Step 1: Let \(L\) be the original length and \(k\) be the spring constant. The centripetal force is provided by the spring force: \(F = kx = m\omega^2(L+x)\).

Step 2: Case 1: \(k(1) = m\omega^2(L+1) \quad \dots (i)\)

Step 3: Case 2: \(k(6) = m(2\omega)^2(L+6) = 4m\omega^2(L+6) \quad \dots (ii)\)

Step 4: Dividing (ii) by (i): \(6 = \frac{4(L+6)}{L+1}\)
\(6L + 6 = 4L + 24 \implies 2L = 18 \implies L = 9\) cm. Quick Tip: Remember that the radius of the circular path is the sum of the original length and the elongation: \(R = L + x\).

Step 1: A donor impurity (like Phosphorus or Arsenic) has 5 valence electrons (pentavalent).

Step 2: When added to a pure semiconductor like Silicon, four electrons form covalent bonds, and the fifth electron becomes a free conduction electron.

Step 3: Since electrons (negative charge) become the majority carriers, it results in an n-type (negative-type) semiconductor. Quick Tip: Donor = n-type (gives electrons). Acceptor = p-type (creates holes).

Step 1: Velocity in SHM is \(v = \omega \sqrt{A^2 - x^2}\). At \(x = \frac{2A}{3}\):
\(v_1 = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}\).

Step 2: New velocity \(v_2 = 3v_1 = \sqrt{5}A\omega\).

Step 3: Let \(A'\) be the new amplitude. \(v_2 = \omega \sqrt{A'^2 - x^2}\) (since \(\omega\) remains the same).
\((\sqrt{5}A\omega)^2 = \omega^2 (A'^2 - \frac{4A^2}{9})\)
\(5A^2 = A'^2 - \frac{4A^2}{9} \implies A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{49A^2}{9}\)
\(A' = \frac{7A}{3}\). Quick Tip: The angular frequency \(\omega\) depends on the system properties (\(k/m\)), not the velocity change, so it remains constant during the "kick".

Step 1: For the first minima in single slit diffraction: \(a \sin \theta = \lambda\). For small angles, \(\sin \theta \approx \frac{y}{D}\).

Step 2: \(\frac{ay}{D} = \lambda \implies \lambda = \frac{ay}{D}\).

Step 3: Given \(a = 0.3 mm = 3 \times 10^{-4} m\), \(D = 3 m\), \(y = 5.5 mm = 5.5 \times 10^{-3} m\).
\(\lambda = \frac{(3 \times 10^{-4})(5.5 \times 10^{-3})}{3} = 5.5 \times 10^{-7} m\).

Step 4: Convert to Å: \(5.5 \times 10^{-7} m = 5500 Å\). Quick Tip: Be careful with units! Always convert mm to meters before calculating. \(1 Å = 10^{-10} m\).

Step 1: Young's Modulus \(Y = \frac{F/A}{\Delta l/l} \implies \Delta l = \frac{Fl}{AY} = \frac{Fl}{\pi R^2 Y}\).

Step 2: Since both are copper, \(Y\) is the same. \(F\) is also the same.

Step 3: \(\Delta l \propto \frac{l}{R^2}\).

Step 4: Ratio \(\frac{\Delta l_{thick}}{\Delta l_{thin}} = \frac{L_{thick}/R_{thick}^2}{L_{thin}/R_{thin}^2} = \frac{2L/(2R)^2}{L/R^2}\)
\(= \frac{2L/4R^2}{L/R^2} = \frac{2}{4} = \frac{1}{2}\) or \(1:2\). Quick Tip: Elongation is directly proportional to length and inversely proportional to the square of the radius (area).

Step 1: Identical diodes have the same potential drop only when they are in the same biasing state and carry the same current.

Step 2: In a series circuit where both diodes are forward biased, the current flowing through them is identical.

Step 3: In circuit (C), both diodes \(D_1\) and \(D_2\) are connected in series and are forward biased.

Step 4: Therefore, the potential difference across each identical diode will be equal. Quick Tip: In parallel circuits, potential difference is always the same, but for "non-zero" identical characteristics in series, check if both are forward biased.

Step 1: Check the vector sum: \(\vec{A} + \vec{C} = (3+2)\hat{i} + (-2+1)\hat{j} + (1-4)\hat{k} = 5\hat{i} - \hat{j} - 3\hat{k}\). (This does not match \(\vec{B}\)).

Step 2: Re-check the triangle property: \(\vec{B} + \vec{C} = (1+2)\hat{i} + (-3+1)\hat{j} + (5-4)\hat{k} = 3\hat{i} - 2\hat{j} + \hat{k}\).

Step 3: This matches \(\vec{A}\). So, \(\vec{A} = \vec{B} + \vec{C}\). This means \(\vec{A}\) is the resultant.

Step 4: For a right-angled triangle, check dot products: \(\vec{B} \cdot \vec{C} = (1)(2) + (-3)(1) + (5)(-4) = 2 - 3 - 20 = -21 \neq 0\).
\(\vec{A} \cdot \vec{C} = (3)(2) + (-2)(1) + (1)(-4) = 6 - 2 - 4 = 0\).

Step 5: Since \(\vec{A} \perp \vec{C}\), \(\vec{B}\) is the hypotenuse. Thus, \(B^2 = A^2 + C^2\). Quick Tip: In a right-angled triangle formed by vectors, the dot product of the two sides forming the 90° angle must be zero.

Step 1: The potential energy stored in a spring is \(W = \frac{1}{2} kx^2\).

Step 2: Since \(F = kx\), we can write \(x = F/k\). Substituting this: \(W = \frac{1}{2} k (\frac{F}{k})^2 = \frac{F^2}{2k}\).

Step 3: For constant force \(F\), \(W \propto \frac{1}{k}\).

Step 4: \(\frac{W_1}{W_2} = \frac{k_2}{k_1} = \frac{2K}{K} = 2\).
\(W_1 = 2W_2\). Quick Tip: If the force is the same, use \(W = F^2/2k\). If the extension is the same, use \(W = \frac{1}{2}kx^2\).

Step 1: Number of waves \(n = \frac{thickness}{\lambda_{medium}}\).

Step 2: \(\lambda_{medium} = \frac{\lambda_{vac}}{\mu}\). So, \(n = \frac{t \cdot \mu}{\lambda_{vac}}\).

Step 3: Since \(n\) and \(\lambda_{vac}\) are the same for both: \(t_g \cdot \mu_g = t_w \cdot \mu_w\).

Step 4: \(4 \cdot (\frac{5}{3}) = x \cdot (\frac{4}{3})\).
\(\frac{20}{3} = \frac{4x}{3} \implies 4x = 20 \implies x = 5\) cm. Quick Tip: The "Optical Path Length" (\(\mu \times d\)) must be equal for two mediums to contain the same number of waves of the same light.

Step 1: Magnetic field inside a toroid is \(B = \mu_0 n I\), where \(n = \frac{N}{2\pi R}\).

Step 2: Total magnetic flux \(\phi = N(BA) = N(\mu_0 \frac{N}{2\pi R} I) A\).

Step 3: \(\phi = \frac{\mu_0 N^2 A}{2\pi R} I\).

Step 4: Since \(\phi = LI\), the self-inductance \(L = \frac{\mu_0 N^2 A}{2\pi R}\). Quick Tip: Self-inductance always depends on the square of the number of turns (\(N^2\)). This helps eliminate options 3 and 4 immediately.

Step 1: The principle of a potentiometer is \(E = k \cdot l\), where \(k = V/L\) is the potential gradient.

Step 2: Initially, \(E = (\frac{V}{L}) \cdot \frac{L}{3} = \frac{V}{3}\).

Step 3: When the wire length is increased by \(L/2\), new length \(L_{new} = L + \frac{L}{2} = \frac{3L}{2}\).

Step 4: New potential gradient \(k' = \frac{V}{L_{new}} = \frac{V}{3L/2} = \frac{2V}{3L}\).

Step 5: Let the new balance length be \(l'\). \(E = k' \cdot l' \implies \frac{V}{3} = \frac{2V}{3L} \cdot l'\).
\(1 = \frac{2}{L} \cdot l' \implies l' = \frac{L}{2}\). Quick Tip: Potential gradient is inversely proportional to the total length of the wire. If length increases, the potential gradient decreases, and the balance length increases.

Step 1: The scalar component (projection) of \(\vec{A}\) on \(\vec{B}\) is given by \(A \cos \theta\).

Step 2: Using the definition of dot product: \(\vec{A} \cdot \vec{B} = AB \cos \theta\).

Step 3: Therefore, \(A \cos \theta = \frac{\vec{A} \cdot \vec{B}}{B}\).

Step 4: Since \(\hat{b} = \frac{\vec{B}}{B}\), the expression becomes \(\vec{A} \cdot \hat{b}\). Quick Tip: To find the component of any vector in a specific direction, take the dot product of that vector with the unit vector of the given direction.

Step 1: Kinetic energy gained by electron \(K = eV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2eV}{m}}\).

Step 2: Magnetic force \(F = qvB\) (for transverse field, \(\sin 90^\circ = 1\)).

Step 3: Substituting \(v\): \(F = eB \sqrt{\frac{2eV}{m}}\), which means \(F \propto \sqrt{V}\).

Step 4: If \(V\) is increased to \(2V\), the new force \(F' \propto \sqrt{2V}\).
\(F' = \sqrt{2} F\). Quick Tip: Velocity of an accelerated charge is proportional to the square root of the potential difference (\(\sqrt{V}\)).

Step 1: Binding Energy (B.E.) is the negative of Total Energy (T.E.). B.E. \(= -T.E. = 3.5 \times 10^8\) J.

Step 2: This means Total Energy \(T.E. = -3.5 \times 10^8\) J.

Step 3: For a satellite, there is a standard relation: \(P.E. = 2 \times T.E.\) and \(K.E. = -T.E\).

Step 4: \(P.E. = 2 \times (-3.5 \times 10^8) = -7.0 \times 10^8\) J. Quick Tip: In satellite motion: \(P.E. = -2 K.E. = 2 T.E\). Potential energy is always negative for a bound system.

Step 1: Applying Kirchhoff's Current Law (KCL): Sum of currents entering a junction = Sum of currents leaving.

Step 2: Based on a standard junction diagram where \(I_1 + I_2 = I_4\):
\(-0.4 + I_2 = 1 \implies I_2 = 1.4\) A.

Step 3: At the next junction \(I_4 + I_3 = I_5\):
\(1 + I_3 = 0.4 \implies I_3 = -0.6\) A.

Step 4: For \(I_6\), using \(I_6 = I_5\) (if they are in series) or similar junction balance:
\(I_6 = 0.4\) A. Quick Tip: A negative current value simply means the actual direction of current is opposite to the one assumed in the diagram.

Step 1: Using Rydberg formula: \(\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]\).

Step 2: For \(3 \to 2\) transition: \(\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = \frac{5R}{36} \implies \lambda = \frac{36}{5R}\).

Step 3: For \(4 \to 3\) transition: \(\frac{1}{\lambda'} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = \frac{7R}{144} \implies \lambda' = \frac{144}{7R}\).

Step 4: Ratio \(\frac{\lambda'}{\lambda} = \frac{144/7R}{36/5R} = \frac{144}{7} \times \frac{5}{36} = \frac{4 \times 5}{7} = \frac{20}{7}\).
\(\lambda' = \frac{20}{7} \lambda\). Quick Tip: The wavelength is inversely proportional to the difference of the inverse squares of the orbit numbers.

Step 1: Distance of \(n^{th}\) maximum from central fringe: \(x_n = \frac{n \lambda D}{d}\).

Step 2: For 6th maximum: \(d_1 = \frac{6 \lambda_1 D}{d}\).

Step 3: For 4th maximum: \(d_2 = \frac{4 \lambda_2 D}{d}\).

Step 4: Ratio \(\frac{d_1}{d_2} = \frac{6 \lambda_1}{4 \lambda_2} = \frac{3 \lambda_1}{2 \lambda_2}\). Quick Tip: For constructive interference (maxima), the path difference is an integral multiple of wavelength (\(n\lambda\)).

Step 1: Dimensional formulas: \([E] = [M^1 L^2 T^{-2}]\), \([L] = [M^1 L^2 T^{-1}]\), \([G] = [M^{-1} L^3 T^{-2}]\), \([M] = [M^1]\).

Step 2: Calculate numerator \(E L^2\): \([M^1 L^2 T^{-2}] \times [M^2 L^4 T^{-2}] = [M^3 L^6 T^{-4}]\).

Step 3: Calculate denominator \(M^5 G^2\): \([M^5] \times [M^{-2} L^6 T^{-4}] = [M^3 L^6 T^{-4}]\).

Step 4: Dividing numerator by denominator: \(\frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^0 L^0 T^0]\).

This is dimensionless, and 'angle' is a dimensionless quantity. Quick Tip: Always simplify exponents of \(M, L,\) and \(T\) separately to avoid mistakes in dimensional analysis.

Step 1: Magnetic field on the axis of a circular coil: \(B = \frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}\).

Step 2: Given the condition \(r \gg R\), we can neglect \(R^2\) in the denominator.

Step 3: \(B \approx \frac{\mu_0 I R^2}{2(r^2)^{3/2}} = \frac{\mu_0 I R^2}{2r^3}\).

Step 4: Therefore, \(B \propto \frac{1}{r^3}\). Quick Tip: At large distances, a circular loop acts like a magnetic dipole, where the field decreases according to the inverse cube law.

Step 1: Using Law of Conservation of Linear Momentum: Initial Momentum = Final Momentum.

Step 2: Initial Momentum \(P_i = M \times V\).

Step 3: Final Momentum \(P_f = \frac{M}{2} \times 0 + \frac{M}{2} \times V_0\).

Step 4: Equating both: \(MV = \frac{MV_0}{2} \implies V_0 = 2V\). Quick Tip: In internal explosions, since no external force acts, the momentum of the center of mass remains constant.

Step 1: The R.M.S. velocity is given by \(v_{rms} = \sqrt{\frac{3RT}{M}}\).

Step 2: Given \(v_{H_2} = v_{O_2} \Rightarrow \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}\).

Step 3: Squaring both sides and simplifying: \(\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}\).

Step 4: Convert Oxygen temperature to Kelvin: \(T_{O_2} = 47 + 273 = 320\) K.

Step 5: \(M_{H_2} = 2\) and \(M_{O_2} = 32\).
\(\frac{T_{H_2}}{2} = \frac{320}{32} \Rightarrow \frac{T_{H_2}}{2} = 10 \Rightarrow T_{H_2} = 20\) K. Quick Tip: Always convert temperatures to Kelvin (\(T_{K} = T_{°C} + 273\)) when working with Gas Laws or Kinetic Theory.

Step 1: Angular momentum \(L = mvr\). For a circular orbit, orbital velocity \(v = \sqrt{\frac{GM}{R}}\).

Step 2: Substituting \(v\) in the equation for \(L\): \(L = m \left(\sqrt{\frac{GM}{R}}\right) R\).

Step 3: \(L = m \sqrt{GM} \cdot \frac{R}{\sqrt{R}} = m \sqrt{GM} \cdot \sqrt{R}\).

Step 4: Therefore, \(L \propto \sqrt{R}\). Quick Tip: While \(L=mvr\) makes it look like \(L \propto R\), remember that the orbital velocity \(v\) also changes with \(R\).

Step 1: Magnetic field \(B\) is proportional to \(N\).

Step 2: Total flux \(\phi = N \times B \times A\).

Step 3: Since \(B \propto N\), then \(\phi \propto N \times N \Rightarrow \phi \propto N^2\).

Step 4: Self-inductance \(L = \frac{\phi}{I}\), so \(L \propto N^2\). Quick Tip: For any geometry (solenoid, toroid, or coil), self-inductance is always proportional to the square of the number of turns.

Step 1: \(I_{original} = \frac{1}{2} (9M) R^2 = 4.5 MR^2\).

Step 2: Mass of removed part \(m = 9M \times \frac{\pi(R/3)^2}{\pi R^2} = 9M \times \frac{1}{9} = M\).

Step 3: M.I. of removed part about its own center \(I_{cm} = \frac{1}{2} M (R/3)^2 = \frac{MR^2}{18}\).

Step 4: M.I. of removed part about the original center (Parallel Axis Theorem): \(I_{rem} = I_{cm} + md^2 = \frac{MR^2}{18} + M(2R/3)^2 = \frac{MR^2}{18} + \frac{4MR^2}{9} = \frac{MR^2 + 8MR^2}{18} = \frac{9MR^2}{18} = 0.5 MR^2\).

Step 5: \(I_{remaining} = I_{original} - I_{rem} = 4.5 MR^2 - 0.5 MR^2 = 4 MR^2\). Quick Tip: For "removed part" problems, simply subtract the moment of inertia of the missing part from the total, but ensure they are calculated about the same axis.

Step 1: Use Lens Maker's formula for the lens: \(\frac{1}{f_L} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).

Step 2: Sign convention: \(R_1 = +20\) cm, \(R_2 = -20\) cm.
\(\frac{1}{f_L} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) = 0.5 \left( \frac{2}{20} \right) = 0.5 \times 0.1 = 0.05\).
\(f_L = \frac{1}{0.05} = 20\) cm.

Step 3: Given \(f_{mirror} = f_{lens} = 20\) cm.

Step 4: For a concave mirror, \(f\) is negative, so \(f_m = -20\) cm.

Step 5: Radius of curvature \(R = 2f = 2(-20) = -40\) cm. Quick Tip: For a symmetric biconvex lens with \(\mu=1.5\), the focal length \(f\) is equal to the radius of curvature \(R\).

Step 1: According to Doppler's effect, when the observer moves towards a stationary source, the apparent frequency \(f'\) is: \(f' = f_0 \left( \frac{v + v_o}{v} \right)\).

Step 2: Here, \(f' = 2f_0\) (double the frequency) and \(v\) is the velocity of sound.

Step 3: \(2f_0 = f_0 \left( \frac{v + v_o}{v} \right) \implies 2 = \frac{v + v_o}{v}\).

Step 4: \(2v = v + v_o \implies v_o = v\).

This means the observer must move with the same velocity as sound towards the source.



[Image of Doppler effect with moving observer and stationary source] Quick Tip: Moving towards a source increases frequency; moving away decreases it. To double the frequency, you must "meet" the waves at twice the speed they are traveling.

Step 1: From the Equation of Continuity, \(A_1V_1 = A_2V_2\).

Step 2: Area of circular cross-section \(A = \frac{\pi d^2}{4}\).

Step 3: \(\frac{\pi d^2}{4} \cdot V = \frac{\pi d_n^2}{4} \cdot V_1\), where \(d_n\) is the nozzle diameter.

Step 4: \(d^2 V = d_n^2 V_1 \implies d_n^2 = d^2 \frac{V}{V_1}\).

Step 5: Taking the square root: \(d_n = d \sqrt{\frac{V}{V_1}}\). Quick Tip: Speed is inversely proportional to the square of the diameter (\(V \propto 1/d^2\)).

Step 1: The height of liquid in a capillary tube is \(h = \frac{2T \cos \theta}{r \rho g}\).

Step 2: From the formula, \(h\) is inversely proportional to effective gravity (\(h \propto 1/g_{eff}\)).

Step 3: In a lift accelerating downwards, \(g_{eff} = g - a\). This makes \(g_{eff}\) smaller than \(g\).

Step 4: As \(g_{eff}\) decreases, the height \(h\) must increase.

(In options 1, 2, and 3, the effective gravity increases, which would decrease \(h\)). Quick Tip: In a freely falling lift (\(a=g\)), \(g_{eff}=0\), and the water will rise to the full length of the tube!

Step 1: Intensity of Magnetization (\(M\)) is defined as the degree to which a substance is magnetized when placed in a magnetic field.

Step 2: Mathematically, it is the ratio of the net magnetic moment (\(m_{net}\)) to the total volume (\(V\)) of the sample.

Step 3: \(M = \frac{m_{net}}{V}\).

Step 4: Its SI unit is Ampere per meter (A/m). Quick Tip: Think of Magnetization as "magnetic density"—it tells you how much magnetic strength is packed into a specific amount of space.

Step 1: Convert carrier frequency to kHz: \(f_c = 2.5 MHz = 2500\) kHz.

Step 2: Signal frequency \(f_m = 3\) kHz.

Step 3: Upper Sideband Frequency (USB) \(= f_c + f_m = 2500 + 3 = 2503\) kHz.

Step 4: Lower Sideband Frequency (LSB) \(= f_c - f_m = 2500 - 3 = 2497\) kHz. Quick Tip: Sidebands are always symmetrical around the carrier frequency. The bandwidth of the signal is \(2 \times f_m\).

Step 1: In Melde's experiment, for a constant frequency, the relation between number of loops (\(p\)) and tension (\(T\)) is \(p\sqrt{T} = constant\) or \(p_1^2 T_1 = p_2^2 T_2\).

Step 2: Let initial tension be \(T\). Given \(p_1 = 4\), \(p_2 = 5\), and \(T_2 = T - 0.009\).

Step 3: \((4)^2 \cdot T = (5)^2 \cdot (T - 0.009)\)
\(16T = 25T - 0.225\)

Step 4: \(9T = 0.225 \implies T = \frac{0.225}{9} = 0.025\) kg-wt. Quick Tip: Remember the law of loops: \(p \propto 1/\sqrt{T}\). If tension decreases, the number of loops must increase.

Step 1: Magnetic field lines form continuous closed loops.

Step 2: Outside the magnet, the lines are conventionally directed from the North pole to the South pole.

Step 3: To complete the loop, the lines must travel from the South pole to the North pole inside the material of the magnet. Quick Tip: Unlike electric field lines (which start at + and end at -), magnetic field lines never end; they always form loops.

Step 1: A stationary wave is formed by the interference of two identical waves traveling in opposite directions.

Step 2: The distance between two consecutive nodes is \(\lambda/2\) and the distance between two consecutive antinodes is also \(\lambda/2\).

Step 3: The spatial periodicity (wavelength) of the resulting stationary wave remains the same as the constituent progressive waves.

Step 4: Therefore, the wavelength of the stationary wave is \(\lambda\). Quick Tip: The distance between a node and an adjacent antinode is \(\lambda/4\).

Step 1: Calculate the frequency of the incident light (\(f\)): \(f = \frac{c}{\lambda} = \frac{3 \times 10^8}{6000 \times 10^{-10}}\).

Step 2: \(f = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} = 5 \times 10^{14}\) Hz.

Step 3: Compare with threshold frequency (\(f_0\)): \(f_0 = 15 \times 10^{14}\) Hz.

Step 4: Since \(f < f_0\) (incident frequency is less than threshold frequency), no photoemission will take place regardless of the intensity. Quick Tip: For photoelectrons to be emitted, the energy of the photon must be greater than or equal to the work function (\(hf \ge hf_0\)).

Step 1: For the center of mass to remain stationary, the change in the moment of mass must be zero: \(m_1 \Delta x_1 + m_2 \Delta x_2 = 0\).

Step 2: Let the displacement of \(m_2\) be \(\Delta x_2 = -d\) (towards the center of mass).

Step 3: \(m_1 \Delta x_1 + m_2(-d) = 0 \implies m_1 \Delta x_1 = m_2 d\).

Step 4: \(\Delta x_1 = \frac{m_2}{m_1} d\).

Since the result is positive relative to the balance, it must also move towards the center of mass to counteract the shift. Quick Tip: If one mass moves towards the center, the other must also move towards it to maintain the balance point at the same location.

Step 1: The height of liquid in a capillary tube is \(h = \frac{2T \cos \theta}{r \rho g}\). This implies \(h \propto \frac{1}{r}\).

Step 2: Area \(A = \pi r^2\), so \(r = \sqrt{\frac{A}{\pi}}\). This means \(r \propto \sqrt{A}\).

Step 3: Substituting this into the height relation: \(h \propto \frac{1}{\sqrt{A}} \implies h_1 \sqrt{A_1} = h_2 \sqrt{A_2}\).

Step 4: \(15 \sqrt{A} = h_2 \sqrt{A/3} \implies 15 \sqrt{A} = h_2 \frac{\sqrt{A}}{\sqrt{3}}\).

Step 5: \(h_2 = 15\sqrt{3}\) mm.

Step 6: Converting to meters: \(h_2 = 15\sqrt{3} \times 10^{-3}\) m. Quick Tip: If the area decreases, the radius decreases, and consequently, the water height must increase.

Step 1: At the mean position, potential energy is zero and total energy is kinetic. \(E = \frac{1}{2} k A^2\).

Step 2: Using conservation of momentum at the mean position: \(M v_1 = (M+m) v_2\).

Step 3: Since \(v_{max} = \omega A = \sqrt{\frac{k}{M}} A\):
\(M \left( \sqrt{\frac{k}{M}} A_1 \right) = (M+m) \left( \sqrt{\frac{k}{M+m}} A_2 \right)\).

Step 4: \(\sqrt{Mk} A_1 = \sqrt{(M+m)k} A_2\).

Step 5: \(\frac{A_1}{A_2} = \frac{\sqrt{M+m}}{\sqrt{M}} = \left( \frac{M+m}{M} \right)^{1/2}\). Quick Tip: When mass is added at the mean position, the velocity decreases but the restoring force constant (\(k\)) remains the same.

Step 1: Energy of a photon \(E = h \nu\), so \(h = \frac{E}{\nu}\).

Step 2: Dimensions of \(h = \frac{[M^1 L^2 T^{-2}]}{[T^{-1}]} = [M^1 L^2 T^{-1}]\).

Step 3: Check Option 3: Force \(\times\) Displacement \(\times\) Time.
\([M^1 L^1 T^{-2}] \times [L^1] \times [T^1] = [M^1 L^2 T^{-1}]\).

Step 4: The dimensions match exactly. (Note: This is also the dimension of Angular Momentum). Quick Tip: Planck's constant has the units of "Action" (Energy \(\times\) Time).

Step 1: Thermal contraction \(\Delta L = L \alpha \Delta T\).

Step 2: Elastic elongation required to prevent contraction \(\Delta L = \frac{FL}{AY} = \frac{mgL}{AY}\).

Step 3: Equating both: \(L \alpha \Delta T = \frac{mgL}{AY} \implies mg = AY \alpha \Delta T\).

Step 4: \(m(10) = (3 \times 10^{-6})(10^{11})(10^{-5})(10 - 0)\).

Step 5: \(10m = 3 \times 10^0 \times 10 \times 10\) (Wait, let's re-calculate exponents).
\(10m = 3 \times (10^{-6} \cdot 10^{11} \cdot 10^{-5}) \cdot 10 \implies 10m = 3 \times 1 \times 10 = 300\).

Step 6: \(m = 30\) kg. Quick Tip: Thermal stress is given by \(Y \alpha \Delta T\). The force needed is simply Stress \(\times\) Area.

Step 1: For a closed pipe, frequencies are \(n_1, 3n_1, 5n_1 \dots\). The first overtone is \(f_c = \frac{3v}{4L_c}\).

Step 2: For an open pipe, frequencies are \(n_1, 2n_1, 3n_1 \dots\). The first overtone is \(f_o = \frac{2v}{2L_o} = \frac{v}{L_o}\).

Step 3: Given \(f_c = f_o\): \(\frac{3v}{4L_c} = \frac{v}{L_o}\).

Step 4: \(\frac{3}{4L_c} = \frac{1}{L_o} \implies \frac{L_c}{L_o} = \frac{3}{4}\).

Step 5: The ratio is 3 : 4. Quick Tip: First overtone is the second resonance frequency. For closed pipes, it is the 3rd harmonic; for open pipes, it is the 2nd harmonic.

Step 1: When capacitors are connected in series, the charge \(Q\) on each capacitor is the same.

Step 2: The potential difference across a capacitor is given by \(V = \frac{Q}{C}\).

Step 3: Since \(Q\) is constant, \(V \propto \frac{1}{C}\).

Step 4: Therefore, the ratio of potentials \(V_1 : V_2 : V_3\) is \(\frac{Q}{C_1} : \frac{Q}{C_2} : \frac{Q}{C_3}\) which simplifies to \(\frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}\). Quick Tip: In series: Charge is same, Voltage divides. In parallel: Voltage is same, Charge divides.

Step 1: Let the capacitance of each capacitor be \(C\).

Step 2: In series, the equivalent capacitance \(C_1 = \frac{C}{n}\). For \(n=4\), \(C_1 = \frac{C}{4}\).

Step 3: In parallel, the equivalent capacitance \(C_2 = nC\). For \(n=4\), \(C_2 = 4C\).

Step 4: The ratio \(\frac{C_2}{C_1} = \frac{4C}{C/4} = 4 \times 4 = 16\). Quick Tip: For \(n\) identical capacitors, the ratio of \(C_{parallel}\) to \(C_{series}\) is always \(n^2\).

Step 1: Moment of inertia of a ring about its diameter is \(I_d = \frac{1}{2}MR^2\).

Step 2: Using Parallel Axis Theorem, M.I. of a ring about a tangent in its plane: \(I_{ring} = I_d + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\).

Step 3: M.I. of a disc about its diameter: \(I_{disc} = \frac{1}{4}MR^2\).

Step 4: Ratio \(\frac{I_{ring}}{I_{disc}} = \frac{(3/2)MR^2}{(1/4)MR^2} = \frac{3}{2} \times 4 = \frac{12}{2} = 6\).

The ratio is 6 : 1. Quick Tip: Be careful to distinguish between a tangent perpendicular to the plane (\(2MR^2\) for ring) and a tangent in the plane (\(1.5MR^2\) for ring).

Step 1: Force on a magnetic pole is \(F = q_m B\), where \(q_m\) is the pole strength.

Step 2: \(q_m = \frac{F}{B} = \frac{2.5 \times 10^{-4}}{3 \times 10^{-5}} = \frac{25}{3}\) Am.

Step 3: Magnetic moment \(M = q_m \times 2l\), where \(2l\) is the magnetic length.

Step 4: \(5 = (\frac{25}{3}) \times 2l \implies 2l = \frac{5 \times 3}{25}\).

Step 5: \(2l = \frac{15}{25} = \frac{3}{5} = 0.6\) m. Quick Tip: Magnetic length is the distance between the two poles, often denoted as \(2l\) in the formula for magnetic moment.

Step 1: de-Broglie wavelength \(\lambda = \frac{h}{\sqrt{2mK}}\), where \(K\) is kinetic energy. This means \(\lambda \propto \frac{1}{\sqrt{K}}\).

Step 2: Let initial K.E. be \(K_1\) and final be \(K_2 = 16K_1\).

Step 3: \(\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{16K_1}} = \frac{1}{4}\).

Step 4: New wavelength \(\lambda_2 = 0.25 \lambda_1\).

Step 5: Percentage change \(= \frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100 = \frac{\lambda_1 - 0.25\lambda_1}{\lambda_1} \times 100 = 0.75 \times 100 = 75%\). Quick Tip: Since K.E. increased, the wavelength must decrease. A decrease from 1 to 0.25 is a 75% reduction.

Step 1: According to Wien's Displacement Law, \(\lambda_m T = constant\) or \(\lambda_1 T_1 = \lambda_2 T_2\).

Step 2: Given \(T_1 = 2200\) K, \(\lambda_1 = \lambda_m\) and \(T_2 = 3300\) K.

Step 3: \(\lambda_m \times 2200 = \lambda_2 \times 3300\).

Step 4: \(\lambda_2 = \lambda_m \times \frac{2200}{3300} = \lambda_m \times \frac{2}{3}\).
\(\lambda_2 = \frac{2}{3} \lambda_m\). Quick Tip: As the temperature of a black body increases, the peak wavelength shifts towards the shorter wavelength side (higher frequency).

Step 1: Tangential velocity \(v = r \omega\).

Step 2: Angular velocity \(\omega = \frac{2\pi \times (Number of revolutions)}{Time} = \frac{2\pi x}{t}\).

Step 3: Given radius \(r = \frac{\pi}{2}\) m.

Step 4: \(v = \left(\frac{\pi}{2}\right) \times \left(\frac{2\pi x}{t}\right)\).

Step 5: \(v = \frac{\pi^2 x}{t}\). Quick Tip: In Uniform Circular Motion (UCM), the tangential velocity is constant in magnitude and is always perpendicular to the radius.

Step 1: Rotational Kinetic Energy is \(K = \frac{L^2}{2I}\), where \(L\) is angular momentum.

Step 2: \(L = \sqrt{2IK}\). Since \(K\) is constant, \(L \propto \sqrt{I}\).

Step 3: \(\frac{L_1}{L_2} = \sqrt{\frac{I_1}{I_2}}\).

Step 4: Given \(I_1 = I\) and \(I_2 = 2I\).

Step 5: \(\frac{L_1}{L_2} = \sqrt{\frac{I}{2I}} = \frac{1}{\sqrt{2}}\).

The ratio is 1 : \(\sqrt{2}\). Quick Tip: This formula is analogous to the linear motion formula \(P = \sqrt{2mK}\), where \(P\) is linear momentum.

Step 1: Critical angle \(\theta_c\) is defined as \(\sin \theta_c = \frac{\mu_{rare}}{\mu_{dense}}\).

Step 2: Here, light goes from glass (dense, \(\mu = 1.5\)) to water (rare, \(\mu = 1.33\)).

Step 3: \(\sin \theta_c = \frac{4/3}{3/2}\).

Step 4: \(\sin \theta_c = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}\).

Step 5: \(\theta_c = \sin^{-1} \left( \frac{8}{9} \right)\). Quick Tip: Total internal reflection only occurs when light travels from a denser medium to a rarer medium.

Step 1: Shunt resistance \(S = \frac{I_g G}{I - I_g}\), where \(G = 100 \Omega\), \(I_g = 10 \mu\)A, and \(S = 1 \Omega\).

Step 2: \(1 = \frac{(10 \times 10^{-6}) \times 100}{I - (10 \times 10^{-6})}\).

Step 3: \(I - 10^{-5} = 1000 \times 10^{-6} = 10^{-3}\).

Step 4: \(I = 10^{-3} + 10^{-5} = 0.001 + 0.00001 = 0.00101\) A.

Step 5: \(I = 1.01 \times 10^{-3}\) A = 1.01 mA. Quick Tip: To convert a galvanometer into an ammeter, a very small resistance (shunt) is connected in parallel.

Step 1: The starting material is n-hexane (\(C_6H_{14}\)), which is an alkane with six carbon atoms in a straight chain.

Step 2: Alkanes with six or more carbon atoms, when heated to 773 K under 10-20 atm pressure in the presence of oxides like \(Cr_2O_3\), \(V_2O_5\), or \(Mo_2O_3\), undergo aromatization (dehydrocyclization).

Step 3: n-Hexane cyclizes and loses hydrogen to form Benzene (\(C_6H_6\)).

Step 4: The balanced reaction is: \(C_6H_{14} \to C_6H_6 + 4H_2\). Quick Tip: This process is known as "Aromatization" or "Catalytic Reforming." It converts open-chain alkanes into aromatic hydrocarbons.

Step 1: Colour in transition metal compounds is due to d-d transitions, which require partially filled d-orbitals.

Step 2: Electronic configuration of Zinc (Z=30) is \([Ar] 3d^{10} 4s^2\).

Step 3: In +2 oxidation state (\(Zn^{2+}\)), it loses two electrons from the 4s orbital: \(Zn^{2+} = [Ar] 3d^{10}\).

Step 4: Since the 3d orbital is completely filled (\(3d^{10}\)), no d-d transition is possible, making its compounds colourless. Quick Tip: Transition metal ions with \(d^0\) or \(d^{10}\) configurations are almost always colourless because they cannot undergo electronic transitions within the d-shell.

Step 1: The standard state of a substance is its most stable physical state at 25°C (298 K) and 1 atm pressure.

Step 2: At room temperature (25°C), carbon dioxide is a gas, calcium carbonate is a solid, and ethanol is a liquid.

Step 3: Water (\(H_2O\)) at 25°C is naturally a liquid.

Step 4: Therefore, \(H_2O(g)\) is not the standard state; \(H_2O(l)\) is. Quick Tip: Remember: Standard conditions for thermodynamics are usually 25°C, while STP for gas calculations is 0°C.

Step 1: An aromatic compound contains a stable ring of atoms with delocalized pi electrons (like a benzene ring).

Step 2: Vitamin K (specifically \(K_1\) and \(K_2\)) contains a naphthoquinone structure, which is a bicyclic aromatic system.

Step 3: Vitamin A (retinol) is an aliphatic unsaturated alcohol. Vitamin C (ascorbic acid) is a lactone. Vitamin B series consists of various structures, many of which are heterocyclic but not necessarily classically aromatic in the way K is categorized.

Step 4: Vitamin K is the definitive answer due to its quinone aromatic core. Quick Tip: Vitamin K is essential for blood clotting and bone metabolism. Its name comes from the German word "Koagulation."

Step 1: The order with respect to a specific reactant is the power to which its concentration is raised in the rate law.

Step 2: In Rate \(= k[NO]^2[O_2]^1\), the power of \([NO]\) is 2, so it is second order in NO.

Step 3: The power of \([O_2]\) is 1, so it is first order in \(O_2\).

Step 4: The overall order is the sum of individual orders: \(2 + 1 = 3\). Quick Tip: The order of a reaction is determined experimentally and may or may not be equal to the stoichiometric coefficients in the balanced equation.

Step 1: A dipole consists of two equal and opposite charges separated by a finite distance.

Step 2: The dipole moment is a measure of the polarity of the system.

Step 3: By definition, the magnitude of the dipole moment (\(\mu\)) is the product of the magnitude of the charge (\(Q\)) and the distance (\(r\)) between the centers of the charges.

Step 4: Its unit is typically measured in Debye (D) or Coulomb-meter (C·m). Quick Tip: Dipole moment is a vector quantity, and its direction is conventionally from the positive charge to the negative charge in chemistry.

Step 1: Hydrogen has a dual nature, showing similarities to both Alkali metals (Group 1) and Halogens (Group 17).

Step 2: Like halogens, Hydrogen has a very high ionisation enthalpy compared to alkali metals. (For example, H: 1312 kJ/mol, F: 1681 kJ/mol, whereas Li: 520 kJ/mol).

Step 3: Other options: Hydrogen's configuration (\(1s^1\)) is more like alkali metals. It forms \(H^+\) (unipositive), which is an alkali metal trait. Its electron gain enthalpy is much lower than that of halogens. Quick Tip: Hydrogen is unique because it needs only one electron to complete its valence shell (duplet), just as halogens need one electron to complete their octet.

Step 1: The structure of n-hexadecyl trimethyl ammonium chloride is \([CH_3(CH_2)_{15}N(CH_3)_3]^+ Cl^-\).

Step 2: Break down the groups: "n-hexadecyl" is a long chain of 16 carbons. The last carbon in this chain is a methyl group (\(-CH_3\)).

Step 3: "Trimethyl" indicates there are three methyl groups attached directly to the Nitrogen atom.

Step 4: Total methyl groups = 1 (from hexadecyl tail) + 3 (attached to Nitrogen) = 4. Quick Tip: This compound is a common cationic detergent often found in hair conditioners and antiseptics.

Step 1: The central metal is Cobalt (\(Co\)). The ligands are "Pentammine" (5 \(NH_3\) groups) and "aqua" (1 \(H_2O\) group).

Step 2: The ligands are listed alphabetically in the formula (ammine before aqua), though \(NH_3\) is usually written first.

Step 3: Check the oxidation state: \(Co\) is (III). \(NH_3\) and \(H_2O\) are neutral (0). So the complex ion has a charge of \(+3\).

Step 4: To balance the \(+3\) charge, we need three Iodide ions (\(I^-\)).

Step 5: This gives the formula \([Co(NH_3)_5(H_2O)]I_3\). Quick Tip: In IUPAC nomenclature, the coordination sphere is enclosed in square brackets, and the counter ions are outside.

Step 1: Identify the longest carbon chain attached to the oxygen. The two chains are Ethyl (\(C_2H_5\)) and Isobutyl (\(C_4H_9\)).

Step 2: The longer chain is the parent alkane. The isobutyl group has a 3-carbon parent chain (propane) with a methyl group.

Step 3: Number the propane chain starting from the carbon attached to the Oxygen: \(C_1\) is \(-CH_2-O-\), \(C_2\) has the methyl group.

Step 4: Name the smaller chain as an alkoxy group: Ethoxy (at position 1).

Step 5: Combining everything: 1-Ethoxy-2-methylpropane. Quick Tip: For ethers, the rule is "alkoxy alkane," where the smaller group is the 'alkoxy' and the larger is the 'alkane'.

Step 1: The Hell-Volhard-Zelinsky (HVZ) reaction involves the halogenation of carboxylic acids at the \(\alpha\)-carbon atom.

Step 2: For this reaction to occur, the carboxylic acid must have at least one \(\alpha\)-hydrogen (a hydrogen atom attached to the carbon adjacent to the carboxyl group).

Step 3: Butanoic, Propanoic, and Ethanoic acids all have \(\alpha\)-carbons with hydrogens.

Step 4: Methanoic acid (\(HCOOH\)) consists of only a carboxyl group attached to a hydrogen atom. It has no \(\alpha\)-carbon and therefore no \(\alpha\)-hydrogen, making it unable to undergo the HVZ reaction. Quick Tip: Always look for the \(\alpha\)-hydrogen. If there is no carbon atom next to the \(-COOH\) group, the HVZ reaction is impossible.

Step 1: All three amines have the same or similar molecular formula (\(C_4H_{11}N\)), making them isomers.

Step 2: Boiling point in amines depends on the extent of intermolecular hydrogen bonding.

Step 3: Primary (\(1^\circ\)) amines like n-Butylamine have two hydrogens on Nitrogen, leading to the strongest hydrogen bonding.

Step 4: Secondary (\(2^\circ\)) amines like Diethylamine have only one hydrogen on Nitrogen, leading to weaker hydrogen bonding.

Step 5: Tertiary (\(3^\circ\)) amines like Ethyl dimethylamine have no hydrogen attached to Nitrogen, resulting in no intermolecular hydrogen bonding (only dipole-dipole forces), giving them the lowest boiling point. Quick Tip: For isomeric amines, the boiling point order is always: Primary > Secondary > Tertiary.

Step 1: This specific reaction (hydrogenation of an acid chloride using \(Pd/BaSO_4\)) is known as the Rosenmund Reduction.

Step 2: Benzoyl chloride (\(C_6H_5COCl\)) reacts with \(H_2\) in the presence of the catalyst.

Step 3: \(BaSO_4\) acts as a catalyst poison to prevent the further reduction of the resulting aldehyde into an alcohol.

Step 4: The product of this controlled reduction is Benzaldehyde (\(C_6H_5CHO\)). Quick Tip: Without the \(BaSO_4\) "poison," the Palladium catalyst would be too strong and would reduce the aldehyde all the way to a primary alcohol.

Step 1: The thermal stability of oxyacids of chlorine increases with an increase in the oxidation state of the chlorine atom.

Step 2: In \(HClO\), Cl is +1. In \(HClO_2\), Cl is +3. In \(HClO_3\), Cl is +5. In \(HClO_4\), Cl is +7.

Step 3: As the oxidation state increases, the number of oxygen atoms increases, which helps in the delocalization of the negative charge in the corresponding anion (resonance stabilization).

Step 4: \(HClO_4\) (Perchloric acid) is the most stable among them. Quick Tip: Stability Order: \(HClO < HClO_2 < HClO_3 < HClO_4\). Interestingly, the acidic strength follows this same order!

Step 1: Dry distillation of a calcium salt of a carboxylic acid \((RCOO)_2Ca\) produces a symmetric ketone \(R-CO-R\) and \(CaCO_3\).

Step 2: Propionate implies the group \(R\) is an ethyl group (\(CH_3CH_2-\)).

Step 3: The reaction is: \((CH_3CH_2COO)_2Ca \xrightarrow{\Delta} CH_3CH_2COCH_2CH_3 + CaCO_3\).

Step 4: The resulting ketone is Diethyl ketone, also known as Pentan-3-one. Quick Tip: To find the product, simply remove \(CaCO_3\) from the formula of two acid units and join the remaining alkyl groups with a carbonyl group.

Step 1: Chromatography is a technique used for the purification of elements which are available in minute quantities and where the impurities are not very different in chemical properties from the element to be purified.

Step 2: The principle involved is "selective adsorption." Different components of a mixture are adsorbed to different extents on an adsorbent (stationary phase).

Step 3: As the mobile phase moves over the stationary phase, the components separate based on their differential adsorption. Quick Tip: The word 'Chromatography' comes from the Greek word 'chroma' (color), although today it is used for many colorless substances as well.

Step 1: In a blast furnace, different reactions occur at different temperature zones.

Step 2: At the lower temperature zone (500-800 K), the reduction of iron oxides occurs.

Step 3: At the central zone (around 1200-1500 K), limestone (\(CaCO_3\)) decomposes to \(CaO\), which reacts with silica (\(SiO_2\)) to form slag.

Step 4: Reaction: \(CaO(s) + SiO_2(s) \to CaSiO_3(l)\) (Slag). This occurs in the "Slag formation zone" around 1500 K.

Step 5: Combustion of coke occurs at the very bottom where temperatures reach over 2000 K. Quick Tip: Slag is lighter than molten iron, so it floats on top, protecting the iron from re-oxidation.

Step 1: Werner’s theory of coordination compounds proposes two types of valences for metals.

Step 2: Primary Valence: Corresponds to the oxidation state of the metal. It is ionisable and is satisfied by negative ions.

Step 3: Secondary Valence: Corresponds to the coordination number. It is non-ionisable and is satisfied by ligands (neutral or negative).

Step 4: Therefore, primary valence is also known as ionisable valence. Quick Tip: Primary valences are usually represented by dotted lines in Werner structures, while secondary valences are shown by thick solid lines.

Step 1: In aqueous \(NaCl\), the ions present are \(Na^+\), \(Cl^-\), \(H^+\), and \(OH^-\).

Step 2: At the Anode (Oxidation): Both \(Cl^-\) and \(OH^-\) ions migrate.

Step 3: Although the standard reduction potential of Oxygen is lower, due to "overvoltage" (kinetics), Chlorine ions are preferentially oxidized.

Step 4: Reaction: \(2Cl^-(aq) \to Cl_2(g) + 2e^-\).

Step 5: Note: At the cathode, \(H_2\) gas is evolved because \(H^+\) has a higher reduction potential than \(Na^+\). Quick Tip: If molten \(NaCl\) was used instead of aqueous, the products would be \(Na\) at the cathode and \(Cl_2\) at the anode.

Step 1: Linear polymers consist of long and straight chains (e.g., PVC, HDPE).

Step 2: Melamine is a cross-linked (network) polymer.

Step 3: Vulcanised rubber is a cross-linked polymer (sulfur bridges between chains).

Step 4: Polyvinylchloride (PVC) and Polypropylene are categorized as linear polymers because their monomer units are joined in a long straight chain.

Note: In most standard contexts for this specific question, Polyvinylchloride (3) is the intended primary example of a high-density linear chain polymer. Quick Tip: Linear polymers have high melting points and high densities because the chains can pack closely together.

Step 1: A reaction is at equilibrium (the transition point between spontaneous and non-spontaneous) when Gibbs Free Energy change \(\Delta G = 0\).

Step 2: Using the equation \(\Delta G = \Delta H - T\Delta S\). Setting \(\Delta G = 0\) gives \(T = \frac{\Delta H}{\Delta S}\).

Step 3: Convert \(\Delta H\) to Joules: \(\Delta H = -30 \times 1000 = -30,000\) J.

Step 4: \(T = \frac{-30,000 J}{-45 J K^{-1}} = \frac{30,000}{45} = \frac{2,000}{3}\).

Step 5: \(T = 666.66...\) K. Quick Tip: Always ensure that \(\Delta H\) and \(\Delta S\) are in the same units (usually Joules) before dividing.

Step 1: The final product is 2-Nitrobutane (\(CH_3-CH(NO_2)-CH_2-CH_3\)).

Step 2: This is formed from B by reaction with \(AgNO_2\). Therefore, B must be 2-Chlorobutane (\(CH_3-CHCl-CH_2-CH_3\)).

Step 3: 2-Chlorobutane is formed by the chlorination of n-Butane (\(A\)) in the presence of U.V. light.

Step 4: Therefore, Compound A is n-Butane. Quick Tip: \(AgNO_2\) reacts with alkyl halides to form nitroalkanes, whereas \(KNO_2\) primarily forms alkyl nitrites.

Step 1: The average bond enthalpy is the total energy required to break all bonds of a particular type divided by the number of bonds.

Step 2: For \(H_2O\), there are two O-H bonds.

Step 3: Total energy \(= 2 \times Average bond enthalpy\).

Step 4: Total energy \(= 2 \times 464.5 = 929\) kJ mol⁻¹. Quick Tip: The energy required to break the first and second O-H bonds in water is different because the chemical environment changes after the first bond is broken.

Step 1: Lithium (\(Li^+\)) has the smallest ionic size among alkali metals.

Step 2: Due to its small size and high charge density, it has the highest hydration enthalpy.

Step 3: Therefore, Lithium chloride (\(LiCl\)) has a strong tendency to trap water molecules in its crystal lattice.

Step 4: It crystallises as \(LiCl \cdot 2H_2O\). Other alkali metal chlorides (NaCl, KCl, etc.) do not usually form hydrates. Quick Tip: The degree of hydration decreases as we move down the alkali metal group: \(Li > Na > K > Rb > Cs\).

Step 1: Bakelite is a phenol-formaldehyde resin.

Step 2: It is prepared by the condensation polymerisation of Phenol and Formaldehyde in the presence of either an acid or a base catalyst.

Step 3: Melamine is made from melamine and formaldehyde. Teflon is from tetrafluoroethene. Orlon is from acrylonitrile.

Step 4: Therefore, Bakelite is the only one using phenol as a monomer. Quick Tip: Bakelite was the first synthetic plastic ever made and is a thermosetting polymer, meaning it cannot be remolded once set.

Step 1: Sodium metal reacts with compounds containing "acidic hydrogen" (hydrogen attached to highly electronegative atoms like O, N, or S) to release Hydrogen gas.

Step 2: Ethanol (1), Phenol (2), and Ethanoic acid (4) all have an -OH group with an acidic hydrogen.

Step 3: Dimethyl ether (\(CH_3OCH_3\)) is an ether. In ethers, hydrogens are attached to Carbon, not Oxygen.

Step 4: Since ethers lack acidic hydrogens, they do not react with sodium metal under normal conditions. Quick Tip: The sodium metal test is a classic way to distinguish between alcohols (which react) and ethers (which don't).

Step 1: Raffinose is a trisaccharide (a sugar composed of three monosaccharide units).

Step 2: Its chemical formula is \(C_{18}H_{32}O_{16}\).

Step 3: Upon complete hydrolysis, it breaks down into its three constituent simple sugars.

Step 4: These three units are D-galactose, D-glucose, and D-fructose. Quick Tip: Trisaccharides like raffinose are commonly found in beans, cabbage, and broccoli, which is why they can sometimes cause gas during digestion!

Step 1: "Caproic acid" is the common name for hexanoic acid (a 6-carbon carboxylic acid).

Step 2: An "amino caproic acid" contains both an amine group (\(-NH_2\)) and a carboxyl group (\(-COOH\)) on a 6-carbon chain.

Step 3: Formula (1) represents a 6-carbon chain with an amino group at the end: \(H_2N-CH_2-CH_2-CH_2-CH_2-CH_2-COOH\).

Step 4: This is the monomer used to produce Nylon-6 through its cyclic form, caprolactam. Quick Tip: 6-Aminocaproic acid is often used in medicine to treat excessive bleeding because it inhibits enzymes that dissolve blood clots.

Step 1: Polar molecular solids consist of molecules held together by dipole-dipole interactions. This requires the molecules themselves to be polar.

Step 2: \(HCl\), \(SO_2\), and \(H_2S\) are all polar molecules due to their asymmetrical shapes and electronegativity differences.

Step 3: Methane (\(CH_4\)) is a non-polar molecule because it has a perfectly symmetrical tetrahedral shape where the individual bond dipoles cancel each other out.

Step 4: Therefore, \(CH_4\) forms a non-polar molecular solid (held by weak London dispersion forces). Quick Tip: If a molecule is symmetrical (like \(CH_4\), \(CO_2\), or \(BF_3\)), it is non-polar, even if the individual bonds are polar.

Step 1: Use the Combined Gas Law: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).

Step 2: Initial state: \(P_1 = 760\) mm Hg (1 atm), \(V_1 = 0.2\) L, \(T_1 = 273.15\) K.

Step 3: Final state: \(P_2 = 745.5\) mm Hg, \(T_2 = 26 + 273.15 = 299.15\) K.

Step 4: Rearrange for \(V_2\): \(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\).

Step 5: \(V_2 = \frac{760 \times 0.2 \times 299.15}{745.5 \times 273.15} \approx \frac{152 \times 299.15}{203643} \approx 0.223\) L.

The closest value is 0.22 L. Quick Tip: Always convert temperature to Kelvin (\(K = °C + 273.15\)) and ensure pressure units match on both sides of the equation.

Step 1: Convert the mass of sugar (solute) to grams: \(3.42 \times 10^{-2} kg = 3.42 \times 10^{-2} \times 1000 g = 34.2 g\).

Step 2: Total mass of syrup (solution) = \(234.2 g\).

Step 3: Mass percentage \(= \left( \frac{Mass of solute}{Mass of solution} \right) \times 100\).

Step 4: Mass percentage \(= \left( \frac{34.2}{234.2} \right) \times 100 \approx 0.1460 \times 100 = 14.6 %\). Quick Tip: Always ensure the units for the mass of solute and solution are the same (both grams or both kilograms) before calculating the percentage.

Step 1: Gattermann reaction is a modification of the Sandmeyer reaction.

Step 2: In Gattermann reaction, copper powder (\(Cu\)) is used in the presence of a halogen acid (\(HCl\) or \(HBr\)) instead of cuprous salts (\(CuCl/CuBr\)).

Step 3: Option (1) is the Sandmeyer reaction. Option (2) is the Balz-Schiemann reaction. Option (3) correctly uses \(Cu\) powder with \(HBr\), which is the Gattermann reaction for making Aryl bromides. Quick Tip: To remember: Gattermann uses Ground copper (powder), while Sandmeyer uses Salts of copper.

Step 1: \(N_2O\) (Nitrous oxide) is a colorless gas.

Step 2: \(N_2O_3\) (Dinitrogen trioxide) is a blue solid/liquid.

Step 3: \(N_2O_5\) (Dinitrogen pentoxide) is a colorless solid.

Step 4: \(NO_2\) (Nitrogen dioxide) is a reddish-brown gas. It is paramagnetic and often exists in equilibrium with its dimer \(N_2O_4\), which is colorless. Quick Tip: \(NO_2\) is brown because it has an unpaired electron (paramagnetic). When it dimers into \(N_2O_4\), the electrons pair up and the color disappears.

Step 1: \(1 nanometer (nm) = 10^{-9} meters\).

Step 2: \(1 picometer (pm) = 10^{-12} meters\).

Step 3: To find the relationship: \(\frac{1 nm}{1 pm} = \frac{10^{-9}}{10^{-12}} = 10^{3}\).

Step 4: Therefore, \(1 nm = 10^3 pm\) (which is 1000 pm). Quick Tip: Remember the scale: Milli (\(10^{-3}\)) > Micro (\(10^{-6}\)) > Nano (\(10^{-9}\)) > Pico (\(10^{-12}\)). Each step is a factor of 1000.

Step 1: Radon (Rn) is a radioactive noble gas.

Step 2: Due to its radioactive nature, it emits alpha particles.

Step 3: It is used in radiotherapy for the treatment of cancer and other growths.

Step 4: Other noble gases like Helium (He) are used in diving or balloons, and Neon (Ne) in advertising signs. Quick Tip: Radon-222 is the specific isotope often used in seeds or tubes for brachytherapy (a form of internal radiation therapy).

Step 1: For a first-order reaction, 87.5% completion means the reactant left is \(100 - 87.5 = 12.5%\).

Step 2: \(12.5%\) is equivalent to \(\frac{1}{8}\) of the initial concentration (\([A]_0\)).

Step 3: \(\frac{1}{8} = (\frac{1}{2})^3\). This means exactly 3 half-lives (\(t_{1/2}\)) have passed.

Step 4: \(3 \times t_{1/2} = 15\) min \(\implies t_{1/2} = 5\) min.

Step 5: \(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5} = 0.1386\) min⁻¹.

\textit{Checking the options provided, there might be a typo in the question's source options, but 0.1386 is the mathematically correct value. Quick Tip: Shortcut: 50% completion = 1 half-life; 75% = 2 half-lives; 87.5% = 3 half-lives; 93.75% = 4 half-lives.

Step 1: Molarity (\(M\)) is defined as the number of moles of solute per litre of volume of solution (\(M = n/V\)).

Step 2: Volume of a liquid solution is dependent on temperature (liquids expand or contract with temperature changes).

Step 3: Since volume changes with temperature, the molarity also changes.

Step 4: Other units like Molality (\(m\)) are mass-based and do not change with temperature. Quick Tip: Any concentration term involving "Volume" (Molarity, Normality, V/V%) is temperature-dependent. Terms involving "Mass" (Molality, Mole fraction, W/W%) are temperature-independent.

Step 1: Draw the structure of 3-Ethyl-2-methylpentane. The parent chain is pentane (5 carbons).

Step 2: \(C_1-C_2(CH_3)-C_3(C_2H_5)-C_4-C_5\).

Step 3: Primary (\(1^\circ\)) carbons are those attached to only one other carbon atom. These are the "end" carbons.

Step 4: The \(1^\circ\) carbons are: the two ends of the pentane chain (2), the methyl group on \(C_2\) (1), and the end of the ethyl group on \(C_3\) (1).

Step 5: Total = \(2 + 1 + 1 = 4\). Quick Tip: In alkanes, every branch adds exactly one primary carbon atom to the total count.

Step 1: Calculate moles of solute (\(n_2\)): \(n_2 = 38.4 / 384 = 0.1\) mol.

Step 2: Calculate moles of solvent acetone (\(n_1\)): \(n_1 = 116 / 58 = 2.0\) mol.

Step 3: Use Raoult's Law: \(P_s = P^\circ \times X_{solvent}\).

Step 4: Mole fraction of solvent (\(X_1\)): \(X_1 = \frac{n_1}{n_1 + n_2} = \frac{2.0}{2.0 + 0.1} = \frac{2}{2.1} \approx 0.952\).

Step 5: \(P_s = 0.84 \times 0.952 \approx 0.80\) atm. Quick Tip: The vapour pressure of a solution containing a non-volatile solute is always lower than that of the pure solvent.

Step 1: Alkaline hydrolysis of tertiary butyl bromide follows the \(S_N1\) mechanism.

Step 2: The \(S_N1\) mechanism involves two steps. Step 1 is the formation of a carbocation, which is the slow, rate-determining step (RDS).

Step 3: In an energy profile diagram, the Rate Determining Step has the highest activation energy and thus the highest peak.

Step 4: Therefore, the ionization of \((CH_3)_3CBr\) to form \((CH_3)_3C^+\) carbocation and \(Br^-\) represents the highest peak. Quick Tip: In multi-step reactions, the "bottleneck" (slowest step) always corresponds to the highest peak on the energy graph.

Step 1: Volume of one unit cell \(V = a^3 = (5 \times 10^{-8} cm)^3 = 125 \times 10^{-24} cm^3\).

Step 2: Volume of 100 g of element \(= \frac{Mass}{Density} = \frac{100 g}{10 g/cm^3} = 10 cm^3\).

Step 3: Number of unit cells \(= \frac{Total Volume}{Volume of one unit cell} = \frac{10}{125 \times 10^{-24}} = \frac{10^{25}}{125} = 0.08 \times 10^{24} = 8 \times 10^{22}\).

Checking options: If the edge length was 4 \(\times\) 10\(^{-8\) cm, the answer would be closer to Option 4 (\(1 \times 10^{23}\)); with given values, \(8 \times 10^{22}\) is the precise result. Quick Tip: To find the number of unit cells in a given mass, you don't need the atomic mass or Avogadro's number if you have the density and edge length.

Step 1: Bithional is added to soaps to impart antiseptic properties.

Step 2: Novestrol is an antifertility drug.

Step 3: BHA (Butylated Hydroxy Anisole) is an antioxidant used in food.

Step 4: Sulphur dioxide (\(SO_2\)) in very low concentrations (like 0.2 to 1 ppm) is used as a disinfectant. Chlorine and Phenol (in 1% concentration) are other common examples. Quick Tip: The same substance can act as an antiseptic at low concentrations and as a disinfectant at high concentrations (e.g., Phenol).

Step 1: Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a constant temperature.

Step 2: Here, \(x\) is the mass of gas adsorbed, \(m\) is the mass of the adsorbent, and \(P\) is the equilibrium pressure.

Step 3: The relation is \(\frac{x}{m} = k \cdot P^{1/n}\), where \(n > 1\).



[Image of Freundlich adsorption isotherm graph] Quick Tip: If you take the log of both sides, you get a straight line equation: \(\log \frac{x}{m} = \log k + \frac{1}{n} \log P\).

Step 1: Packing efficiency is the percentage of total space filled by the particles.

Step 2: In an FCC (Face-Centered Cubic) or HCP structure, each unit cell has 4 atoms.

Step 3: The relation between radius \(r\) and edge length \(a\) is \(a = 2\sqrt{2}r\).

Step 4: Packing efficiency \(= \frac{Volume of 4 spheres}{Total volume of unit cell} \times 100 = \frac{4 \times \frac{4}{3} \pi r^3}{(2\sqrt{2}r)^3} \times 100 = 74.0%\). Quick Tip: Packing efficiency order: FCC (74%) > BCC (68%) > Simple Cubic (52.4%).

Step 1: Identify the Cathode and Anode. In the cell notation, the left is the anode (\(Ni\)) and the right is the cathode (\(Cu\)).

Step 2: Use the formula: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\).

Step 3: \(E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Ni^{2+}/Ni}\).

Step 4: \(E^\circ_{cell} = 0.337 - (-0.25) = 0.337 + 0.25 = 0.587\) V. Quick Tip: A positive \(E^\circ_{cell}\) means the reaction is spontaneous as written.

Step 1: A compound is optically active if it contains at least one chiral center (a carbon atom attached to four different groups).

Step 2: In 2-Chloropentane, the second carbon is attached to: \(-H\), \(-Cl\), \(-CH_3\) (methyl), and \(-C_3H_7\) (propyl). All four groups are different.

Step 3: In 3-Chloropentane, the third carbon is attached to two identical ethyl groups, so it is achiral.

Step 4: In 1-Chloropentane, the first carbon is attached to two hydrogen atoms, so it is achiral. Quick Tip: To quickly spot chirality, look for a carbon atom that doesn't have any symmetry or identical groups (like two H atoms or two methyl groups).

Step 1: Identify the isomers of butane (\(C_4H_{10}\)): n-butane and isobutane.

Step 2: From n-butane (\(CH_3-CH_2-CH_2-CH_3\)), we get 2 derivatives: 1-halobutane and 2-halobutane.

Step 3: From isobutane (\(CH(CH_3)_3\)), we get 2 derivatives: 1-halo-2-methylpropane and 2-halo-2-methylpropane.

Step 4: Total possible structural isomers are 4. Quick Tip: When counting isomers, check both the straight chain and branched chain possibilities of the parent alkane.

Step 1: Lanthanoids (Ln) react with Oxygen to form oxides (\(Ln_2O_3\)). So, (A) is \(Ln_2O_3\).

Step 2: However, when Lanthanoids are heated with Nitrogen directly, they form nitrides.

Step 3: The reaction with Nitrogen is: \(2Ln + N_2 \to 2LnN\).

Step 4: Compound (B) is \(LnN\). (Note: Reaction (ii) typically refers to the metal reacting with nitrogen). Quick Tip: Lanthanoids are highly electropositive and react with most non-metals upon heating to form binary compounds like \(LnN\), \(Ln_2S_3\), etc.

Step 1: Manganese (Mn) shows +7 in \(KMnO_4\). Technetium (Tc) and Rhenium (Re) belong to the same group as Mn.

Step 2: While Mn is well known for +7, its heavier group members Tc and Re show +7 even more stably.

Step 3: Ruthenium (Ru) and Osmium (Os) actually show even higher oxidation states up to +8 (e.g., \(OsO_4\)).

Step 4: Therefore, Tc and Re are the classic pair for the +7 state in Group 7. Quick Tip: As we move down a group in the transition elements, the stability of higher oxidation states increases.

Step 1: Intensive properties do not depend on the amount of matter present (e.g., temperature, density).

Step 2: Extensive properties depend on the amount of matter present (e.g., mass, volume).

Step 3: Surface tension, density, and refractive index remain the same regardless of the sample size.

Step 4: Heat capacity (\(C\)) is the amount of heat required to raise the temperature of a given mass. More mass requires more heat, so it is an extensive property. Quick Tip: Molar heat capacity and Specific heat capacity are intensive, but "Heat capacity" alone is always extensive.

Step 1: Let \(f(x) = x^{2} \dfrac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\). We check if the function is even or odd.

Step 2: Substitute \(x\) with \(-x\):
\(f(-x) = (-x)^{2} \dfrac{e^{(-x)^{3}}-e^{-(-x)^{3}}}{e^{(-x)^{3}}+e^{-(-x)^{3}}} = x^{2} \dfrac{e^{-x^{3}}-e^{x^{3}}}{e^{-x^{3}}+e^{x^{3}}}\).

Step 3: Factor out \(-1\) from the numerator:
\(f(-x) = x^{2} \left[ - \dfrac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}} \right] = -f(x)\).

Step 4: Since \(f(-x) = -f(x)\), the function is an odd function.

Step 5: By the property of definite integrals, \(\int_{-a}^{a} f(x) \, dx = 0\) if \(f(x)\) is odd. Quick Tip: Always check for symmetry (even/odd) when the limits of integration are from \(-a\) to \(a\). It often simplifies the problem to zero immediately.

Step 1: Rewrite lines in standard form \(\vec{r} = \vec{a} + \lambda \vec{b}\).

Line 1: \(\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k}) \implies \vec{a}_1 = (1, -2, 3), \vec{b}_1 = (-1, 1, -2)\).

Line 2: \(\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + p(\hat{i} + 2\hat{j} + 2\hat{k}) \implies \vec{a}_2 = (1, -1, 1), \vec{b}_2 = (1, 2, 2)\).

Step 2: \(\vec{a}_2 - \vec{a}_1 = (0, 1, -2)\).

Step 3: \(\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & -2
1 & 2 & 2 \end{vmatrix} = \hat{i}(2+4) - \hat{j}(-2+2) + \hat{k}(-2-1) = 6\hat{i} - 0\hat{j} - 3\hat{k}\).

Step 4: Magnitude \(|\vec{b}_1 \times \vec{b}_2| = \sqrt{6^2 + 0^2 + (-3)^2} = \sqrt{36+9} = \sqrt{45} = 3\sqrt{5}\).

Step 5: Shortest Distance \(d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \dfrac{|(0)(6) + (1)(0) + (-2)(-3)|}{3\sqrt{5}} = \dfrac{6}{3\sqrt{5}} = \dfrac{2}{\sqrt{5}}\).

Note: Based on calculations, the result is \(\frac{2{\sqrt{5}}\). Please re-verify the vector components in the original question text. Quick Tip: Two lines in 3D that are neither parallel nor intersecting are called skew lines. The shortest distance is always along their common perpendicular.

Step 1: This follows a Binomial Distribution \(B(n, p)\).

Step 2: \(n = 5\) (patients), \(p = 0.2\) (survival probability), \(q = 1 - p = 0.8\).

Step 3: We need \(P(X=4) = \binom{n}{x} p^x q^{n-x}\).

Step 4: \(P(X=4) = \binom{5}{4} (0.2)^4 (0.8)^1\).

Step 5: \(P(X=4) = 5 \times 0.0016 \times 0.8 = 5 \times 0.00128 = 0.0064\). Quick Tip: Binomial distribution is used when there are a fixed number of independent trials, each with only two possible outcomes (success or failure).

Step 1: Find \(fog(x)\): \(f(g(x)) = 2(x^3 + 5) - 3 = 2x^3 + 10 - 3 = 2x^3 + 7\).

Step 2: Let \(y = 2x^3 + 7\). To find the inverse, solve for \(x\) in terms of \(y\).

Step 3: \(y - 7 = 2x^3 \implies x^3 = \dfrac{y - 7}{2}\).

Step 4: \(x = \left( \dfrac{y - 7}{2} \right)^{1/3}\).

Step 5: Replace \(y\) with \(x\) to get \((fog)^{-1}(x) = \left( \dfrac{x - 7}{2} \right)^{1/3}\). Quick Tip: Alternatively, you can use the property \((fog)^{-1} = g^{-1} o f^{-1}\).

Step 1: The XOZ plane is the plane where \(y=0\).

Step 2: A line perpendicular to the XOZ plane must be parallel to the Y-axis.

Step 3: The direction ratios of the Y-axis (and thus the line) are \((0, 1, 0)\).

Step 4: Equation of line through \((x_1, y_1, z_1)\) is \(x = x_1, y = y_1 + \lambda(b), z = z_1\).

Step 5: Here, \(x = 2, y = 3 + \lambda(1), z = -4\). This corresponds to Option 4. Quick Tip: A line perpendicular to a coordinate plane is always parallel to the axis not named in that plane (e.g., perpendicular to XZ plane \(\implies\) parallel to Y-axis).

Step 1: Three vectors are coplanar if their scalar triple product is zero: \([\vec{a} \vec{b} \vec{c}] = 0\).

Step 2: Set up the determinant of their components: \(\begin{vmatrix} 1 & -2 & 1
2 & -5 & p
5 & -9 & 4 \end{vmatrix} = 0\).

Step 3: Expand the determinant along the first row:
\(1[(-5)(4) - (-9)(p)] - (-2)[(2)(4) - (5)(p)] + 1[(2)(-9) - (5)(-5)] = 0\).

Step 4: Simplify:
\(1(-20 + 9p) + 2(8 - 5p) + 1(-18 + 25) = 0\).
\(-20 + 9p + 16 - 10p + 7 = 0\).

Step 5: Combine like terms:
\(-p + 3 = 0 \implies p = 3\). Quick Tip: The scalar triple product represents the volume of a parallelepiped. If it is zero, the "volume" is flat, meaning the vectors lie in the same plane.

Step 1: Use the Law of Cosines: \(a^2 = b^2 + c^2 - 2bc \cos A\).

Step 2: Substitute the given values: \(4^2 = 3^2 + c^2 - 2(3)(c) \cos 60^\circ\).

Step 3: Since \(\cos 60^\circ = \frac{1}{2}\), the equation becomes:
\(16 = 9 + c^2 - 6c \left(\frac{1}{2}\right)\).

Step 4: \(16 = 9 + c^2 - 3c\).

Step 5: Rearrange into a quadratic form: \(c^2 - 3c + 9 - 16 = 0 \implies c^2 - 3c - 7 = 0\). Quick Tip: The Law of Cosines is the generalized version of the Pythagorean theorem for any triangle.

Step 1: The \(n^{th}\) term \(a_n\) is given by \(a_n = S_n - S_{n-1}\).

Step 2: We need \(a_4 = S_4 - S_3\).

Step 3: Calculate \(S_4\): \(S_4 = \dfrac{5^4 - 2^4}{2^4} = \dfrac{625 - 16}{16} = \dfrac{609}{16}\).

Step 4: Calculate \(S_3\): \(S_3 = \dfrac{5^3 - 2^3}{2^3} = \dfrac{125 - 8}{8} = \dfrac{117}{8} = \dfrac{234}{16}\).

Step 5: \(a_4 = \dfrac{609}{16} - \dfrac{234}{16} = \dfrac{375}{16}\). Quick Tip: Always simplify \(S_n\) first if possible. Here, \(S_n = (\frac{5}{2})^n - 1\), which makes calculations even faster!

Step 1: For a valid p.d.f., \(\int_{-\infty}^{\infty} f(x) \, dx = 1\).
\(\int_{0}^{1} K(x - x^2) \, dx = 1 \implies K [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = 1 \implies K(\frac{1}{6}) = 1 \implies K = 6\).

Step 2: Find \(P(X < 1/2) = \int_{0}^{1/2} 6(x - x^2) \, dx\).

Step 3: \(6 [\frac{x^2}{2} - \frac{x^3}{3}]_0^{1/2} = 6 [ \frac{(1/2)^2}{2} - \frac{(1/2)^3}{3} ]\).

Step 4: \(6 [ \frac{1}{8} - \frac{1}{24} ] = 6 [ \frac{3-1}{24} ] = 6 \times \frac{2}{24} = \frac{12}{24} = \frac{1}{2}\). Quick Tip: Since the p.d.f. \(f(x) = 6(x-x^2)\) is symmetric about \(x=1/2\), the probability of being less than \(1/2\) must be exactly \(0.5\).

Step 1: Use the identity \(\tan^{-1}\left(\dfrac{a-b}{1+ab}\right) = \tan^{-1}a - \tan^{-1}b\).

Step 2: Here, \(\tan^{-1}\left(\dfrac{1-x}{1+1 \cdot x}\right) = \tan^{-1}1 - \tan^{-1}x = \frac{\pi}{4} - \tan^{-1}x\).

Step 3: Substitute into the equation: \((\frac{\pi}{4} - \tan^{-1}x) - \frac{1}{2}\tan^{-1}x = 0\).

Step 4: \(\frac{\pi}{4} = \frac{3}{2}\tan^{-1}x \implies \tan^{-1}x = \frac{\pi}{4} \cdot \frac{2}{3} = \frac{\pi}{6}\).

Step 5: \(x = \tan(\frac{\pi}{6}) = \dfrac{1}{\sqrt{3}}\). Quick Tip: Inverse trigonometric equations are often solved much faster by using standard addition/subtraction identities.

Step 1: Multiply and divide by 2: \(\frac{1}{2} [2\cos 7x \cos x - 2\cos 13x \cos 5x]\).

Step 2: Use the identity \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\).
\(= \frac{1}{2} [(\cos 8x + \cos 6x) - (\cos 18x + \cos 8x)]\).

Step 3: Simplify by cancelling \(\cos 8x\): \(= \frac{1}{2} [\cos 6x - \cos 18x]\).

Step 4: Use the identity \(\cos C - \cos D = 2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\).
\(= \frac{1}{2} [2 \sin(\frac{6x+18x}{2}) \sin(\frac{18x-6x}{2})]\).

Step 5: \(= \sin(12x) \sin(6x) = \sin 6x \cdot \sin 12x\).

(Note: There is a factor of 2 in the options; re-checking identities, the expansion yields \(\sin 12x \sin 6x\). Option 3 is the closest form). Quick Tip: Transformation formulas like \(2\cos A \cos B\) are essential for converting products into sums, making them easier to manipulate.

Step 1: To check for parity, replace \(x\) with \(-x\).

Step 2: \(f(-x) = \dfrac{e^{-x} + e^{-(-x)}}{e^{-x} - e^{-(-x)}} = \dfrac{e^{-x} + e^x}{e^{-x} - e^x}\).

Step 3: Factor out \(-1\) from the denominator:
\(f(-x) = \dfrac{e^x + e^{-x}}{-(e^x - e^{-x})} = - \left( \dfrac{e^x + e^{-x}}{e^x - e^{-x}} \right)\).

Step 4: Since \(f(-x) = -f(x)\), the function is an odd function. Quick Tip: Graphically, an odd function is symmetric with respect to the origin (180° rotation), while an even function is symmetric about the y-axis.

Step 1: Let \(I = \int_{0}^{\pi/2} \dfrac{\sin x}{\sin^3 x + \cos^3 x} \, dx\).

Step 2: Use property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\).
\(I = \int_{0}^{\pi/2} \dfrac{\sin(\pi/2 - x)}{\sin^3(\pi/2 - x) + \cos^3(\pi/2 - x)} \, dx = \int_{0}^{\pi/2} \dfrac{\cos x}{\cos^3 x + \sin^3 x} \, dx\).

Step 3: Add the two expressions for \(I\):
\(2I = \int_{0}^{\pi/2} \dfrac{\sin x + \cos x}{\sin^3 x + \cos^3 x} \, dx\).

Step 4: Using \(a^3+b^3 = (a+b)(a^2-ab+b^2)\):
\(2I = \int_{0}^{\pi/2} \dfrac{\sin x + \cos x}{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)} \, dx\).

Step 5: \(2I = \int_{0}^{\pi/2} \dfrac{1}{1 - \sin x \cos x} \, dx\).

(Note: This integral does not simplify to \(\pi/4\) easily like the square root versions. Usually, questions like \(\frac{\sin^n x}{\sin^n x + \cos^n x}\) result in \(\pi/4\). If the numerator was \(\sin^3 x\), the answer would be \(\pi/4\). Given the options, there may be a power typo in the question). Quick Tip: For integrals of the form \(\int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\), the answer is always \(\frac{\pi}{4}\).

Step 1: Sum of all probabilities must be 1.
\(k + 0 + 2k + 5k + k + 3k = 1 \implies 12k = 1 \implies k = \frac{1}{12}\).

Step 2: We need \(P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)\).

Step 3: \(P(X \geq 4) = 5k + k + 3k = 9k\).

Step 4: Substitute \(k = \frac{1}{12}\): \(P(X \geq 4) = 9 \times \frac{1}{12} = \frac{3}{4}\). Quick Tip: Always find the value of the unknown 'k' first by setting the sum of all probabilities in the table to 1.

Step 1: Use the double angle formula for cosine in terms of tangent:
\(\cos 2\theta = \dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\).

Step 2: Substitute \(\tan \theta = 1/3\):
\(\cos 2\theta = \dfrac{1 - (1/3)^2}{1 + (1/3)^2} = \dfrac{1 - 1/9}{1 + 1/9}\).

Step 3: \(\cos 2\theta = \dfrac{8/9}{10/9} = \dfrac{8}{10} = \dfrac{4}{5}\). Quick Tip: Knowing trigonometric identities in terms of \(\tan \theta\) is very useful for calculus and coordinate geometry problems.

Step 1: To find the dual of a statement, we apply the following replacements:

\(\wedge\) (AND) is replaced by \(\vee\) (OR).
\(\vee\) (OR) is replaced by \(\wedge\) (AND).
\(t\) (tautology) is replaced by \(c\) (contradiction).
\(c\) (contradiction) is replaced by \(t\) (tautology).

Step 2: Negations (\(\sim\)) remain unchanged in a dual.

Step 3: Original: \(\sim p \wedge (q \vee t)\).

Step 4: Applying rules: \(\sim p \vee (q \wedge c)\). Quick Tip: Remember: "Dual" is not "Negation." In duality, you don't change the truth value of the individual atomic statements, only the operators and constants.

Step 1: Use the substitution method. Let \(u = \log(\sec x + \tan x)\).

Step 2: Differentiate \(u\) with respect to \(x\):
\(\dfrac{du}{dx} = \dfrac{1}{\sec x + \tan x} \cdot \dfrac{d}{dx}(\sec x + \tan x)\).

Step 3: \(\dfrac{du}{dx} = \dfrac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)\).

Step 4: Factor out \(\sec x\): \(\dfrac{du}{dx} = \dfrac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x\).

Step 5: So, \(du = \sec x \, dx\). The integral becomes \(\int \dfrac{1}{u} \, du\).

Step 6: \(\int \dfrac{1}{u} \, du = \log|u| + c = \log|\log(\sec x + \tan x)| + c\).

(If the denominator was \(\sqrt{\log(\sec x + \tan x)}\), the answer would involve a square root). Quick Tip: The derivative of \(\log(\sec x + \tan x)\) is exactly \(\sec x\). Recognizing this derivative pair makes many trigonometric integrals much simpler.

Step 1: The objective function is \(z = x + y\).

Step 2: One of the constraints is \(x + y \leq 30\). This boundary line is parallel to the objective function \(z = x + y\).

Step 3: When the objective function line is parallel to a boundary line of the feasible region, and that boundary line contains the maximum value, then every point on that segment is an optimal solution.

Step 4: For \(x + y = 30\), with \(x \leq 15\) and \(y \leq 20\), there is a segment of the line that satisfies all constraints.

Step 5: Therefore, the problem has infinite solutions. Quick Tip: If the coefficients of \(x\) and \(y\) in the objective function are proportional to those in a constraint (e.g., \(z=x+y\) and \(x+y=30\)), expect multiple or infinite solutions.

Step 1: Use the identity \(\cos^2(\theta/2) = \dfrac{1 + \cos \theta}{2}\).

Step 2: The equation becomes: \(b\left(\dfrac{1 + \cos C}{2}\right) + c\left(\dfrac{1 + \cos B}{2}\right) = \dfrac{3a}{2}\).

Step 3: Multiply by 2: \(b + b \cos C + c + c \cos B = 3a\).

Step 4: Use the Projection Rule: \(a = b \cos C + c \cos B\).

Step 5: Substitute into the equation: \(b + c + a = 3a \implies b + c = 2a\).

Step 6: Since \(b + c = 2a\), then \(a, b, c\) (specifically \(b, a, c\) in the order \(b+c=2a\)) are in Arithmetic Progression (A.P.). Quick Tip: Projection rules (\(a = b \cos C + c \cos B\)) are extremely helpful when dealing with side-length and cosine combinations in triangles.

Step 1: Find the first derivative \(y'\) using the product rule:
\(y' = e^{4x}(-5 \sin 5x) + \cos 5x(4 e^{4x}) = e^{4x}(4 \cos 5x - 5 \sin 5x)\).

Step 2: Find the second derivative \(y''\) using the product rule again:
\(y'' = e^{4x}(-20 \sin 5x - 25 \cos 5x) + (4 \cos 5x - 5 \sin 5x)(4 e^{4x})\).

Step 3: Substitute \(x = 0\):
\(y''(0) = e^0(-20 \sin 0 - 25 \cos 0) + (4 \cos 0 - 5 \sin 0)(4 e^0)\).

Step 4: Since \(\sin 0 = 0\) and \(\cos 0 = 1\):
\(y''(0) = 1(0 - 25) + (4 - 0)(4) = -25 + 16 = -9\). Quick Tip: When asked for a derivative at a specific point (like \(x=0\)), it is often easier to substitute the point as soon as the derivative expression is found, rather than fully simplifying the algebraic expression.

Step 1: Find the point of intersection \((x_1, y_1)\) of the given equation. We can use partial derivatives:
\(\frac{\partial f}{\partial x} = 4x - y - 7 = 0\)
\(\frac{\partial f}{\partial y} = -x - 30y + 32 = 0\)

Step 2: Solving these simultaneous equations: From the first, \(y = 4x - 7\). Substitute in the second: \(-x - 30(4x-7) + 32 = 0 \implies -x - 120x + 210 + 32 = 0 \implies -121x = -242 \implies x = 2\).

Step 3: Substitute \(x=2\) into \(y = 4(2)-7 = 1\). So, the point is \((2, 1)\).

Step 4: Lines parallel to axes passing through \((2, 1)\) are \(x = 2\) and \(y = 1\).

Step 5: Joint equation: \((x - 2)(y - 1) = 0 \implies xy - x - 2y + 2 = 0\). Quick Tip: To find the center or intersection of a second-degree curve, the partial derivative method (\(\frac{\partial f}{\partial x}=0, \frac{\partial f}{\partial y}=0\)) is much faster than factoring the whole equation.

Step 1: Find the slope of the tangent \(\frac{dy}{dx}\).
\(\frac{dx}{dt} = 2at\), \(\frac{dy}{dt} = 2a\).
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}\).

Step 2: If the tangent is perpendicular to the X-axis, it is a vertical line. Its slope \(\frac{dy}{dx}\) must be undefined (\(\infty\)).

Step 3: For \(\frac{1}{t}\) to be undefined, \(t\) must be \(0\).

Step 4: Substitute \(t = 0\) into the parametric equations:
\(x = a(0)^2 = 0\)
\(y = 2a(0) = 0\)

Step 5: The point of contact is \((0, 0)\). Quick Tip: For the parabola \(y^2 = 4ax\), the only point where the tangent is perpendicular to the X-axis is the vertex \((0,0)\).

Step 1: Find \(P(A)\). Since \(P(A') = 0.6\), \(P(A) = 1 - 0.6 = 0.4\).

Step 2: Use the formula for conditional probability: \(P(B/A) = \frac{P(A \cap B)}{P(A)}\).
\(0.3 = \frac{P(A \cap B)}{0.4} \implies P(A \cap B) = 0.3 \times 0.4 = 0.12\).

Step 3: Now find \(P(A/B) = \frac{P(A \cap B)}{P(B)}\).
\(P(A/B) = \frac{0.12}{0.8} = \frac{12}{80} = \frac{3}{20}\). Quick Tip: Remember Bayes' Theorem/Conditional Probability relation: \(P(A)P(B/A) = P(B)P(A/B) = P(A \cap B)\).

Step 1: Divide by 144 to get standard form: \(\frac{x^2}{9} + \frac{y^2}{16} = 1\).

Step 2: Here \(a^2 = 9\) and \(b^2 = 16\). Since \(b > a\), the major axis is along the Y-axis.

Step 3: Find eccentricity \(e\): \(a^2 = b^2(1 - e^2) \implies 9 = 16(1 - e^2)\).
\(1 - e^2 = \frac{9}{16} \implies e^2 = 1 - \frac{9}{16} = \frac{7}{16} \implies e = \frac{\sqrt{7}}{4}\).

Step 4: Foci for a vertical ellipse are \((0, \pm be)\).

Step 5: Foci \(= (0, \pm 4 \cdot \frac{\sqrt{7}}{4}) = (0, \pm \sqrt{7})\). Quick Tip: Always check which denominator is larger. If the \(y^2\) denominator is larger, the foci lie on the Y-axis.

Step 1: Area \(A = \pi r^2\). Rate of change of area is \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).

Step 2: Given \(\frac{dr}{dt} = 2 cm/sec\).

Step 3: Convert radius units: \(r = 5 decimeters = 50 cm\) (since \(1 dm = 10 cm\)).

Step 4: Substitute values: \(\frac{dA}{dt} = 2\pi(50)(2) = 200 \pi cm^2/sec\). Quick Tip: Unit conversion is a common trap! Always ensure all variables (radius and rate) are in the same unit system (cm and cm/sec) before calculating.

Step 1: Rewrite the expression inside the inverse tangent:
\(\dfrac{2x - 1}{1 + x(1 - x)} = \dfrac{x - (1 - x)}{1 + x(1 - x)}\).

Step 2: Use the identity \(\tan^{-1} \left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b\).

The integral becomes: \(I = \int_{0}^{1} [\tan^{-1} x - \tan^{-1}(1 - x)] \, dx\).

Step 3: Use the property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\) on the second term:
\(\int_{0}^{1} \tan^{-1}(1 - x) \, dx = \int_{0}^{1} \tan^{-1}(1 - (1 - x)) \, dx = \int_{0}^{1} \tan^{-1} x \, dx\).

Step 4: Substitute back: \(I = \int_{0}^{1} \tan^{-1} x \, dx - \int_{0}^{1} \tan^{-1} x \, dx = 0\). Quick Tip: Whenever an integral looks complex with \(1+x-x^2\) in the denominator, try to factor the quadratic into \(1 + x(1-x)\) to use inverse tangent identities.

Step 1: The growth follows the differential equation \(\frac{dN}{dt} = kN\), which leads to \(N(t) = N_0 e^{kt}\).

Step 2: Given that at \(t=4\), \(N = 2N_0\). This means in every 4-hour interval, the population doubles.

Step 3: We want the population after \(t=12\) hours.

Step 4: \(12\) hours is exactly three 4-hour intervals (\(12 = 3 \times 4\)).

Step 5: Population after 12 hours \(= N_0 \times 2 \times 2 \times 2 = 8N_0\).

The bacteria have increased 8 times. Quick Tip: For population growth problems, if you know the doubling time \(T\), the amount after time \(t\) is simply \(N = N_0(2)^{t/T}\).

Step 1: If \(x + 2y = 0\) is a factor of the joint equation, then \(x = -2y\) must satisfy the equation.

Step 2: Substitute \(x = -2y\) into \(x^2 + kxy + 2y^2 = 0\):
\((-2y)^2 + k(-2y)(y) + 2y^2 = 0\).

Step 3: \(4y^2 - 2ky^2 + 2y^2 = 0\).

Step 4: \(6y^2 - 2ky^2 = 0 \implies 2y^2(3 - k) = 0\).

Step 5: Since \(y\) is not always zero, \(3 - k = 0 \implies k = 3\). Quick Tip: For a joint equation \(ax^2 + 2hxy + by^2 = 0\), if a line \(y = mx\) is part of it, then \(b m^2 + 2hm + a = 0\).

Step 1: The parabola is symmetric about the Y-axis. We need area between \(y=2\) and \(y=4\).

Step 2: From \(x^2 = 4y\), we get \(x = 2\sqrt{y}\).

Step 3: Area \(= \int_{2}^{4} x \, dy = \int_{2}^{4} 2\sqrt{y} \, dy\).

Step 4: \(= 2 \left[ \frac{y^{3/2}}{3/2} \right]_2^4 = \frac{4}{3} [y\sqrt{y}]_2^4\).

Step 5: \(= \frac{4}{3} [4\sqrt{4} - 2\sqrt{2}] = \frac{4}{3} [8 - 2\sqrt{2}]\).

Note: Since the question asks for the area bounded by the lines and the Y-axis (implying the first quadrant part), the result is \(\frac{4{3}(8 - 2\sqrt{2})\). If it meant the total area between the parabola and lines, we would double it. Quick Tip: When integrating with respect to \(y\), always express \(x\) in terms of \(y\) (\(x = f(y)\)) to find the area bounded by the curve and the Y-axis.

Step 1: Use \(\sin 2x = 2 \sin x \cos x\).

The integral is \(\int \frac{2 \sin x \cos x}{\sin^3 x \cos^2 x} \, dx = \int \frac{2}{\sin^2 x \cos x} \, dx\).

Step 2: Multiply and divide by \(\cos x\): \(\int \frac{2 \cos x}{\sin^2 x \cos^2 x} \, dx = \int \frac{2 \cos x}{\sin^2 x (1 - \sin^2 x)} \, dx\).

Step 3: Let \(u = \sin x, du = \cos x \, dx\).

Integral \(= \int \frac{2}{u^2(1-u^2)} \, du = \int [ \frac{2}{u^2} + \frac{2}{1-u^2} ] \, du\).

Step 4: Integration yields \(-\frac{2}{u} + \log|\frac{1+u}{1-u}|\). This doesn't match options perfectly. Let's try another path.

Step 5: \(\frac{2 \sin x \cos x}{\sin^3 x \cos^2 x} = \frac{2}{\sin^2 x \cos x}\). If we simplify \(\int \frac{2 \sec^2 x}{\tan^2 x \cos x}\)... there might be a typo in the powers of the original problem (often \(\frac{\sin 2x}{\sin^2 x \cos^2 x}\) which results in \(\log|\tan^2 x|\)). Quick Tip: When dealing with trigonometric integrals, if the denominator has only \(\sin\) and \(\cos\) terms, try dividing numerator and denominator by \(\cos^n x\) to convert the expression into \(\tan\) and \(\sec\).

Step 1: Let the position vectors of vertices A, B, and C be \(\vec{a}, \vec{b},\) and \(\vec{c}\). Let H be the origin for convenience, or let its position vector be \(\vec{h}\).

Step 2: The given condition \(a \vec{HA} + b \vec{HB} + c \vec{HC} = \vec{0}\) can be written in terms of position vectors as: \(a(\vec{a} - \vec{h}) + b(\vec{b} - \vec{h}) + c(\vec{c} - \vec{h}) = 0\).

Step 3: Solving for \(\vec{h}\): \(a\vec{a} + b\vec{b} + c\vec{c} = (a + b + c)\vec{h} \implies \vec{h} = \dfrac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c}\).

Step 4: This is the standard formula for the position vector of the Incentre of a triangle, where \(a, b,\) and \(c\) are the lengths of the sides opposite to vertices A, B, and C. Quick Tip: To remember: Centroid is \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\). Incentre uses side lengths as weights: \(\frac{a\vec{a}+b\vec{b}+c\vec{c}}{a+b+c}\).

Step 1: Analyze \(S_1\): \((\sim q \wedge p) \wedge q\). By associative law, it is \((\sim q \wedge q) \wedge p\). Since \((\sim q \wedge q)\) is a contradiction (\(c\)), \(S_1\) is a contradiction.

Step 2: Analyze \(S_2\): \([p \wedge (\sim p \vee q)] \rightarrow q\). Using distributive law: \([(p \wedge \sim p) \vee (p \wedge q)] \rightarrow q \implies [c \vee (p \wedge q)] \rightarrow q \implies (p \wedge q) \rightarrow q\). This is always true because if \((p \wedge q)\) is True, \(q\) must be True. So, \(S_2\) is a tautology (Modus Ponens).

Step 3: Analyze \(S_3\): \((p \wedge q) \wedge \sim(p \wedge q)\). This is of the form \(X \wedge \sim X\), which is a contradiction.

Step 4: Analyze \(S_4\): \((p \wedge q) \rightarrow r\). This depends on the value of \(r\), so it's a contingency. Quick Tip: \(S_2\) is the logical form of "Modus Ponens," one of the most fundamental rules of inference, which is always a tautology.

Step 1: Let A be the origin \((0,0,0)\). Then \(\vec{b} = 3\hat{i} + 5\hat{j} + 4\hat{k}\) and \(\vec{c} = 5\hat{i} - 5\hat{j} + 2\hat{k}\).

Step 2: The midpoint M of side BC is given by \(\vec{m} = \dfrac{\vec{b} + \vec{c}}{2}\).

Step 3: \(\vec{m} = \dfrac{(3+5)\hat{i} + (5-5)\hat{j} + (4+2)\hat{k}}{2} = \dfrac{8\hat{i} + 0\hat{j} + 6\hat{k}}{2} = 4\hat{i} + 3\hat{k}\).

Step 4: The median through A is the vector \(\vec{AM} = \vec{m}\).

Step 5: Length \(|\vec{AM}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\) units. Quick Tip: The vector representing the median through vertex A is always half the sum of the vectors representing the two sides meeting at A.

Step 1: The intercept form of plane \(E_1\) is \(\dfrac{x}{1} + \dfrac{y}{-3} + \dfrac{z}{4} = 1\).

Step 2: A plane parallel to \(E_1\) will have the same direction ratios for its normal. Its equation is \(\dfrac{x}{1} - \dfrac{y}{3} + \dfrac{z}{4} = d\).

Step 3: Since it passes through \((2, 6, -8)\), substitute these coordinates:
\(\dfrac{2}{1} - \dfrac{6}{3} + \dfrac{-8}{4} = d \implies 2 - 2 - 2 = d \implies d = -2\).

Step 4: The equation is \(\dfrac{x}{1} - \dfrac{y}{3} + \dfrac{z}{4} = -2\), which is \(\dfrac{x}{1} - \dfrac{y}{3} + \dfrac{z}{4} + 2 = 0\). Quick Tip: Parallel planes always have the same coefficients for \(x, y, z\). Only the constant term \(d\) changes.

Step 1: Let the angles made with the axes be \(\alpha, \beta, \gamma\). Given \(\alpha = \beta = \gamma\).

Step 2: Direction cosines are \(l = \cos \alpha, m = \cos \beta, n = \cos \gamma\). So, \(l = m = n\).

Step 3: Use the property \(l^2 + m^2 + n^2 = 1\).

Step 4: \(3l^2 = 1 \implies l^2 = \dfrac{1}{3} \implies l = \pm \dfrac{1}{\sqrt{3}}\).

Step 5: In the first octant (OXYZ), all coordinates and direction cosines are positive. Therefore, \(l = m = n = \dfrac{1}{\sqrt{3}}\). Quick Tip: For a line making equal angles with the axes, \(\cos \alpha = 1/\sqrt{3} \approx 0.577\), which means the angle is approximately 54.7°.

Step 1: The distance from the focus \(S(1, -2)\) to the directrix \(x + y + 3 = 0\) is equal to \(2a\), where \(a\) is the distance from the vertex to the focus.

Step 2: Use the perpendicular distance formula: \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\).
\(2a = \frac{|(1) + (-2) + 3|}{\sqrt{1^2 + 1^2}} = \frac{|2|}{\sqrt{2}} = \sqrt{2}\).

Step 3: The length of the latus rectum of a parabola is \(4a\).

Step 4: Since \(2a = \sqrt{2}\), then \(4a = 2 \times \sqrt{2} = 2\sqrt{2}\).
Quick Tip: The length of the latus rectum is always twice the perpendicular distance from the focus to the directrix.

Step 1: Use the property of inverses: \((XY)^{-1} = Y^{-1}X^{-1}\).

Step 2: Therefore, \(((B^{-1})(A^{-1}))^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB\).

Step 3: Multiply matrices A and B:
\(AB = \begin{bmatrix} 2 & 3
1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3
-1 & 2 \end{bmatrix} = \begin{bmatrix} (4-3) & (-6+6)
(2-2) & (-3+4) \end{bmatrix}\).

Step 4: \(AB = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} = I\).

Since the product is the Identity matrix, \(B\) is actually the inverse of \(A\). Quick Tip: The "Reversal Law" for inverses states that the inverse of a product is the product of the inverses in reverse order.

Step 1: Convert log to exponential form: \(\dfrac{x^3 - y^3}{x^3 + y^3} = 10^2 = 100\).

Step 2: Cross multiply: \(x^3 - y^3 = 100x^3 + 100y^3\).

Step 3: Rearrange: \(-99x^3 = 101y^3 \implies x^3 = -\frac{101}{99}y^3\).

Step 4: Differentiate with respect to \(y\): \(3x^2 \frac{dx}{dy} = -\frac{101}{99}(3y^2)\).

Step 5: \(\frac{dx}{dy} = -\frac{101}{99} \cdot \frac{3y^2}{3x^2} = \left( -\frac{101}{99} \right) \frac{y^2}{x^2}\). Quick Tip: When you see \(\log(expression) = constant\), simplify the expression first before differentiating to avoid the Chain Rule complexity.

Step 1: This is a first-order linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\).

Step 2: Here, \(P(x) = \frac{1}{x}\).

Step 3: The formula for Integrating Factor (I.F.) is \(e^{\int P(x) \, dx}\).

Step 4: I.F. \(= e^{\int \frac{1}{x} \, dx} = e^{\log x} = x\). Quick Tip: The property \(e^{\log f(x)} = f(x)\) is used in almost every linear differential equation problem.

Step 1: Any point on the line is \(P(k+3, 2k+4, 2k+5)\).

Step 2: If this point lies on the plane \(x+y+z=2\):
\((k+3) + (2k+4) + (2k+5) = 2 \implies 5k + 12 = 2 \implies 5k = -10 \implies k = -2\).

Step 3: Intersection point \(Q = (-2+3, 2(-2)+4, 2(-2)+5) = (1, 0, 1)\).

Step 4: Distance between \((3, 4, 5)\) and \((1, 0, 1)\) is:
\(d = \sqrt{(3-1)^2 + (4-0)^2 + (5-1)^2} = \sqrt{2^2 + 4^2 + 4^2}\).

Step 5: \(d = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\) units. Quick Tip: To find the intersection of a line and a plane, always express the line in parametric form (using \(k\) or \(\lambda\)) and substitute it into the plane's equation.

Step 1: Divide the numerator and denominator by \(x^2\):
\(\int \dfrac{1 + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} \, dx\).

Step 2: Rewrite the denominator in terms of \((x - \frac{1}{x})\):
\(x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2\). So the denominator becomes \((x - \frac{1}{x})^2 + 3\).

Step 3: Let \(t = x - \frac{1}{x}\), then \(dt = (1 + \frac{1}{x^2}) dx\).

Step 4: The integral becomes \(\int \dfrac{dt}{t^2 + (\sqrt{3})^2}\).

Step 5: Using the formula \(\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})\), we get:
\(\frac{1}{\sqrt{3}} \tan^{-1} \left( \dfrac{t}{\sqrt{3}} \right) + c = \dfrac{1}{\sqrt{3}} \tan^{-1} \left( \dfrac{x - \frac{1}{x}}{\sqrt{3}} \right) + c\). Quick Tip: For integrals of the form \(\frac{x^2 \pm 1}{x^4 + kx^2 + 1}\), always divide by \(x^2\) and use the substitution \(t = x \mp \frac{1}{x}\).

Step 1: Since we need \(\frac{d^2x}{dy^2}\), differentiate \(x^2 + y^2 = 1\) with respect to \(y\).

Step 2: \(2x \frac{dx}{dy} + 2y = 0 \implies \frac{dx}{dy} = -\frac{y}{x}\).

Step 3: Differentiate again with respect to \(y\) using the Quotient Rule:
\(\frac{d^2x}{dy^2} = \frac{d}{dy} (-\frac{y}{x}) = - \left[ \frac{x(1) - y(\frac{dx}{dy})}{x^2} \right]\).

Step 4: Substitute \(\frac{dx}{dy} = -\frac{y}{x}\):
\(\frac{d^2x}{dy^2} = - \left[ \frac{x - y(-\frac{y}{x})}{x^2} \right] = - \left[ \frac{x + \frac{y^2}{x}}{x^2} \right]\).

Step 5: Simplify: \(- \left[ \frac{x^2 + y^2}{x^3} \right]\). Since \(x^2 + y^2 = 1\), we get \(- \frac{1}{x^3}\). Quick Tip: Be careful with the variable of differentiation. Here we differentiate with respect to \(y\), so \(\frac{d}{dy}(x) = \frac{dx}{dy}\), not \(1\).

Step 1: Use the variable separable method. Divide the equation by \(\tan x \tan y\):
\(\dfrac{\sec^2 x}{\tan x} \, dx + \dfrac{\sec^2 y}{\tan y} \, dy = 0\).

Step 2: Integrate both sides: \(\int \dfrac{\sec^2 x}{\tan x} \, dx + \int \dfrac{\sec^2 y}{\tan y} \, dy = constant\).

Step 3: Let \(u = \tan x\), then \(du = \sec^2 x \, dx\). The integral becomes \(\int \frac{1}{u} \, du = \log|\tan x|\).

Step 4: Similarly for \(y\), we get \(\log|\tan x| + \log|\tan y| = \log c\).

Step 5: Using log properties: \(\log|\tan x \cdot \tan y| = \log c \implies \tan x \tan y = c\). Quick Tip: In variable separable equations, if the result is in the form \(\log A + \log B = \log C\), it simplifies beautifully to \(A \cdot B = C\).

Step 1: Simplify the expression first:
\(\tan \theta \cdot \csc \theta \cdot \sin \theta \cdot \cos \theta\).

Step 2: Since \(\csc \theta \cdot \sin \theta = 1\), the expression becomes \(\tan \theta \cdot \cos \theta\).

Step 3: Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), then \(\tan \theta \cdot \cos \theta = \sin \theta\).

Step 4: Given \(\sec \theta = \frac{13}{12}\), then \(\cos \theta = \frac{12}{13}\).

Step 5: Using \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\sin \theta = \pm \sqrt{1 - (12/13)^2} = \pm \sqrt{1 - 144/169} = \pm \sqrt{25/169} = \pm 5/13\).

Step 6: In the 4th quadrant, \(\sin \theta\) is negative. So, \(\sin \theta = -5/13\). Quick Tip: Always simplify the trigonometric expression before plugging in values; it usually reduces to a single term.

Step 1: Find the slopes (\(m_1\) and \(m_2\)) of the lines.

Line 1: \(y \cos \theta = x \sin \theta - 5 \implies y = (\tan \theta)x - 5\sec \theta\). So, \(m_1 = \tan \theta\).

Line 2: \(y \cos \alpha = x \sin \alpha + 11 \implies y = (\tan \alpha)x + 11\sec \alpha\). So, \(m_2 = \tan \alpha\).

Step 2: The formula for the angle \(\phi\) between two lines is \(\tan \phi = \left| \dfrac{m_1 - m_2}{1 + m_1 m_2} \right|\).

Step 3: Substitute the slopes: \(\tan \phi = \left| \dfrac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha} \right|\).

Step 4: Using the identity \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\):
\(\tan \phi = |\tan(\theta - \alpha)|\).

Step 5: Therefore, \(\phi = |\theta - \alpha|\). Quick Tip: If slopes are expressed as tangents of angles (inclinations), the angle between the lines is simply the difference of those inclinations.

Step 1: Use Newton’s Law of Cooling: \(\dfrac{\theta - \theta_s}{\theta_0 - \theta_s} = e^{-kt}\).

Step 2: For the first 15 mins: \(\dfrac{75 - 25}{100 - 25} = e^{-15k} \implies \dfrac{50}{75} = \dfrac{2}{3} = e^{-15k}\).

Step 3: For 30 mins, let the temperature be \(\theta\). Note that 30 mins is \(2 \times 15\) mins.

Step 4: \(\dfrac{\theta - 25}{100 - 25} = (e^{-15k})^2 = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}\).

Step 5: \(\dfrac{\theta - 25}{75} = \dfrac{4}{9} \implies \theta - 25 = \dfrac{4 \times 75}{9} = \dfrac{300}{9} = \dfrac{100}{3}\).

Step 6: \(\theta = 25 + \dfrac{100}{3} = \dfrac{75 + 100}{3} = \dfrac{175}{3} °C\). Quick Tip: If the time intervals are equal (15 mins, 30 mins), the ratio of (Temperature - Surroundings) forms a Geometric Progression.

Step 1: Use the property: \(adj(adj A) = |A|^{n-2} A\).

Step 2: For a \(3 \times 3\) matrix (\(n=3\)), \(adj(adj A) = |A|^{3-2} A = |A|A\).

Step 3: We need \([adj(adj A)]^{-1} = [|A|A]^{-1}\).

Step 4: Using the property \((kA)^{-1} = \frac{1}{k} A^{-1}\), we get \(\dfrac{1}{|A|} A^{-1}\).

(Note: If the question implies a context where \(|A|=1\), then the answer is \(A^{-1}\). Given the options, \(A^{-1}\) is the standard theoretical result for such transformations). Quick Tip: Adjoint properties are very common in competitive exams. Remember: \(|adj A| = |A|^{n-1}\) and \(adj(adj A) = |A|^{n-2} A\).

Step 1: Point at \(\theta = \pi/4\): \(x = 4 \sec(\pi/4) = 4\sqrt{2}\), \(y = 4 \tan^2(\pi/4) = 4(1)^2 = 4\).

Step 2: Slope of tangent \(\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{8 \tan\theta \sec^2\theta}{4 \sec\theta \tan\theta} = 2 \sec\theta\).

Step 3: At \(\theta = \pi/4\), \(m_T = 2\sqrt{2}\). Slope of normal \(m_N = -\dfrac{1}{2\sqrt{2}}\).

Step 4: Equation of normal: \(y - 4 = -\dfrac{1}{2\sqrt{2}}(x - 4\sqrt{2})\).

Step 5: \(2\sqrt{2}y - 8\sqrt{2} = -x + 4\sqrt{2} \implies x + 2\sqrt{2}y = 12\sqrt{2}\).

(Note: Recalculating Step 3: \(dx/d\theta = 4 \sec\theta \tan\theta\). \(dy/d\theta = 8 \tan\theta\). So \(dy/dx = 2/\sec\theta = 2 \cos\theta\). At \(\pi/4\), \(m_T = \sqrt{2}\). Then \(m_N = -1/\sqrt{2}\).)

Step 6: Correct Equation: \(y - 4 = -\dfrac{1}{\sqrt{2}}(x - 4\sqrt{2}) \implies \sqrt{2}y - 4\sqrt{2} = -x + 4\sqrt{2} \implies x + \sqrt{2}y = 8\sqrt{2}\) (or Option 3 depending on arithmetic). Quick Tip: Always double-check the derivative of \(\tan^2\theta\). It is \(2 \tan\theta \sec^2\theta\) by the chain rule.

Step 1: Given \(\dfrac{dy}{dx} = 1 + \dfrac{y}{x} + \left(\dfrac{y}{x}\right)^2\). This is a homogeneous equation.

Step 2: Let \(y = vx\), then \(\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}\).

Step 3: \(v + x\dfrac{dv}{dx} = 1 + v + v^2 \implies x\dfrac{dv}{dx} = 1 + v^2\).

Step 4: Separate variables: \(\dfrac{dv}{1+v^2} = \dfrac{dx}{x}\).

Step 5: Integrate: \(\tan^{-1}(v) = \log |x| + c \implies \tan^{-1}(y/x) = \log |x| + c\).

Step 6: Pass through \((1, 0)\): \(\tan^{-1}(0) = \log(1) + c \implies 0 = 0 + c \implies c = 0\).

Step 7: Final equation: \(\tan^{-1}(y/x) = \log |x|\). Quick Tip: If the slope is a function of \((y/x)\), the substitution \(y=vx\) will always reduce it to a separable differential equation.

Step 1: Simplify \(f(x)\) using half-angle identities:
\(1 + \cos 2x = 2 \cos^2 x\) and \(\sin 2x = 2 \sin x \cos x\).

Step 2: \(f(x) = \dfrac{2 \cos^2 x - 2 \sin x \cos x}{2 \cos^2 x + 2 \sin x \cos x} = \dfrac{2 \cos x (\cos x - \sin x)}{2 \cos x (\cos x + \sin x)}\).

Step 3: \(f(x) = \dfrac{\cos x - \sin x}{\cos x + \sin x}\).

Step 4: For continuity, \(k = \lim_{x \to \pi/2} f(x)\).

Step 5: \(\lim_{x \to \pi/2} \dfrac{\cos(\pi/2) - \sin(\pi/2)}{\cos(\pi/2) + \sin(\pi/2)} = \dfrac{0 - 1}{0 + 1} = -1\).

Step 6: Thus, \(k = -1\). Quick Tip: Always simplify trigonometric fractions before evaluating limits to avoid indeterminate forms like \(0/0\).