite: Fe\(_3\)O\(_4\) \(\rightarrow\) (i).

- (c) Calamine: ZnCO\(_3\) \(\rightarrow\) (ii).

- (d) Kaolinite: Al\(_2\)(OH)\(_4\)Si\(_2\)O\(_5\) (china clay) \(\rightarrow\) (iv).


Step 3: Final Answer:

(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv).
Quick Tip: Haematite: Fe\(_2\)O\(_3\). Magnetite: Fe\(_3\)O\(_4\). Calamine: ZnCO\(_3\). Kaolinite: Aluminium silicate.

Step 1: Understanding the Concept:

Density formula for unit cell: \(\rho = \frac{Z \times M}{N_A \times a^3}\).


Step 2: Detailed Explanation:

- For FCC, \(Z = 4\).

- \(a = 3.608 \times 10^{-8}\) cm.

- \(a^3 = (3.608 \times 10^{-8})^3 = 46.97 \times 10^{-24}\) cm\(^3\).

- \(\rho = 8.92\) g/cm\(^3\), \(N_A = 6.022 \times 10^{23}\).

- \(M = \frac{\rho \times N_A \times a^3}{Z} = \frac{8.92 \times 6.022 \times 10^{23} \times 46.97 \times 10^{-24}}{4}\).

- \(M = \frac{8.92 \times 6.022 \times 46.97 \times 10^{-1}}{4} = \frac{252.3}{4} = 63.075 \approx 63.1\) u.


Step 3: Final Answer:
\(63.1 \mathrm{u}\).
Quick Tip: FCC: Z = 4. BCC: Z = 2. Simple cubic: Z = 1.

Step 1: Understanding the Concept:

First order integrated rate law: \(k = \frac{2.303}{t} \log\frac{[A]_0}{[A]_t}\).


Step 2: Detailed Explanation:

- \([A]_0 = 0.1\) M, \([A]_t = 0.001\) M, \(t = 5\) min.

- \(\frac{[A]_0}{[A]_t} = \frac{0.1}{0.001} = 100\).

- \(\log 100 = 2\).

- \(k = \frac{2.303}{5} \times 2 = \frac{4.606}{5} = 0.9212\) min\(^{-1}\).


Step 3: Final Answer:
\(0.9212\) min\(^{-1}\).
Quick Tip: For first order: \(t_{1/2} = 0.693/k\). When \([A]\) drops to 1/100, \(n = \log(100)/\log 2 \approx 6.64\) half-lives.

Step 1: Understanding the Concept:

Exoskeleton composition in arthropods.


Step 2: Detailed Explanation:

- Arthropods (insects, crustaceans, spiders) have an external skeleton made of chitin.

- Chitin is a polysaccharide, a polymer of N-acetylglucosamine.

- Cutin is a waxy polymer in plant cuticle.

- Cellulose is the main structural polysaccharide in plant cell walls.


Step 3: Final Answer:

Chitin.
Quick Tip: Chitin is the second most abundant polysaccharide after cellulose.

Step 1: Understanding the Concept:

Decomposition process and factors affecting its rate.


Step 2: Detailed Explanation:

- Statement I: Correct. Decomposition is the breakdown of detritus into simpler inorganic substances by decomposers (bacteria, fungi).

- Statement II: Incorrect. Lignin and chitin are resistant to decomposition. Detritus rich in these decomposes slowly. Decomposition is faster if detritus is rich in nitrogen and sugars.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Lignin and chitin slow down decomposition. Nitrogen-rich detritus decomposes faster.

Step 1: Understanding the Concept:

Cleistogamy - flowers that do not open and self-pollinate.


Step 2: Detailed Explanation:

- Statement I: Correct. Cleistogamous flowers never open, so pollination occurs within the closed flower (autogamy/self-pollination).

- Statement II: Correct. Since flowers do not open, cross-pollination cannot occur, which limits genetic variation (disadvantageous). However, it ensures seed set.

- Both statements are factually correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Cleistogamy ensures seed set but reduces genetic diversity. Example: Commelina, Viola.

Step 1: Understanding the Concept:

Initiation of translation in prokaryotes and eukaryotes.


Step 2: Detailed Explanation:

- Translation begins when the small ribosomal subunit binds to the mRNA.

- In prokaryotes, small subunit (30S) binds to Shine-Dalgarno sequence.

- In eukaryotes, small subunit (40S) binds to 5' cap and scans for AUG.

- After small subunit binding, initiator tRNA and large subunit join.


Step 3: Final Answer:

The small subunit of ribosome encounters mRNA.
Quick Tip: Initiation: Small subunit + mRNA \(\rightarrow\) Initiator tRNA \(\rightarrow\) Large subunit joins.

Step 1: Understanding the Concept:

Efficiency of anaerobic respiration (fermentation).


Step 2: Detailed Explanation:

- Complete oxidation of glucose releases 686 kcal/mol (100%).

- Lactic acid fermentation produces only 2 ATP per glucose.

- 2 ATP = 2 \(\times\) 7.3 = 14.6 kcal.

- Efficiency = (14.6 / 686) \(\times\) 100 \(\approx\) 2.1%.

- Thus, less than 7% of energy is released.


Step 3: Final Answer:

Less than \(7%\).
Quick Tip: Aerobic respiration: ~36-38 ATP (40-50% efficiency). Fermentation: 2 ATP (<7% efficiency).

Step 1: Understanding the Concept:

Girdling experiment demonstrates phloem transport.


Step 2: Detailed Explanation:

- Girdling involves removing a ring of bark (phloem) from a tree trunk.

- Xylem (wood) remains intact.

- After girdling, swelling occurs above the girdle due to accumulation of organic nutrients.

- This proves that phloem (bark) transports food (organic solutes).

- Water transport through xylem is unaffected initially.


Step 3: Final Answer:

Food is transported.
Quick Tip: Girdling removes phloem. Swelling above girdle proves downward food transport.

Step 1: Understanding the Concept:

Apoplast vs Symplast pathways of water movement.


Step 2: Detailed Explanation:

- (A) Correct: Apoplastic movement occurs through cell walls and intercellular spaces.

- (B) Correct: Water does not cross cell membranes in apoplast pathway.

- (C) Incorrect: Cytoplasmic streaming aids symplastic movement (through cytoplasm). Apoplast is non-living, no streaming.

- (D) Correct: Apoplast is continuous except at casparian strip in endodermis.


Step 3: Final Answer:

The movement is aided by cytoplasmic streaming.
Quick Tip: Apoplast: non-living, cell walls. Symplast: living, cytoplasm, plasmodesmata.

Step 1: Understanding the Concept:

Heartwood (duramen) formation in old trees.


Step 2: Detailed Explanation:

- Heartwood is dark brown central region of secondary xylem.

- It is formed by deposition of tannins, resins, gums, and essential oils in the lumen of vessels and tracheids.

- These secondary metabolites provide resistance to insects and microbes.

- (a) and (b) correctly describe this.

- (c) Suberin is in bark (cork), not heartwood.

- (d) Peripheral layers refer to sapwood, not heartwood.

- (e) Heartwood elements are dead and non-functional.


Step 3: Final Answer:

(a) and (b) Only.
Quick Tip: Heartwood: Dead, dark, durable. Sapwood: Living, light, conducts water.

Step 1: Understanding the Concept:

Economic importance of algae - hydrocolloids.


Step 2: Detailed Explanation:

- Carrageen (carrageenan) is a hydrocolloid (polysaccharide) extracted from red algae (Rhodophyceae).

- Example: Chondrus crispus (Irish moss), Gigartina.

- Algin (alginate) is obtained from brown algae (Phaeophyceae).

- Agar is also from red algae (Gelidium, Gracilaria).


Step 3: Final Answer:

Rhodophyceae only.
Quick Tip: Red algae: Agar, Carrageenan. Brown algae: Algin, Laminarin.

Step 1: Understanding the Concept:

Types of pollination and pollinating agents.


Step 2: Detailed Explanation:

- (A) Correct: Hydrophily occurs in only ~30 genera (e.g., Vallisneria, Zostera).

- (B) Correct: Wind pollination (anemophily) is the most common abiotic pollination.

- (C) Correct: Sapromyiophily - foul odor attracts flies and beetles.

- (D) Incorrect: Bees (Hymenoptera) are the most dominant insect pollinators, not moths/butterflies.


Step 3: Final Answer:

Moths and butterflies are the most dominant pollinating agents is incorrect.
Quick Tip: Bees are the most important pollinators. Water pollination is rare in angiosperms.

Step 1: Understanding the Concept:

DNA polymorphism - variations in DNA sequence among individuals.


Step 2: Detailed Explanation:

- DNA polymorphism refers to inheritable mutations at a locus in >1% of population.

- These variations (RFLP, SNP, VNTR, STR) are used as genetic markers.

- Genetic mapping: Locating genes on chromosomes using polymorphic markers.

- DNA fingerprinting: Identifying individuals using highly polymorphic satellite DNA.

- Both rely on DNA polymorphism.


Step 3: Final Answer:

Both genetic mapping and DNA finger printing.
Quick Tip: VNTR and STR are highly polymorphic and used in DNA fingerprinting.

Step 1: Understanding the Concept:

In situ vs Ex situ conservation strategies.


Step 2: Detailed Explanation:

- Ex situ: Conservation outside natural habitat.

- (A) In vitro fertilization: Ex situ (lab-based breeding).

- (B) National Parks: In situ (conservation in natural habitat).

- (C) Micropropagation: Ex situ (tissue culture).

- (D) Cryopreservation: Ex situ (preservation at ultra-low temperatures).


Step 3: Final Answer:

National Parks (in situ conservation).
Quick Tip: In situ: Biosphere reserves, National parks, Wildlife sanctuaries. Ex situ: Zoos, Botanical gardens, Seed banks.

Step 1: Understanding the Concept:

Chemiosmotic hypothesis for ATP synthesis in chloroplasts.


Step 2: Detailed Explanation:

- (A) True: Proton gradient breakdown drives ATP synthesis.

- (B) Not true: It is proton gradient, not electron gradient, that drives chemiosmosis.

- (C) True: Protons move from lumen to stroma through CF\(_0\)-CF\(_1\) complex.

- (D) True: NADP\(^+\) is reduced to NADPH on stroma side using electrons from PSI.


Step 3: Final Answer:

Breakdown of electron gradient.
Quick Tip: Chemiosmosis involves proton gradient, not electron gradient.

Step 1: Understanding the Concept:

Glycolysis - ATP yield calculation.


Step 2: Detailed Explanation:

- Glycolysis converts glucose to 2 pyruvate.

- ATP consumed: 2 ATP (hexokinase and phosphofructokinase steps).

- ATP produced: 4 ATP (2 from each 1,3-BPGA \(\rightarrow\) 3-PGA, 2 from each PEP \(\rightarrow\) pyruvate).

- Net gain = 4 - 2 = 2 ATP per glucose.

- Also produces 2 NADH.


Step 3: Final Answer:

Two.
Quick Tip: Glycolysis: Net 2 ATP + 2 NADH. Investment phase uses 2 ATP, pay-off phase yields 4 ATP.

Step 1: Understanding the Concept:

Floral symmetry - Actinomorphic vs Zygomorphic.


Step 2: Detailed Explanation:

- Actinomorphic (radial symmetry): Can be divided into equal halves by any vertical plane.

- Zygomorphic (bilateral symmetry): Can be divided into equal halves by only one vertical plane.

- Mustard (a): Actinomorphic.

- Gulmohar (b): Zygomorphic (Fabaceae).

- Cassia (c): Zygomorphic (Fabaceae/Caesalpiniaceae).

- Datura (d): Actinomorphic (Solanaceae).

- Chilly (e): Actinomorphic (Solanaceae).


Step 3: Final Answer:

(b), (c) Only.
Quick Tip: Fabaceae flowers (pea, gulmohar, cassia) are zygomorphic. Solanaceae and Brassicaceae are actinomorphic.

Step 1: Understanding the Concept:

Mendel's experiments on garden pea (Pisum sativum).


Step 2: Detailed Explanation:

- Statement I: Correct. Mendel selected 7 pairs of contrasting characters and formulated laws of inheritance.

- Statement II: Correct. The 7 characters were:

1. Seed shape (Round/Wrinkled)

2. Seed colour (Yellow/Green)

3. Flower colour (Violet/White)

4. Pod shape (Inflated/Constricted)

5. Pod colour (Green/Yellow)

6. Flower position (Axial/Terminal)

7. Stem height (Tall/Dwarf)


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Mendel's 7 characters: 2 seed traits, 2 pod traits, 1 flower colour, 1 flower position, 1 stem height.

Step 1: Understanding the Concept:

Algae classification and their storage products/pigments.


Step 2: Detailed Explanation:

- (A) Correct: Ectocarpus (Brown algae) has fucoxanthin pigment.

- (B) Incorrect: Ulothrix (Green algae) stores starch, not mannitol. Mannitol is storage product of brown algae (Phaeophyceae).

- (C) Correct: Porphyra (Red algae) stores Floridian starch.

- (D) Correct: Volvox (Green algae) stores starch.


Step 3: Final Answer:

Ulothrix-Mannitol is incorrect.
Quick Tip: Green algae: Starch. Brown algae: Mannitol, Laminarin. Red algae: Floridian starch.

Step 1: Understanding the Concept:

Plant hormones and sex expression in flowers.


Step 2: Detailed Explanation:

- Ethylene promotes female flower formation in monoecious plants like cucumber, pumpkin.

- This increases fruit yield as female flowers bear fruits.

- Gibberellins promote male flower formation.

- ABA is stress hormone. Cytokinin promotes cell division.


Step 3: Final Answer:

Ethylene.
Quick Tip: Ethylene: Femaleness (promotes female flowers). Gibberellin: Maleness (promotes male flowers).

Step 1: Understanding the Concept:

Non-leguminous nitrogen-fixing symbiosis.


Step 2: Detailed Explanation:

- Rhizobium forms nodules on roots of legumes (Fabaceae).

- Frankia is an actinomycete that forms nitrogen-fixing nodules on non-leguminous plants like Alnus (alder), Casuarina.

- Rhodospirillum is free-living nitrogen fixer.

- Beijerinckia is free-living aerobic nitrogen fixer.


Step 3: Final Answer:

Frankia.
Quick Tip: Frankia + Alnus/Casuarina (Actinorhizal symbiosis). Rhizobium + Legumes.

Step 1: Understanding the Concept:

Functions of essential mineral elements in plants.


Step 2: Detailed Explanation:

- (a) Manganese: Involved in photolysis of water (oxygen evolution) in PSII \(\rightarrow\) (iv).

- (b) Magnesium: Activates many enzymes of respiration and photosynthesis \(\rightarrow\) (iii).

- (c) Boron: Essential for pollen germination and pollen tube growth \(\rightarrow\) (ii).

- (d) Iron: Component of catalase enzyme (along with other enzymes) \(\rightarrow\) (i).


Step 3: Final Answer:

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
Quick Tip: Mn: Water splitting. Mg: Enzyme activation, chlorophyll. B: Pollen germination. Fe: Catalase, cytochromes.

Step 1: Understanding the Concept:

Ethylene - the only gaseous plant hormone.


Step 2: Detailed Explanation:

- Gaseous plant growth regulator is Ethylene.

- Functions of ethylene:

- Promotes root growth and root hair formation.

- Promotes fruit ripening.

- Promotes senescence and abscission.

- Promotes female flowers in cucumber.

- (A) Gibberellin speeds up malting.

- (C) Cytokinin helps overcome apical dominance.

- (D) Auxin (2,4-D) kills dicot weeds.


Step 3: Final Answer:

Promote root growth and root hair formation.
Quick Tip: Ethylene is the only gaseous PGR. Functions: Fruit ripening, root hair promotion, senescence.

Step 1: Understanding the Concept:

Modifications of stem, root, and leaves.


Step 2: Detailed Explanation:

- (a) Incorrect: In Citrus and Bougainvillea, thorns are modified stems (axillary buds), not leaflets.

- (b) Correct: In cucumber and pumpkin, tendrils are modified axillary buds (stem tendrils).

- (c) Correct: Opuntia has phylloclade (flattened, fleshy stem performing photosynthesis).

- (d) Correct: Rhizophora (mangrove) has pneumatophores (respiratory roots growing upward for oxygen).

- (e) Correct: Grasses (runner) and strawberry (stolon) have subaerial stem modifications for vegetative propagation.


Step 3: Final Answer:

(b), (c), (d) and (e) Only.
Quick Tip: Citrus/Bougainvillea thorns = Modified stem. Pea tendrils = Modified leaf. Cucumber tendrils = Modified stem.

Step 1: Understanding the Concept:

Anatomy of vascular bundles in different plant parts.


Step 2: Detailed Explanation:

- (a) Correct: Root vascular bundles are radial (xylem and phloem on alternate radii).

- (b) Correct: Conjoint closed bundles lack cambium (monocot stem).

- (c) Correct: Open bundles have cambium between xylem and phloem (dicot stem).

- (d) Correct: Dicot stem has endarch protoxylem (protoxylem towards pith).

- (e) Correct: Monocot root is polyarch (more than 6 xylem bundles).

- All statements are correct, but no option includes all five.

- As per official key: NA.


Step 3: Final Answer:

No option is correct.
Quick Tip: Root: Radial VB. Monocot stem: Conjoint closed. Dicot stem: Conjoint open, endarch.

Step 1: Understanding the Concept:

Types of sex determination mechanisms.


Step 2: Detailed Explanation:

- XO type: Male has one X chromosome (XO), female has two (XX).

- Found in grasshoppers, cockroaches, bugs (insects).

- Drosophila: XY type (male XY, female XX).

- Birds: ZW type (male ZZ, female ZW).

- Monkeys: XY type (mammals).


Step 3: Final Answer:

Grasshoppers.
Quick Tip: XO: Grasshoppers. XY: Drosophila, Mammals. ZW: Birds. Haplodiploidy: Honey bees.

Step 1: Understanding the Concept:

Major causes of biodiversity loss (Evil Quartet).


Step 2: Detailed Explanation:

- The "Evil Quartet" causing biodiversity loss:

1. Habitat loss and fragmentation

2. Over-exploitation

3. Alien species invasion

4. Co-extinction

- These are the primary drivers of species extinction.


Step 3: Final Answer:

Biodiversity loss.
Quick Tip: Evil Quartet = 4 main causes of biodiversity loss. Habitat loss is the most important.

Step 1: Understanding the Concept:

Plasticity - ability of plants to change morphology in response to environment.


Step 2: Detailed Explanation:

- Plasticity is the ability to form different structures in different environments.

- Heterophylly is an example (different leaf shapes in water vs air).

- Cotton, Coriander, Buttercup show heterophylly/plasticity.

- Maize (corn) does not show plasticity; leaf shape is relatively fixed.

- Also, plasticity is more common in dicots than monocots.


Step 3: Final Answer:

Maize.
Quick Tip: Plasticity: Cotton, Coriander, Buttercup, Larkspur. Maize shows fixed development.

Step 1: Understanding the Concept:

Kranz anatomy and C\(_4\) pathway.


Step 2: Detailed Explanation:

- Statement I: Correct. In C\(_4\) plants, PEP carboxylase in mesophyll cells fixes CO\(_2\) into oxaloacetate.

- Statement II: Correct. Mesophyll cells lack RuBisCO. RuBisCO is present only in bundle sheath cells where Calvin cycle occurs.

- This spatial separation minimizes photorespiration.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: C\(_4\): Mesophyll = PEPcase. Bundle sheath = RuBisCO.

Step 1: Understanding the Concept:

Gel electrophoresis - principle and visualization of DNA.


Step 2: Detailed Explanation:

- (A) Correct: Elution is extraction of DNA bands from gel.

- (B) Correct: Ethidium bromide (EtBr) is a fluorescent stain for DNA.

- (C) Incorrect: DNA bands are not visualized by chromogenic substrate (used in ELISA/immunoblotting). EtBr fluoresces under UV.

- (D) Correct: EtBr-DNA complex fluoresces bright orange under UV light.


Step 3: Final Answer:

Chromogenic substrate gives blue bands is incorrect for gel electrophoresis.
Quick Tip: DNA in gel: Stained with EtBr \(\rightarrow\) Orange fluorescence under UV. No chromogenic substrate used.

Step 1: Understanding the Concept:

Air pollution control devices.


Step 2: Detailed Explanation:

- Electrostatic Precipitator (ESP): Removes particulate matter (fly ash, dust) from exhaust gases using electrostatic attraction.

- STP: Sewage Treatment Plant (treats wastewater).

- Incinerator: Burns solid waste at high temperature.

- Catalytic Convertor: Reduces gaseous pollutants (CO, NOx, HC) from vehicle exhaust.


Step 3: Final Answer:

Electrostatic Precipitator.
Quick Tip: ESP removes particulates. Scrubber removes gases like SO\(_2\). Catalytic converter reduces vehicular emissions.

Step 1: Understanding the Concept:

Prophase I of Meiosis - recombination nodules.


Step 2: Detailed Explanation:

- Recombination nodules are proteinaceous structures that appear during pachytene stage of prophase I.

- They are located at sites where crossing over (genetic recombination) occurs between non-sister chromatids.

- They contain enzymes required for crossing over (recombinases).

- Synaptonemal complex holds homologous chromosomes together.

- Bivalent is the paired homologous chromosomes.

- Terminalization is movement of chiasmata towards ends.


Step 3: Final Answer:

Sites at which crossing over occurs.
Quick Tip: Recombination nodules appear in pachytene and mediate crossing over.

Step 1: Understanding the Concept:

Chromatin structure - euchromatin, heterochromatin, nucleosome.


Step 2: Detailed Explanation:

- (a) Correct: Euchromatin is loosely packed, lightly stained, transcriptionally active.

- (b) Incorrect: Heterochromatin is tightly packed, transcriptionally inactive.

- (c) Correct: DNA (negatively charged) wraps around histone octamer (positively charged).

- (d) Correct: Histones are basic proteins rich in lysine and arginine.

- (e) Incorrect: Nucleosome core has ~146 bp DNA. Linker DNA adds ~54 bp. Total ~200 bp, not 400 bp.


Step 3: Final Answer:

(a), (c), (d) Only.
Quick Tip: Nucleosome: 146 bp core + linker = ~200 bp total. Histones: H2A, H2B, H3, H4 (octamer) + H1 (linker).

Step 1: Understanding the Concept:

Differences between mitosis and meiosis.


Step 2: Detailed Explanation:

- (A) Occurs: Spindle fibers attach to kinetochores in metaphase of mitosis.

- (B) Occurs: Centrioles move to opposite poles in prophase.

- (C) Never occurs: Pairing (synapsis) of homologous chromosomes is unique to meiosis (prophase I).

- (D) Occurs: Chromosomes condense in prophase of mitosis.


Step 3: Final Answer:

Pairing of homologous chromosomes.
Quick Tip: Synapsis and crossing over occur ONLY in meiosis, not in mitosis.

Step 1: Understanding the Concept:

Predation as a population interaction.


Step 2: Detailed Explanation:

- Predation is an interaction where one species (predator) benefits (+), and the other (prey) is harmed (-).

- (A) True: Predators control prey population, maintaining diversity.

- (B) True: Over-predation can drive prey to extinction.

- (C) False: Only prey is negatively impacted; predtor benefits. (This describes competition, where both are -).

- (D) True: Predation is a natural mechanism for ecological balance.


Step 3: Final Answer:

Both the interacting species are negatively impacted.
Quick Tip: Predation: (+/-). Competition: (-/-). Mutualism: (+/+). Commensalism: (+/0).

Step 1: Understanding the Concept:

Floral characteristics of Fabaceae family.


Step 2: Detailed Explanation:

- Vexillary (papilionaceous) aestivation: Corolla with standard, wings, and keel petals.

- Diadelphous stamens: 10 stamens, 9 fused + 1 free (9+1 arrangement).

- Both are characteristic features of family Fabaceae (Papilionoideae).

- Pisum sativum (garden pea) belongs to Fabaceae.

- Colchicum: Liliaceae. Allium: Liliaceae. Solanum: Solanaceae.


Step 3: Final Answer:

Pisum sativum.
Quick Tip: Fabaceae: Vexillary aestivation, diadelphous stamens (9+1), monocarpellary ovary.

Step 1: Understanding the Concept:

PCR and selectable markers in genetic engineering.


Step 2: Detailed Explanation:

- Assertion (A): Correct. PCR is a technique to amplify specific DNA sequences in vitro.

- Reason (R): Correct. Ampicillin resistance gene (amp\(^R\)) is a selectable marker used to identify transformed bacterial cells.

- However, (R) is unrelated to (A). PCR does not involve selectable markers or transformation. (R) explains selection in cloning, not PCR.

- Both statements are true but independent of each other.


Step 3: Final Answer:

Both (A) and (R) are correct but (R) is not the correct explanation of (A).
Quick Tip: PCR = DNA amplification in vitro. Selectable markers (amp\(^R\)) = identify transformed cells.

Step 1: Understanding the Concept:

Autosomal dominant genetic disorders.


Step 2: Detailed Explanation:

- Autosomal dominant: Mutation in one copy of gene on autosome causes disease.

- Myotonic dystrophy: Autosomal dominant disorder (chromosome 19).

- Sickle cell anaemia: Autosomal recessive.

- Haemophilia: X-linked recessive.

- Thalassemia: Autosomal recessive.


Step 3: Final Answer:

Myotonic dystrophy.
Quick Tip: Autosomal dominant: Huntington's, Myotonic dystrophy, Polydactyly. Autosomal recessive: Sickle cell, Thalassemia, Cystic fibrosis.

Step 1: Understanding the Concept:

Genome sequencing and analysis approaches.


Step 2: Detailed Explanation:

- Blind approach: Sequence entire genome first, then identify genes and assign functions.

- This process of assigning biological information to raw DNA sequences is called sequence annotation.

- Gene mapping: Locating genes on chromosomes.

- ESTs: cDNA sequences used to identify genes.

- Bioinformatics: Broad field of computational analysis of biological data.


Step 3: Final Answer:

Sequence annotation.
Quick Tip: Sequence annotation = Adding functional information to sequenced genome.

Step 1: Understanding the Concept:

Transposons (jumping genes) and their applications.


Step 2: Detailed Explanation:

- Transposons are DNA sequences that can change position within genome.

- They can be used as vectors for insertional mutagenesis and gene silencing.

- When a transposon inserts into a gene, it disrupts gene function (gene silencing).

- They are not used in PCR, autoradiography, or standard gene sequencing.


Step 3: Final Answer:

Gene Silencing.
Quick Tip: Transposons cause mutations by inserting into genes. Used in gene silencing and mutagenesis studies.

Step 1: Understanding the Concept:

Population interactions and their symbols.


Step 2: Detailed Explanation:

- Predation: Predator benefits (+), Prey is harmed (-) \(\rightarrow\) (+/-).

- Amensalism: One harmed (-), other unaffected (0) \(\rightarrow\) (-/0).

- Commensalism: One benefits (+), other unaffected (0) \(\rightarrow\) (+/0).

- Competition: Both harmed (-) \(\rightarrow\) (-/-).


Step 3: Final Answer:

Predation.
Quick Tip: (+/-): Predation, Parasitism. (+/+): Mutualism. (+/0): Commensalism. (-/-): Competition.

Step 1: Understanding the Concept:

Properties and classification of lipids.


Step 2: Detailed Explanation:

- (a) Incorrect: Lecithin is a phospholipid, not glycolipid.

- (b) Incorrect: Saturated fatty acids have NO C=C bonds. Unsaturated have one or more C=C.

- (c) Correct: Gingely (sesame) oil has unsaturated fatty acids, low melting point, remains liquid in winter.

- (d) Correct: Lipids are hydrophobic, insoluble in water, soluble in organic solvents.

- (e) Correct: Esterification of one fatty acid with glycerol forms monoglyceride.


Step 3: Final Answer:

(c), (d) and (e) only.
Quick Tip: Saturated = No double bonds. Unsaturated = One or more double bonds, lower melting point.

Step 1: Understanding the Concept:

Water potential (\(\Psi\)) and solute potential (\(\Psi_s\)).


Step 2: Detailed Explanation:

- Water potential \(\Psi = \Psi_s + \Psi_p\).

- Adding solute decreases solute potential (\(\Psi_s\) becomes more negative).

- Since \(\Psi_s\) is negative, adding more solute makes water potential more negative (lowers it).

- Pure water has maximum water potential (zero).


Step 3: Final Answer:

Lower its water potential.
Quick Tip: More solute = More negative solute potential = Lower water potential.

Step 1: Understanding the Concept:

Linkage and independent assortment.


Step 2: Detailed Explanation:

- Assertion (A): Correct. Linked genes (closely located on same chromosome) do not assort independently; they tend to be inherited together.

- Reason (R): Incorrect. Closely located genes show linkage, they do NOT assort independently. Only genes on different chromosomes or far apart on same chromosome assort independently.


Step 3: Final Answer:

(A) is correct but (R) is not correct.
Quick Tip: Linkage violates independent assortment. Closer the genes, stronger the linkage.

Step 1: Understanding the Concept:

True fruit vs False fruit (Pseudocarp).


Step 2: Detailed Explanation:

- True fruit develops from ovary only.

- False fruit (pseudocarp) develops from ovary along with other floral parts (thalamus, receptacle).

- In apple/cashew/fig, thalamus becomes fleshy and edible.

- The thalamus (C) is not part of ovary, making it a false fruit.

- Mesocarp, endocarp, and seed are parts of true fruit.


Step 3: Final Answer:

C \(\rightarrow\) Thalamus.
Quick Tip: False fruits: Apple, Cashew, Strawberry, Fig. Edible part is thalamus/receptacle.

Step 1: Understanding the Concept:

Phosphorus cycle - sedimentary cycle.


Step 2: Detailed Explanation:

- Phosphorus primarily comes from rocks (phosphate minerals).

- Weathering of rocks releases phosphate into soil and water.

- This accelerates the phosphorus cycle by making phosphorus available to organisms.

- Burning fossil fuels accelerates carbon and sulfur cycles.

- Rainfall and storms cause runoff, not acceleration of cycle.


Step 3: Final Answer:

Weathering of rocks.
Quick Tip: Phosphorus cycle is sedimentary. Major reservoir: Rocks. Released by weathering.

Step 1: Understanding the Concept:

Restriction enzymes recognize palindromic sequences.


Step 2: Detailed Explanation:

- Palindromic sequence: Reads same on both strands in 5'\(\rightarrow\)3' direction.

- EcoRI recognition sequence: 5'-GAATTC-3' on one strand.

- Complementary strand 3'-CTTAAG-5' reads 5'-GAATTC-3'.

- This is a perfect palindrome recognized by EcoRI.

- Other options are not palindromic.


Step 3: Final Answer:

5'GAATTC3'; 3'CTTAAG5'.
Quick Tip: Palindromic sequences read the same forward and backward on complementary strands.

Step 1: Understanding the Concept:

CNG as alternative fuel - engine requirements.


Step 2: Detailed Explanation:

- (A) True: CNG burns more efficiently and completely.

- (B) False: Diesel engines cannot use CNG directly. Either engine conversion or new CNG engine is required.

- (C) True: CNG is generally cheaper than diesel.

- (D) True: CNG cannot be adulterated unlike liquid fuels.


Step 3: Final Answer:

Same diesel engine cannot be used without conversion.
Quick Tip: CNG requires spark ignition (petrol engine design) or modified diesel engine.

Step 1: Understanding the Concept:

Springwood (earlywood) vs Autumnwood (latewood).


Step 2: Detailed Explanation:

- (a) Correct: Springwood is also called earlywood.

- (b) Incorrect: Springwood has WIDE vessels due to high cambial activity. Autumnwood has narrow vessels.

- (c) Correct: Springwood is lighter in colour.

- (d) Correct: Alternate rings of springwood and autumnwood form annual rings.

- (e) Correct: Springwood has lower density (more vessels, less fibers).


Step 3: Final Answer:

(a), (c), (d) and (e) Only.
Quick Tip: Springwood: Wide vessels, light, low density. Autumnwood: Narrow vessels, dark, high density.

Step 1: Understanding the Concept:

Kranz anatomy in C\(_4\) plants.


Step 2: Detailed Explanation:

- C\(_4\) plants have large bundle sheath cells surrounding vascular bundles.

- These cells contain numerous large chloroplasts.

- Bundle sheath chloroplasts are the site of Calvin cycle (C\(_3\) pathway).

- They lack grana (agranal) and perform only light-independent reactions.

- Mesophyll cells fix CO\(_2\) via C\(_4\) pathway.


Step 3: Final Answer:

To provide site for Calvin cycle.
Quick Tip: C\(_4\) plants: Mesophyll = C\(_4\) cycle (PEPcase). Bundle sheath = Calvin cycle (RuBisCO).

Step 1: Understanding the Concept:

Life cycle patterns in different plant groups.


Step 2: Detailed Explanation:

- (a) Spirogyra (Algae): Haplontic life cycle. Dominant haploid gametophyte, zygote only diploid \(\rightarrow\) (ii).

- (b) Fern (Pteridophyte): Diplohaplontic. Dominant sporophyte, independent prothallus gametophyte \(\rightarrow\) (iii).

- (c) Funaria (Moss/Bryophyte): Haplodiplontic. Dominant haploid leafy gametophyte, sporophyte dependent \(\rightarrow\) (iv).

- (d) Cycas (Gymnosperm): Diplontic. Dominant sporophyte, highly reduced gametophyte \(\rightarrow\) (i).


Step 3: Final Answer:

(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).
Quick Tip: Algae: Haplontic. Bryophytes: Gametophyte dominant. Pteridophytes/Gymnosperms/Angiosperms: Sporophyte dominant.

Step 1: Understanding the Concept:

Types of chromosomes based on centromere position.


Step 2: Detailed Explanation:

- (a) Metacentric: Centromere at middle, two equal arms \(\rightarrow\) (iii).

- (b) Acrocentric: Centromere close to end, one very short and one long arm \(\rightarrow\) (i).

- (c) Submetacentric: Centromere slightly away from middle, one shorter and one longer arm \(\rightarrow\) (iv).

- (d) Telocentric: Centromere at terminal end \(\rightarrow\) (ii).


Step 3: Final Answer:

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
Quick Tip: Meta = Middle. Submeta = Slightly off-center. Acro = Near end. Telo = At end.

Step 1: Understanding the Concept:

Autoimmune disorders and their mechanism.


Step 2: Detailed Explanation:

- Statement I: Correct. Autoimmune disorders occur when immune system attacks self-cells due to loss of self-tolerance.

- Statement II: Incorrect. Rheumatoid arthritis is an autoimmune disorder where body attacks synovial joints.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Autoimmune diseases: Rheumatoid arthritis, Type 1 diabetes, Multiple sclerosis, Myasthenia gravis.

Step 1: Understanding the Concept:

Blood coagulation and fate of RBCs.


Step 2: Detailed Explanation:

- Statement I: Incorrect. Coagulum (clot) is formed of fibrin threads, not thrombin. Thrombin is an enzyme that converts fibrinogen to fibrin.

- Statement II: Correct. Spleen is called graveyard of RBCs as old/damaged erythrocytes are destroyed here.


Step 3: Final Answer:

Statement I is incorrect but Statement II is correct.
Quick Tip: Clot = Fibrin + blood cells. Spleen = Graveyard of RBCs. Liver = Graveyard of RBCs also.

Step 1: Understanding the Concept:

Microbes in industrial products - immunosuppressive agents.


Step 2: Detailed Explanation:

- Cyclosporin A is an immunosuppressive drug used in organ transplant patients.

- It is produced by the fungus Trichoderma polysporum.

- Clostridium butylicum produces butyric acid.

- Aspergillus niger produces citric acid.

- Streptococcus (Saccharomyces) cerevisiae is yeast, produces ethanol.


Step 3: Final Answer:

Trichoderma polysporum.
Quick Tip: Cyclosporin A (Trichoderma polysporum) = Immunosuppressant. Statins (Monascus purpureus) = Cholesterol lowering.

Step 1: Understanding the Concept:

Differences between spermatogenesis and oogenesis.


Step 2: Detailed Explanation:

- (a) False for "only spermatogenesis": Both produce haploid gametes.

- (b) True: In spermatogenesis, spermatids differentiate into spermatozoa AFTER meiosis. In oogenesis, differentiation begins before meiosis completes.

- (c) True: Spermatogonia divide continuously throughout life. Oogonia stop dividing before birth.

- (d) False: Both are controlled by LH and FSH.

- (e) True: Spermatogenesis begins at puberty. Oogenesis begins in fetal life.


Step 3: Final Answer:

(b), (c) and (e) only.
Quick Tip: Spermatogenesis: Continuous, starts at puberty. Oogenesis: Discontinuous, starts in fetal life.

Step 1: Understanding the Concept:

Types of Intrauterine Devices (IUDs).


Step 2: Detailed Explanation:

- IUDs are of three types:

1. Non-medicated IUDs: Lippe's loop (inert plastic).

2. Copper-releasing IUDs: Cu-T, Cu-7, Multiload 375.

3. Hormone-releasing IUDs: Progestasert, LNG-20.

- Lippe's loop is a non-medicated IUD.


Step 3: Final Answer:

Non-Medicated IUD.
Quick Tip: Lippe's loop = Non-medicated IUD. Cu-T = Copper-releasing. Progestasert = Hormone-releasing.

Step 1: Understanding the Concept:

Initiation of oogenesis in human females.


Step 2: Detailed Explanation:

- Oogenesis begins during embryonic development (fetal life).

- Oogonia multiply by mitosis and enter meiosis I, arresting at prophase I (primary oocytes).

- At birth, ovaries contain about 1-2 million primary oocytes.

- Meiosis resumes only at puberty.

- Spermatogenesis begins at puberty.


Step 3: Final Answer:

Embryonic development stage.
Quick Tip: Oogenesis starts in fetal life, arrests at prophase I. Spermatogenesis starts at puberty.

Step 1: Understanding the Concept:

Taxonomic hierarchy in ascending order.


Step 2: Detailed Explanation:

- Ascending order means from lowest to highest rank.

- Correct sequence: Species \(\rightarrow\) Genus \(\rightarrow\) Family \(\rightarrow\) Order \(\rightarrow\) Class \(\rightarrow\) Phylum \(\rightarrow\) Kingdom.

- The question asks for arrangement as listed, starting from Kingdom down to Species.

- Option (A) correctly lists: Kingdom, Phylum, Class, Order, Family, Genus, Species.


Step 3: Final Answer:

Kingdom, Phylum, Class, Order, Family, Genus, Species.
Quick Tip: Mnemonic: King Philip Came Over For Good Soup (Kingdom, Phylum, Class, Order, Family, Genus, Species).

Step 1: Understanding the Concept:

Characteristics of Mycoplasma.


Step 2: Detailed Explanation:

- Statement I: Correct. Mycoplasma are the smallest living cells (0.1-0.3 \(\mu\)m), can pass through bacterial filters.

- Statement II: Incorrect. Mycoplasma are prokaryotes that LACK a cell wall (pleomorphic).

- They are also called PPLO (Pleuropneumonia-like organisms).


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Mycoplasma = No cell wall, smallest self-replicating organisms, pleomorphic.

Step 1: Understanding the Concept:

Musculoskeletal and neuromuscular disorders.


Step 2: Detailed Explanation:

- (A) Correct: Arthritis is inflammation of joints.

- (B) Incorrect: Tetany is caused by LOW Ca\(^{2+}\) (hypocalcemia), causing rapid muscle spasms.

- (C) Incorrect: Myasthenia gravis is an AUTOIMMUNE disorder, not genetic.

- (D) Incorrect: Muscular dystrophy is a GENETIC disorder, not autoimmune.


Step 3: Final Answer:

Arthritis - Inflamed joints.
Quick Tip: Tetany = Low Ca\(^{2+}\). Myasthenia gravis = Autoimmune. Muscular dystrophy = Genetic.

Step 1: Understanding the Concept:

Lac operon regulation and i gene mutation.


Step 2: Detailed Explanation:

- i gene codes for repressor protein.

- Mutation in i gene: Repressor cannot bind inducer (allolactose/lactose).

- Repressor remains bound to operator permanently.

- RNA polymerase cannot transcribe structural genes (z, y, a).

- No transcription \(\rightarrow\) No translation of z, y, a genes.


Step 3: Final Answer:

z, y, a genes will not be translated.
Quick Tip: i gene mutation (repressor cannot bind inducer) = Super-repressed state (always OFF).

Step 1: Understanding the Concept:

Conducting part vs Respiratory part of respiratory system.


Step 2: Detailed Explanation:

- Conducting part: Nose, pharynx, larynx, trachea, bronchi, bronchioles (up to terminal bronchioles).

- Functions: Filtration, humidification, temperature adjustment of air.

- Respiratory part: Respiratory bronchioles, alveolar ducts, alveoli.

- Gas exchange (diffusion of O\(_2\) and CO\(_2\)) occurs ONLY in respiratory part.


Step 3: Final Answer:

Provides surface for diffusion of O\(_2\) and CO\(_2\).
Quick Tip: Conducting part = No gas exchange. Respiratory part = Gas exchange occurs.

Step 1: Understanding the Concept:

Strategies for enhancing food quality.


Step 2: Detailed Explanation:

- Biofortification: Breeding crops to increase nutritional value (vitamins, minerals, proteins).

- Examples: Golden rice (Vitamin A), Iron-fortified rice.

- Biomagnification: Increase in toxin concentration up food chain.

- Bioremediation: Using microbes to clean pollutants.

- Bioaccumulation: Accumulation of substances in organism.


Step 3: Final Answer:

Bio-fortification.
Quick Tip: Biofortification = Enhancing nutritional quality of crops. Golden rice is best example.

Step 1: Understanding the Concept:

Functions of saliva and salivary enzymes.


Step 2: Detailed Explanation:

- Saliva contains salivary amylase (ptyalin) which digests STARCH (polysaccharide) into maltose.

- It does NOT digest disaccharides. Disaccharides are digested by disaccharidases in small intestine.

- Lysozyme in saliva controls bacterial population.

- Mucus lubricates food.


Step 3: Final Answer:

Digestion of disaccharides.
Quick Tip: Salivary amylase acts on starch (polysaccharide), not on disaccharides.

Step 1: Understanding the Concept:

Wings of cockroach - morphology.


Step 2: Detailed Explanation:

- Cockroach has two pairs of wings:

1. Forewings (Tegmina): Arise from MESOTHORAX. They are thick, leathery, protective.

2. Hindwings: Arise from METATHORAX. They are thin, membranous, used in flight.

- Prothorax bears no wings.


Step 3: Final Answer:

Mesothorax.
Quick Tip: Tegmina = Mesothorax (forewings). Hindwings = Metathorax. Prothorax = No wings.

Step 1: Understanding the Concept:

Endoplasmic Reticulum in eukaryotic cells.


Step 2: Detailed Explanation:

- (A) Correct: Rough ER has ribosomes on cytoplasmic surface.

- (B) Correct: Smooth ER lacks ribosomes.

- (C) Incorrect: Prokaryotes DO NOT have membrane-bound organelles including ER.

- (D) Correct: SER synthesizes lipids and steroids.


Step 3: Final Answer:

Prokaryotes lack ER entirely.
Quick Tip: Prokaryotes = No membrane-bound organelles. ER is present only in eukaryotes.

Step 1: Understanding the Concept:

Spermatogenesis terminology.


Step 2: Detailed Explanation:

- Statement I: Correct. Release of spermatozoa from Sertoli cells into lumen of seminiferous tubules is called spermiation. (Note: The question says "spermat" but means spermiation).

- Statement II: Incorrect. Spermiogenesis is transformation of spermatids into spermatozoa. Formation of sperms from spermatogonia is spermatogenesis.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Spermatogenesis: Spermatogonia \(\rightarrow\) Sperms. Spermiogenesis: Spermatids \(\rightarrow\) Spermatozoa. Spermiation: Release of sperms.

Step 1: Understanding the Concept:

Joints of vertebral column.


Step 2: Detailed Explanation:

- Adjacent vertebrae are separated by intervertebral discs.

- Intervertebral discs are made of fibrocartilage.

- They act as shock absorbers and allow slight movement.

- Intercalated discs are found in cardiac muscle.

- Areolar tissue is loose connective tissue.


Step 3: Final Answer:

Cartilage.
Quick Tip: Intervertebral discs = Fibrocartilage. Intercalated discs = Cardiac muscle.

Step 1: Understanding the Concept:

Digestive system of birds (Aves).


Step 2: Detailed Explanation:

- Crop and gizzard are specialized chambers in digestive tract of birds.

- Crop: Stores and softens food.

- Gizzard: Grinds food (muscular stomach).

- Pavo (peacock), Psittacula (parrot), Corvus (crow) are all birds.

- Other options contain reptiles, amphibians, mammals, fish which lack crop and gizzard.


Step 3: Final Answer:

Pavo, Psittacula, Corvus.
Quick Tip: Crop and gizzard are characteristic of birds (Aves) for mechanical digestion.

Step 1: Understanding the Concept:

Asexual reproduction in fungi.


Step 2: Detailed Explanation:

- Penicillium is a fungus (Ascomycetes).

- It reproduces asexually by conidia (non-motile spores).

- Conidia are borne on conidiophores (brush-like appearance).

- Zoospores: Motile spores in some algae and fungi (Phycomycetes).

- Gemmules: Asexual reproductive structures in sponges.

- Buds: Asexual reproduction in yeast and Hydra.


Step 3: Final Answer:

Conidia.
Quick Tip: Penicillium = Conidia (brush-like conidiophores). Yeast = Budding.

Step 1: Understanding the Concept:

Absorption of lipids in small intestine.


Step 2: Detailed Explanation:

- Statement I: Correct. Fatty acids and glycerol are insoluble in blood. They are absorbed into lacteals (lymph).

- Statement II: Correct. Fatty acids and glycerol reassemble into chylomicrons in intestinal cells, enter lacteals (lymphatic capillaries), and eventually reach blood via thoracic duct.

- Both statements are correct and related.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Lipids absorbed via lacteals (lymph). Carbohydrates and amino acids absorbed via blood capillaries.

Step 1: Understanding the Concept:

In situ vs Ex situ conservation.


Step 2: Detailed Explanation:

- In situ conservation: Protecting species in their natural habitat along with the entire ecosystem.

- Examples: National parks, wildlife sanctuaries, biosphere reserves.

- Ex situ conservation: Protecting species outside natural habitat (zoos, botanical gardens, seed banks).

- In situ protects entire ecosystem, not just individual species.


Step 3: Final Answer:

Protect and conserve the whole ecosystem.
Quick Tip: In situ = On-site conservation (protect habitat). Ex situ = Off-site conservation.

Step 1: Understanding the Concept:

Death rate (mortality rate) calculation.


Step 2: Detailed Explanation:

- Death rate = Number of deaths / Total population.

- Number of deaths = 8.

- Total population = 80.

- Death rate = 8 / 80 = 0.1 per individual per week.


Step 3: Final Answer:

0.1.
Quick Tip: Death rate = Deaths / Total population. Per capita death rate.

Step 1: Understanding the Concept:

Classification of animal tissues.


Step 2: Detailed Explanation:

- Connective tissues: Blood, bone, cartilage, adipose tissue, areolar tissue.

- Neuroglia (glial cells) are supporting cells of NERVOUS TISSUE, not connective tissue.

- Blood is fluid connective tissue.

- Adipose is fat-storing connective tissue.

- Cartilage is skeletal connective tissue.


Step 3: Final Answer:

Neuroglia.
Quick Tip: Neuroglia = Nervous tissue. All others = Connective tissue. Blood is connective tissue.

Step 1: Understanding the Concept:

Restriction enzymes and their recognition sites.


Step 2: Detailed Explanation:

- Statement I: Correct. Restriction endonucleases recognize specific palindromic sequences (e.g., GAATTC for EcoRI).

- Statement II: Correct. Most restriction enzymes cut DNA asymmetrically, slightly away from center, creating sticky ends.

- EcoRI cuts between G and A, creating overhangs.

- Some cut at center creating blunt ends (e.g., SmaI).


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: EcoRI: Cuts between G and A, produces sticky ends. Palindromic sequence recognized.

Step 1: Understanding the Concept:

Types of natural selection.


Step 2: Detailed Explanation:

- Directional selection: Favors one extreme phenotype, shifting mean character value.

- Stabilizing selection: Favors mean phenotype, reduces variation.

- Disruptive selection: Favors both extremes, may lead to speciation.

- Question describes shift towards specific character value (other than mean) = Directional change.


Step 3: Final Answer:

Directional change.
Quick Tip: Directional: Mean shifts. Stabilizing: Mean remains, variation reduces. Disruptive: Two peaks form.

Step 1: Understanding the Concept:

Excretory products and their forms in different animals.


Step 2: Detailed Explanation:

- Birds (e.g., Pavo - peacock) excrete uric acid (uricotelic).

- Uric acid is insoluble, excreted as semi-solid paste/pellet to conserve water.

- Ornithorhynchus (platypus): Mammal, ureotelic (liquid urine).

- Salamandra (salamander): Amphibian, ureotelic/ammonotelic.

- Hippocampus (seahorse): Fish, ammonotelic.


Step 3: Final Answer:

Pavo.
Quick Tip: Birds and reptiles: Uricotelic (pellet/paste). Mammals: Ureotelic (liquid). Fish: Ammonotelic.

Step 1: Understanding the Concept:

Osteoporosis - causes and characteristics.


Step 2: Detailed Explanation:

- Assertion (A): Correct. Osteoporosis is reduction in bone density leading to fragile bones and fractures.

- Reason (R): Incorrect. Osteoporosis is caused by DECREASED estrogen levels (post-menopausal), not increased.

- Low estrogen increases osteoclast activity, causing bone resorption.


Step 3: Final Answer:

(A) is correct but (R) is not correct.
Quick Tip: Osteoporosis = Low estrogen (post-menopause), decreased bone density.

Step 1: Understanding the Concept:

Oxygen transport and delivery by blood.


Step 2: Detailed Explanation:

- 100 ml of oxygenated blood contains ~20 ml of O\(_2\).

- Venous blood contains ~15 ml of O\(_2\) per 100 ml.

- Amount delivered to tissues = Arterial O\(_2\) - Venous O\(_2\).

- = 20 ml - 15 ml = 5 ml per 100 ml of blood.


Step 3: Final Answer:

5 ml.
Quick Tip: Oxygen delivery = 5 ml per 100 ml blood. CO\(_2\) delivery = 4 ml per 100 ml blood.

Step 1: Understanding the Concept:

Dehydration synthesis of disaccharides.


Step 2: Detailed Explanation:

- Two glucose molecules: C\(_6\)H\(_{12}\)O\(_6\) + C\(_6\)H\(_{12}\)O\(_6\) = C\(_{12}\)H\(_{24}\)O\(_{12}\).

- Dehydration removes ONE water molecule (H\(_2\)O).

- Maltose formula = C\(_{12}\)H\(_{24}\)O\(_{12}\) - H\(_2\)O = C\(_{12}\)H\(_{22}\)O\(_{11}\).

- General formula for disaccharide: C\(_n\)H\(_{2n-2}\)O\(_{n-1}\) where n = total carbons.


Step 3: Final Answer:
\(\mathrm{C_{12}H_{22}O_{11}}\).
Quick Tip: Disaccharide = Monosaccharide \(\times\) 2 - H\(_2\)O. Maltose, Sucrose, Lactose all have C\(_{12}\)H\(_{22}\)O\(_{11}\).

Step 1: Understanding the Concept:

DNA length to base pair conversion.


Step 2: Detailed Explanation:

- Distance between two base pairs = 0.34 nm = 0.34 \(\times\) 10\(^{-9}\) m.

- Total length = 1.1 m.

- Number of base pairs = Total length / Distance per bp.

- = 1.1 / (0.34 \(\times\) 10\(^{-9}\)) = (1.1 / 0.34) \(\times\) 10\(^9\).

- \(\approx\) 3.23 \(\times\) 10\(^9\) bp \(\approx\) 3.3 \(\times\) 10\(^9\) bp.

- This matches the approximate size of human genome.


Step 3: Final Answer:
\(3.3\times 10^{9}\mathrm{bp}\).
Quick Tip: 1 bp = 0.34 nm. Human haploid genome = 3.3 \(\times\) 10\(^9\) bp \(\approx\) 1.1 m DNA.

Step 1: Understanding the Concept:

Stages of mitosis and spindle fiber attachment.


Step 2: Detailed Explanation:

- (A) Correct: Chromosomes align at metaphase plate.

- (B) Incorrect: Spindle fibers attach to KINETOCHORE (protein structure at centromere), not directly to centromere.

- (C) Correct: Chromosomes decondense in telophase.

- (D) Correct: Centromere splits in anaphase, sister chromatids separate.


Step 3: Final Answer:

Spindle fibers attach to kinetochore, not centromere directly.
Quick Tip: Kinetochore = Protein disc at centromere where spindle fibers attach.

Step 1: Understanding the Concept:

Chordate classification and vertebrate characteristics.


Step 2: Detailed Explanation:

- Assertion (A): Correct. Vertebrates are a subphylum of chordates. Protochordates (Urochordates, Cephalochordates) are chordates but not vertebrates.

- Reason (R): Correct. In vertebrates, notochord is replaced by vertebral column during development. In protochordates, notochord persists.

- (R) correctly explains why some chordates are not vertebrates.


Step 3: Final Answer:

Both (A) and (R) are correct and (R) is the correct explanation of (A).
Quick Tip: All vertebrates are chordates. Protochordates are chordates but not vertebrates (notochord persists).

Step 1: Understanding the Concept:

Stages of meiosis and DNA replication.


Step 2: Detailed Explanation:

- (A) Correct: Meiosis has two divisions: Meiosis I and Meiosis II.

- (B) Incorrect: DNA replication occurs ONLY before Meiosis I (in S phase of interphase). There is NO S phase between Meiosis I and II.

- (C) Correct: Synapsis and crossing over occur in Prophase I.

- (D) Correct: Meiosis produces four haploid cells.


Step 3: Final Answer:

No DNA replication in Meiosis II.
Quick Tip: DNA replicates only once before Meiosis I. Interkinesis has NO S phase.

Step 1: Understanding the Concept:

Gene therapy for ADA deficiency.


Step 2: Detailed Explanation:

- ADA deficiency is treated by introducing functional ADA gene into lymphocytes.

- Lymphocytes are short-lived (not immortal), so periodic infusion is required.

- If stem cells (from bone marrow) are genetically engineered, it can be a permanent cure.

- Retroviral vector is used to introduce gene into lymphocytes cultured outside.


Step 3: Final Answer:

Genetically engineered lymphocytes are not immortal cells.
Quick Tip: Lymphocyte-based gene therapy = Temporary (repeat infusions). Stem cell-based = Permanent cure.

Step 1: Understanding the Concept:

Steps of decomposition process.


Step 2: Detailed Explanation:

- Decomposition steps:

1. Fragmentation: Breakdown of detritus into smaller particles by detritivores (earthworms, termites).

2. Leaching: Water-soluble nutrients percolate into soil.

3. Catabolism: Enzymatic degradation by decomposers.

4. Humification: Formation of humus.

5. Mineralization: Release of inorganic nutrients.


Step 3: Final Answer:

Fragmentation.
Quick Tip: Fragmentation = Detritivores break detritus. Decomposition = Microbial enzymatic breakdown.

Step 1: Understanding the Concept:

Contraceptive methods and their mechanisms of action.


Step 2: Detailed Explanation:

- (a) Diaphragms: Barrier method, cover cervix blocking sperm entry \(\rightarrow\) (iv).

- (b) Contraceptive Pills: Hormonal method, inhibit ovulation and implantation \(\rightarrow\) (i).

- (c) Intra Uterine Devices (IUDs): Increase phagocytosis of sperm in uterus \(\rightarrow\) (ii).

- (d) Lactational Amenorrhea: Absence of menstruation and ovulation during breastfeeding \(\rightarrow\) (iii).


Step 3: Final Answer:

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).
Quick Tip: Barrier methods: Block sperm entry. Hormonal: Inhibit ovulation. IUDs: Phagocytosis of sperm.

Step 1: Understanding the Concept:

Homologous vs Analogous organs.


Step 2: Detailed Explanation:

- (A) Correct: Analogous structures evolve independently due to similar function (convergent evolution).

- (B) Correct: Sweet potato (root) and potato (stem) are analogous (different origin, same function).

- (C) Correct: Homologous structures indicate common ancestry.

- (D) Incorrect: Penguin flipper (bird) and dolphin flipper (mammal) have different origins, similar function \(\rightarrow\) ANALOGOUS, not homologous.


Step 3: Final Answer:

They are analogous organs, not homologous.
Quick Tip: Homologous: Same origin, different function. Analogous: Different origin, same function.

Step 1: Understanding the Concept:

Acquired immunity vs Innate immunity.


Step 2: Detailed Explanation:

- (A) Correct: Primary immune response occurs on first exposure.

- (B) Correct: Secondary (anamnestic) response on subsequent exposure.

- (C) Correct: Memory cells from first encounter enable faster secondary response.

- (D) Incorrect: Acquired immunity is SPECIFIC and develops AFTER birth. Innate immunity is non-specific and present from birth.


Step 3: Final Answer:

Acquired immunity is specific and develops after birth.
Quick Tip: Innate: Non-specific, present at birth. Acquired: Specific, develops after exposure.

Step 1: Understanding the Concept:

Gene mapping using recombination frequencies.


Step 2: Detailed Explanation:

- Recombination frequency \(\propto\) distance between genes.

- Smallest distance: a \& c = 5% (closest).

- Next: b \& d = 9% (close).

- Next: b \& c = 15%.

- Next: a \& b = 20%.

- Next: c \& d = 24%.

- Largest: a \& d = 29% (farthest).

- Arranging: a - (5) - c - (15) - b - (9) - d.

- Check a to d = 5 + 15 + 9 = 29 (matches).

- Sequence: a, c, b, d.


Step 3: Final Answer:

a, c, b, d.
Quick Tip: 1% recombination frequency = 1 centiMorgan (cM) map distance.

Step 1: Understanding the Concept:

Biological molecules and their functions.


Step 2: Detailed Explanation:

- (a) Glycogen: Polysaccharide, storage product in animals \(\rightarrow\) (iv).

- (b) Globulin: Protein, functions as antibody (immunoglobulins) \(\rightarrow\) (iii).

- (c) Steroids: Lipids, many function as hormones (e.g., testosterone, estrogen) \(\rightarrow\) (i).

- (d) Thrombin: Enzyme, biocatalyst in blood coagulation \(\rightarrow\) (ii).


Step 3: Final Answer:

(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).
Quick Tip: Glycogen = Animal starch (storage). Globulins = Antibodies. Steroids = Hormones. Thrombin = Enzyme.

Step 1: Understanding the Concept:

Comparison of electrical and chemical synapses.


Step 2: Detailed Explanation:

- (A) Correct: Electrical synapses have gap junctions with close membrane proximity.

- (B) Correct: Ions flow directly through gap junctions in electrical synapses.

- (C) Correct: Chemical synapses release neurotransmitters.

- (D) Incorrect: Electrical synapses are FASTER than chemical synapses (no synaptic delay). Chemical synapses have ~0.5 ms delay.


Step 3: Final Answer:

Electrical synapses are faster than chemical synapses.
Quick Tip: Electrical synapse: Faster, bidirectional. Chemical synapse: Slower (synaptic delay), unidirectional.

Step 1: Understanding the Concept:

Genetically engineered insulin is produced using recombinant DNA technology.


Step 2: Detailed Explanation:

(a) Incorrect: This is a general property of pro-insulin, not specific to genetically engineered insulin.

(b) Correct: A and B chains were produced separately in \textit{E. coli and later combined by forming disulphide bonds.

(c) Incorrect: This refers to insulin obtained from animals, not genetically engineered insulin.

(d) Incorrect: This is a general biological process, not specific to recombinant insulin production.

(e) Incorrect: Allergic reactions were associated with animal insulin.


Step 3: Final Answer:

(b) only.

Step 1: Understanding the Concept:

Meselson-Stahl experiment and DNA replication.


Step 2: Detailed Explanation:

- E. coli division time = 20 minutes.

- Total time = 60 minutes = 3 generations.

- Initial: 10 cells with \(^{15}\mathrm{N}/^{15}\mathrm{N}\) DNA.

- After 1st generation (20 min): 20 cells, all \(^{15}\mathrm{N}/^{14}\mathrm{N}\) (hybrid).

- After 2nd generation (40 min): 40 cells, 20 hybrid + 20 light (\(^{14}\mathrm{N}/^{14}\mathrm{N}\)).

- After 3rd generation (60 min): 80 cells total.

- Hybrid: 20 cells (from previous hybrid division).

- Light (totally free from \(^{15}\mathrm{N}\)): 80 - 20 = 60 cells.


Step 3: Final Answer:

60 cells.
Quick Tip: After n generations: Light DNA = Total cells - Initial cells \(\times\) 2. Total cells = Initial \(\times\) 2\(^n\).

Step 1: Understanding the Concept:

Cardiac cycle and valve actions.


Step 2: Detailed Explanation:

- (A) Incorrect: SA node generates action potential for atrial contraction, AV node relays to ventricles.

- (B) Incorrect: AV valves open due to atrial pressure > ventricular pressure, not necessarily contraction.

- (C) Correct: During joint diastole, both atria and ventricles relaxed, blood flows passively from atria to ventricles.

- (D) Incorrect: Increased ventricular pressure OPENS semilunar valves.


Step 3: Final Answer:

Blood moves freely from atrium to ventricle during joint diastole.
Quick Tip: Joint diastole: Both chambers relaxed, AV valves open, semilunar closed.

Step 1: Understanding the Concept:

Characteristics of ideal cloning vector.


Step 2: Detailed Explanation:

- (A) Desirable: Ori allows replication of vector.

- (B) Desirable: Marker gene (e.g., antibiotic resistance) helps select transformed cells.

- (C) Desirable: Single restriction site for easy insertion of foreign DNA.

- (D) Not desirable: Multiple recognition sites for same enzyme cause vector to be cut into fragments, complicating cloning.


Step 3: Final Answer:

Presence of two or more recognition sites.
Quick Tip: Ideal vector: Single restriction site, ori, selectable marker, small size.