Let the potential at the center \(P\) be \(V_c\) and the electric field be \(E\). The potential at a point on the surface is given by \(V = V_c - \vec{E} \cdot \vec{r}\).


The maximum potential occurs when \(\vec{r}\) is opposite to \(\vec{E}\), and the minimum when \(\vec{r}\) is along \(\vec{E}\).

\(V_{max} = V_c + Er = 589.8\) V

\(V_{min} = V_c - Er = 589.0\) V


The potential at the center \(V_c\) is the average of the maximum and minimum potentials:

\(V_c = \frac{589.8 + 589.0}{2} = 589.4\) V


The difference \(2Er = 589.8 - 589.0 = 0.8\) V, so \(Er = 0.4\) V.


We need the potential at an angle \(\theta = 60^{\circ}\) with the field direction:

\(V = V_c - Er \cos(60^{\circ})\)

\(V = 589.4 - 0.4 \times \frac{1}{2}\)

\(V = 589.4 - 0.2 = 589.2\) V
Quick Tip: For a uniform field, the potential varies linearly with position. The potential at any point on a sphere is \(V(\theta) = V_{center} - \Delta V_{max} \cos \theta\), where \(\Delta V_{max}\) is the maximum potential difference from the center to the surface.

For the net force on the test charge to be zero, the electric force must cancel the magnetic force: \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) = 0\).


This implies \(\vec{E} = - (\vec{v} \times \vec{B}) = \vec{B} \times \vec{v}\).


Let's calculate the cross product \(\vec{v} \times \vec{B}\):

\(\vec{v} \times \vec{B} = v_0 B_0 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -1 & 2
1 & 2 & -4 \end{vmatrix}\)

\(= v_0 B_0 [ \hat{i}((-1)(-4) - (2)(2)) - \hat{j}((3)(-4) - (2)(1)) + \hat{k}((3)(2) - (-1)(1)) ]\)

\(= v_0 B_0 [ \hat{i}(4 - 4) - \hat{j}(-12 - 2) + \hat{k}(6 + 1) ]\)

\(= v_0 B_0 [ 0\hat{i} + 14\hat{j} + 7\hat{k} ]\)


Since \(\vec{E} = - (\vec{v} \times \vec{B})\), we have:

\(\vec{E} = - v_0 B_0 (14\hat{j} + 7\hat{k})\)
Quick Tip: This is the principle behind a velocity selector. For zero deflection, \(\vec{E}\) must be equal and opposite to \(\vec{v} \times \vec{B}\).

The rod experiences expansion due to heat and compression due to the force F.


The change in length due to thermal expansion is \(\Delta l_{th} = l \alpha \Delta T\).


The change in length due to mechanical compression is \(\Delta l_{mech} = \frac{Fl}{AY}\).


Since the net change in length is zero, the magnitude of expansion must equal the magnitude of compression:

\(l \alpha \Delta T = \frac{Fl}{AY}\)


The length \(l\) cancels out. Solving for F:

\(F = AY \alpha \Delta T\)
Quick Tip: Thermal stress is generated when expansion is restricted. The force required to prevent expansion is independent of the length of the rod.

The potential energy \(U\) of a magnetic dipole \(\vec{m}\) in a field \(\vec{B}\) is given by \(U = -\vec{m} \cdot \vec{B} = -mB \cos \theta\).


The torque \(\tau\) is given by \(\vec{\tau} = \vec{m} \times \vec{B}\), with magnitude \(\tau = mB \sin \theta\).


The torque is maximum when \(\sin \theta = 1\), i.e., \(\theta = 90^{\circ}\).


At \(\theta = 90^{\circ}\), the potential energy is \(U = -mB \cos(90^{\circ}) = 0\).


Thus, when the torque is maximum, the potential energy is zero.
Quick Tip: Remember: Torque relates to sine (max at \(90^{\circ}\)), and Potential Energy relates to cosine (zero at \(90^{\circ}\)).

Write the dimensional formulas for the given quantities:

\([T] = [T]\)

\([C] = [L T^{-1}]\)

\([h] = [M L^2 T^{-1}]\) (angular momentum \(mvr\))


Let \([M] = [T]^x [C]^y [h]^z\).


Substitute the dimensions:

\([M] = [T]^x [L T^{-1}]^y [M L^2 T^{-1}]^z\)

\([M] = [M]^z [L]^{y+2z} [T]^{x-y-z}\)


Comparing powers of M: \(z = 1\).


Comparing powers of L: \(y + 2z = 0 \Rightarrow y + 2(1) = 0 \Rightarrow y = -2\).


Comparing powers of T: \(x - y - z = 0 \Rightarrow x - (-2) - 1 = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1\).


Substituting the values back:

\([M] = [T]^{-1} [C]^{-2} [h]^1 = [T^{-1} C^{-2} h]\).
Quick Tip: Use the method of dimensional homogeneity. Set up equations for the exponents of Mass, Length, and Time and solve the system of linear equations.

The magnetic field \(B\) produced by the large loop at its center is \(B = \frac{\mu_0 I}{2b}\).


Since \(b \gg a\), we assume this field is constant over the area of the small loop.


The magnetic flux through the small loop is \(\Phi = B \cdot (\pi a^2) = \frac{\mu_0 I \pi a^2}{2b}\).


Substitute \(I = I_0 \cos(\omega t)\):

\(\Phi = \frac{\mu_0 \pi a^2 I_0}{2b} \cos(\omega t)\).


The induced emf is \(\varepsilon = -\frac{d\Phi}{dt}\).

\(\varepsilon = - \frac{\mu_0 \pi a^2 I_0}{2b} \frac{d}{dt}(\cos(\omega t))\).

\(\varepsilon = - \frac{\mu_0 \pi a^2 I_0}{2b} (-\omega \sin(\omega t))\).

\(\varepsilon = \frac{\pi \mu_0 I_0 a^2 \omega}{2b} \sin(\omega t)\).
Quick Tip: For mutual inductance problems where one loop is much smaller, assume the field from the larger loop is uniform across the smaller loop's area.

The reaction is \(_1^2H + _1^2H \rightarrow _2^4He\).


The energy released (Q-value) is the difference between the total binding energy of the products and the reactants.


Total Binding Energy of reactants (2 Deuterons):

Each deuteron (\(^2H\)) has 2 nucleons. Binding energy per nucleon is 1.1 MeV.
\(BE_{reactants} = 2 \times (2 \times 1.1) = 4.4\) MeV.


Total Binding Energy of product (1 Helium):

Helium (\(^4He\)) has 4 nucleons. Binding energy per nucleon is 7.0 MeV.
\(BE_{product} = 4 \times 7.0 = 28.0\) MeV.


Energy Released = \(BE_{product} - BE_{reactants}\).


Energy Released = \(28.0 - 4.4 = 23.6\) MeV.
Quick Tip: Energy released in fusion = (Binding Energy of Product) - (Sum of Binding Energies of Reactants). Higher binding energy implies a more stable nucleus.

Heat lost by Aluminium = Heat gained by Water + Heat gained by Calorimeter.


Mass of Aluminium \(m_{Al} = 0.20\) kg.

Change in temp of Al \(\Delta T_{Al} = 150 - 40 = 110^{\circ}\)C.


Mass of water \(m_w = 150 cc \approx 0.150\) kg (density 1 g/cc).

Water equivalent of calorimeter \(w = 0.025\) kg.

Total equivalent mass of water \(M = m_w + w = 0.150 + 0.025 = 0.175\) kg.

Change in temp of water \(\Delta T_w = 40 - 27 = 13^{\circ}\)C.

Specific heat of water \(c_w = 4200\) J/kg-\(^{\circ}\)C (since 4.2 J = 1 cal and 1 cal/g\(^{\circ}\)C = 4200 J/kg\(^{\circ}\)C).


Equation:
\(m_{Al} \cdot c_{Al} \cdot \Delta T_{Al} = M \cdot c_w \cdot \Delta T_w\)

\(0.20 \cdot c_{Al} \cdot 110 = 0.175 \cdot 4200 \cdot 13\)

\(22 \cdot c_{Al} = 735 \cdot 13\)

\(22 \cdot c_{Al} = 9555\)

\(c_{Al} = \frac{9555}{22} \approx 434.3\) J/kg-\(^{\circ}\)C.


Matching the closest option, we get 434 J/kg-\(^{\circ}\)C.
Quick Tip: When using water equivalent for a calorimeter, simply add it to the mass of the water and treat the calorimeter as that much extra water.

Let the mass of the empty pans be \(m\). Let the length of the left arm be \(L_1\) and the right arm be \(L_2\).


Initially, without the extra weight, the balance was not horizontal. The fact that adding weight to the left makes it horizontal means the right side moment was initially greater than the left side moment (\(mgL_2 > mgL_1\)).


When 5 mg (\(w\)) is added to the left pan, the beam is balanced.


Taking moments about the pivot:
\((m + w)g L_1 = m g L_2\)


Dividing by \(g\):
\(m L_1 + w L_1 = m L_2\)

\(m (L_2 - L_1) = w L_1\)


Since mass \(m\), weight \(w\), and length \(L_1\) are all positive values, the term \((L_2 - L_1)\) must be positive.

\(L_2 - L_1 > 0 \Rightarrow L_2 > L_1\)


Therefore, the right arm (\(L_2\)) is longer than the left arm (\(L_1\)), or the left arm is shorter than the right arm.
Quick Tip: In a physical balance, equilibrium is achieved when the net torque is zero: \(m_{left} L_{left} = m_{right} L_{right}\). If adding mass to one side balances it, that side's initial moment was smaller.

For Simple Harmonic Motion (SHM):

Maximum velocity \(v_{max} = \omega A\)

Maximum acceleration \(a_{max} = \omega^2 A\)


The ratio is given as:
\(\frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} = \omega = 10 s^{-1}\).


The displacement equation is \(x = A \sin(\omega t + \phi)\).

Given that at \(t=0\), \(x=5\) m and \(\phi = \frac{\pi}{4}\).


Substitute these values to find Amplitude \(A\):
\(5 = A \sin(0 + \frac{\pi}{4})\)
\(5 = A \frac{1}{\sqrt{2}}\)
\(A = 5\sqrt{2}\) m.


Now calculate the maximum acceleration:
\(a_{max} = \omega^2 A\)
\(a_{max} = (10)^2 \times (5\sqrt{2})\)
\(a_{max} = 100 \times 5\sqrt{2} = 500\sqrt{2} m/s^2\).
Quick Tip: Key SHM formulas: \(v_{max} = \omega A\), \(a_{max} = \omega^2 A\). The ratio \(a_{max}/v_{max}\) gives the angular frequency \(\omega\).

The degrees of freedom of the gas molecules are given as \(f = 5\).


The molar specific heat at constant volume is given by \(C_v = \frac{f}{2}R\).
\(C_v = \frac{5}{2}R\)


The molar specific heat at constant pressure is given by Mayer's relation \(C_p = C_v + R\).
\(C_p = \frac{5}{2}R + R = \frac{7}{2}R\)


The ratio of specific heats is \(\gamma = \frac{C_p}{C_v}\).
\(\gamma = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}\)
Quick Tip: For an ideal gas with \(f\) degrees of freedom, the adiabatic index is given directly by \(\gamma = 1 + \frac{2}{f}\). For \(f=5\), \(\gamma = 1 + 0.4 = 1.4 = 7/5\).

The energy stored in the electric field of a charged sphere is given by the potential energy formula:
\(U = \frac{Q^2}{2C}\)


For an isolated conducting sphere of radius \(R\), the capacitance is \(C = 4\pi\epsilon_0 R\).

Therefore, \(U = \frac{Q^2}{2(4\pi\epsilon_0 R)} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{2R}\).


Given:
\(U = 4.5\) J
\(Q = 4 \muC = 4 \times 10^{-6}\) C
\(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) N-m\(^2\)/C\(^2\)


Substitute these values into the energy equation:
\(4.5 = (9 \times 10^9) \frac{(4 \times 10^{-6})^2}{2R}\)

\(4.5 = \frac{9 \times 10^9 \times 16 \times 10^{-12}}{2R}\)

\(4.5 = \frac{144 \times 10^{-3}}{2R}\)

\(9.0 R = 0.144\)

\(R = \frac{0.144}{9} = 0.016\) m


Converting to millimeters:
\(R = 16\) mm
Quick Tip: The self-energy of a conducting sphere is \(\frac{kQ^2}{2R}\). Do not confuse this with the interaction energy between two charges.

Step 1: Interpreting the given condition

The question states that the object loses \(50%\) of its kinetic energy after each collision with the ground.

Kinetic energy \(K \propto v^2\).

If kinetic energy is reduced to \(50%\), then: \[ \frac{1}{2}mv'^2 = \frac{1}{2} \left( \frac{1}{2}mv^2 \right) \Rightarrow v'^2 = \frac{1}{2}v^2 \Rightarrow v' = \frac{v}{\sqrt{2}} \]
This would give the coefficient of restitution \(e = \frac{1}{\sqrt{2}}\), leading to a total distance of \(3h\), which is not among the options.


Hence, the intended (and option-consistent) interpretation is that the object loses \(50%\) of its velocity after each collision, i.e., \[ e = \frac{1}{2} \]

Step 2: Heights reached after successive bounces

When an object rebounds with coefficient of restitution \(e\), the maximum height reached after each bounce reduces by a factor \(e^2\).

Thus, the heights reached are: \[ h_1 = e^2 h,\quad h_2 = e^4 h,\quad h_3 = e^6 h,\ \ldots \]

Step 3: Expression for total distance travelled

The object first falls a distance \(h\).

After that, for each bounce, it travels upward and downward through the same height.

Hence, total distance \(D\) is: \[ D = h + 2(h_1 + h_2 + h_3 + \cdots) \] \[ D = h + 2h(e^2 + e^4 + e^6 + \cdots) \]

Step 4: Evaluating the geometric series

The series \(e^2 + e^4 + e^6 + \cdots\) is an infinite geometric series with: \[ a = e^2,\quad r = e^2 \] \[ Sum = \frac{e^2}{1 - e^2} \]

Step 5: Substitute the value of \(e\)

Given \(e = \frac{1}{2}\), so \(e^2 = \frac{1}{4}\): \[ D = h + 2h \left( \frac{\frac{1}{4}}{1 - \frac{1}{4}} \right) = h + 2h \left( \frac{\frac{1}{4}}{\frac{3}{4}} \right) \] \[ D = h + 2h \left( \frac{1}{3} \right) = h + \frac{2}{3}h \]

Step 6: Final Answer
\[ D = \frac{5}{3}h \]
\[ \boxed{D = \frac{5}{3}h} \] Quick Tip: The total distance traveled by a bouncing object is \(H \left(\frac{1+e^2}{1-e^2}\right)\), where \(H\) is the initial drop height and \(e\) is the coefficient of restitution.

The electron moving in an orbit is equivalent to a current loop.

The effective current \(I\) is charge divided by time period: \(I = \frac{e}{T} = \frac{e v}{2\pi r}\).


The magnetic field at the center of a circular loop is \(B = \frac{\mu_0 I}{2r}\).

Substituting \(I\):
\(B = \frac{\mu_0 (e v / 2\pi r)}{2r} \propto \frac{v}{r^2}\).


From Bohr's theory:

Velocity \(v \propto \frac{Z}{n} \propto \frac{1}{n}\).

Radius \(r \propto \frac{n^2}{Z} \propto n^2\).


Substitute these proportionalities into the expression for B:
\(B \propto \frac{(1/n)}{(n^2)^2}\)
\(B \propto \frac{1}{n \cdot n^4}\)
\(B \propto \frac{1}{n^5}\) or \(n^{-5}\).
Quick Tip: In the Bohr model, remember the key dependencies: \(r \propto n^2\), \(v \propto 1/n\), \(T \propto n^3\), and \(I \propto n^{-3}\).

Let the refractive index of the denser medium be \(\mu_1\) and the rarer medium be \(\mu_2\).

The condition given (reflected \(\perp\) refracted) is the condition for Brewster's angle (polarizing angle).


Using Snell's law at the interface: \(\mu_1 \sin A = \mu_2 \sin r\).

Since the reflected and refracted rays are perpendicular: \(r + 90^{\circ} + A = 180^{\circ} \Rightarrow r = 90^{\circ} - A\).


Substitute \(r\) in Snell's law:
\(\mu_1 \sin A = \mu_2 \sin(90^{\circ} - A) = \mu_2 \cos A\).
\(\tan A = \frac{\mu_2}{\mu_1}\).


We are given the critical angle \(\theta_C\). The formula for critical angle is:
\(\sin \theta_C = \frac{\mu_{rare}}{\mu_{dense}} = \frac{\mu_2}{\mu_1}\).


Comparing the two equations:
\(\tan A = \sin \theta_C\).
\(A = \tan^{-1} (\sin \theta_C)\).
Quick Tip: When the reflected and refracted rays are perpendicular, the angle of incidence satisfies \(\tan i_p = \frac{n_{refracted}}{n_{incident}}\). This is Brewster's Law.

For single slit diffraction, the position of the \(n\)-th minimum is given by \(y_n = \frac{n \lambda D}{b}\).


Given:

Distance of 2nd minimum (\(n=2\)): \(y_2 = 3\) cm.

Distance of 4th minimum (\(n=4\)): \(y_4 = 6\) cm.

This is consistent since \(y_4 = 2y_2\).


We need to find the width of the central maximum. The central maximum extends from the first minimum on one side (\(n=1\)) to the first minimum on the other side (\(n=-1\)).


First, find the position of the 1st minimum (\(y_1\)):

Since \(y_n \propto n\), \(y_1 = \frac{y_2}{2} = \frac{3 cm}{2} = 1.5\) cm.


The width of the central maximum is \(2 y_1\).

Width \(= 2 \times 1.5\) cm \(= 3.0\) cm.
Quick Tip: The width of the central maximum in single slit diffraction is twice the width of any other secondary fringe (distance between consecutive minima). \(W_{central} = \frac{2\lambda D}{b}\).

Since the joint O is a node and the ends are nodes (bridges), both wires vibrate in their respective harmonic modes with the same frequency \(f\).


The frequency of vibration for a wire of length \(L\) with \(n\) antinodes (loops) is \(f = \frac{n}{2L} v\), where \(v = \sqrt{\frac{T}{\mu}}\).
\(\mu = \rho \pi r^2\) is the linear mass density.


Given that lengths are equal (\(L_1 = L_2\)), Tension \(T\) is same, and radius \(r\) is same.
\(v \propto \frac{1}{\sqrt{\rho}}\).


For wire 1: \(f = \frac{n_1}{2L} \sqrt{\frac{T}{\rho_1 \pi r^2}}\).

For wire 2: \(f = \frac{n_2}{2L} \sqrt{\frac{T}{\rho_2 \pi r^2}}\).


Equating the frequencies:
\(n_1 \sqrt{\frac{1}{\rho_1}} = n_2 \sqrt{\frac{1}{\rho_2}}\).

\(\frac{n_1}{n_2} = \sqrt{\frac{\rho_1}{\rho_2}}\).


Given \(\rho_2 = 4\rho_1\):
\(\frac{n_1}{n_2} = \sqrt{\frac{\rho_1}{4\rho_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\).


Ratio \(n_1 : n_2 = 1 : 2\).
Quick Tip: For composite wires in series with the same tension and frequency, the number of loops \(n\) is inversely proportional to the wave velocity \(v\). Since \(v \propto 1/\sqrt{\rho}\), \(n \propto \sqrt{\rho}\).

Let \(T\) be the tension in the string and \(a\) be the acceleration of mass \(m\).

Equation of motion for mass \(m\):
\(mg - T = ma\) --- (1)


Equation of torque for the disc:
\(\tau = T \cdot R = I \alpha\)

Since the string does not slip, \(a = R \alpha\).

Moment of inertia of disc \(I = \frac{1}{2}MR^2\).
\(T \cdot R = (\frac{1}{2}MR^2) (\frac{a}{R})\)
\(T = \frac{1}{2} M a\) --- (2)


Substitute (2) into (1):
\(mg - \frac{1}{2} M a = ma\)
\(mg = ma + \frac{1}{2} M a\)
\(mg = a (m + \frac{M}{2})\)
\(mg = a (\frac{2m + M}{2})\)

\(a = \frac{2mg}{2m + M}\)
Quick Tip: For "falling mass unwinding a pulley" problems, the acceleration is always \(a = \frac{g}{1 + I/(mR^2)}\). Substituting \(I_{disc} = MR^2/2\) gives the result quickly.

Step 1: Understanding the physical condition

The object has:

Positive velocity \(\Rightarrow v > 0\) (motion in the positive direction).
Constant negative acceleration \(\Rightarrow a < 0\) (uniform retardation).

Hence, the speed of the object continuously decreases but remains positive until it eventually becomes zero.


Step 2: Identifying the type of graph

From the given options, Graphs 2 and 3 are velocity (\(v\)) vs distance (\(x\)) graphs, since the velocity decreases to zero at a finite distance.


Step 3: Using the equation of motion

For motion with constant acceleration, the relation between velocity, acceleration, and distance is: \[ v^2 = u^2 + 2ax \]
Since the acceleration is negative, let \(a = -k\) where \(k > 0\): \[ v^2 = u^2 - 2kx \]

Step 4: Nature of the \(v\)--\(x\) graph

The above equation shows that \(v^2\) varies linearly with \(x\), which implies that the \(v\) vs \(x\) graph is a parabola.

At: \[ x = 0,\quad v = u \ (positive) \]
As \(x\) increases, the value of \(v\) decreases continuously and finally becomes zero at a finite distance.


Step 5: Slope of the \(v\)--\(x\) graph

Differentiate \(v^2 = u^2 - 2kx\) with respect to \(x\): \[ 2v \frac{dv}{dx} = -2k \] \[ \frac{dv}{dx} = -\frac{k}{v} \]

As \(v \rightarrow 0\), the magnitude of the slope: \[ \left| \frac{dv}{dx} \right| \rightarrow \infty \]
This means the graph must meet the distance axis with a vertical tangent.


Step 6: Matching with the given graphs


Graph 2 approaches the axis asymptotically, which contradicts the requirement of an infinite slope.
Graph 3 starts with a positive velocity, decreases smoothly, and meets the distance axis with a vertical tangent.


Therefore, Graph 3 correctly represents motion with constant negative acceleration and positive velocity.

\[ \boxed{Correct option: (C) Graph 3} \] Quick Tip: For constant acceleration, the \(v\)--\(x\) graph is parabolic (\(v^2 \propto x\)). If an object comes to rest, the curve must hit the \(x\)-axis perpendicularly because \(\displaystyle \frac{dv}{dx} = \frac{a}{v}\) becomes infinite at \(v = 0\).

We need to determine the dynamic resistance in forward bias and the resistance in reverse bias.


Forward Bias Resistance (\(r_f\)):

From the graph, in the linear region (around the operating points shown):
\(V_1 = 0.7\) V, \(I_1 = 10\) mA
\(V_2 = 0.8\) V, \(I_2 = 20\) mA
\(r_f = \frac{\Delta V}{\Delta I} = \frac{0.8 - 0.7}{(20 - 10) \times 10^{-3}} = \frac{0.1}{10 \times 10^{-3}} = \frac{0.1}{0.01} = 10 \, \Omega\).


Reverse Bias Resistance (\(R_r\)):

From the graph, at \(V = -10\) V, the current is \(I = -1 \, \mu\)A.

Since the curve is flat, we take the static resistance:
\(R_r = \frac{V}{I} = \frac{10}{1 \times 10^{-6}} = 10^7 \, \Omega\).


Ratio of forward to reverse resistance:

Ratio \(= \frac{r_f}{R_r} = \frac{10}{10^7} = 10^{-6}\).
Quick Tip: Forward resistance of a diode is typically low (\(1-100 \Omega\)), while reverse resistance is very high (\(M\Omega\)). The ratio \(R_{fwd}/R_{rev}\) is extremely small.

Using Einstein's photoelectric equation:
\(K_{max} = h\nu - \Phi\)


Case 1: Frequency is \(n\). Maximum velocity is \(v\).
\(\frac{1}{2}mv^2 = hn - \Phi\) \quad ...(1)


Case 2: Frequency is \(3n\). Maximum velocity is \(v'\).
\(\frac{1}{2}mv'^2 = h(3n) - \Phi\) \quad ...(2)


Substitute \(hn\) from equation (1) into equation (2):
\(hn = \frac{1}{2}mv^2 + \Phi\)
\(\frac{1}{2}mv'^2 = 3(\frac{1}{2}mv^2 + \Phi) - \Phi\)

\(\frac{1}{2}mv'^2 = \frac{3}{2}mv^2 + 3\Phi - \Phi\)
\(\frac{1}{2}mv'^2 = \frac{3}{2}mv^2 + 2\Phi\)


Since the work function \(\Phi\) is positive (\(\Phi > 0\)):
\(\frac{1}{2}mv'^2 > \frac{3}{2}mv^2\)

\(v'^2 > 3v^2\)
\(v' > \sqrt{3}v\)
Quick Tip: When the frequency of incident light is increased by a factor of \(x\), the maximum kinetic energy increases by a factor greater than \(x\) because the constant work function subtraction becomes relatively smaller.

Let \(k\) be the potential gradient along the potentiometer wire. The potential drop across a length \(l\) is \(V = kl\).


When the key \(K_3\) connects terminals 1 and 2, the galvanometer measures the potential drop across resistor \(R_1\). The balancing length is \(l_1\).
\(V_1 = I R_1 = k l_1\) \quad ...(1)


When the key \(K_3\) connects terminals 1 and 3, the galvanometer measures the potential drop across the series combination of \(R_1\) and \(R_2\). The balancing length is \(l_2\).
\(V_2 = I (R_1 + R_2) = k l_2\) \quad ...(2)


Divide equation (1) by equation (2):
\(\frac{I R_1}{I (R_1 + R_2)} = \frac{k l_1}{k l_2}\)

\(\frac{R_1}{R_1 + R_2} = \frac{l_1}{l_2}\)


Invert the equation:
\(\frac{R_1 + R_2}{R_1} = \frac{l_2}{l_1}\)

\(1 + \frac{R_2}{R_1} = \frac{l_2}{l_1}\)

\(\frac{R_2}{R_1} = \frac{l_2}{l_1} - 1 = \frac{l_2 - l_1}{l_1}\)


Taking the reciprocal to find \(\frac{R_1}{R_2}\):
\(\frac{R_1}{R_2} = \frac{l_1}{l_2 - l_1}\)
Quick Tip: In a potentiometer, the balancing length is directly proportional to the potential difference being measured. \(l_1 \propto V_1\) and \(l_2 \propto V_2\).

Given values:

Modulating signal: \(A_m = 5\) V, \(f_m = 20\) kHz.

Carrier wave: \(A_c = 25\) V, \(f_c = 1.2\) MHz = 1200 kHz.


Modulation Index \(\mu\):
\(\mu = \frac{A_m}{A_c} = \frac{5}{25} = 0.2\)


Side band frequencies are given by \(f_c \pm f_m\).

Upper Side Band (USB) = \(f_c + f_m = 1200 + 20 = 1220\) kHz.

Lower Side Band (LSB) = \(f_c - f_m = 1200 - 20 = 1180\) kHz.


Comparing with the options, (A) is correct.
Quick Tip: The modulation index determines the quality of transmission and should be less than or equal to 1 to avoid distortion. Sidebands are located symmetrically around the carrier frequency.

Let the mass of the original large triangle ABC be \(M\) and side length be \(L\).

The moment of inertia is \(I_0 \propto ML^2\).


The removed triangle DEF is formed by the midpoints. Its side length is \(L/2\).

Since the lamina is uniform, Mass \(\propto\) Area \(\propto (side)^2\).

Mass of removed triangle \(m = M (\frac{L/2}{L})^2 = \frac{M}{4}\).


The removed triangle DEF is geometrically similar and shares the same centroid O.

Its moment of inertia \(I_{removed}\) about O follows the same scaling law:
\(I_{removed} \propto m (L/2)^2 \propto (\frac{M}{4}) (\frac{L}{2})^2 = \frac{1}{16} (ML^2)\).


Thus, \(I_{removed} = \frac{I_0}{16}\).


The moment of inertia of the remaining part is:
\(I_{remaining} = I_{total} - I_{removed}\)
\(I_{remaining} = I_0 - \frac{I_0}{16} = \frac{15 I_0}{16}\).
Quick Tip: For a self-similar part cut from a laminar body (where both share the same axis of rotation), if the linear dimension scales by \(k\), the Mass scales by \(k^2\) and the Moment of Inertia scales by \(k^4\). Here \(k=1/2\), so \(I' = (1/2)^4 I = I/16\).

The effective gravity at the equator due to rotation is \(g' = g - \omega^2 R\).

The weight is \(W' = m g'\). The original weight is \(W = mg\).


Given \(W' = \frac{3}{4} W\).
\(m(g - \omega^2 R) = \frac{3}{4} mg\)
\(g - \omega^2 R = \frac{3}{4} g\)
\(\omega^2 R = \frac{1}{4} g\)

\(\omega = \sqrt{\frac{g}{4R}} = \frac{1}{2} \sqrt{\frac{g}{R}}\)


Substitute values (\(g=10\), \(R=6400 \times 10^3\) m):
\(\omega = \frac{1}{2} \sqrt{\frac{10}{6400000}} = \frac{1}{2} \sqrt{\frac{1}{640000}}\)
\(\omega = \frac{1}{2} \times \frac{1}{800} = \frac{1}{1600}\) rad/s.

\(\omega = 0.000625\) rad/s.
\(\omega = 0.625 \times 10^{-3}\) rad/s \(\approx 0.63 \times 10^{-3}\) rad/s.
Quick Tip: The centrifugal force \(m\omega^2 R\) acts opposite to gravity at the equator, reducing the apparent weight. For weightlessness, \(\omega = \sqrt{g/R}\).

The wave propagation direction is determined by the argument \((kx + \omega t)\). Since it is \((+x, +t)\), the wave travels in the \(-\hat{i}\) direction.


The direction of propagation is given by the vector \(\vec{E} \times \vec{B}\).

Given \(\vec{B}\) is in the \(\hat{j}\) direction.

We need \(dir(\vec{E}) \times \hat{j} = -\hat{i}\).


Using the cyclic properties of unit vectors:
\(\hat{k} \times \hat{j} = -\hat{i}\).

So, the electric field must be in the \(\hat{k}\) direction.


The magnitude of the electric field is related to the magnetic field by \(E_0 = c B_0\).


Therefore, \(\vec{E} = c B_0 \sin (kx + \omega t) \hat{k}\).
Quick Tip: For EM waves, \(\hat{v} = \hat{E} \times \hat{B}\) where \(\hat{v}\) is the direction of propagation. Also \(E = cB\). Always check the phase \((kx \pm \omega t)\) to determine direction.

Work done \(W\) is the area enclosed by the cycle:
\(W = (2V_0 - V_0)(2P_0 - P_0) = P_0 V_0\).


Heat is absorbed (\(Q_{in}\)) during processes where Temperature increases.
\(T_A = \frac{P_0 V_0}{nR}\), \(T_B = \frac{2P_0 V_0}{nR} = 2T_A\), \(T_C = \frac{2P_0 (2V_0)}{nR} = 4T_A\), \(T_D = 2T_A\).


Process A \(\rightarrow\) B (Isochoric Heating):
\(Q_{AB} = n C_v \Delta T = n (1.5 R) (T_B - T_A) = 1.5 n R (T_A) = 1.5 P_0 V_0\).


Process B \(\rightarrow\) C (Isobaric Expansion):
\(Q_{BC} = n C_p \Delta T\). Since \(C_v = 1.5R\), \(C_p = 2.5R\).
\(Q_{BC} = n (2.5 R) (T_C - T_B) = 2.5 n R (2T_A) = 5 n R T_A = 5 P_0 V_0\).


Total Heat Input \(Q_{in} = 1.5 P_0 V_0 + 5 P_0 V_0 = 6.5 P_0 V_0\).


Efficiency \(\eta = \frac{W}{Q_{in}} = \frac{P_0 V_0}{6.5 P_0 V_0} = \frac{1}{6.5}\).
\(\eta = \frac{10}{65} = \frac{2}{13} \approx 0.1538\).


Rounding to two decimal places gives 0.15.
Quick Tip: Efficiency \(\eta = \frac{Work Done}{Heat Absorbed}\). Be careful to only sum the positive heat inputs (where temperature increases in isochoric/isobaric steps) for \(Q_{in}\).

Frequency of a spring-mass system is \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\).


Case 1: \(m_1 = 1\) kg, \(k_1 = k\).
\(f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{1}} = 1\) Hz.

So, \(\frac{\sqrt{k}}{2\pi} = 1\).


Case 2: \(m_2 = 8\) kg. Two springs in parallel, so equivalent spring constant \(k_2 = k + k = 2k\).
\(f_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m_2}} = \frac{1}{2\pi} \sqrt{\frac{2k}{8}} = \frac{1}{2\pi} \sqrt{\frac{k}{4}}\).

\(f_2 = \frac{1}{2} \left( \frac{1}{2\pi} \sqrt{\frac{k}{1}} \right)\).


Substitute \(f_1\):
\(f_2 = \frac{1}{2} \times 1 = 0.5\) Hz.
Quick Tip: Springs in parallel add their stiffness (\(k_{eq} = k_1 + k_2\)). Frequency scales as \(\sqrt{k/m}\). Here \(k\) doubles and \(m\) becomes 8 times, so ratio is \(\sqrt{2/8} = 1/2\).

Let the equivalent resistance of the infinite ladder network be \(R_{eq}\).

The network consists of repeating units of series resistors (\(1\Omega + 1\Omega\)) and a shunt resistor (\(4\Omega\)).

Since it is infinite, adding one more section to the front does not change the equivalent resistance.

\(R_{eq} = (1 + 1) + (4 || R_{eq})\)
\(R_{eq} = 2 + \frac{4 R_{eq}}{4 + R_{eq}}\)
\(R_{eq} (4 + R_{eq}) = 2(4 + R_{eq}) + 4 R_{eq}\)
\(4 R_{eq} + R_{eq}^2 = 8 + 2 R_{eq} + 4 R_{eq}\)
\(R_{eq}^2 - 2 R_{eq} - 8 = 0\)
\((R_{eq} - 4)(R_{eq} + 2) = 0\)

Since resistance must be positive, \(R_{eq} = 4 \Omega\).


Now consider the whole circuit. The battery (\(E=9\) V, \(r=0.5 \Omega\)) is connected to \(R_{eq} = 4 \Omega\).

Total current \(I = \frac{E}{r + R_{eq}} = \frac{9}{0.5 + 4} = \frac{9}{4.5} = 2\) A.


The ammeter \(A_1\) is in series with the input line (top branch), so it reads the total current.

Reading of \(A_1 = 2\) A.


Let's check the voltmeter reading. V is connected across the input of the network.
\(V = E - I r = 9 - 2(0.5) = 8\) V. (So Options A and D are incorrect).
Quick Tip: For infinite ladder networks, assume the resistance of the infinite chain is \(R\) and set up the equation \(R = R_{series} + (R_{shunt} || R)\).

The conductivity \(\sigma\) of a semiconductor is given by:
\(\sigma = e (n_e \mu_e + n_h \mu_h)\)


Given values:
\(e = 1.6 \times 10^{-19}\) C
\(n_e = 5 \times 10^{18}\) m\(^{-3}\)
\(n_h = 5 \times 10^{19}\) m\(^{-3}\)
\(\mu_e = 2.0\) m\(^2\)/Vs
\(\mu_h = 0.01\) m\(^2\)/Vs


Substitute these values:
\(\sigma = 1.6 \times 10^{-19} [ (5 \times 10^{18} \times 2.0) + (5 \times 10^{19} \times 0.01) ]\)


Calculate the terms inside the bracket:
\(n_e \mu_e = 10 \times 10^{18} = 10^{19}\)
\(n_h \mu_h = 0.05 \times 10^{19} = 5 \times 10^{17} = 0.5 \times 10^{18}\)


Let's keep powers consistent (\(10^{18}\)):

Sum \(= 10 \times 10^{18} + 0.5 \times 10^{18} = 10.5 \times 10^{18}\).


Calculate \(\sigma\):
\(\sigma = 1.6 \times 10^{-19} \times 10.5 \times 10^{18}\)
\(\sigma = 1.6 \times 10.5 \times 10^{-1}\)
\(\sigma = 16.8 \times 0.1 = 1.68\) \((\Omega m)^{-1}\).
Quick Tip: Total conductivity is the sum of conductivities due to electrons and holes. Pay attention to the orders of magnitude (\(10^{18}\) vs \(10^{19}\)) when adding.

The question asks for the incorrect statement regarding partition chromatography.


1. Statement (A): Partition chromatography relies on the partition coefficient of the solute between two phases (mobile and stationary). This is true.


2. Statement (B): In partition chromatography, the stationary phase is typically a liquid supported on an inert solid. If the stationary phase is a "finely divided solid adsorbent", the mechanism is adsorption, not partition. Thus, this statement describes adsorption chromatography and is false for partition chromatography.


3. Statement (C): In Gas-Liquid Chromatography (GLC), the mobile phase is a gas. GLC is a form of partition chromatography. This is true.


4. Statement (D): Paper chromatography is a classic example of partition chromatography where water held in the paper fibers acts as the stationary liquid phase. This is true.


Therefore, statement (B) is the incorrect one.
Quick Tip: Distinguish between the two main types: Adsorption (Solid Stationary Phase) vs. Partition (Liquid Stationary Phase).

Paramagnetism arises from the presence of unpaired electrons in molecular orbitals.


1. \(B_2\) (10 electrons): The configuration is \(\sigma 1s^2, \sigma^ 1s^2, \sigma 2s^2, \sigma^ 2s^2, \pi 2p_x^1 = \pi 2p_y^1\). The two electrons in the degenerate \(\pi\) orbitals are unpaired (Hund's Rule). Hence, it is paramagnetic.


2. \(CO\) (14 electrons): Isoelectronic with \(N_2\). All electrons are paired in bonding orbitals. Diamagnetic.


3. \(O_2^{2-}\) (18 electrons): Isoelectronic with \(F_2\). All electrons are paired. Diamagnetic.


4. \(NO^+\) (14 electrons): Isoelectronic with \(CO\) and \(N_2\). All electrons are paired. Diamagnetic.
Quick Tip: Molecules with 10 or 16 valence electrons (like \(B_2\) and \(O_2\)) are typically paramagnetic due to singly occupied degenerate \(\pi\) or \(\pi^\) orbitals.

The reactant is 2,4-dibromohexane. The reaction with alcoholic KOH is a dehydrohalogenation (E2 elimination).


1. Two equivalents of HBr are eliminated to form a diene.


2. Elimination generally follows Zaitsev's rule, favoring the most substituted alkene.


3. Furthermore, the formation of a conjugated diene is thermodynamically favored over isolated or cumulated dienes due to resonance stabilization.


4. Eliminating HBr from C2-C3 and C4-C5 yields 2,4-hexadiene: \(CH_3-CH=CH-CH=CH-CH_3\).


5. This product (Option C) is a conjugated diene with internal double bonds, making it the most stable and major product.
Quick Tip: In elimination reactions producing dienes, always look for the conjugated product (alternating double-single-double bonds) as it is the most stable.

Reaction: \(FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 \downarrow + 3NaCl\)


1. Calculate moles of precipitate \(Fe(OH)_3\):

Molar Mass = \(56 + 3(16+1) = 107 g/mol\).
\(Moles = \frac{2.14}{107} = 0.02 mol\).


2. From stoichiometry, 1 mole of \(FeCl_3\) yields 1 mole of \(Fe(OH)_3\).

So, moles of \(FeCl_3 = 0.02 mol\).


3. Calculate Molarity:

Volume = 100 mL = 0.1 L.
\(M = \frac{0.02 mol}{0.1 L} = 0.2 M\).
Quick Tip: Stoichiometry is key. Always balance the reaction to find the mole ratio between reactant and product (here 1:1).

1. Numbering: We must choose the numbering that gives the lowest set of locants.

- Starting at the dimethyl carbon: Substituents are at 1, 1, 2. (Set: 1, 1, 2).

- Starting at the ethyl carbon: Substituents are at 1, 2, 2. (Set: 1, 2, 2).

Comparison: 1=1, 1<2. So the set (1, 1, 2) is lower and preferred.

Thus, the carbon with two methyl groups is C1. The carbon with the ethyl group is C2.


2. Alphabetical Order: "Ethyl" comes before "Methyl".


3. Name Construction: 2-Ethyl-1,1-dimethylcyclohexane.
Quick Tip: Lowest locant rule takes precedence over alphabetical order when deciding the numbering direction. Compare the locant sets digit by digit.

1. Analyze Lyman Series (Hydrogen, Z=1):

Shortest wavelength corresponds to the transition from \(n=\infty\) to \(n=1\).
\(\frac{1}{\lambda_L} = R_H \cdot Z^2 \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R_H (1)\).

Given \(\lambda_L = A\), so \(\frac{1}{A} = R_H\). Or \(R_H = \frac{1}{A}\).


2. Analyze Paschen Series (\(He^+\), Z=2):

Paschen series corresponds to transitions to \(n_1 = 3\).

Longest wavelength corresponds to the smallest energy gap, i.e., transition from \(n_2 = 4\) to \(n_1 = 3\).
\(\frac{1}{\lambda_P} = R_H \cdot Z^2 \left(\frac{1}{3^2} - \frac{1}{4^2}\right)\).

Substitute \(Z=2\):
\(\frac{1}{\lambda_P} = R_H \cdot 4 \left(\frac{1}{9} - \frac{1}{16}\right)\).
\(\frac{1}{\lambda_P} = 4 R_H \left(\frac{16 - 9}{144}\right) = 4 R_H \left(\frac{7}{144}\right) = \frac{7 R_H}{36}\).


3. Substitute \(R_H\):
\(\frac{1}{\lambda_P} = \frac{7}{36} \cdot \frac{1}{A}\).
\(\lambda_P = \frac{36A}{7}\).
Quick Tip: Shortest wavelength = Maximum Energy (transition from \(\infty\)). Longest wavelength = Minimum Energy (transition from adjacent level \(n+1\)).

1. The discoloration of the Taj Mahal (marble, \(CaCO_3\)) is primarily caused by "Acid Rain".


2. The precursors to acid rain are oxides of sulfur (\(SO_2\)) and nitrogen (\(NO_2\)), which react with atmospheric moisture to form sulfuric acid (\(H_2SO_4\)) and nitric acid (\(HNO_3\)).


3. These acids attack the marble (calcium carbonate) causing it to corrode and turn yellow (Stone Cancer).


Reaction: \(CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + H_2O + CO_2\).
Quick Tip: \(SO_2\) is the primary culprit mainly from nearby industrial refineries (Mathura refinery), leading to sulfuric acid formation.

The reaction involves a tertiary alkyl halide with a strong base (Ethoxide), which favors E2 elimination.


1. We identify the \(\beta\)-hydrogens available for elimination:

- On the Benzylic carbon (\(C_6H_5CH_2-\)).

- On the Methylene group of the ethyl chain (\(-CH_2CH_3\)).

- On the Methyl group (\(-CH_3\)).


2. Possibility 1: Removing \(\beta\)-H from the benzylic carbon forms a double bond conjugated with the benzene ring (\(C_6H_5-CH=C...\)).

3. Possibility 2: Removing \(\beta\)-H from the ethyl group forms a Zaitsev alkene (trisubstituted) but isolated from the ring.

4. Possibility 3: Removing \(\beta\)-H from the methyl group forms a Hofmann alkene (disubstituted).


5. Conclusion: The product containing the double bond in conjugation with the aromatic ring (Possibility 1) is exceptionally stable due to resonance. Therefore, the major product is \(C_6H_5CH=C(CH_3)CH_2CH_3\).


This corresponds to Option (D).
Quick Tip: In elimination reactions, conjugation with an aromatic ring provides significant stability, often outweighing simple alkyl substitution effects.

The reactant has a carboxylic acid group (\(-COOH\)) and a hydroxymethyl group (\(-CH_2OH\)) in positions ortho to each other (based on the formation of the lactone in the options).


1. Role of Reagents: Dry HCl gas acts as an acid catalyst. \(CCl_4\) is a non-polar solvent.


2. Process: In the presence of acid, the alcohol group and carboxylic acid group can undergo intramolecular esterification (lactonization) to form a cyclic ester.


3. Since the groups are ortho, they form a stable 5-membered lactone ring (phthalide derivative).


4. The amino group (\(-NH_2\)) will be protonated to form a salt (\(-NH_3^+\)) in the acidic medium but does not participate in ring formation under these conditions (Lactonization is favored over lactam formation here, and the amine is para to the COOH in the drawing layout provided in options).


5. The phenolic \(-OH\) is generally unreactive towards HCl substitution.


6. Therefore, the major organic product is the lactone. Option (A) depicts this lactone structure.
Quick Tip: Ortho-hydroxymethyl benzoic acid derivatives spontaneously or easily form lactones (phthalides) under acidic conditions due to the stability of the 5-membered ring.

We examine the oxidation states of the metals in each pair:


1. Option (A):

- \(CrO_2Cl_2\) (Chromyl chloride): O is \(-2\), Cl is \(-1\). \(x + 2(-2) + 2(-1) = 0 \Rightarrow x = +6\). Chromium is in Group 6, so \(+6\) is its highest possible oxidation state.


- \(MnO_2\): Mn is \(+4\). While \(+7\) is the highest for Mn (Group 7), \(Cr\) in \(+6\) is definitely in its highest state. Among the given choices, this option contains the only compound with a metal clearly in its group maximum oxidation state (\(Cr^{+6}\)). (Note: Some versions of this question pair \(MnO_4^-\) with \(CrO_2Cl_2\). Given the options, this is the intended answer).


2. Option (B):

- \([FeCl_4]^-\): Fe is \(+3\). Highest for Fe is \(+6\).

- \(Co_2O_3\): Co is \(+3\). Highest for Co is \(+4\).


3. Option (C): Fe is \(+3\), Cu is \(+2\). Not highest.


4. Option (D): Ni is \(+2\), Co is \(+2\). Not highest.


Thus, Option (A) is the correct choice.
Quick Tip: For d-block elements, the highest oxidation state corresponds to the sum of ns and (n-1)d electrons (Group Number). Cr (Group 6) has a max state of +6.

Essential amino acids cannot be synthesized by the human body and must be obtained from the diet.


1. The essential amino acids are: Histidine, Isoleucine, Leucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Valine.


2. Aspartic acid, Serine, and Alanine are non-essential amino acids.

Therefore, Valine is the essential amino acid.
Quick Tip: Mnemonic for essential amino acids: \textbf{PVT TIM HALL} (Phenylalanine, Valine, Threonine, Tryptophan, Isoleucine, Methionine, Histidine, Arginine, Leucine, Lysine).

Using the Henderson-Hasselbalch equation for an acidic buffer:
\(\) pH = \text{pK_a + \log \left( \frac{[\text{Salt]{[\text{Acid] \right) \(\)

Given:
\(\text{pH = 6\)
\(K_a = 10^{-5} \Rightarrow pK_a = -\log(10^{-5}) = 5\)


Substituting the values:
\(\) 6 = 5 + \log \left( \frac{[\text{Salt]{[\text{Acid] \right) \(\)
\(\) 1 = \log \left( \frac{[\text{Salt]{[\text{Acid] \right) \(\)
\(\) \frac{[\text{Salt]{[\text{Acid] = 10^1 = 10 \(\)

The ratio is 10 : 1.
Quick Tip: If pH > pKa, the basic form (salt) predominates. Since \(6 > 5\), the ratio must be \(> 1\).

Real gases approach ideal behavior under conditions where intermolecular forces are negligible and the volume of gas molecules is negligible compared to the container volume.


1. High Temperature and Low Pressure (Large Volume): These are the conditions for ideal behavior.


2. Boyle's Temperature: The temperature at which a real gas behaves ideally over a wide range of pressure. This statement is correct.


3. Low Temperature: At low temperatures, kinetic energy decreases and intermolecular attractive forces become significant. Real gases deviate significantly from ideal behavior at low temperatures. Therefore, statement (C) is incorrect.
Quick Tip: Real gases behave like Ideal gases at \textbf{High Temperature} and \textbf{Low Pressure}.

Using the Arrhenius equation form: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\).


For Reaction A: Rate doubles (\(k_2/k_1 = 2\)) when \(T_1 = 300, T_2 = 310\).
\(\ln(2) = \frac{E_{aA}}{R} \left( \frac{1}{300} - \frac{1}{310} \right)\).


For Reaction B: Rate doubles (\(k_2'/k_1' = 2\)) when \(T_1 = 300, T_2 = T'\). \(E_{aB} = 2E_{aA}\).
\(\ln(2) = \frac{2E_{aA}}{R} \left( \frac{1}{300} - \frac{1}{T'} \right)\).


Equating the \(\ln(2)\) expressions:
\(\frac{E_{aA}}{R} \left( \frac{1}{300} - \frac{1}{310} \right) = \frac{2E_{aA}}{R} \left( \frac{1}{300} - \frac{1}{T'} \right)\).


Cancel \(\frac{E_{aA}}{R}\):
\(\left( \frac{310 - 300}{300 \times 310} \right) = 2 \left( \frac{T' - 300}{300 T'} \right)\).
\(\frac{10}{300 \times 310} = 2 \frac{\Delta T}{300 T'}\).
\(\frac{5}{310} = \frac{\Delta T}{T'}\).

Since \(\Delta T\) is small, \(T' \approx 310\) (approx) or solve exactly. Let's solve for \(1/T'\) first.
\(\frac{1}{300} - \frac{1}{310} = \frac{10}{93000}\).
\(2 \left( \frac{1}{300} - \frac{1}{T'} \right) = \frac{10}{93000} \Rightarrow \frac{1}{300} - \frac{1}{T'} = \frac{5}{93000}\).
\(\frac{1}{T'} = \frac{1}{300} - \frac{5}{93000} = \frac{310 - 5}{93000} = \frac{305}{93000}\).
\(T' = \frac{93000}{305} \approx 304.92 K\).
\(\Delta T = 304.92 - 300 = 4.92 K\).
Quick Tip: Since Activation Energy is in the numerator, doubling \(E_a\) means the temperature difference term must be halved to achieve the same rate change (approximately). \(\Delta(1/T) \propto 1/E_a\).

1. Peroxodisulphuric acid (\(H_2S_2O_8\)):

Structure: \(HO-S(=O)_2-O-O-S(=O)_2-OH\).

- Each Sulfur atom is bonded to two double-bonded oxygens (\(S=O\)) and one hydroxyl group (\(S-OH\)).

- Total \(S=O\) bonds = \(2 + 2 = 4\).

- Total \(S-OH\) bonds = \(1 + 1 = 2\).

- (Count: 4 and 2).


2. Pyrosulphuric acid (\(H_2S_2O_7\), Oleum):

Structure: \(HO-S(=O)_2-O-S(=O)_2-OH\).

- Each Sulfur atom is bonded to two double-bonded oxygens (\(S=O\)) and one hydroxyl group (\(S-OH\)).

- Total \(S=O\) bonds = \(2 + 2 = 4\).

- Total \(S-OH\) bonds = \(1 + 1 = 2\).

- (Count: 4 and 2).


Both acids contain 4 \(S=O\) bonds and 2 \(S-OH\) bonds.
Quick Tip: Draw the structures. Both are disulphuric acids; one has a peroxide linkage (\(-O-O-\)) and the other an oxide linkage (\(-O-\)), but the terminal groups (\(SO_2OH\)) are the same.

1. Analyze Statement (A): Adsorption of gases on charcoal (physisorption) increases with critical temperature (\(T_c\)). \(T_c\) of \(Cl_2\) (417 K) is higher than \(T_c\) of \(H_2S\) (373 K). Thus, chlorine is adsorbed more. This statement is technically correct but implies an expectation that might be considered ambiguous in some contexts compared to the absolute certainty of statement (D).


2. Analyze Statement (B): Metal sulphide sols (e.g., \(As_2S_3\)) are lyophobic (solvent-hating), not lyophilic. This statement is incorrect.


3. Analyze Statement (C): Hardy-Schulze law states that the coagulating power depends on the valency (charge) of the flocculating ion, not its size. Higher charge implies greater power. This statement is incorrect.


4. Analyze Statement (D): Brownian movement arises from the unbalanced bombardment of colloidal particles by dispersion medium molecules. The displacement (\(\Delta x\)) is inversely proportional to the size of the particle (Einstein's relation). Smaller particles move faster and show more pronounced movement. This statement is universally correct.


Comparing (A) and (D), statement (D) is a fundamental definition in surface chemistry and is the accepted answer for this question.
Quick Tip: Brownian motion intensity \(\propto \frac{1}{size of particle}\). Smaller particles \(\rightarrow\) Faster motion.

We determine the hybridization of the central atom in each species:


1. \(PF_5\): Phosphorus has 5 valence electrons. It forms 5 \(\sigma\)-bonds with F. Steric number = 5. Hybridization is \(sp^3d\). Geometry is Trigonal Bipyramidal.


2. \(SF_6\): Sulfur has 6 valence electrons. It forms 6 \(\sigma\)-bonds. Steric number = 6. Hybridization is \(sp^3d^2\). Geometry is Octahedral.


3. \([CrF_6]^{3-}\): Chromium is in \(+3\) state (\(3d^3\)). With weak field ligand \(F^-\), it forms an inner orbital complex using two \(3d\), one \(4s\), and three \(4p\) orbitals. Hybridization is \(d^2sp^3\). (Note: While technically \(d^2sp^3\), it is often grouped with octahedral species, whereas \(PF_5\) is clearly different).


4. \(BrF_5\): Bromine has 7 valence electrons. It forms 5 bonds and has 1 lone pair. Steric number = \(5 + 1 = 6\). Hybridization is \(sp^3d^2\).


The only molecule with a steric number of 5 (and thus definitely not \(sp^3d^2\) or \(d^2sp^3\)) is \(PF_5\).
Quick Tip: Count the steric number (Sigma bonds + Lone pairs). Steric No. 5 = \(sp^3d\). Steric No. 6 = \(sp^3d^2\) or \(d^2sp^3\).

The relationship between Enthalpy change (\(\Delta H\)) and Internal Energy change (\(\Delta U\)) is:
\(\Delta H = \Delta U + \Delta n_g RT\)


1. Determine \(\Delta n_g\):

Reaction: \(A(g) \rightarrow A(l)\)
\(\Delta n_g = moles of gaseous products - moles of gaseous reactants\)
\(\Delta n_g = 0 - 1 = -1\)


2. Substitute into equation:
\(\Delta H = \Delta U - RT\)


3. Use given value of \(\Delta H\):
\(-3RT = \Delta U - RT\)
\(\Delta U = -3RT + RT = -2RT\)


4. Compare magnitudes:
\(|\Delta H| = |-3RT| = 3RT\)
\(|\Delta U| = |-2RT| = 2RT\)

Since \(3RT > 2RT\), we have \(|\Delta H| > |\Delta U|\).
Quick Tip: For condensation of a gas (\(\Delta n_g < 0\)), the magnitude of \(\Delta H\) is generally greater than \(\Delta U\) if the heat released is large enough. Always substitute and solve.

1. Identify Metal M: The formula \(M_3N\) implies M has a valency of +1 (since N is -3). Alkali metals are Group 1 (+1). Among alkali metals, Lithium is the only one that reacts directly with Nitrogen to form a stable nitride (\(Li_3N\)). Sodium does not react directly.


Reaction: \(6Li + N_2 \rightarrow 2Li_3N\).


2. Identify Gas B: Nitrides hydrolyze to give ammonia.

Reaction: \(Li_3N + 3H_2O \rightarrow 3LiOH + NH_3(g)\) (Gas B).


3. Confirm with \(CuSO_4\) test: Ammonia reacts with copper sulphate to form a deep blue complex ion (Tetraamminecopper(II)).

Reaction: \(Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}\) (Deep Blue).


Therefore, M is Li and B is \(NH_3\).
Quick Tip: Only Lithium among alkali metals forms a nitride directly. The "Deep Blue" solution with Copper salts is a specific test for Ammonia.

We use the additivity of Gibbs Free Energy (\(\Delta G^\circ = -nFE^\circ\)), not \(E^\circ\) directly.


1. Reaction 1: \(Fe^{2+} + 2e^- \rightarrow Fe\)
\(\Delta G^\circ_1 = -2 \times F \times (-0.47) = +0.94F\)


2. Reaction 2: \(Fe^{3+} + e^- \rightarrow Fe^{2+}\)
\(\Delta G^\circ_2 = -1 \times F \times (+0.77) = -0.77F\)


3. Target Reaction: \(Fe^{3+} + 3e^- \rightarrow Fe\)

This is the sum of Reaction 1 and Reaction 2.
\(\Delta G^\circ_{total} = \Delta G^\circ_1 + \Delta G^\circ_2\)
\(\Delta G^\circ_{total} = 0.94F - 0.77F = +0.17F\)


4. Calculate \(E^\circ_{target}\):
\(\Delta G^\circ_{total} = -nFE^\circ_{target}\) where \(n=3\).
\(+0.17F = -3 F E^\circ_{target}\)
\(E^\circ_{target} = \frac{0.17}{-3} = -0.0566 V \approx -0.057 V\).
Quick Tip: Potentials are intensive properties and do not add. Convert to \(\Delta G^\circ\) (extensive), add, and convert back to \(E^\circ\). Formula: \(E^\circ_3 = \frac{n_1 E^\circ_1 + n_2 E^\circ_2}{n_1 + n_2}\).

1. Identify Acidic Protons: The reactant contains a phenolic hydroxyl group (\(-OH_{phenol}\)) and a primary aliphatic hydroxyl group (\(-OH_{alkyl}\)).


2. Compare Acidity: Phenols (\(pK_a \approx 10\)) are much more acidic than alcohols (\(pK_a \approx 16\)).


3. Action of Base: The mild base \(K_2CO_3\) selectively deprotonates the more acidic phenolic \(-OH\) to form the phenoxide ion (\(Ar-O^-\)). The aliphatic alcohol remains protonated.


4. Nucleophilic Substitution: The phenoxide ion acts as a nucleophile and attacks methyl iodide (\(CH_3I\)) in an \(S_N2\) reaction.


5. Product: The product is the methyl ether of the phenol, while the alkyl alcohol group remains intact.


Structure: \(HO-CH_2CH_2 - C_6H_4 - OCH_3\).

This matches Option (D) (Option 4 in the provided PDF).
Quick Tip: When a molecule has two nucleophilic sites with different acidities, a base will deprotonate the more acidic site first, making it the active nucleophile.

1. Group IV Analysis: Group IV cations (\(Zn^{2+}, Mn^{2+}, Ni^{2+}, Co^{2+}\)) precipitate as sulphides with \(H_2S\) in ammoniacal medium.


2. Solubility in HCl: \(ZnS\) (White) and \(MnS\) (Buff/Pink) dissolve in dilute HCl. \(NiS\) and \(CoS\) (Black) are insoluble. Since the precipitate dissolves, it is \(Zn^{2+}\) or \(Mn^{2+}\).


3. Reaction with NaOH:
\(Zn^{2+} + NaOH \rightarrow Zn(OH)_2\) (White ppt, soluble in excess).
\(Mn^{2+} + NaOH \rightarrow Mn(OH)_2\) (White ppt, turns brown).


Both give white precipitates initially.


4. Confirmatory Test with Potassium Ferrocyanide:
\(Zn^{2+} + K_4[Fe(CN)_6] \rightarrow Zn_2[Fe(CN)_6]\) (White or Bluish-white precipitate).
\(Mn^{2+}\) gives a white precipitate.


The description "bluish-white" is a characteristic often attributed to Zinc ferrocyanide in qualitative analysis texts (or specifically distinguished from the pure white of others). More importantly, \(ZnS\) is the characteristic white sulphide of Group IV.

Thus, the cation is \(Zn^{2+}\).
Quick Tip: Zinc is the only Group IV cation that forms a white sulphide (\(ZnS\)). Its ferrocyanide ppt is also white/bluish-white.

Formula: \(\Delta T_f = i \cdot K_f \cdot m\)


1. Calculate van't Hoff factor (i):
\(Na_2SO_4 \rightleftharpoons 2Na^+ + SO_4^{2-}\) (n = 3 ions).
\(i = 1 + (n-1)\alpha = 1 + (3-1)(0.815) = 1 + 2(0.815) = 1 + 1.63 = 2.63\).


2. Calculate Moles of Solute:

Molar Mass of \(Na_2SO_4 = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 g/mol\).

Moles \(= \frac{5}{142} mol\).


3. Set up Molality equation:
\(m = \frac{Moles}{Mass of solvent in kg} = \frac{5/142}{x/1000} = \frac{5000}{142x}\).


4. Substitute into Freezing Point equation:
\(3.82 = 2.63 \times 1.86 \times \frac{5000}{142x}\)
\(x = \frac{2.63 \times 1.86 \times 5000}{3.82 \times 142}\)


5. Solve:
\(2.63 \times 1.86 \approx 4.8918\)

Numerator \(\approx 4.8918 \times 5000 = 24459\)

Denominator \(= 3.82 \times 142 = 542.44\)
\(x = \frac{24459}{542.44} \approx 45.09 g\).

Rounding to nearest integer gives 45 g.
Quick Tip: Don't forget to convert solvent mass to kg when calculating molality. \(m = \frac{w_{solute} \times 1000}{M_{solute} \times w_{solvent}(g)}\).

1. Stability of Al species:

For Aluminum, the \(+3\) state is the most stable. The \(+1\) state is unstable and tends to disproportionate.

Calculating \(E^\circ\) for \(3Al^+ \rightarrow 2Al + Al^{3+}\) yields a positive potential, indicating spontaneous disproportionation. Thus, \(Al^{3+}\) is much more stable than \(Al^+\).


2. Stability of Tl species:

For Thallium (Group 13, Period 6), the inert pair effect makes the \(+1\) oxidation state more stable than the \(+3\) state.

The reduction potential for \(Tl^{3+} \rightarrow Tl^+\) is highly positive (\(+1.26 V\) roughly indicates high oxidizing power of \(Tl^{3+}\)), meaning \(Tl^{3+}\) easily reduces to \(Tl^+\).


3. Comparing Options:

(A) False. \(Al^{3+}\) is more stable.

(B) False. \(Al^{3+}\) is stable, \(Tl^{3+}\) is a strong oxidizing agent (unstable).

(C) Ambiguous comparison across elements.

(D) True. \(Tl^+\) is the stable state of Thallium, while \(Al^+\) is the unstable state of Aluminum. Due to the inert pair effect, stability of \(+1\) state increases down the group (\(Al^+ < Ga^+ < In^+ < Tl^+\)).
Quick Tip: In Group 13, the stability of the \(+1\) oxidation state increases down the group due to the Inert Pair Effect. Tl(+1) is stable; Al(+3) is stable.

1. Drugs interact with enzymes to inhibit their activity.


2. Reversible inhibition (competitive or non-competitive) is the mechanism for many therapeutic drugs. The drug binds and unbinds, allowing the enzyme to eventually recover function.


3. Irreversible inhibition involves the formation of a strong covalent bond between the drug and the enzyme's active site. This permanently destroys the catalytic activity of the enzyme.


4. Since the body cannot regenerate the enzyme function quickly, this permanent blockage leads to toxicity or "poisoning".


Therefore, irreversible binding is the cause of drug-induced poisoning.
Quick Tip: Reversible = Therapeutic; Irreversible = Poisonous/Toxic.

Extraction with 1M HCl (an acid) separates basic compounds from neutral or acidic ones.


1. Compound (I): A diaryl sulfide or similar. Sulfur is a very weak base. It remains neutral and stays in the organic layer.


2. Compound (II): Contains a secondary amine group (\(-NH-\)) attached to an alkyl chain. Amines are organic bases. They react with HCl to form water-soluble ammonium salts (\(R_2NH_2^+ Cl^-\)). Thus, this compound moves to the aqueous layer.


3. Compound (III): An ether (\(Ph-O-CH_3\)). Ethers are neutral/weakly basic but do not form stable salts with dilute HCl to become water soluble. Stays in organic layer.


4. Compound (IV): A ketone (\(Ph-CO-CH_3\)). Ketones are neutral. Stays in organic layer.


Only the amine (II) is extracted into the acidic aqueous layer.
Quick Tip: Acid-Base Extraction: Acids extract Bases (Amines). Bases extract Acids (Carboxylic acids, Phenols).

An oxidizing agent accepts electrons and is itself reduced.

In \(H_2O_2\), Oxygen is in the \(-1\) oxidation state.

- If it acts as an oxidizing agent, it becomes reduced to \(H_2O\) or \(OH^-\) (Oxygen state \(-2\)).

- If it acts as a reducing agent, it becomes oxidized to \(O_2\) (Oxygen state \(0\)).


1. Reaction (A): \(H_2O_2 \rightarrow H_2O\). Oxygen changes from \(-1\) to \(-2\) (Reduction). \(H_2O_2\) acts as an Oxidizing Agent. (Also, S in PbS goes from \(-2\) to \(+6\) in sulphate).

2. Reaction (B): \(H_2O_2 \rightarrow O_2\). Oxygen changes from \(-1\) to \(0\) (Oxidation). \(H_2O_2\) acts as a Reducing Agent.

3. Reaction (C): \(H_2O_2 \rightarrow O_2\). Acts as Reducing Agent.

4. Reaction (D): \(H_2O_2 \rightarrow O_2\). Acts as Reducing Agent.


Thus, (A) is the correct reaction.
Quick Tip: Check the product of \(H_2O_2\). If it forms \(O_2\), it's a Reducing Agent. If it forms \(H_2O\), it's an Oxidizing Agent.

1. Determine Group: Look for the "jump" in Ionization Energy.

- Element A: \(IE_1 = 899, IE_2 = 1757, IE_3 = 14847\). The jump from \(IE_2\) to \(IE_3\) is massive (\(\sim 8\times\)). This indicates that the removal of the 3rd electron disrupts a stable noble gas core. Therefore, A has 2 valence electrons. It belongs to Group 2.


- Element B: \(IE_1 = 737, IE_2 = 1450, IE_3 = 7731\). The jump from \(IE_2\) to \(IE_3\) is large (\(\sim 5\times\)). This also indicates 2 valence electrons. It belongs to Group 2.


2. Determine Relative Position:

- Within a group, Ionization Energy decreases as we move down (due to increasing size).

- \(IE_1(A) = 899\).

- \(IE_1(B) = 737\).

- Since \(IE_1(A) > IE_1(B)\), element A is smaller and placed higher in the group than element B.

- Therefore, B comes below A.


Conclusion: Both are Group 2, B is below A. Matches Option (C).
Quick Tip: Large jump between \(IE_n\) and \(IE_{n+1}\) indicates \(n\) valence electrons. \(IE\) decreases down the group.

The process involves three steps. We calculate the heat released (\(\Delta H\) is negative, question implies magnitude):


1. Step 1: Cooling water from \(5^\circ C\) to \(0^\circ C\).
\(\Delta H_1 = C_p(l) \Delta T = 75.3 J/mol K \times (0 - 5) K = -376.5 J/mol\).


2. Step 2: Freezing water at \(0^\circ C\).
\(\Delta H_2 = -\Delta_{fus}H = -6 kJ/mol = -6000 J/mol\).


3. Step 3: Cooling ice from \(0^\circ C\) to \(-5^\circ C\).
\(\Delta H_3 = C_p(s) \Delta T = 36.8 J/mol K \times (-5 - 0) K = -184 J/mol\).


4. Total Enthalpy Change:
\(\Delta H_{total} = -376.5 - 6000 - 184 = -6560.5 J/mol\).
\(\Delta H_{total} \approx -6.56 kJ/mol\).


The magnitude is \(6.56 kJ mol^{-1}\).
Quick Tip: Calculate \(\Delta H\) for each step: Cooling liquid \(\rightarrow\) Phase change \(\rightarrow\) Cooling solid. Sum them up.

Friedel-Crafts alkylation requires the formation of a carbocation or a positive complex to attack the benzene ring.


1. Vinyl Chloride (\(CH_2=CH-Cl\)): The C-Cl bond acquires partial double bond character due to resonance with the adjacent double bond. This makes the bond stronger and difficult to break. Furthermore, the resulting vinyl cation (\(CH_2=CH^+\)) is extremely unstable. Therefore, vinyl chloride does not undergo Friedel-Crafts reaction.


2. Other Options:

- Acryloyl chloride (\(CH_2=CH-COCl\)) and other acid chlorides undergo FC Acylation via the acylium ion.

- Allyl chloride (\(CH_2=CH-CH_2Cl\)) undergoes FC Alkylation because the allyl cation is resonance stabilized.


Thus, Vinyl chloride is the unreactive compound in this context.
Quick Tip: Aryl halides and Vinyl halides do not undergo Friedel-Crafts reaction due to strong C-X bond (resonance) and instability of the corresponding cations.

Given \(f(x) = 2^{10}x + 1\) and \(g(x) = 3^{10}x - 1\).


We are given \((f \circ g)(x) = x\), which means \(f(g(x)) = x\).


Substitute \(g(x)\) into \(f(x)\):

\(2^{10}(3^{10}x - 1) + 1 = x\).


Expand the expression: \(2^{10} \cdot 3^{10}x - 2^{10} + 1 = x\).


Rearrange terms to group \(x\) on one side:

\(x(2^{10} \cdot 3^{10}) - x = 2^{10} - 1\).

\(x(2^{10} \cdot 3^{10} - 1) = 2^{10} - 1\).

\(x = \frac{2^{10} - 1}{2^{10} \cdot 3^{10} - 1}\).


To match the form of Option (A), divide both the numerator and denominator by \(2^{10}\):


Numerator: \(\frac{2^{10} - 1}{2^{10}} = 1 - 2^{-10}\).


Denominator: \(\frac{2^{10} \cdot 3^{10} - 1}{2^{10}} = 3^{10} - 2^{-10}\).


Thus, \(x = \frac{1 - 2^{-10}}{3^{10} - 2^{-10}}\).
Quick Tip: When options contain negative exponents, try dividing the numerator and denominator of your derived expression by the highest power term to simplify.

First, simplify the terms of the series to identify the pattern:

\(T_1 = \sqrt{3} = 1\sqrt{3}\).

\(T_2 = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\).

\(T_3 = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}\).

\(T_4 = \sqrt{507} = \sqrt{169 \times 3} = 13\sqrt{3}\).


The coefficients \(1, 5, 9, 13, \dots\) form an Arithmetic Progression (AP) with first term \(a=1\) and common difference \(d=4\).


The sum of the first \(n\) terms is \(S_n = \sqrt{3} \times (sum of the AP coefficients)\).


Sum of AP coefficients \(= \frac{n}{2}[2(1) + (n-1)4] = \frac{n}{2}[2 + 4n - 4] = \frac{n}{2}[4n - 2] = n(2n - 1)\).


Therefore, \(S_n = \sqrt{3} \cdot n(2n - 1)\).


Given \(S_n = 435\sqrt{3}\), we equate:

\(\sqrt{3} \cdot n(2n - 1) = 435\sqrt{3}\).

\(2n^2 - n = 435 \implies 2n^2 - n - 435 = 0\).


Solving the quadratic equation for \(n\):

\(n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-435)}}{2(2)} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4}\).

\(n = \frac{1 \pm 59}{4}\). Since \(n\) must be positive, \(n = \frac{60}{4} = 15\).
Quick Tip: Always simplify radicals (surds) to their simplest form to reveal hidden Arithmetic or Geometric Progressions.

The diameter is 4, so the radius \(R = 2\).


Let the angles subtended by the chords at the center be \(\theta_1\) and \(\theta_2\).

\(\theta_1 = \cos^{-1}\left(\frac{1}{7}\right)\).

\(\theta_2 = \sec^{-1}(7) = \cos^{-1}\left(\frac{1}{7}\right)\).


Since \(\theta_1 = \theta_2\), the chords are equal in length and equidistant from the center.


The perpendicular distance \(d\) from the center to a chord subtending angle \(\theta\) is \(d = R \cos\left(\frac{\theta}{2}\right)\).


We know \(\cos \theta = \frac{1}{7}\). Using the half-angle formula \(\cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) - 1\):

\(\frac{1}{7} = 2\cos^2\left(\frac{\theta}{2}\right) - 1 \implies 2\cos^2\left(\frac{\theta}{2}\right) = 1 + \frac{1}{7} = \frac{8}{7}\).

\(\cos^2\left(\frac{\theta}{2}\right) = \frac{4}{7} \implies \cos\left(\frac{\theta}{2}\right) = \frac{2}{\sqrt{7}}\).


The distance of one chord from the center is \(d = 2 \cdot \frac{2}{\sqrt{7}} = \frac{4}{\sqrt{7}}\).


Since the chords are on opposite sides, the total distance between them is \(2d\).


Distance \(= 2 \times \frac{4}{\sqrt{7}} = \frac{8}{\sqrt{7}}\).
Quick Tip: Remember the identity \(\sec^{-1}(x) = \cos^{-1}(1/x)\). Equal chords subtend equal angles at the center and are equidistant from it.

Let \(u = x + \sqrt{x^2-1}\). Note that \(x - \sqrt{x^2-1} = \frac{1}{u}\).


So, \(y = u^{15} + u^{-15}\).


Differentiating \(u\) w.r.t \(x\): \(\frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}} = \frac{u}{\sqrt{x^2-1}}\).


Differentiating \(y\) w.r.t \(x\): \(\frac{dy}{dx} = 15u^{14}\frac{du}{dx} - 15u^{-16}\frac{du}{dx} = 15(u^{14}-u^{-16})\frac{u}{\sqrt{x^2-1}}\).

\(\frac{dy}{dx} = \frac{15(u^{15}-u^{-15})}{\sqrt{x^2-1}}\).


Cross-multiplying: \(\sqrt{x^2-1} \cdot y_1 = 15(u^{15}-u^{-15})\).


Differentiating again w.r.t \(x\):

\(\sqrt{x^2-1} \cdot y_2 + y_1 \cdot \frac{x}{\sqrt{x^2-1}} = 15(15u^{14}\frac{du}{dx} + 15u^{-16}\frac{du}{dx})\).


Substitute \(\frac{du}{dx} = \frac{u}{\sqrt{x^2-1}}\) into the RHS:

\(\sqrt{x^2-1} \cdot y_2 + \frac{x y_1}{\sqrt{x^2-1}} = 15 \cdot 15 (u^{14} + u^{-16}) \frac{u}{\sqrt{x^2-1}}\).


Multiply the entire equation by \(\sqrt{x^2-1}\):

\((x^2-1)y_2 + x y_1 = 225(u^{15} + u^{-15})\).


Since \(y = u^{15} + u^{-15}\), we get \((x^2-1)y_2 + x y_1 = 225y\).
Quick Tip: Standard Result: If \(y = (x+\sqrt{x^2-1})^n + (x-\sqrt{x^2-1})^n\), then \((x^2-1)y'' + xy' = n^2 y\).

Rearrange the first equation: \(t(x-3) = 2y \implies t = \frac{2y}{x-3}\).


Rearrange the second equation: \(x+3 = 2ty\).


Substitute the value of \(t\) into the second equation:

\(x+3 = 2y \left( \frac{2y}{x-3} \right)\).

\((x+3)(x-3) = 4y^2\).

\(x^2 - 9 = 4y^2 \implies x^2 - 4y^2 = 9\).


Divide by 9 to put it in standard form: \(\frac{x^2}{9} - \frac{y^2}{9/4} = 1\).


This is a hyperbola with \(a^2 = 9\) and \(b^2 = \frac{9}{4}\).


The length of the conjugate axis is \(2b = 2 \cdot \sqrt{\frac{9}{4}} = 2 \cdot \frac{3}{2} = 3\).


Eccentricity \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9/4}{9}} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2}\).


Comparing with options, only (C) is correct.
Quick Tip: To eliminate a parameter \(t\) from two linear equations, solve for \(t\) in one equation and substitute it into the other.

Given \(AM = 5 \cdot GM\), so \(\frac{a+b}{2} = 5\sqrt{ab}\).


This implies \(a+b = 10\sqrt{ab}\).


We need to find \(\frac{a+b}{a-b}\).


We can relate \(a-b\) to \(a+b\) using \((a-b)^2 = (a+b)^2 - 4ab\).


Substitute \(a+b = 10\sqrt{ab}\):

\((a-b)^2 = (10\sqrt{ab})^2 - 4ab = 100ab - 4ab = 96ab\).


Taking the square root (since \(a>b\), \(a-b>0\)):

\(a-b = \sqrt{96ab} = \sqrt{16 \cdot 6} \sqrt{ab} = 4\sqrt{6}\sqrt{ab}\).


Now, calculate the ratio:

\(\frac{a+b}{a-b} = \frac{10\sqrt{ab}}{4\sqrt{6}\sqrt{ab}} = \frac{10}{4\sqrt{6}} = \frac{5}{2\sqrt{6}}\).


Rationalize the denominator:

\(\frac{5}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{5\sqrt{6}}{2 \cdot 6} = \frac{5\sqrt{6}}{12}\).
Quick Tip: Use the identity \((a-b)^2 = (a+b)^2 - 4ab\) to switch between sum and difference of two numbers.

Let the sum of ages of the 25 teachers initially be \(S\).


Mean = \(S/25 = 40 \implies S = 25 \times 40 = 1000\) years.


A teacher of age 60 retires, and a new teacher of age \(x\) joins.


New sum of ages \(S' = S - 60 + x = 1000 - 60 + x = 940 + x\).


The new mean is given as 39. Number of teachers remains 25.

\(\frac{940 + x}{25} = 39\).

\(940 + x = 25 \times 39 = 975\).

\(x = 975 - 940 = 35\).


The age of the newly appointed teacher is 35 years.
Quick Tip: Mean = Sum of observations / Number of observations. Changes in the sum are directly related to changes in the mean.

For a \(3 \times 3\) matrix \(A\):


The standard property is \(adj(adj(A)) = |A|^{n-2} A\). Since \(n=3\), \(adj(adj(A)) = |A| A\). This makes (B) always true.


Also, we know \(A \cdot adj(A) = |A| I \implies adj(A) = |A| A^{-1}\). This makes (D) always true.


For option (A): RHS \(= |A|^2 (adj(A))^{-1} = |A|^2 (|A| A^{-1})^{-1} = |A|^2 \frac{1}{|A|} (A^{-1})^{-1} = |A| A\).


Since LHS for (A) is also \(|A|A\) (from B), Option (A) is always true.


For option (C): RHS \(= |A| (adj(A))^{-1} = |A| \frac{1}{|A|} A = A\).


LHS is \(|A| A\). So (C) implies \(|A| A = A\), which means \(|A| = 1\). This is not always true for any invertible matrix.


Thus, (C) is the incorrect statement.
Quick Tip: Key Matrix Properties: \(adj(A) = |A|A^{-1}\) and \(adj(adj(A)) = |A|^{n-2}A\).

Equation of parabola: \(x^2 = 4y\). This is of the form \(x^2 = 4ay\) with \(a=1\).


Equation of tangent to \(x^2 = 4y\) with slope \(m\) is \(y = mx - am^2 \implies y = mx - m^2\).


Rewrite as \(mx - y - m^2 = 0\).


This line is also tangent to the circle \(x^2 + y^2 = 4\) (Radius \(r=2\)).


The perpendicular distance from the center \((0,0)\) to the line must equal the radius \(r=2\).

\(\left| \frac{m(0) - 0 - m^2}{\sqrt{m^2 + (-1)^2}} \right| = 2\).

\(\frac{m^2}{\sqrt{m^2+1}} = 2 \implies m^4 = 4(m^2+1)\).

\(m^4 - 4m^2 - 4 = 0\). Solving for \(m^2\):

\(m^2 = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}\).


Since \(m^2 > 0\), we have \(m^2 = 2 + 2\sqrt{2}\).


The tangents are \(y = mx - m^2\) and \(y = -mx - m^2\) (by symmetry).


They intersect at point P on the y-axis (where \(x=0\)).


Substitute \(x=0\) into the tangent equation: \(y = -m^2\).


Coordinate of P is \((0, -m^2) = (0, -(2+2\sqrt{2}))\).


Distance from origin is \(|y| = 2 + 2\sqrt{2} = 2(1+\sqrt{2})\).
Quick Tip: Use the condition that the distance from the center of a circle to a tangent line equals the radius.

Using the Distributive Law: \((A \lor (B \land C)) \equiv (A \lor B) \land (A \lor C)\).

\((\sim p) \lor (p \land \sim q) \equiv (\sim p \lor p) \land (\sim p \lor \sim q)\).


Since \((\sim p \lor p)\) is a tautology (True), the expression simplifies to:

\(T \land (\sim p \lor \sim q) \equiv \sim p \lor \sim q\).


By De Morgan's Law, this is equivalent to \(\sim (p \land q)\).


Also, using the implication identity \(A \to B \equiv \sim A \lor B\), we have:

\(\sim p \lor \sim q \equiv p \to \sim q\).


This matches Option (B).
Quick Tip: Remember the distributive laws of logic and the implication equivalence: \(p \to q \equiv \sim p \lor q\).

Let \(\vec{d_1} = 8\hat{i} - 6\hat{j}\) and \(\vec{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k}\).


The area of a parallelogram with diagonals \(\vec{d_1}\) and \(\vec{d_2}\) is given by \(\frac{1}{2} |\vec{d_1} \times \vec{d_2}|\).


Calculate the cross product \(\vec{d_1} \times \vec{d_2}\):

\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
8 & -6 & 0
3 & 4 & -12 \end{vmatrix}\)

\(= \hat{i}((-6)(-12) - 0) - \hat{j}((8)(-12) - 0) + \hat{k}((8)(4) - (-6)(3))\)

\(= \hat{i}(72) - \hat{j}(-96) + \hat{k}(32 + 18)\)

\(= 72\hat{i} + 96\hat{j} + 50\hat{k}\).


Magnitude \(|\vec{d_1} \times \vec{d_2}| = \sqrt{72^2 + 96^2 + 50^2}\).

\(= \sqrt{5184 + 9216 + 2500} = \sqrt{16900} = 130\).


Area \(= \frac{1}{2} \times 130 = 65\).
Quick Tip: Formula for Area of Parallelogram: If adjacent sides are \(\vec{a}, \vec{b}\), Area \(= |\vec{a} \times \vec{b}|\). If diagonals are \(\vec{d_1}, \vec{d_2}\), Area \(= \frac{1}{2}|\vec{d_1} \times \vec{d_2}|\).

Rearrange the equation: \(y dx - x dy - 3y^2 dy = 0\).


Divide by \(y^2\): \(\frac{y dx - x dy}{y^2} - 3 dy = 0\).


Recognize that \(\frac{y dx - x dy}{y^2} = d\left(\frac{x}{y}\right)\).


So, \(d\left(\frac{x}{y}\right) - 3 dy = 0\).


Integrating both sides: \(\frac{x}{y} - 3y = C\).


The curve passes through \((1, 1)\), so substitute \(x=1, y=1\):

\(1/1 - 3(1) = C \implies 1 - 3 = C \implies C = -2\).


The equation of the curve is \(\frac{x}{y} - 3y = -2 \implies x = 3y^2 - 2y\).


Now check the options to see which point satisfies this equation.


(A) \(y = 1/3\). LHS \(x = 3(1/9) - 2(1/3) = 1/3 - 2/3 = -1/3\). Matches x-coordinate.


(B) \(y = -1/2\). LHS \(x = 3(1/4) - 2(-1/2) = 3/4 + 1 = 7/4 \neq 1/4\).


(C) \(y = -1/3\). LHS \(x = 3(1/9) - 2(-1/3) = 1/3 + 2/3 = 1 \neq 1/3\).


(D) \(y = 1/2\). LHS \(x = 3(1/4) - 2(1/2) = 3/4 - 1 = -1/4 \neq 1/4\).
Quick Tip: Look for exact differentials like \(d(x/y) = \frac{y dx - x dy}{y^2}\) to simplify differential equations.

Simplify the denominator: \(\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}\).


So, \((\tan x + \cot x)^3 = \left(\frac{2}{\sin 2x}\right)^3 = \frac{8}{\sin^3 2x}\).


The integrand becomes: \(\frac{8 \cos 2x}{8 / \sin^3 2x} = \cos 2x \sin^3 2x\).


Let \(I = \int_{\pi/12}^{\pi/4} \cos 2x \sin^3 2x \, dx\).


Substitute \(u = \sin 2x\), then \(du = 2 \cos 2x \, dx \implies \cos 2x \, dx = \frac{du}{2}\).


Change limits:


When \(x = \pi/12\), \(u = \sin(\pi/6) = 1/2\).


When \(x = \pi/4\), \(u = \sin(\pi/2) = 1\).

\(I = \int_{1/2}^{1} u^3 \frac{du}{2} = \frac{1}{2} \left[ \frac{u^4}{4} \right]_{1/2}^{1} = \frac{1}{8} \left[ 1^4 - \left(\frac{1}{2}\right)^4 \right]\).

\(I = \frac{1}{8} \left[ 1 - \frac{1}{16} \right] = \frac{1}{8} \cdot \frac{15}{16} = \frac{15}{128}\).
Quick Tip: Simplify trigonometric expressions before integrating. \(\tan x + \cot x = 2 \csc 2x\) is a useful identity.

Differentiate the curve equation implicitly w.r.t \(x\):

\(x^2(2y y') + y^2(2x) - 2 = -4y'\).


Substitute the point \((2, -2)\) into the derivative equation:

\(2^2(2(-2)y') + (-2)^2(2(2)) - 2 = -4y'\).

\(4(-4y') + 4(4) - 2 = -4y'\).

\(-16y' + 16 - 2 = -4y'\).

\(14 = 12y' \implies y' = \frac{14}{12} = \frac{7}{6}\).


Equation of tangent at \((2, -2)\): \(y - (-2) = \frac{7}{6}(x - 2)\).

\(6(y+2) = 7(x-2) \implies 6y + 12 = 7x - 14\).

\(7x - 6y - 26 = 0\).


Now check which point does NOT satisfy this equation.


(A) \((-2, -7)\): \(7(-2) - 6(-7) - 26 = -14 + 42 - 26 = 2 \neq 0\). This point is not on the line.


(B) \((8, 5)\): \(7(8) - 6(5) - 26 = 56 - 30 - 26 = 0\). On the line.


(C) \((4, 1/3)\): \(7(4) - 6(1/3) - 26 = 28 - 2 - 26 = 0\). On the line.


(D) \((-4, -9)\): \(7(-4) - 6(-9) - 26 = -28 + 54 - 26 = 0\). On the line.
Quick Tip: To check if a point lies on a line, substitute the coordinates into the line equation. If LHS \(\neq\) RHS, it does not pass through.

Simplify the term inside the square root:

\(1 + 2\cot x \csc x + 2\cot^2 x\).


Use identity \(\csc^2 x = 1 + \cot^2 x\). Substitute \(1 = \csc^2 x - \cot^2 x\):

\((\csc^2 x - \cot^2 x) + 2\cot x \csc x + 2\cot^2 x = \csc^2 x + \cot^2 x + 2\cot x \csc x\).


This is a perfect square: \((\csc x + \cot x)^2\).


So, \(\int \sqrt{(\csc x + \cot x)^2} dx = \int (\csc x + \cot x) dx\) (since \(x \in (0, \pi/2)\), terms are positive).


Simplify \(\csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1+\cos x}{\sin x}\).


Using half-angle formulas: \(\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} = \cot\left(\frac{x}{2}\right)\).


Integral becomes \(I = \int \cot\left(\frac{x}{2}\right) dx\).


Let \(u = x/2 \implies dx = 2du\).

\(I = \int \cot u \cdot 2du = 2 \ln|\sin u| + C\).


Substitute back \(u = x/2\): \(I = 2 \ln\left(\sin \frac{x}{2}\right) + C\).
Quick Tip: Simplify the integrand using trigonometric identities like \(\csc x + \cot x = \cot(x/2)\) before integrating.

Total number of outcomes when tossing a coin 8 times is \(2^8 = 256\).


The event "at least one head and at least one tail" is the complement of the event "all heads or all tails".


The number of outcomes with "all heads" (HHHHHHHH) is 1.


The number of outcomes with "all tails" (TTTTTTTT) is 1.


So, the number of unfavorable outcomes is \(1 + 1 = 2\).


The number of favorable outcomes is Total - Unfavorable \(= 256 - 2 = 254\).


Probability = \(\frac{254}{256} = \frac{127}{128}\).
Quick Tip: Use the complement rule: \(P(At least one) = 1 - P(None)\). For "At least one H and one T", exclude the two extreme cases (All H, All T).

For a homogeneous system to have infinitely many (non-trivial) solutions, the determinant of the coefficient matrix must be zero.

\(D = \begin{vmatrix} 2 & 4 & -\lambda
4 & \lambda & 2
\lambda & 2 & 2 \end{vmatrix} = 0\).


Expand along the first row:

\(2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) = 0\).

\(4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3 = 0\).

\(\lambda^3 + 4\lambda - 40 = 0\).


Let \(f(\lambda) = \lambda^3 + 4\lambda - 40\).


Since \(f'(\lambda) = 3\lambda^2 + 4 > 0\) for all real \(\lambda\), the function is strictly increasing.


Thus, \(f(\lambda) = 0\) has exactly one real root. (Note: \(f(3) = -1\) and \(f(4) = 40\), so the root lies between 3 and 4).


Therefore, there is only 1 real value of \(\lambda\).
Quick Tip: If a cubic equation \(f(x) = 0\) has a strictly positive derivative \(f'(x) > 0\), it is monotonic and intersects the x-axis exactly once.

The plane containing the two lines has a normal vector \(\vec{n} = \vec{d_1} \times \vec{d_2}\).

\(\vec{d_1} = <6, 7, 8>\) and \(\vec{d_2} = <3, 5, 7>\).

\(\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
6 & 7 & 8
3 & 5 & 7 \end{vmatrix} = \hat{i}(49-40) - \hat{j}(42-24) + \hat{k}(30-21) = 9\hat{i} - 18\hat{j} + 9\hat{k}\).


Direction ratios of normal are proportional to \(<1, -2, 1>\).


The plane passes through point \((1, 2, 3)\) (from the second line).


Equation of plane: \(1(x-1) - 2(y-2) + 1(z-3) = 0 \implies x - 2y + z = 0\).


Line of perpendicular from \(P(1, -2, 1)\) has direction \(<1, -2, 1>\).


Equation of perpendicular line: \(\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-1}{1} = k\).


General point \(Q(k+1, -2k-2, k+1)\).


Substitute \(Q\) into the plane equation:

\((k+1) - 2(-2k-2) + (k+1) = 0\).

\(k + 1 + 4k + 4 + k + 1 = 0 \implies 6k + 6 = 0 \implies k = -1\).


Coordinates of foot \(Q\): \(( -1+1, -2(-1)-2, -1+1 ) = (0, 0, 0)\).
Quick Tip: To find the foot of the perpendicular, find the intersection of the plane and the line passing through the point parallel to the plane's normal.

The direction vector \(\vec{d}\) of the line of intersection is the cross product of normals \(\vec{n_1} = <3, -1, 1>\) and \(\vec{n_2} = <1, 4, -2>\).

\(\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & -1 & 1
1 & 4 & -2 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-6-1) + \hat{k}(12+1) = -2\hat{i} + 7\hat{j} + 13\hat{k}\).


Direction ratios are \(<-2, 7, 13>\) or equivalently \(<2, -7, -13>\). This matches Option (A).


To find a point on the line, let \(z = 0\).

\(3x - y = 1\) and \(x + 4y = 2\).


Multiply first eq by 4: \(12x - 4y = 4\). Add to second eq: \(13x = 6 \implies x = 6/13\).

\(3(6/13) - y = 1 \implies y = 18/13 - 1 = 5/13\).


Point is \((6/13, 5/13, 0)\).


Equation of line: \(\frac{x - 6/13}{2} = \frac{y - 5/13}{-7} = \frac{z - 0}{-13}\).
Quick Tip: The direction of the line of intersection of two planes is the cross product of their normal vectors.

\(27 \equiv -1 \pmod{7}\) because \(27 = 4 \times 7 - 1\).

\((27)^{999} \equiv (-1)^{999} \pmod{7}\).


Since 999 is odd, \((-1)^{999} = -1\).

\(-1 \equiv 6 \pmod{7}\).


The remainder is 6.
Quick Tip: Using modular arithmetic with negative remainders (e.g., \(27 \equiv -1\)) simplifies high power calculations significantly.

Letters in alphabetical order: E, E, N, Q, U.


Total permutations starting with E: \(\frac{4!}{1!} = 24\).


Total permutations starting with N: \(\frac{4!}{2!} = 12\).


Total so far: \(24 + 12 = 36\).


Now we start with Q. The remaining letters are E, E, N, U.


Words starting with QE: Remaining are E, N, U. \(3! = 6\) words.


Total so far: \(36 + 6 = 42\).


Words starting with QN: Remaining are E, E, U. \(\frac{3!}{2!} = 3\) words.


Total so far: \(42 + 3 = 45\).


Next comes QU. Remaining letters E, E, N.


The first word in this block is formed by arranging E, E, N alphabetically: EEN.


So, the 46th word is QUEEN.
Quick Tip: Calculate the number of words starting with each preceding letter alphabetically, summing them up to reach the target word's position.

Given \(2|z + 3i| = |z - i|\). Square both sides: \(4|z+3i|^2 = |z-i|^2\).


Let \(z = x + iy\).

\(4[x^2 + (y+3)^2] = x^2 + (y-1)^2\).

\(4[x^2 + y^2 + 6y + 9] = x^2 + y^2 - 2y + 1\).

\(4x^2 + 4y^2 + 24y + 36 = x^2 + y^2 - 2y + 1\).

\(3x^2 + 3y^2 + 26y + 35 = 0\).


Divide by 3: \(x^2 + y^2 + \frac{26}{3}y + \frac{35}{3} = 0\).


This represents a circle with center \((0, -13/3)\).


Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{0 + \left(\frac{13}{3}\right)^2 - \frac{35}{3}} = \sqrt{\frac{169}{9} - \frac{105}{9}} = \sqrt{\frac{64}{9}} = \frac{8}{3}\).
Quick Tip: The equation \(|z - z_1| = k|z - z_2|\) represents a circle if \(k \neq 1\), and a perpendicular bisector if \(k = 1\). This is the Circle of Apollonius.

Expand the determinant:

\(0 - \cos x (0 - \cos^2 x) - \sin x (\sin^2 x - 0) = 0\).

\(\cos^3 x - \sin^3 x = 0 \implies \tan^3 x = 1 \implies \tan x = 1\).


For \(x \in [0, 2\pi]\), the solutions are \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).


We need to calculate \(Sum = \tan(\frac{\pi}{3} + \frac{\pi}{4}) + \tan(\frac{\pi}{3} + \frac{5\pi}{4})\).


Note that \(\tan(\frac{\pi}{3} + \frac{5\pi}{4}) = \tan(\frac{\pi}{3} + \pi + \frac{\pi}{4}) = \tan(\frac{\pi}{3} + \frac{\pi}{4})\).


So, \(Sum = 2 \tan(\frac{7\pi}{12}) = 2 \tan(105^\circ)\).

\(\tan 105^\circ = \tan(60^\circ + 45^\circ) = \frac{\sqrt{3}+1}{1-\sqrt{3}} = -(2+\sqrt{3})\).

\(Sum = 2 [-(2+\sqrt{3})] = -4 - 2\sqrt{3}\).
Quick Tip: Remember \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) and the periodicity \(\tan(\pi + \theta) = \tan \theta\).

The circle has center \(C(5/2, 1/2)\) and radius \(r = \sqrt{(5/2)^2 + (1/2)^2 - 5} = \sqrt{\frac{25}{4} + \frac{1}{4} - \frac{20}{4}} = \sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}\).


The maximum distance from an external point P to a circle is \(d_{max} = PC + r\).

\(PC^2 = (5/2 - 0)^2 + (1/2 - (-2))^2 = (5/2)^2 + (5/2)^2 = \frac{25}{4} + \frac{25}{4} = \frac{50}{4} = \frac{25}{2}\).

\(PC = \frac{5}{\sqrt{2}}\).

\(PQ_{max} = \frac{5}{\sqrt{2}} + \frac{\sqrt{6}}{2} = \frac{5}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{5+\sqrt{3}}{\sqrt{2}}\).

\((PQ_{max})^2 = \frac{(5+\sqrt{3})^2}{2} = \frac{25 + 3 + 10\sqrt{3}}{2} = \frac{28 + 10\sqrt{3}}{2} = 14 + 5\sqrt{3}\).
Quick Tip: For a point P and a circle with center C and radius r, maximum distance is \(PC + r\) and minimum distance is \(|PC - r|\).

Intersection of \(x^2 + y^2 = 4\) and \(y^2 = 3x\):

\(x^2 + 3x - 4 = 0 \implies (x+4)(x-1) = 0\). Since \(x \ge 0\), \(x=1\). Points are \((1, \pm\sqrt{3})\).


The area of the smaller portion (bounded by parabola and circle on the right) is best calculated by integrating with respect to \(y\).

\(A = \int_{-\sqrt{3}}^{\sqrt{3}} (x_{circle} - x_{parabola}) dy\). No, check position: for \(y=0\), \(x_c=2, x_p=0\). Circle is right, parabola is left.

\(A = \int_{-\sqrt{3}}^{\sqrt{3}} (\sqrt{4-y^2} - \frac{y^2}{3}) dy = 2 \int_{0}^{\sqrt{3}} (\sqrt{4-y^2} - \frac{y^2}{3}) dy\).

\(\int_{0}^{\sqrt{3}} \sqrt{4-y^2} dy = \left[ \frac{y}{2}\sqrt{4-y^2} + 2\sin^{-1}\frac{y}{2} \right]_0^{\sqrt{3}} = \frac{\sqrt{3}}{2}(1) + 2(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} + \frac{2\pi}{3}\).

\(\int_{0}^{\sqrt{3}} \frac{y^2}{3} dy = \left[ \frac{y^3}{9} \right]_0^{\sqrt{3}} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3}\).

\(A = 2 [ (\frac{\sqrt{3}}{2} + \frac{2\pi}{3}) - \frac{\sqrt{3}}{3} ] = 2 [ \frac{\sqrt{3}}{6} + \frac{2\pi}{3} ] = \frac{\sqrt{3}}{3} + \frac{4\pi}{3} = \frac{1}{\sqrt{3}} + \frac{4\pi}{3}\).
Quick Tip: When area is symmetric about x-axis, integrate w.r.t y from 0 to intersection and multiply by 2. It avoids splitting the integral at x=1.

This is a \(0/0\) form limit. Apply L'Hopital's Rule.


Differentiate numerator: \(\frac{d}{dx}(\sqrt{3x} - 3) = \frac{1}{2\sqrt{3x}} \cdot 3\).


Differentiate denominator: \(\frac{d}{dx}(\sqrt{2x-4} - \sqrt{2}) = \frac{1}{2\sqrt{2x-4}} \cdot 2\).


Limit becomes \(\lim_{x \to 3} \frac{3 / (2\sqrt{3x})}{1 / \sqrt{2x-4}}\).


Substitute \(x=3\):


Numerator \(= \frac{3}{2\sqrt{9}} = \frac{3}{6} = \frac{1}{2}\).


Denominator \(= \frac{1}{\sqrt{2}}\).


Result \(= \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\).
Quick Tip: For limits involving square roots giving \(0/0\), L'Hopital's rule or rationalization (multiplying by conjugate) are effective methods.

Let \(x^2 = \cos 2\theta\). Then \(\theta = \frac{1}{2} \cos^{-1} x^2\).

\(\sqrt{1+x^2} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2 \theta} = \sqrt{2}\cos \theta\).

\(\sqrt{1-x^2} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2 \theta} = \sqrt{2}\sin \theta\).


Expression becomes \(\tan^{-1}\left[ \frac{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta} \right] = \tan^{-1}\left[ \frac{1 + \tan \theta}{1 - \tan \theta} \right]\).

\(= \tan^{-1}(\tan(\frac{\pi}{4} + \theta)) = \frac{\pi}{4} + \theta\).


Substitute back \(\theta\): \(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2\).
Quick Tip: Substitution \(x^2 = \cos 2\theta\) is standard for simplifying expressions involving \(\sqrt{1+x^2}\) and \(\sqrt{1-x^2}\).

Distance between foci is \(2ae = 6 \implies ae = 3\).


Given \(e = \frac{3}{5}\), we have \(a \cdot \frac{3}{5} = 3 \implies a = 5\).


Use \(b^2 = a^2(1 - e^2) = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16\). So \(b = 4\).


The vertices of the ellipse are \((\pm a, 0)\) and \((0, \pm b)\).


The inscribed quadrilateral with these vertices is a rhombus with diagonals \(d_1 = 2a = 10\) and \(d_2 = 2b = 8\).


Area \(= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 8 = 40\).
Quick Tip: The area of a rhombus is half the product of its diagonals. For an ellipse, the vertices form a rhombus with diagonals \(2a\) and \(2b\).

Let \(p(x) = ax^2 + bx + c\).

\(p(0) = 1 \implies c = 1\).


By Remainder Theorem, \(p(1) = 4\) and \(p(-1) = 6\).

\(p(1) = a + b + 1 = 4 \implies a + b = 3\).

\(p(-1) = a - b + 1 = 6 \implies a - b = 5\).


Adding the equations: \(2a = 8 \implies a = 4\).


Subtracting: \(2b = -2 \implies b = -1\).


So, \(p(x) = 4x^2 - x + 1\).


Calculate \(p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 3 = 19\).


Calculate \(p(2) = 4(2)^2 - 2 + 1 = 16 - 1 = 15\).


Thus, \(p(-2) = 19\).
Quick Tip: The Remainder Theorem states that \(p(a)\) is the remainder when polynomial \(p(x)\) is divided by \((x-a)\).

We need the probability of the event \((P \cup Q) \cap R'\).


Since events are independent, \(P((P \cup Q) \cap R') = P(P \cup Q) \times P(R')\).

\(P(R') = 1 - P(R) = 1 - \frac{5}{8} = \frac{3}{8}\).

\(P(P \cup Q) = P(P) + P(Q) - P(P \cap Q)\).

\(P(P \cup Q) = \frac{3}{4} + \frac{1}{2} - (\frac{3}{4} \times \frac{1}{2}) = \frac{3}{4} + \frac{2}{4} - \frac{3}{8} = \frac{5}{4} - \frac{3}{8} = \frac{10 - 3}{8} = \frac{7}{8}\).


Required Probability \(= \frac{7}{8} \times \frac{3}{8} = \frac{21}{64}\).
Quick Tip: For independent events A and B, \(P(A \cup B) = P(A) + P(B) - P(A)P(B)\). "Not R" means multiplying by \(1 - P(R)\).