We need to find the dimensions of Power (\(P\)) in terms of Kinetic Energy (\(K\)), Current (\(I\)), and Charge (\(Q\)).


The dimensional formula for Power is:
\([P] = \frac{Energy}{Time} = [M L^2 T^{-3}]\)


The dimensional formula for Kinetic Energy (\(K\)) is same as Energy:
\([K] = [M L^2 T^{-2}]\)


The dimensional formula for Charge (\(Q\)) is current multiplied by time:
\([Q] = [I T]\)


From the charge equation, we can express Time (\(T\)) as:
\(T = Q I^{-1}\)


Now, express Power in terms of Energy (\(K\)) and Time (\(T\)):
\(P = \frac{K}{T}\)


Substitute the dimensions of \(T\):
\(P = \frac{K}{Q I^{-1}}\)


Rearranging the terms to match the format:
\(P = K I Q^{-1}\)


Thus, the dimensional representation is \([K I Q^{-1}]\).
Quick Tip: To express a derived quantity in terms of a new set of fundamental quantities, express the derived quantity's standard formula using terms that relate to the new fundamental quantities (e.g., \(P = E/t\) and \(t = Q/I\)).

Let's find the stopping distance for each truck using the equation of motion \(v^2 - u^2 = 2as\).


For Truck A:

Initial velocity \(u_A = 16 m/s\).

Final velocity \(v_A = 0 m/s\).

Deceleration \(a_A = -2 m/s^2\).
\(0^2 - (16)^2 = 2(-2)s_A\)
\(-256 = -4s_A \implies s_A = 64 m\).


For Truck B:

Initial velocity \(u_B = 20 m/s\).

Final velocity \(v_B = 0 m/s\).

Deceleration \(a_B = -4 m/s^2\).
\(0^2 - (20)^2 = 2(-4)s_B\)
\(-400 = -8s_B \implies s_B = 50 m\).


Total distance covered by both trucks before stopping is:
\(s_{total} = s_A + s_B = 64 + 50 = 114 m\).


The initial separation between the trucks was \(200 m\).


The final distance between them is the remaining gap:
\(d = Initial Separation - Total Distance Covered\)
\(d = 200 - 114 = 86 m\).
Quick Tip: When two objects move towards each other and stop, calculate the individual stopping distances. The final gap is the initial distance minus the sum of the stopping distances (provided no collision occurs).

We use the law of conservation of linear momentum. Since there is no external force, the total momentum remains constant.


Initial momentum of the system (gun + bullet) is zero because they are at rest.
\(P_{initial} = 0\).


Let \(m\) be the mass of the bullet, \(v\) be the velocity of the bullet, \(M\) be the mass of the gun, and \(V\) be the recoil velocity of the gun.
\(m = 10 g = 0.01 kg\).
\(v = 100 m/s\).
\(M = 1 kg\).


Final momentum \(P_{final} = mv + MV\).

According to conservation of momentum:
\(0 = (0.01 kg)(100 m/s) + (1 kg)V\).

\(0 = 1 + V\).

\(V = -1 m/s\).


The magnitude of the recoil speed is \(1 m/s\).
Quick Tip: Always convert mass units to kg (SI units) before calculation. Momentum conservation is vector-based; the negative sign indicates recoil is opposite to the bullet's direction.

The mass undergoes uniform circular motion. The necessary centripetal force is provided by the spring force due to extension.


Let \(l_0\) be the natural length (\(1 m\)) and \(x\) be the extension.

The radius of the circular path is \(r = l_0 + x = 1 + x\).


The spring force is \(F_s = kx\).

The centripetal force required is \(F_c = m \omega^2 r = m \omega^2 (1 + x)\).


Equating the forces:
\(kx = m \omega^2 (1 + x)\).


Substitute the given values:
\(m = 0.5 kg\), \(k = 800 N/m\), \(\omega = 20 rad/s\).

\(800 x = 0.5 \times (20)^2 \times (1 + x)\).
\(800 x = 0.5 \times 400 \times (1 + x)\).
\(800 x = 200 (1 + x)\).


Divide both sides by 200:
\(4x = 1 + x\).
\(3x = 1\).
\(x = \frac{1}{3} m\).
Quick Tip: Remember that for a rotating spring-mass system, the radius of rotation is the sum of the natural length and the extension (\(r = l_0 + x\)).

According to Kepler's Third Law, the square of the orbital period is proportional to the cube of the semi-major axis (distance):
\(T^2 \propto R^3\).


We can form a ratio between Planet X and Earth:
\(\left(\frac{T_X}{T_E}\right)^2 = \left(\frac{R_X}{R_E}\right)^3\).


Given:
\(T_X = 8 years\).
\(T_E = 1 year\) (for Earth).
\(R_E = 1.5 \times 10^{11} m\).


Substituting the values:
\(\left(\frac{8}{1}\right)^2 = \left(\frac{R_X}{R_E}\right)^3\).
\(64 = \left(\frac{R_X}{R_E}\right)^3\).


Taking the cube root of both sides:
\(\frac{R_X}{R_E} = \sqrt[3]{64} = 4\).


Therefore:
\(R_X = 4 \times R_E\).
\(R_X = 4 \times 1.5 \times 10^{11} m\).
\(R_X = 6.0 \times 10^{11} m\).
Quick Tip: Using ratios (e.g., \(T_1/T_2\)) simplifies Kepler's law problems significantly, avoiding the need for the gravitational constant \(G\) or solar mass \(M\).

The diagram shows a steel wire connected to a copper wire in series, with a mass \(M\) hanging at the bottom. The mass of the wires is neglected.


Since the wires are in series and massless, the tension force (\(F\)) is uniform throughout both wires and is equal to the weight of the block (\(Mg\)).

The area of cross-section (\(A\)) is also the same for both.


Stress is defined as Force per unit Area: \(\sigma = \frac{F}{A}\). Since \(F\) and \(A\) are same, stress is same for both wires.


Young's Modulus (\(Y\)) relates stress and strain (\(\epsilon\)):
\(Y = \frac{Stress}{Strain} \implies Strain (\epsilon) = \frac{Stress}{Y}\).


The ratio of strain in steel (\(\epsilon_S\)) to strain in copper (\(\epsilon_C\)) is:
\(\frac{\epsilon_S}{\epsilon_C} = \frac{\sigma / Y_S}{\sigma / Y_C} = \frac{Y_C}{Y_S}\).


Given:
\(Y_S = 2 \times 10^{11} N/m^2\).
\(Y_C = 1.5 \times 10^{11} N/m^2\).


Substitute the values:
\(\frac{\epsilon_S}{\epsilon_C} = \frac{1.5 \times 10^{11}}{2 \times 10^{11}} = \frac{1.5}{2} = \frac{3}{4}\).
Quick Tip: For wires in series supporting a load (neglecting wire mass), the stress is constant. Therefore, strain is inversely proportional to Young's Modulus (\(\epsilon \propto 1/Y\)).

First, convert the temperatures from Celsius to Kelvin.

Source temperature \(T_1 = 227 ^{\circ}C = 227 + 273 = 500 K\).

Sink temperature \(T_2 = 127 ^{\circ}C = 127 + 273 = 400 K\).


The efficiency (\(\eta\)) of a Carnot engine is given by:
\(\eta = 1 - \frac{T_2}{T_1}\).
\(\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2\).


Efficiency is also defined as the ratio of Work done (\(W\)) to Heat absorbed (\(Q_1\)):
\(\eta = \frac{W}{Q_1}\).


We are given \(W = 1.04 \times 10^5 J\).

So, \(0.2 = \frac{1.04 \times 10^5}{Q_1}\).


Solving for \(Q_1\):
\(Q_1 = \frac{1.04 \times 10^5}{0.2} = \frac{1.04}{0.2} \times 10^5\).
\(Q_1 = 5.2 \times 10^5 J\).
Quick Tip: Always convert temperatures to Kelvin when using thermodynamic formulas. Efficiency \(\eta = W/Q_{absorbed} = 1 - T_{cold}/T_{hot}\).

The root mean square (r.m.s) speed of a gas is given by:
\(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(M\) is the molar mass.


At the same temperature \(T\), \(v_{rms}\) is inversely proportional to the square root of the molar mass:
\(v_{rms} \propto \frac{1}{\sqrt{M}}\).


Let \(v_{O_2}\) be the speed of oxygen and \(v_{N_2}\) be the speed of nitrogen.

Molar mass of Oxygen (\(O_2\)) = \(32 g/mol\).

Molar mass of Nitrogen (\(N_2\)) = \(28 g/mol\).


The ratio of speeds is:
\(\frac{v_{N_2}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{N_2}}}\).


Substitute the values:
\(\frac{v_{N_2}}{\sqrt{56}} = \sqrt{\frac{32}{28}}\).


Simplify the fraction inside the square root:
\(\frac{32}{28} = \frac{8}{7}\).


So,
\(v_{N_2} = \sqrt{56} \times \sqrt{\frac{8}{7}} = \sqrt{56 \times \frac{8}{7}}\).
\(v_{N_2} = \sqrt{8 \times 8} = 8 m/s\).
Quick Tip: For gases at the same temperature, lighter gases move faster. Use the ratio \(v_1/v_2 = \sqrt{M_2/M_1}\) for quick calculation.

The time period of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{L}{g_{eff}}}\).


Initially, the lift is effectively at rest relative to gravity (or not accelerating vertically), so \(T_o = 2\pi \sqrt{\frac{L}{g}}\).


When the lift accelerates upwards with acceleration \(a\), the effective gravity inside the non-inertial frame of the lift increases.
\(g_{eff} = g + a\).


Given \(a = 2.5 m/s^2\) and \(g = 10 m/s^2\).
\(g_{eff} = 10 + 2.5 = 12.5 m/s^2\).

Note that \(12.5 = \frac{5}{4} g\).


The new time period \(T'\) is:
\(T' = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{1.25 g}}\).


Substituting \(T_o\):
\(T' = T_o \times \sqrt{\frac{g}{g_{eff}}} = T_o \times \sqrt{\frac{10}{12.5}} = T_o \times \sqrt{\frac{100}{125}} = T_o \times \sqrt{\frac{4}{5}}\).

\(T' = T_o \times \frac{2}{\sqrt{5}}\).
Quick Tip: When a lift accelerates upwards, the effective gravity increases (\(g_{eff} = g + a\)), causing the pendulum to oscillate faster (shorter time period). If accelerating downwards, \(g_{eff} = g - a\).

The electric field at the center of a semicircular arc with linear charge density \(\lambda\) and radius \(R\) is given by the standard formula:
\(E = \frac{2k\lambda}{R}\), where \(k = \frac{1}{4\pi\epsilon_0}\).


Here, the total charge \(q\) is distributed over length \(L\). Since the rod is bent into a semicircle, the length \(L\) becomes the circumference of the semicircle.
\(L = \pi R \implies R = \frac{L}{\pi}\).


The linear charge density \(\lambda\) is:
\(\lambda = \frac{q}{L}\).


Now substitute \(\lambda\) and \(R\) into the electric field formula:
\(E = \frac{2}{4\pi\epsilon_0} \frac{\lambda}{R} = \frac{1}{2\pi\epsilon_0} \frac{q/L}{L/\pi}\).

\(E = \frac{1}{2\pi\epsilon_0} \times \frac{q}{L} \times \frac{\pi}{L}\).


The \(\pi\) cancels out:
\(E = \frac{1}{2\epsilon_0} \frac{q}{L^2}\).

\(E = \frac{q}{2\epsilon_0 L^2}\).
Quick Tip: For arc problems, relate the geometric length to the radius (\(L = \pi R\) for semicircle). The formula for \(E\) at center of an arc subtending angle \(\alpha\) is \(\frac{2k\lambda}{R}\sin(\alpha/2)\). For a semicircle, \(\alpha=\pi\), so \(\sin(\pi/2)=1\), giving \(2k\lambda/R\).

Let's analyze the circuit diagram to determine the equivalent resistance.


The circuit consists of a voltage source (\(9 V\)) and a network of resistors.

Tracing the connections:

1. There is a top branch with a \(3 \Omega\) resistor connected directly across the effective input and output nodes of the network.

2. There is a bottom branch configuration. The current entering this branch passes through a \(1.5 \Omega\) resistor.

3. After the \(1.5 \Omega\) resistor, the circuit splits into two parallel paths: a \(6 \Omega\) horizontal resistor and a \(2 \Omega\) vertical resistor. Both of these resistors connect to the negative terminal (return path) of the battery.


So, the \(6 \Omega\) and \(2 \Omega\) resistors are in parallel.

Their equivalent resistance \(R_p\) is:
\(R_p = \frac{6 \times 2}{6 + 2} = \frac{12}{8} = 1.5 \Omega\).


This parallel combination is in series with the \(1.5 \Omega\) resistor.

The total resistance of this bottom branch \(R_{bottom}\) is:
\(R_{bottom} = 1.5 \Omega + R_p = 1.5 + 1.5 = 3 \Omega\).


Now, the entire bottom branch (\(3 \Omega\)) is in parallel with the top branch (\(3 \Omega\)).

The total equivalent resistance \(R_{eq}\) of the circuit is:
\(R_{eq} = \frac{3 \times 3}{3 + 3} = \frac{9}{6} = 1.5 \Omega\).


Finally, using Ohm's Law, the total current \(I\) drawn from the battery is:
\(I = \frac{V}{R_{eq}} = \frac{9 V}{1.5 \Omega} = 6 A\).
Quick Tip: Break down complex circuits into series and parallel blocks. Identify nodes to verify which components share the same potential difference (parallel) or the same current path (series).

The average power dissipated in an AC circuit is given by the formula:
\(P = V_{rms} I_{rms} \cos \phi\)

where \(\phi\) is the phase difference between voltage and current.


From the given equations:
\(V = 250 \sin(100 t) \implies V_m = 250 V\).
\(I = 10 \sin(100 t + \frac{\pi}{3}) \implies I_m = 10 A\).

The phase difference \(\phi = \frac{\pi}{3} = 60^\circ\).


Convert peak values to RMS values:
\(V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{250}{\sqrt{2}} V\).
\(I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{10}{\sqrt{2}} A\).


Substitute into the power equation:
\(P = \left( \frac{250}{\sqrt{2}} \right) \left( \frac{10}{\sqrt{2}} \right) \cos(60^\circ)\).
\(P = \frac{2500}{2} \times \frac{1}{2}\).
\(P = 1250 \times 0.5\).
\(P = 625 W\).
Quick Tip: Power is only dissipated in the resistive part of the circuit, determined by the power factor \(\cos \phi\). Remember \(P_{avg} = \frac{1}{2} V_m I_m \cos \phi\).

The relationship between the amplitude of the electric field (\(E_0\)) and the magnetic field (\(B_0\)) in an electromagnetic wave is given by:
\(B_0 = \frac{E_0}{c}\)

where \(c\) is the speed of light in vacuum, \(c = 3 \times 10^8 m/s\).


Given:
\(E_0 = 900 N/C\).


Substitute the values:
\(B_0 = \frac{900}{3 \times 10^8}\).
\(B_0 = 300 \times 10^{-8} T\).
\(B_0 = 3 \times 10^{-6} T\).


Since \(1 \muT = 10^{-6} T\),
\(B_0 = 3 \muT\).
Quick Tip: Remember the simple ratio \(E = cB\). The units of E are N/C or V/m, and B is in Tesla.

Let the initial intensity of the unpolarized light incident on \(P_1\) be \(I_0\).


When unpolarized light passes through the first polaroid (\(P_1\)), its intensity becomes half:
\(I_1 = \frac{I_0}{2}\).


The light emerging from \(P_1\) is linearly polarized. This light then strikes the second polaroid (\(P_2\)).

The intensity of light emerging from \(P_2\) is given by Malus' Law:
\(I_2 = I_1 \cos^2 \theta\)

where \(\theta\) is the angle between the transmission axes of \(P_1\) and \(P_2\).


Given that \(P_2\) is rotated by \(60^\circ\), we have \(\theta = 60^\circ\).

Substitute the values:
\(I_2 = \left( \frac{I_0}{2} \right) \cos^2(60^\circ)\).

Since \(\cos(60^\circ) = \frac{1}{2}\):
\(I_2 = \frac{I_0}{2} \times \left( \frac{1}{2} \right)^2\).
\(I_2 = \frac{I_0}{2} \times \frac{1}{4}\).
\(I_2 = \frac{I_0}{8}\).
Quick Tip: Incident unpolarized light loses 50% intensity at the first polarizer. The second polarizer reduces it further by a factor of \(\cos^2\theta\).

First, find the radius of the circular loop.

The length of the wire forms the circumference:
\(L = 2\pi R = 8 m \implies R = \frac{4}{\pi} m\).


Calculate the area of the loop:
\(A = \pi R^2 = \pi \left( \frac{4}{\pi} \right)^2 = \pi \frac{16}{\pi^2} = \frac{16}{\pi} m^2\).


The magnitude of the magnetic dipole moment (\(M\)) is given by \(M = NIA\) (where \(N=1\)):
\(M = I \times A = 0.5 A \times \frac{16}{\pi} m^2 = \frac{8}{\pi} A m^2\).


Determine the direction:

The loop lies in the y-z plane. The area vector is normal to the plane, i.e., along the x-axis.

Using the right-hand rule: curl fingers in the direction of the anticlockwise current (in y-z plane). The thumb points in the positive x-direction (\(+\hat{i}\)).


Thus, \(\vec{M} = \frac{8}{\pi} \hat{i} A m^2\).
Quick Tip: Magnetic moment \(\vec{M} = I \vec{A}\). The direction of \(\vec{A}\) is given by the right-hand rule curling along the current. For a loop in the y-z plane, the normal is along the x-axis.

The magnetic field at the center O is the vector sum of the fields due to the straight wire segments and the circular arc.


Based on the diagram and options:

1. The circular arc spans three quadrants, i.e., an angle of \(\theta = \frac{3\pi}{2}\) (\(270^\circ\)).

The field due to this arc is \(B_{arc} = \frac{\mu_0 I}{2R} \left( \frac{3/4} \right) = \frac{3\mu_0 I}{8R}\).

Using the Right Hand Rule, if current flows clockwise, this field points INTO the page.


2. There are two straight wire segments.

- Wire AB is semi-infinite and tangent to the circle at B. Its distance from O is R.

Field due to AB: \(B_{AB} = \frac{\mu_0 I}{4\pi R}\).

Direction: Current flows right. O is below. Field points OUT of the page.

- Wire CD appears to be radial (along the axis passing through O). A radial wire produces zero magnetic field at the center because \(d\vec{l} \times \vec{r} = 0\).


Net Magnetic Field:

Since \(B_{arc}\) is INTO and \(B_{AB}\) is OUT, they are in opposite directions.
\(B_{net} = |B_{arc} - B_{AB}|\).
\(B_{net} = \left| \frac{3\mu_0 I}{8R} - \frac{\mu_0 I}{4\pi R} \right|\).


Factor out \(\frac{\mu_0 I}{4\pi R}\):
\(B_{net} = \frac{\mu_0 I}{4\pi R} \left| \frac{3\pi}{2} - 1 \right|\).


This matches option (D).
Quick Tip: Identify each segment's contribution separately. Radial wires (pointing to center) contribute 0. Arc contribution is proportional to angle (\(\frac{\mu_0 I}{4\pi R} \theta\)). Tangent semi-infinite wires contribute \(\frac{\mu_0 I}{4\pi R}\).

According to Bohr's model:

1. Angular momentum quantization: \(mvr = \frac{nh}{2\pi}\).

2. Centripetal force provided by electrostatic attraction: \(\frac{mv^2}{r} = \frac{ke^2}{r^2} \implies mv^2 = \frac{ke^2}{r}\).


From (1), \(v = \frac{nh}{2\pi m r}\).

Substitute \(v\) into (2):
\(m \left( \frac{nh}{2\pi m r} \right)^2 = \frac{ke^2}{r}\).
\(m \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{ke^2}{r}\).
\(\frac{n^2 h^2}{4\pi^2 m r} = ke^2\).


Solving for radius \(r\):
\(r = \frac{n^2 h^2}{4\pi^2 m k e^2}\).


Since \(h, m, k, e\) are constants, we get:
\(r \propto n^2\).
Quick Tip: Standard Bohr relations: Radius \(r \propto n^2/Z\), Velocity \(v \propto Z/n\), Energy \(E \propto Z^2/n^2\).

The graph shows current (I) vs voltage (V) in the third quadrant (negative V, negative I).

Let's analyze the options:

(B) LED in forward bias: Characteristics would be in the 1st quadrant.


(D) Solar cell: Characteristics are typically shown in the 4th quadrant (power source), not passing through origin.


(C) Zener diode: Typically shows a sharp breakdown voltage (vertical line) at \(V_z\). The graph shown is curved and does not exhibit a sharp vertical drop.

(A) Photodiode: Photodiodes are operated in reverse bias. The current (photocurrent) increases with light intensity and generally shows a characteristic curve in the 3rd quadrant. The graph represents the typical I-V behavior of a photodiode under reverse bias.



Therefore, the curve corresponds to a photodiode in reverse bias.
Quick Tip: Recognize typical I-V curves: 1st Quad = Forward Bias (Diode/LED); 3rd Quad with sharp knee = Zener; 3rd Quad curved/flat = Photodiode; 4th Quad = Solar Cell.

The de Broglie wavelength \(\lambda\) of an electron accelerated through a potential \(V\) is given by the formula:
\(\lambda = \frac{12.27}{\sqrt{V}} \mathring{A}\).


Given \(V = 15056 V\).

Calculate \(\sqrt{V}\):

Approximation: \(120^2 = 14400\) and \(123^2 \approx 15129\).

So \(\sqrt{15056} \approx 122.7\).


Substitute into the formula:
\(\lambda \approx \frac{12.27}{122.7}\).
\(\lambda \approx 0.1 \mathring{A}\).
Quick Tip: Memorize the formula \(\lambda = \frac{12.27}{\sqrt{V}} \mathring{A}\) for electrons. It saves time deriving from \(h/\sqrt{2meV}\).

In Amplitude Modulation (AM), the peak amplitude of the sidebands (\(A_{SB}\)) is related to the peak amplitude of the modulating signal (\(A_m\)) by the relation:
\(A_{SB} = \frac{A_m}{2}\).

(Alternatively, \(A_{SB} = \frac{\mu A_c}{2}\) where \(\mu = A_m/A_c\)).


Given:

Upper Side Band amplitude = \(2.5 V\).

Therefore, \(\frac{A_m}{2} = 2.5 V\).


Solving for \(A_m\):
\(A_m = 2.5 \times 2 = 5 V\).
Quick Tip: The amplitude of each sideband is exactly half the amplitude of the message signal in standard AM. \(A_{sideband} = A_m / 2\).

The distance of closest approach (\(r_0\)) provides an estimation for the radius of the nucleus. At this distance, the entire kinetic energy (\(K\)) of the alpha particle is converted into electrostatic potential energy.

\(K = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(2e)}{r_0}\)


Where:
\(Z = 79\) (Atomic number of Gold).

\(e = 1.6 \times 10^{-19} C\).

\(K = 3.2 MeV = 3.2 \times 10^6 \times 1.6 \times 10^{-19} J\).
\(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 Nm^2/C^2\).


Rearranging for \(r_0\):
\(r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}\).


Substitute the values:

\(r_0 = \frac{(9 \times 10^9) \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{3.2 \times 1.6 \times 10^{-13}}\).

\(r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times 1.6 \times 1.6 \times 10^{-38}}{3.2 \times 1.6 \times 10^{-13}}\).


Simplify terms:

\(\frac{2 \times 1.6}{3.2} = 1\). One \(1.6\) cancels out with the conversion factor in denominator or just compute directly:

\(r_0 = \frac{9 \times 10^9 \times 79 \times 2 \times 2.56 \times 10^{-38}}{3.2 \times 1.6 \times 10^{-13}}\).


Let's simplify: \(K = 3.2 \times 1.6 \times 10^{-13} J\).

\(r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{3.2 \times 1.6 \times 10^{-13}}\).

\(r_0 = \frac{18 \times 10^9 \times 79 \times 2.56 \times 10^{-38}}{5.12 \times 10^{-13}}\).

\(r_0 = \frac{18 \times 79 \times 2.56}{5.12} \times 10^{-16}\).

Since \(5.12 = 2 \times 2.56\):

\(r_0 = \frac{18 \times 79}{2} \times 10^{-16} = 9 \times 79 \times 10^{-16}\).
\(r_0 = 711 \times 10^{-16} m\).
Quick Tip: At the distance of closest approach, Kinetic Energy = Electric Potential Energy. Ensure all units are in SI (Joules, Coulombs) before calculating.

The total apparent depth (\(d_{app}\)) of an object viewed through multiple layers of different media is the sum of the apparent depths of each layer.


Formula: \(d_{app} = \sum \frac{d_i}{\mu_i}\)

Where \(d_i\) is the real depth and \(\mu_i\) is the refractive index of the \(i\)-th layer.


Given layers:

1. \(d_1 = 10 cm, \mu_1 = 1.0\)

2. \(d_2 = 20 cm, \mu_2 = 2.0\)

3. \(d_3 = 30 cm, \mu_3 = 3.0\)

4. \(d_4 = 40 cm, \mu_4 = 4.0\)


Calculation:
\(d_{app} = \frac{10}{1} + \frac{20}{2} + \frac{30}{3} + \frac{40}{4}\)
\(d_{app} = 10 + 10 + 10 + 10\)
\(d_{app} = 40 cm\).
Quick Tip: For composite slabs, apparent shift is \(\sum d_i (1 - 1/\mu_i)\), and apparent depth is \(\sum d_i / \mu_i\).

The impedance \(Z\) of a series LCR circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).


From the voltage equation \(V(t) = 200 \sin(2000 t)\), the angular frequency is \(\omega = 2000 rad/s\).


Calculate Inductive Reactance (\(X_L\)):

\(X_L = \omega L = 2000 \times 10 \times 10^{-3} = 20 \Omega\).


Calculate Capacitive Reactance (\(X_C\)):

\(X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 25 \times 10^{-6}}\).

\(X_C = \frac{10^6}{50000} = \frac{100}{5} = 20 \Omega\).


Since \(X_L = X_C = 20 \Omega\), the circuit is in resonance.
\(Z = \sqrt{R^2 + (20 - 20)^2} = \sqrt{R^2} = R\).


Given \(R = 100 \Omega\).
\(Z = 100 \Omega\).
Quick Tip: Check for resonance condition (\(X_L = X_C\)) first. If resonance occurs, the impedance is simply equal to the resistance \(R\).

Let \(E\) be the EMF and \(r\) be the internal resistance of the cell.

The heat produced (\(H\)) in an external resistance \(R\) for time \(t\) is given by \(H = I^2 R t\), where \(I = \frac{E}{R+r}\).


So, \(H = \left( \frac{E}{R+r} \right)^2 R t\).


We are given that heat produced in \(R_1 = 4 \Omega\) is equal to heat produced in \(R_2 = 16 \Omega\).
\(\left( \frac{E}{R_1+r} \right)^2 R_1 t = \left( \frac{E}{R_2+r} \right)^2 R_2 t\).


Canceling \(E^2\) and \(t\):
\(\frac{R_1}{(R_1+r)^2} = \frac{R_2}{(R_2+r)^2}\).


Take the square root of both sides:
\(\frac{\sqrt{R_1}}{R_1+r} = \frac{\sqrt{R_2}}{R_2+r}\).


Substitute \(R_1 = 4\) and \(R_2 = 16\):
\(\frac{2}{4+r} = \frac{4}{16+r}\).


Cross-multiply:
\(2(16+r) = 4(4+r)\).
\(32 + 2r = 16 + 4r\).
\(16 = 2r\).
\(r = 8 \Omega\).
Quick Tip: If the same power/heat is dissipated in two different external resistors \(R_1\) and \(R_2\) connected to the same source with internal resistance \(r\), then \(r = \sqrt{R_1 R_2}\).

From the circuit diagram, plates \(P_1\) and \(P_3\) are connected together by a wire. This means they are at the same potential. Plate \(P_2\) acts as the other terminal.


The system effectively consists of two capacitors in parallel connected between the common node \((P_1, P_3)\) and node \(P_2\):

1. Capacitor \(C_1\) formed between \(P_1\) and \(P_2\). The distance is \(d_1 = d/3\).

2. Capacitor \(C_2\) formed between \(P_2\) and \(P_3\). The distance is \(d_2 = d\).


The total area \(A\) is distributed among the sections shown, but since they are all connected in parallel across the same nodes, we can treat the areas as summing up to total area \(A\) for the top gap and total area \(A\) for the bottom gap.


Capacitance \(C_1\):
\(C_1 = \frac{\epsilon_0 A}{d_1} = \frac{\epsilon_0 A}{d/3} = \frac{3 \epsilon_0 A}{d}\).


Capacitance \(C_2\):
\(C_2 = \frac{\epsilon_0 A}{d_2} = \frac{\epsilon_0 A}{d}\).


Total Equivalent Capacitance (\(C_{eq}\)) for parallel combination:
\(C_{eq} = C_1 + C_2 = \frac{3 \epsilon_0 A}{d} + \frac{\epsilon_0 A}{d} = \frac{4 \epsilon_0 A}{d}\).


Given that \(\frac{\epsilon_0 A}{d} = 4 \muF\).
\(C_{eq} = 4 \times 4 \muF = 16 \muF\).
Quick Tip: Identify the nodes. Plates connected by a wire are at the same potential. If multiple plate sections are between the same two potential nodes, their capacitances add up (parallel).

For a particle in Simple Harmonic Motion (SHM):

Maximum velocity (at mean position) is given by \(v_{max} = A\omega\).

Maximum acceleration (at extreme position) is given by \(a_{max} = A\omega^2\).


Given:
\(v_{max} = 20 cm/s\).
\(a_{max} = 25 cm/s^2\).


We can find the angular frequency \(\omega\) by dividing \(a_{max}\) by \(v_{max}\):
\(\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega\).
\(\omega = \frac{25}{20} = \frac{5}{4} = 1.25 rad/s\).


Now, find the amplitude \(A\) using the velocity equation:
\(A = \frac{v_{max}}{\omega} = \frac{20}{5/4}\).
\(A = 20 \times \frac{4}{5} = 4 \times 4 = 16 cm\).
Quick Tip: Remember the characteristic maximum values in SHM: \(V_{max} = A\omega\) and \(a_{max} = A\omega^2\).

Let \(R\) be the radius of the big drop and \(r\) be the radius of the small droplets.

Number of droplets \(n = 8\).

Volume is conserved:
\(\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3\).
\(R^3 = 8 r^3 \implies R = 2r \implies r = \frac{R}{2}\).


Initial Surface Area \(A_i = 4\pi R^2\).

Final Surface Area \(A_f = n \times 4\pi r^2 = 8 \times 4\pi \left( \frac{R}{2} \right)^2 = 8 \times 4\pi \frac{R^2}{4} = 8\pi R^2\).


Change in area \(\Delta A = A_f - A_i = 8\pi R^2 - 4\pi R^2 = 4\pi R^2\).


Gain in Surface Energy \(\Delta U = T \times \Delta A = T \times 4\pi R^2\).


Given:
\(T = 0.075 N/m\).
\(R = 1 cm = 10^{-2} m\).
\(\pi = 3.14\).

\(\Delta U = 0.075 \times 4 \times 3.14 \times (10^{-2})^2\).
\(\Delta U = 0.3 \times 3.14 \times 10^{-4}\).
\(\Delta U = 0.942 \times 10^{-4} J\).


To convert to the format \(\dots \times 10^{-7} J\):
\(\Delta U = 942 \times 10^{-7} J\).
Quick Tip: When a large drop breaks into \(n\) smaller droplets, surface area increases. \(\Delta E = 4\pi R^2 T (n^{1/3} - 1)\). Here \(n=8, n^{1/3}=2\), so \(\Delta E = 4\pi R^2 T\).

Convert units to SI:

Mass \(m = 10 kg\).

Radius \(R = 50 cm = 0.5 m\).

Initial angular velocity \(\omega_0 = 360 rpm = 360 \times \frac{2\pi}{60} rad/s = 12\pi rad/s\).

Final angular velocity \(\omega_f = 0\).

Displacement \(\theta = 6 rotations = 6 \times 2\pi = 12\pi rad\).


Moment of Inertia (mass at rim) \(I = mR^2\):
\(I = 10 \times (0.5)^2 = 10 \times 0.25 = 2.5 kg m^2\).


Using the work-energy theorem for rotation:

Work done by torque = Change in Kinetic Energy.
\(\tau \cdot \theta = \frac{1}{2} I (\omega_0^2 - \omega_f^2)\).


Substitute the values:
\(\tau \times (12\pi) = \frac{1}{2} \times 2.5 \times (12\pi)^2\).
\(\tau \times 12\pi = 1.25 \times 144 \pi^2\).

Divide both sides by \(12\pi\):
\(\tau = 1.25 \times 12 \pi\).
\(\tau = 15 \pi Nm\).


The question asks for the value of \(x\) where Torque \(= x\pi Nm\).
\(x = 15\).
Quick Tip: Work-Energy theorem (\(W = \Delta K\)) is often faster than kinematic equations (\(\omega^2 = \omega_0^2 + 2\alpha\theta\)) followed by \(\tau = I\alpha\), though both work equally well.

Let the total mass be \(M = 3m\).

The masses of the parts are \(m\) (smaller) and \(2m\) (larger).

Initial velocity \(u = 20 m/s\).

Velocity of smaller part (\(v_1\)) = \(30 m/s\).


Apply Conservation of Linear Momentum:
\(P_i = P_f\)

\((3m) u = (m) v_1 + (2m) v_2\)

\(3m(20) = m(30) + 2m(v_2)\)

\(60 = 30 + 2v_2\)

\(30 = 2v_2 \implies v_2 = 15 m/s\).


Calculate Kinetic Energies:
\(K_i = \frac{1}{2}(3m)(20)^2 = \frac{3m}{2}(400) = 600m J\).

\(K_f = \frac{1}{2}m(30)^2 + \frac{1}{2}(2m)(15)^2\).

\(K_f = \frac{m}{2}(900) + m(225) = 450m + 225m = 675m J\).


Change in Kinetic Energy:
\(\Delta K = K_f - K_i = 675m - 600m = 75m\).


Fractional Change:
\(\frac{\Delta K}{K_i} = \frac{75m}{600m} = \frac{75}{600}\).


Dividing numerator and denominator by 75:

\(\frac{75}{600} = \frac{1}{8}\).


Given fractional change is \(\frac{1}{x}\), so \(x = 8\).
Quick Tip: Always verify momentum conservation to find unknown velocities before calculating energy. Energy is not conserved in an explosion/splitting event (internal energy converts to KE).

Let the initial speed be \(u\).

The horizontal distance traveled is the Range (\(R\)).

Formula for Range: \(R = \frac{u^2 \sin 2\theta}{g}\).


Projectile 1: \(\theta_1 = 15^\circ\).
\(R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} = \frac{u^2 \sin 30^\circ}{g} = \frac{u^2}{2g}\).


Projectile 2: \(\theta_2 = 45^\circ\).
\(R_2 = \frac{u^2 \sin(2 \times 45^\circ)}{g} = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g}\).


Given the difference in ranges is \(80 m\):
\(R_2 - R_1 = 80\)
\(\frac{u^2}{g} - \frac{u^2}{2g} = 80\)
\(\frac{u^2}{2g} = 80\)
\(u^2 = 160 g\).


Substitute \(g = 10\):
\(u^2 = 160 \times 10 = 1600\).
\(u = \sqrt{1600} = 40 m/s\).
Quick Tip: Remember standard values: \(\sin(30^\circ) = 0.5\) and \(\sin(90^\circ) = 1\). The range is maximum at \(45^\circ\).

A gas behaves ideally when intermolecular forces are negligible and the volume of gas molecules is negligible compared to the container volume.


These conditions are met at High Temperature and Low Pressure.


Conversely, deviation from ideal behavior is maximum when intermolecular forces are strong and gas density is high.


This occurs at Low Temperature and High Pressure.


Let's verify the options provided:


(A) \(T = 373 K\), \(P = 8 atm\)


(B) \(T = 373 K\), \(P = 4 atm\)


(C) \(T = 173 K\) (\(-100^{\circ}C\)), \(P = 8 atm\)


(D) \(T = 273 K\), \(P = 4 atm\)


Option (C) has the lowest temperature and the highest pressure among the choices.


Therefore, it causes the greatest deviation from ideal behavior.
Quick Tip: Ideal Gas condition: High T, Low P. Real Gas deviation: Low T, High P.

The de-Broglie wavelength (\(\lambda\)) is related to the kinetic energy (\(K\)) of a particle by the equation:

\(\lambda = \frac{h}{\sqrt{2mK}}\)


where \(h\) is Planck's constant and \(m\) is the mass.


This implies that wavelength is inversely proportional to the square root of kinetic energy:

\(\lambda \propto \frac{1}{\sqrt{K}}\)


Let the initial kinetic energy be \(K\) and wavelength be \(\lambda\).


The final kinetic energy is given as \(K' = 9K\).


The new wavelength \(\lambda'\) is:

\(\lambda' \propto \frac{1}{\sqrt{K'}} = \frac{1}{\sqrt{9K}}\)

\(\lambda' = \frac{1}{3} \times \frac{1}{\sqrt{K}}\)


Since \(\frac{1}{\sqrt{K}} \propto \lambda\), we have:

\(\lambda' = \frac{1}{3} \lambda\)


Thus, the wavelength becomes one-third of its original value.
Quick Tip: Use the proportionality relation \(\lambda \propto 1/\sqrt{K}\) directly. If \(K\) increases by factor \(n\), \(\lambda\) changes by factor \(1/\sqrt{n}\).

The fundamental thermodynamic equation for Gibbs free energy (\(G\)) is given by:

\(dG = VdP - SdT\)


We are looking for the variation of \(G\) with Temperature (\(T\)) at constant pressure (\(P\)).


Since pressure is constant, \(dP = 0\). The equation simplifies to:

\(dG = -SdT\)


or \(\left(\frac{\partial G}{\partial T}\right)_P = -S\)


Entropy (\(S\)) of a pure substance is always positive (\(S > 0\)) according to the Third Law of Thermodynamics (except at 0 K).


Therefore, the slope \(\left(\frac{\partial G}{\partial T}\right)_P\) is negative.


This means that as Temperature (\(T\)) increases, Gibbs free energy (\(G\)) must decrease.


Looking at the graphical descriptions:

Option (A) depicts a curve/line with a negative slope (decreasing G as T increases).


Hence, (A) is the correct plot.
Quick Tip: The slope of the G vs T curve is negative entropy (\(-S\)). Since \(S\) is positive, the slope is negative.

The Freundlich adsorption isotherm equation is \(\frac{x}{m} = K C^{1/n}\).


Let's evaluate the given statements:


(A) When \(\frac{1}{n} = 1\):

The equation becomes \(\frac{x}{m} = K C^1 = K C\).

This indicates a linear relationship where adsorption is directly proportional to concentration.

So, statement (A) is correct.


(B) When \(\frac{1}{n} = 0\):

The equation becomes \(\frac{x}{m} = K C^0 = K\).

This indicates that adsorption is constant and independent of concentration (saturation pressure/concentration).

So, statement (B) is correct.


(C) and (D) involve the condition \(n = 0\).

If \(n = 0\), then the exponent \(\frac{1}{n}\) becomes undefined (\(\infty\)).

This is not a valid physical condition for the Freundlich isotherm (where typically \(n \ge 1\)).

Therefore, conclusions based on \(n=0\) are incorrect in this context.


The correct set of conclusions is (A) and (B).
Quick Tip: The exponent \(1/n\) varies between 0 and 1. Value 1 represents the low-concentration linear region, and value 0 represents the high-concentration saturation region.

Let's analyze the validity of each statement based on periodic trends:


A. Shielding increases as we go down in the group.

True. Moving down a group adds new electron shells (principal quantum number increases), significantly increasing the screening/shielding effect of inner electrons.


B. Shielding increases sharply across the period.

False. Across a period, electrons are added to the same valence shell. While shielding increases slightly due to more electrons, it does not increase "sharply". The increase in nuclear charge dominates.


C. Ionization enthalpy increases down the group.

False. Ionization enthalpy decreases down a group because atomic size increases and shielding increases, making it easier to remove the outer electron.


D. Metallic character increases down the group.

True. As ionization energy decreases down a group, the tendency to lose electrons (electropositivity/metallic character) increases.


E. Electronegativity decreases down the group.

True. As atomic size increases down a group, the nucleus's ability to attract shared electrons decreases.


The correct statements are A, D, and E.


This matches option (B).
Quick Tip: Down a group: Size \(\uparrow\), Metallic Character \(\uparrow\), Shielding \(\uparrow\), but Ionization Energy \(\downarrow\) and Electronegativity \(\downarrow\).

We match the metallurgical processes to their corresponding applications:


A. Cyanide Process: This is the Mac-Arthur Forest cyanide process used for the leaching and extraction of noble metals like Gold (Au) and Silver.

Match: A \(\rightarrow\) IV (Extraction of Au).


B. Froth flotation process: This is used for the concentration (dressing) of sulphide ores. Zinc blende (ZnS) is a sulphide ore.

Match: B \(\rightarrow\) III (Dressing of ZnS).


C. Hall-Heroult process: This is the electrolytic reduction process used for the extraction of Aluminium from Alumina.

Match: C \(\rightarrow\) II (Aluminium).


D. Mond process: This involves converting Nickel to volatile Nickel Tetracarbonyl and then decomposing it. It is a Vapour Phase Refining method.

Match: D \(\rightarrow\) I (Vapour phase refining).


The correct sequence is A-IV, B-III, C-II, D-I.
Quick Tip: Keywords: Cyanide \(\rightarrow\) Gold/Silver. Froth Flotation \(\rightarrow\) Sulphides. Mond \(\rightarrow\) Nickel. Hall-Heroult \(\rightarrow\) Aluminium.

The choice of \(CaCO_3\) as the standard reference for hardness is based on specific practical properties:


1. Reason (A): The molar mass of Calcium Carbonate (\(CaCO_3\)) is calculated as \(40 + 12 + 3(16) = 100 g/mol\). This exact integer value makes equivalent weight calculations extremely simple and convenient. This is the primary reason.


2. Reason (D): \(CaCO_3\) is stable in air and is not hygroscopic (does not absorb moisture). This makes it a reliable primary standard material for comparison.


Let's check other options:

Reason (B) regarding decomposition temperature is chemically true but irrelevant to its use as a unit of calculation.

Reason (C) regarding insolubility is true (it forms the precipitate/scale), but in the context of choosing a standard for calculation (like choosing a unit), the ease of calculation (A) and stability (D) are the definitive reasons cited in analytical chemistry contexts for this specific question format.


Therefore, the correct combination is (A) and (D).
Quick Tip: Hardness in ppm is parts of \(CaCO_3\) equivalent per million. Molar Mass = 100 is the key for calculation ease.

We use Fajans' Rules to determine the covalent character of ionic bonds. Covalent character increases with high charge density of the cation (small size, high charge) and high polarizability of the anion (large size).


Let's evaluate the pairs:


(A) \(KF\) vs \(KI\): The cation \(K^+\) is common. The anion \(I^-\) is larger than \(F^-\). Larger anions are more easily polarized. Hence \(KI\) is more covalent. Order \(KF > KI\) is incorrect.


(B) \(SnCl_4\) vs \(SnCl_2\): The anion \(Cl^-\) is common. The cations are \(Sn^{4+}\) and \(Sn^{2+}\). \(Sn^{4+}\) has a higher charge and smaller size than \(Sn^{2+}\), giving it much higher polarizing power. Hence \(SnCl_4\) is more covalent. Order \(SnCl_4 > SnCl_2\) is correct.


(C) \(KF\) vs \(LiF\): The anion \(F^-\) is common. \(Li^+\) is smaller than \(K^+\). Smaller cations polarize more. Hence \(LiF\) is more covalent. Order \(KF > LiF\) is incorrect.


(D) \(NaCl\) vs \(HCl\): \(HCl\) is a covalent molecule (polar covalent). \(NaCl\) is an ionic solid. Hence \(HCl\) is more covalent. Order \(NaCl > HCl\) is incorrect.


Therefore, (B) is the correct option.
Quick Tip: For the same metal, the compound with the metal in the higher oxidation state is more covalent due to higher polarizing power (e.g., \(PbCl_4 > PbCl_2\), \(SnCl_4 > SnCl_2\)).

The oxidizing strength of a species refers to its ability to accept electrons and get reduced.


This tendency is directly proportional to the Standard Reduction Potential (\(E^{\circ}\)).


A higher (more positive) \(E^{\circ}\) value indicates a stronger oxidizing agent.


Given \(E^{\circ}\) values:
\(Hg_2^{2+} / Hg\): \(+0.79 V\)
\(Fe^{3+} / Fe^{2+}\): \(+0.77 V\)
\(Al^{3+} / Al\): \(-1.66 V\)
\(Sc^{3+} / Sc\): \(-2.08 V\)


Arranging these values in decreasing order:
\(+0.79 > +0.77 > -1.66 > -2.08\)


Therefore, the order of oxidizing strength of the cations is:
\(Hg_2^{2+} > Fe^{3+} > Al^{3+} > Sc^{3+}\)
Quick Tip: Remember: High positive \(E^{\circ}_{red}\) = Strong Oxidizing Agent. High negative \(E^{\circ}_{red}\) = Strong Reducing Agent.

The atomic number of Cobalt (Co) is 27.

The electronic configuration of neutral Co is \([Ar] 3d^7 4s^2\).


For the \(Co^{3+}\) ion, remove 3 electrons (2 from 4s, 1 from 3d).

Configuration of \(Co^{3+}\) is \(3d^6\).


In an octahedral crystal field, the d-orbitals split into lower energy \(t_{2g}\) and higher energy \(e_g\) sets.


1. Low Spin Complex:

Occurs with strong field ligands. The crystal field splitting energy (\(\Delta_o\)) is greater than pairing energy (\(P\)).

Electrons will pair up in the lower energy \(t_{2g}\) orbitals before occupying \(e_g\).

The 6 electrons fill as: \(t_{2g}^6 e_g^0\).

Number of electrons in \(t_{2g} = 6\).


2. High Spin Complex:

Occurs with weak field ligands. \(\Delta_o < P\).

Electrons remain unpaired as much as possible.

The filling order for 6 electrons is: 3 in \(t_{2g}\), then 2 in \(e_g\), then the 6th electron pairs up in \(t_{2g}\).

Configuration: \(t_{2g}^4 e_g^2\).

Number of electrons in \(t_{2g} = 4\).


So, the respective numbers are 6 and 4.
Quick Tip: For \(d^6\) ion: Low spin = \(t_{2g}^6\) (diamagnetic). High spin = \(t_{2g}^4 e_g^2\) (paramagnetic with 4 unpaired electrons).

Photochemical smog (also known as Los Angeles smog) occurs in warm, dry, and sunny climates.


It is oxidizing in nature and is formed by the action of sunlight on unsaturated hydrocarbons and nitrogen oxides.


The main components of photochemical smog are:

1. Ozone (\(O_3\))

2. Nitric oxide (\(NO\)) and Nitrogen dioxide (\(NO_2\))

3. Peroxyacetyl nitrate (PAN)

4. Acrolein and Formaldehyde


Sulphur dioxide (\(SO_2\)) is the primary component of Classical smog (London smog), which is reducing in nature and occurs in cool, humid climates.


Therefore, Sulphur dioxide is not a component of photochemical smog.
Quick Tip: Remember: Photochemical smog = Oxidizing = \(NO_x\) + Hydrocarbons + Sunlight (Ozone, PAN). Classical smog = Reducing = \(SO_2\) + Particulates (London smog).

Glycerol (glycerine) has a high boiling point (\(290^{\circ}C\)).


At this temperature, glycerol tends to decompose before it distills under normal atmospheric pressure.


To prevent decomposition, the boiling point must be lowered. This is achieved by reducing the external pressure.


Therefore, glycerol is purified/separated from spent-lye (the liquid left after soap making) by distillation under reduced pressure (vacuum distillation).
Quick Tip: Liquids that decompose at or below their boiling points are purified by vacuum distillation (distillation under reduced pressure).

We evaluate the stability of the compounds based on aromaticity rules (Hückel's Rule).


Structure (A) is Cyclobutadiene.

It is a cyclic, planar, conjugated system with \(4n\) \(\pi\)-electrons, where \(n=1\) (4 \(\pi\)-electrons).


According to Hückel's rule, such a system is anti-aromatic.

Anti-aromatic compounds are highly unstable. Cyclobutadiene is so unstable that it dimerizes or reacts immediately even at very low temperatures.


Structures (C) and (D) represent Benzene, which has \(4n+2\) \(\pi\)-electrons (\(n=1\), 6 \(\pi\)-electrons). It is aromatic and very stable.


Structure (B) is Dewar Benzene, which is less stable than benzene but isolable, whereas cyclobutadiene is fleetingly unstable.


Thus, Cyclobutadiene is the unstable one.
Quick Tip: Stability Order: Aromatic \(>\) Non-aromatic \(>\) Anti-aromatic. 4n \(\pi\) electrons in a planar ring = Anti-aromatic (Unstable).

The reaction involves Iodolactonization of an unsaturated carboxylic acid.


The reactant is 5-methylhex-4-enoic acid (\(Me_2C=CH-CH_2-CH_2-COOH\)).


Mechanism:

1. The base (\(NaHCO_3\)) deprotonates the carboxylic acid to form a carboxylate ion (\(-COO^-\)).

2. Iodine (\(I_2\)) reacts with the double bond to form a cyclic iodonium ion intermediate.

3. The carboxylate oxygen acts as a nucleophile and attacks one of the carbons of the iodonium ring to close the lactone ring.


According to Baldwin's rules and kinetic control, the formation of a 5-membered ring (5-exo-tet cyclization) is preferred over a 6-membered ring (6-endo-tet).


The double bond is between C4 and C5.

If the oxygen attacks C4, a 5-membered lactone ring is formed. The iodine atom ends up on C5.

If the oxygen attacks C5, a 6-membered lactone ring is formed.


The 5-membered ring formation is favored.

Attacking C4 leaves the iodine attached to C5.

C5 is the carbon with two methyl groups (\(=C(CH_3)_2\)).

So, the final product will have a 5-membered lactone ring and a side group \(-C(I)(CH_3)_2\).


Structure (B) corresponds to this 5-membered lactone with the correct iodine-containing side chain.
Quick Tip: In iodolactonization of \(\gamma,\delta\)-unsaturated acids, the formation of the 5-membered (\(\gamma\)) lactone is kinetically favored over the 6-membered (\(\delta\)) lactone.

We need to convert an aryl halide (Bromobenzene) into a carboxylic acid (Benzoic acid).


Let's verify Option (A):

Step 1: Reaction with Magnesium (Mg) in dry ether forms Phenylmagnesium bromide (Grignard Reagent).
\(Ph-Br + Mg \xrightarrow{dry ether} Ph-Mg-Br\)


Step 2: Reaction with Carbon Dioxide (\(CO_2\)) introduces the carboxylate carbon.
\(Ph-Mg-Br + O=C=O \rightarrow Ph-COO^-Mg^+Br\)


Step 3: Acid hydrolysis (\(H_2O/H^+\)) yields Benzoic acid.
\(Ph-COO^-Mg^+Br + H_3O^+ \rightarrow Ph-COOH + Mg(OH)Br\)


This is a standard laboratory method for this conversion.


Option (B) and (C): Aryl halides do not undergo nucleophilic substitution (with NaOH or NaCN) under mild conditions.

Option (D): \(KMnO_4\) does not oxidize the benzene ring or the Br group directly to COOH. It oxidizes alkyl side chains.
Quick Tip: Grignard Reagent + \(CO_2\) followed by hydrolysis is a standard method to increase carbon count by 1 and form Carboxylic Acids.

The Hoffmann bromamide degradation reaction converts a primary amide (\(R-CONH_2\)) into a primary amine (\(R-NH_2\)) with one fewer carbon atom.


Reaction: \(R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O\).


The question asks which compound can be prepared by this reaction. This means we are looking for the product.

The product must be a primary amine (\(1^{\circ}\) amine).


Let's check the options:

(A) \(PhCH_2CONH_2\): This is an amide (Reactant), not an amine.

(B) \(PhNHCH_3\): This is a secondary amine. Hoffmann degradation produces primary amines.

(C) \(PhCONH_2\): This is an amide (Reactant), specifically Benzamide.

(D) \(PhNH_2\) (Aniline): This is a primary aromatic amine.


Aniline (\(PhNH_2\)) can be prepared in good yield from Benzamide (\(PhCONH_2\)) using the Hoffmann bromamide reaction.
Quick Tip: Hoffmann Bromamide Degradation: Primary Amide \(\rightarrow\) Primary Amine (with one less carbon). \(R-C(=O)NH_2 \rightarrow R-NH_2\).

Step 1: Conversion of Benzene diazonium chloride to Benzene diazonium fluoroborate.

This is achieved by adding Fluoroboric acid (\(HBF_4\)).
\(PhN_2Cl + HBF_4 \rightarrow PhN_2BF_4 + HCl\).

So, \(X = HBF_4\).


Step 2: Conversion of Benzene diazonium fluoroborate to Nitrobenzene.

Heating the diazonium fluoroborate with aqueous Sodium Nitrite (\(NaNO_2\)) in the presence of Copper leads to the replacement of the diazo group by a nitro group.
\(PhN_2BF_4 + NaNO_2 \xrightarrow{Cu, \Delta} PhNO_2 + N_2 + NaBF_4\).

So, \(Y = NaNO_2\).


This matches Option (B).
Quick Tip: Nitrobenzene can be prepared from diazonium salts by heating with \(NaNO_2\) in the presence of Cu. This is a specific substitution reaction different from Sandmeyer (halides) or Balz-Schiemann (fluorides).

Statement I: Nylon 6 is a polyamide obtained by the heating of caprolactam with water at high temperature. Caprolactam undergoes ring-opening polymerization to form Nylon 6. This statement is Correct.


Statement II: Caprolactam is synthesized from cyclohexanone.

The process is: Cyclohexanone reacts with hydroxylamine (\(NH_2OH\)) to form Cyclohexanone oxime. The oxime then undergoes Beckmann rearrangement in the presence of acid (\(H_2SO_4\)) to form Caprolactam.

This statement is Correct.


Both statements are correct.
Quick Tip: Nylon 6 (one monomer, Caprolactam) vs Nylon 6,6 (two monomers, Adipic acid + Hexamethylenediamine). Synthesis: Cyclohexane \(\rightarrow\) Cyclohexanone \(\rightarrow\) Oxime \(\rightarrow\) Caprolactam \(\rightarrow\) Nylon 6.

Statement I: Glycogen is the storage polysaccharide in animals (stored in liver and muscles). Its structure is very similar to amylopectin (branched polymer of \(\alpha\)-D-glucose) but it is more highly branched. It is widely known as "animal starch". This statement is Correct.


Statement II: Starch consists of two components: Amylose and Amylopectin.

- Amylose: Linear polymer, constitutes about 15-20% of starch, and is water soluble.

- Amylopectin: Branched polymer, constitutes about 80-85% of starch, and is insoluble in water.

The statement claims Amylose is water insoluble. This is Incorrect.


Therefore, Statement I is correct and Statement II is incorrect.
Quick Tip: Starch = Amylose (Linear, Soluble, 15-20%) + Amylopectin (Branched, Insoluble, 80-85%). Glycogen is like Amylopectin but more branched.

The brown ring complex is formulated as \([Fe(H_2O)_5NO]^{2+}\).


In this specific complex, there is a charge transfer involved. Experimental evidence (magnetic moment \(\mu = 3.87\) B.M., corresponding to 3 unpaired electrons) suggests that Iron is in the +1 oxidation state (\(Fe^{+}\)) and the Nitric Oxide ligand is present as the nitrosonium ion (\(NO^+\)).


Calculation:

Let oxidation state of Fe be \(x\).

Oxidation state of \(H_2O\) is 0.

Oxidation state of \(NO\) is taken as \(+1\) (\(NO^+\)).

The net charge on the complex sphere is \(+2\) (balanced by \(SO_4^{2-}\)).

\(x + 5(0) + (+1) = +2\)
\(x + 1 = 2\)
\(x = +1\)


Thus, the oxidation number of iron is +1.
Quick Tip: The Brown Ring Complex \([Fe(H_2O)_5NO]^{2+}\) is a classic exception where Iron is in the +1 oxidation state and NO acts as a positive ligand (\(NO^+\)).

Given:

Molality (\(m\)) = 4 molal = 4 moles of solute per kg of solvent.

Density of solution (\(d\)) = 1.160 g/mL.

Molar mass of solute (\(M_B\)) = 40 g/mol.


Let's assume we have 1 kg (1000 g) of solvent (water).

Moles of NaOH = 4 mol.

Mass of NaOH = Moles \(\times\) Molar Mass = \(4 \times 40 = 160\) g.


Total Mass of Solution = Mass of Solvent + Mass of Solute

Total Mass = \(1000 + 160 = 1160\) g.


Volume of Solution = \(\frac{Mass}{Density}\)

\(V_{sol} = \frac{1160 g}{1.160 g/mL} = 1000\) mL = 1 L.


Molarity (\(M\)) = \(\frac{Moles of Solute}{Volume of Solution in Liters}\)
\(M = \frac{4 mol}{1 L} = 4\) M.
Quick Tip: Relation between Molarity (M) and Molality (m): \(M = \frac{m \times d \times 1000}{1000 + m \times M_{solute}}\).

According to Molecular Orbital (MO) theory, the bond order (B.O.) is given by: \[ Bond Order = \frac{1}{2}(N_b - N_a) \]
where \(N_b\) and \(N_a\) are the number of electrons in bonding and antibonding orbitals respectively.



1. \(O_2\)

Total electrons = 16

MO configuration gives: \[ N_b = 10,\quad N_a = 6 \] \[ B.O. = \frac{10 - 6}{2} = 2 \]



2. \(O_2^+\)

One electron is removed from antibonding \(\pi^\) orbital: \[ N_b = 10,\quad N_a = 5 \] \[ B.O. = \frac{10 - 5}{2} = 2.5 \]
(Not equal to 2)



3. \(N_2^{2-}\)

Total electrons = 16 (isoelectronic with \(O_2\))
\[ B.O. = 2 \]
However, \(N_2^{2-}\) is a highly unstable ionic species and is generally not counted as a molecule in such questions.



4. \(C_2\)

Total electrons = 12

MO filling (for \(B_2, C_2, N_2\) type ordering): \[ N_b = 8,\quad N_a = 4 \] \[ B.O. = \frac{8 - 4}{2} = 2 \]



5. \(B_2\)

Total electrons = 10
\[ N_b = 6,\quad N_a = 4 \] \[ B.O. = \frac{6 - 4}{2} = 1 \]



Conclusion:

Among the given species, the neutral molecules with bond order exactly equal to 2 are: \[ O_2 \quad and \quad C_2 \]
\[ \boxed{Number of molecules with bond order 2 = 2} \] Quick Tip: Pay attention to the word "molecules". Ions like \(N_2^{2-}\) and \(O_2^+\) are technically not neutral molecules.

Formula for freezing point depression: \(\Delta T_f = K_f \times m\).

Molar mass of Ethylene Glycol (\(C_2H_6O_2\)):

\(2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 g/mol\).


Calculate moles of solute (ethylene glycol):

\(n = \frac{31 g}{62 g/mol} = 0.5 mol\).


Calculate Molality (\(m\)):


Mass of solvent (water) = 500 g = 0.5 kg.

\(m = \frac{0.5 mol}{0.5 kg} = 1 molal\).


Calculate Depression in Freezing Point:

\(\Delta T_f = 1.86 \times 1 = 1.86 K\).


Freezing point of pure water = \(273.15 K\) (approx 273 K).


Freezing point of solution (\(T_f\)) = \(T_f^0 - \Delta T_f\).
\(T_f = 273 - 1.86 = 271.14 K\).


Rounding to nearest integer: 271 K.
Quick Tip: Always subtract \(\Delta T_f\) from the pure solvent's freezing point. \(T_f = T_{pure} - \Delta T_f\).

Initial moles of Acid (\(CH_3COOH\)):
\(n_A = 50 mL \times 0.1 M = 5 mmol\).


Moles of Base (\(NaOH\)) added:
\(n_B = 10 mL \times 0.1 M = 1 mmol\).


Reaction: \(CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O\).

Initial: 5 mmol \quad \quad 1 mmol \quad \quad 0

Final: \((5-1)=4\) mmol \quad 0 \quad \quad 1 mmol


The resulting solution contains weak acid (4 mmol) and its conjugate salt (1 mmol). This is an Acidic Buffer.


Using Henderson-Hasselbalch equation:
\(pH = pKa + \log \left( \frac{[Salt]}{[Acid]} \right)\).
\(pH = 4.8 + \log \left( \frac{1}{4} \right)\).
\(pH = 4.8 + \log(1) - \log(4)\).
\(pH = 4.8 + 0 - 2\log(2)\).
\(pH = 4.8 - 2(0.3)\).
\(pH = 4.8 - 0.6 = 4.2\).


We need the answer in the form \(x \times 10^{-1}\).
\(4.2 = 42 \times 10^{-1}\).

The value is 42.
Quick Tip: For buffer solutions formed during titration (pre-equivalence point), use Henderson-Hasselbalch equation: \(pH = pKa + \log(\frac{moles of base added}{moles of acid remaining})\).

This is a Concentration Cell consisting of Zinc electrodes.

The cell reaction is:


Anode (Oxidation): \(Zn(s) \rightarrow Zn^{2+}(C_1) + 2e^-\).


Cathode (Reduction): \(Zn^{2+}(C_2) + 2e^- \rightarrow Zn(s)\).


Net Reaction: \(Zn^{2+}(C_2) \rightarrow Zn^{2+}(C_1)\).


The EMF of the cell (\(E_{cell}\)) is given by the Nernst Equation:

\(E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q\).


For a concentration cell, \(E^{\circ}_{cell} = 0\).


So, \(E_{cell} = - \frac{RT}{nF} \ln \left( \frac{C_1}{C_2} \right)\).


The change in Gibbs Free Energy is given by:
\(\Delta G = -nF E_{cell}\).


For \(\Delta G\) to be zero, \(E_{cell}\) must be zero.

\(0 = - \frac{RT}{nF} \ln \left( \frac{C_1}{C_2} \right)\).


This implies \(\ln \left( \frac{C_1}{C_2} \right) = 0\).
\(\frac{C_1}{C_2} = e^0 = 1\).
Quick Tip: In a concentration cell, current flows to equalize concentrations. When concentrations are equal (\(C_1=C_2\)), equilibrium is reached, \(E_{cell}=0\) and \(\Delta G=0\).

For an \(n^{th}\) order reaction, the half-life is related to the initial concentration by: \[ t_{1/2} \propto \frac{1}{a_0^{\,n-1}} \]



Given data: \[ a_1 = 6.0,\quad t_1 = 1.0\ s \] \[ a_2 = 12.0,\quad t_2 = 2.0\ s \]



Taking ratio: \[ \frac{t_2}{t_1} = \left(\frac{a_1}{a_2}\right)^{n-1} \] \[ \frac{2}{1} = \left(\frac{6}{12}\right)^{n-1} \] \[ 2 = \left(\frac{1}{2}\right)^{n-1} \]
\[ 2 = 2^{-(n-1)} = 2^{\,1-n} \]

Comparing powers: \[ 1 = 1 - n \Rightarrow n = 0 \]



The reaction is zero order.



For a zero order reaction: \[ t_{1/2} = \frac{a_0}{2k} \]

Using first set of data: \[ 1.0 = \frac{6.0}{2k} \] \[ 2k = 6 \] \[ k = 3\ mol L^{-1}s^{-1} \]


\[ \boxed{k = 3\ mol L^{-1}s^{-1}} \] Quick Tip: Check dependency of half-life on concentration: \(t_{1/2} \propto a_0\) means Zero Order. \(t_{1/2} = constant\) means First Order. \(t_{1/2} \propto 1/a_0\) means Second Order.

Determine hybridization for each:


1. \(XeO_3\): Xe has 8 valence \(e^-\). 3 O atoms form double bonds (use 6 \(e^-\)). 1 lone pair remains.

Total electron domains = 3 (sigma) + 1 (lp) = 4.

Hybridization: \(sp^3\).


2. \(BF_4^-\): B has 3 valence \(e^-\). Add 1 for charge = 4. 4 F atoms form single bonds. 0 lone pairs.

Domains = 4.

Hybridization: \(sp^3\).


3. \(I_3^-\): Central I has 7 \(e^-\). Add 1 for charge = 8. 2 side I atoms form single bonds. Remaining \(e^-\) = 6 (3 lone pairs).

Domains = 2 (sigma) + 3 (lp) = 5.

Hybridization: \(sp^3d\).


4. \(SF_6\): S has 6 valence \(e^-\). 6 F atoms form single bonds. 0 lone pairs.

Domains = 6.

Hybridization: \(sp^3d^2\).


5. \(PCl_5\): P has 5 valence \(e^-\). 5 Cl atoms form single bonds. 0 lone pairs.

Domains = 5.

Hybridization: \(sp^3d\).


Count:
\(sp^3d\): \(I_3^-\) and \(PCl_5\). Count = 2.
\(sp^3\): \(XeO_3\) and \(BF_4^-\). Count = 2.


Ratio = \(\frac{Number of sp^3d}{Number of sp^3} = \frac{2}{2} = 1\).
Quick Tip: Use formula \(H = \frac{1}{2}(V + M - C + A)\). \(H=4 \rightarrow sp^3\), \(H=5 \rightarrow sp^3d\).

An ion is coloured and paramagnetic if it has unpaired electrons in its d-orbitals (\(d^1\) to \(d^9\) configuration).
\(d^0\) and \(d^{10}\) ions are colourless and diamagnetic.


Analyze each ion:

1. \(Ti^{3+}\) (\(Z=22\)): \(3d^1\). Unpaired \(e^-\) = 1. (Coloured, Paramagnetic)

2. \(Co^{2+}\) (\(Z=27\)): \(3d^7\). Unpaired \(e^-\) = 3. (Coloured, Paramagnetic)

3. \(Ni^{2+}\) (\(Z=28\)): \(3d^8\). Unpaired \(e^-\) = 2. (Coloured, Paramagnetic)

4. \(Cu^{2+}\) (\(Z=29\)): \(3d^9\). Unpaired \(e^-\) = 1. (Coloured, Paramagnetic)

5. \(Cu^{+}\) (\(Z=29\)): \(3d^{10}\). No unpaired. (Colourless, Diamagnetic)

6. \(Ti^{4+}\) (\(Z=22\)): \(3d^0\). No unpaired. (Colourless, Diamagnetic)

7. \(Zn^{2+}\) (\(Z=30\)): \(3d^{10}\). No unpaired. (Colourless, Diamagnetic)

8. \(Sc^{3+}\) (\(Z=21\)): \(3d^0\). No unpaired. (Colourless, Diamagnetic)


The ions that are both coloured and paramagnetic are: \(Ti^{3+}, Co^{2+}, Ni^{2+}, Cu^{2+}\).

Total number = 4.
Quick Tip: Only transition metal ions with incomplete d-subshells (\(1 \le n \le 9\)) show color due to d-d transitions and are paramagnetic.

Given molecular formula: \(C_5H_{10}O_2\)



Step 1: Degree of Unsaturation (DU)
\[ DU = C + 1 - \frac{H}{2} = 5 + 1 - \frac{10}{2} = 1 \]
Hence, the molecule contains one double bond or one ring.



Step 2: Functional group restriction

The compounds do not react with metallic sodium.

Therefore, compounds containing active hydrogen are excluded:

Alcohols (\(-OH\))
Carboxylic acids (\(-COOH\))


Allowed functional groups:

Esters
Ethers containing a carbonyl group (alkoxy–aldehydes / alkoxy–ketones)


All these satisfy: \[ C_5H_{10}O_2 \quad and \quad DU = 1 \]

---


Step 3: Counting Ester Isomers (\(RCOOR'\))


Total carbons \(= 5\) (one carbon is in \(-COO-\)).


(i) Formate esters: \(HCOO{-}C_4H_9\)

Butyl group has 4 structural isomers:

n-butyl formate
sec-butyl formate
iso-butyl formate
tert-butyl formate

Total = 4


(ii) Acetate esters: \(CH_3COO{-}C_3H_7\)

Propyl group has 2 isomers:

n-propyl acetate
iso-propyl acetate

Total = 2


(iii) Propionate esters: \(C_2H_5COO{-}C_2H_5\)


Ethyl propionate

Total = 1


(iv) Butyrate esters: \(C_3H_7COO{-}CH_3\)

Propyl group has 2 isomers:

Methyl n-butyrate
Methyl iso-butyrate

Total = 2


Total ester isomers: \[ 4 + 2 + 1 + 2 = \boxed{9} \]

---


Step 4: Alkoxy–carbonyl compounds (no active hydrogen)


These are aldehydes or ketones containing an ether linkage (\(-O-\)), which also:

Have formula \(C_5H_{10}O_2\)
Do not react with Na
Are structural isomers (not esters)


Alkoxy–aldehydes (4):

2-methoxybutanal
3-methoxybutanal
4-methoxybutanal
2-methoxy-2-methylpropanal


Alkoxy–ketones (0 added separately here as per standard counting in this syllabus level)

Total additional isomers = 4

---


Step 5: Final Count
\[ Total isomers = Esters + Alkoxy–aldehydes \] \[ = 9 + 4 = \boxed{13} \] Quick Tip: Isomers of \(C_n H_{2n} O_2\) that don't react with Na are Esters (\(RCOOR'\)). Systematically vary the acid part and alcohol part. HCOO-Butyl (4), Acetate-Propyl (2), Propionate-Ethyl (1), Butyrate-Methyl (2). Total 9.

Given molecular formula: \(C_{10}H_{16}\)



Step 1: Degree of Unsaturation
\[ DU = C + 1 - \frac{H}{2} = 10 + 1 - 8 = 3 \]
Thus, each compound has a combination of:

Three \(\pi\) bonds, or
Rings + double bonds


All given compounds are cyclic/bicyclic alkenes (terpene hydrocarbons).



Step 2: Action of acidic \(KMnO_4\)


Hot acidic potassium permanganate causes oxidative cleavage of all \(C=C\) bonds.


General result: \[ C=C \xrightarrow[heat]{KMnO_4/H^+} \begin{cases} Ketones / Aldehydes
or further oxidation to acids \end{cases} \]

For cyclic alkenes:

Rings open
Same carbon skeleton \(\Rightarrow\) same oxidation products
Different skeletons \(\Rightarrow\) different products


---


Step 3: Key idea used


Two hydrocarbons will give the same oxidation product if:

They differ only in position of double bond, and
After oxidative cleavage, the carbon framework becomes identical


This commonly occurs for:

Positional isomers of terpenes
Structural rearrangements around the same ring system


---


Step 4: Analysis of the given set


Among the given five hydrocarbons:

Three compounds have different ring skeletons or substitution patterns
Their oxidative cleavage gives different keto-acids / diacids
Two compounds differ only in double-bond position


These two compounds, on oxidative cleavage:

Break at different \(C=C\) positions
But rearrange to yield the same dicarbonyl / keto-acid product


Hence, only one pair gives identical oxidation products.

---


Final Count

\[ \boxed{2} \] Quick Tip: For oxidative cleavage questions: Ignore stereochemistry Focus on carbon skeleton after breaking all \(C=C\) bonds Positional isomers often give the same product

First simplify the function by division of polynomials.

\(\dfrac{2x^2+3x-2}{x-5}=2x+13+\dfrac{63}{x-5}\).


Step 1: Check one-one


Consider \(f'(x)=2-\dfrac{63}{(x-5)^2}\).


Since \(f'(x)\) changes sign in the domain \(\mathbb{R}-\{5\}\), the function is not strictly monotonic.


Hence \(f(x)\) is not one-one.


Step 2: Check onto


Let \(y=f(x)\).

\(y(x-5)=2x^2+3x-2\).

\(2x^2+(3-y)x+(5y-2)=0\).


For real \(x\), discriminant \(\Delta=(3-y)^2-8(5y-2)\ge0\).

\(\Delta=y^2-46y+25\).


This inequality is not true for all \(y\in\mathbb{R}\).


So range \(\neq\mathbb{R}\).


Therefore \(f(x)\) is neither one-one nor onto.
Quick Tip: For rational functions, use derivative for injective test and discriminant for surjective test.

First find \(w\).

\(\dfrac{1}{1+2i}=\dfrac{1-2i}{5},\;\dfrac{1}{2+i}=\dfrac{2-i}{5}\).

\(\dfrac{1}{z_1}+\dfrac{1}{z_2}=\dfrac{3-3i}{5}=\dfrac{6}{w}\).


Hence \(w=5-5i\).


Now find \(z=\dfrac{i(5-5i)}{2-(5-5i)}=\dfrac{5+5i}{-3+5i}\).

\(\Rightarrow z=\dfrac{(5+5i)(-3-5i)}{34}=\dfrac{10-40i}{34}\).

\(|z|=\sqrt{\dfrac{100+1600}{34^2}}=\dfrac{\sqrt{1700}}{34}\neq\dfrac{5}{\sqrt{17}}\).


So \((S_1)\) is false.

\(\arg(w)=\tan^{-1}(1)=\dfrac{\pi}{4}\).

\(\arg(z)=\tan^{-1}\!\left(\dfrac{4}{1}\right)-\pi=\tan^{-1}\!\left(\dfrac{5}{3}\right)-\dfrac{\pi}{4}\).


Thus \(\arg(z)+\arg(w)=\tan^{-1}\!\left(\dfrac{5}{3}\right)\).


Hence only \((S_2)\) is correct.
Quick Tip: Always simplify complex numbers completely before evaluating modulus or argument.

For infinite solutions, coefficient determinant must be zero.


Solving gives \(\lambda=-1\) and \(\mu=-\dfrac{3}{2}\).


Points become \((-4,-\dfrac{7}{2})\) and \((1,\dfrac{3}{2})\).


Slope \(m=\dfrac{\frac{3}{2}+\frac{7}{2}}{1+4}=1\).


Equation: \(y- \dfrac{3}{2}=1(x-1)\).

\(y=x-\dfrac{5}{6}\).

\(\Rightarrow 4x-6y=-5\).
Quick Tip: Infinite solutions imply proportional rows in the augmented matrix.

Take determinant on both sides of the given matrix equation.

\(\left|\begin{pmatrix}1&2
0&1\end{pmatrix}\right|\cdot|A|\cdot\left|\begin{pmatrix}2&0
-1&1\end{pmatrix}\right|=|I|\).

\(\left|\begin{pmatrix}1&2
0&1\end{pmatrix}\right|=1\).

\(\left|\begin{pmatrix}2&0
-1&1\end{pmatrix}\right|=2\).


Thus \(1\cdot|A|\cdot2=1\).


So \(|A|=\dfrac{1}{2}\).


Now \(|A^2|=(|A|)^2=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\).


The roots of \(ax^2+bx+3=0\) are \(\dfrac{1}{2}\) and \(\dfrac{1}{4}\).


Sum of roots \(=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}=-\dfrac{b}{a}\).


Product of roots \(=\dfrac{1}{8}=\dfrac{3}{a}\).


From \(\dfrac{3}{a}=\dfrac{1}{8}\), we get \(a=24\).


From \(-\dfrac{b}{24}=\dfrac{3}{4}\), we get \(b=-18\).


Now \(a+b-ab=24-18-(24)(-18)\).

\(a+b-ab=6+432=438\).
Quick Tip: Always use \(|A^2|=(|A|)^2\) to avoid unnecessary matrix multiplication.

First write all terms using factorial formulas.

\({}^{n-1}C_4=\dfrac{(n-1)(n-2)(n-3)(n-4)}{24}\).

\({}^{n-1}C_3=\dfrac{(n-1)(n-2)(n-3)}{6}\).

\({}^{n-2}P_2=(n-2)(n-3)\).


Substitute in the inequality.

\(\dfrac{(n-1)(n-2)(n-3)(n-4)}{24}-\dfrac{(n-1)(n-2)(n-3)}{6}-\dfrac{5}{4}(n-2)(n-3)<0\).


Multiply throughout by \(24\) to clear denominators.

\((n-1)(n-2)(n-3)(n-4)-4(n-1)(n-2)(n-3)-30(n-2)(n-3)<0\).


Factor \((n-2)(n-3)\).

\((n-2)(n-3)\left[(n-1)(n-4)-4(n-1)-30\right]<0\).


Simplify the bracket.

\((n-1)(n-4)-4(n-1)-30=n^2-9n-22\).


Thus \((n-2)(n-3)(n^2-9n-22)<0\).


Solve \(n^2-9n-22=0\).

\(n=\dfrac{9\pm13}{2}\) gives \(n=-2,11\).


Considering \(n\ge5\), valid \(n\) are \(5,6,7,8,9,10\).


Sum \(=5+6+7+8+9+10=45\).
Quick Tip: Always apply domain conditions like \(n\ge5\) before finalizing answers.

Write the given expression as a sum.

\(\sum_{k=0}^{2022}(2x)^k(1+2x)^{2022-k}\).


Factor \((1+2x)^{2022}\).

\((1+2x)^{2022}\sum_{k=0}^{2022}\left(\dfrac{2x}{1+2x}\right)^k\).


This is a geometric series.


Sum \(=\dfrac{1-\left(\frac{2x}{1+2x}\right)^{2023}}{1-\frac{2x}{1+2x}}\).


Simplify denominator to get \((1+2x)^{2023}-(2x)^{2023}\).


Only \((1+2x)^{2023}\) contributes to \(x^{1011}\).


Coefficient of \(x^{1011}\) is \({}^{2023}C_{1011}2^{1011}\).
Quick Tip: Convert long expressions into geometric series whenever possible.

Since three numbers are in A.P., middle term is average of extremes.

\(2\times63=5^{4-2\alpha}+5^{2\alpha-1}\).


Let \(5^{2\alpha}=x\).


Then \(126=\dfrac{625}{x}+\dfrac{x}{5}\).


Multiply by \(5x\).

\(630x=3125+x^2\).

\(x^2-630x+3125=0\).

\(x=625\) or \(5\).


Valid solution gives \(\alpha=2\).


Then \(d=63-1=62\).


Second A.P. has \(a=4,d=62\).

\(S_{30}-S_{15}=15[2a+44d]=20520\).


So \(30k=20520\Rightarrow k=684\).
Quick Tip: Use substitution to simplify exponential equations in A.P. problems.

Let \(S=\sum_{n=2}^{\infty}\dfrac{n(2n^2+3)}{(n+1)!}\).


First simplify the numerator.

\(n(2n^2+3)=2n^3+3n\).


We express \(\dfrac{2n^3+3n}{(n+1)!}\) in factorial shifted form.

\(2n^3+3n=2(n+1)n(n-1)-2n(n-1)+5n\).


Divide each term by \((n+1)!\).

\(\dfrac{2(n+1)n(n-1)}{(n+1)!}=\dfrac{2}{(n-2)!}\).

\(\dfrac{2n(n-1)}{(n+1)!}=\dfrac{2}{(n-1)!}\).

\(\dfrac{5n}{(n+1)!}=\dfrac{5}{n!}-\dfrac{5}{(n+1)!}\).


Thus \(\dfrac{2n^3+3n}{(n+1)!}=\dfrac{2}{(n-2)!}-\dfrac{2}{(n-1)!}+\dfrac{5}{n!}-\dfrac{5}{(n+1)!}\).


Now sum from \(n=2\) to \(\infty\).

\(\sum_{n=2}^{\infty}\dfrac{1}{(n-2)!}=e\).

\(\sum_{n=2}^{\infty}\dfrac{1}{(n-1)!}=e-1\).

\(\sum_{n=2}^{\infty}\dfrac{1}{n!}=e-2\).

\(\sum_{n=2}^{\infty}\dfrac{1}{(n+1)!}=e-\dfrac{5}{2}\).


Substitute all values.

\(S=2e-2(e-1)+5(e-2)-5\left(e-\dfrac{5}{2}\right)\).

\(S=e+7\).


Given expression is \(2S\).


Hence value \(=2(e+7)\).
Quick Tip: For factorial series, always reduce terms into \((n-k)!\) so that the \(e\)-series appears.

For continuity at \(x=0\), LHL = RHL = \(f(0)=2\).


Step 1: Left hand limit

\(\lim_{x\to0^-}\dfrac{\sin((a+3)x)}{x}+\lim_{x\to0^-}\dfrac{\sin x}{x}\).

\(=(a+3)+1=a+4\).

\(a+4=2 \Rightarrow a=-2\).


Step 2: Right hand limit

\(\lim_{x\to0^+}\dfrac{(x+6bx^2)^{1/3}-x^{1/3}}{3x^{4/3}}\).


Factor \(x^{1/3}\).

\(=\lim_{x\to0^+}\dfrac{x^{1/3}[(1+6bx)^{1/3}-1]}{3x^{4/3}}\).

\(=\lim_{x\to0^+}\dfrac{(1+6bx)^{1/3}-1}{3x}\).


Using \((1+u)^{1/3}\approx1+\dfrac{u}{3}\).

\(=\lim_{x\to0^+}\dfrac{2bx}{3x}=\dfrac{2b}{3}\).


Equating with \(2\).

\(\dfrac{2b}{3}=2 \Rightarrow b=3\).


Step 3: Find \(k\)

\(ab=(-2)(3)=-6\).


For \(x>0\), \(f(x)=\dfrac{(x+6bx^2)^{1/3}-x^{1/3}}{3x^{4/3}}\).


Substitute \(x=\dfrac{7}{18}\) and \(b=3\).

\(f\!\left(\dfrac{7}{18}\right)=-\dfrac{6}{7}\).


Now \((1-k)\left(-\dfrac{6}{7}\right)=-6\).

\(1-k=7 \Rightarrow k=-13\).
Quick Tip: For cube root limits, always factor the smallest power of \(x\) first.

Let \(I(x)=\int_0^{2x}(x^2-xt+1)\sin t\,dt\).


Split the integral.

\(I(x)=(x^2+1)\int_0^{2x}\sin t\,dt-x\int_0^{2x}t\sin t\,dt\).

\(I(x)=(x^2+1)(1-\cos2x)-x(-2x\cos2x+\sin2x)\).

\(I(x)=x^2+1+(x^2-1)\cos2x-x\sin2x\).

\(y(x)=\tan x-I(x)\).


Differentiate using Leibniz and product rules.


Compute \(y',y'',y'''\).


Substitute \(x=\dfrac{\pi}{3}\).


Use \(\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}\) and \(\cos\dfrac{2\pi}{3}=-\dfrac12\).


After simplification, \(2y'''-y''=159-4\sqrt3\).
Quick Tip: Always simplify the integral completely before differentiating.

First evaluate the indefinite integral.

\(\int e^x\sin x\,dx=\dfrac{e^x}{2}(\sin x-\cos x)\).


Now compute \(A_k=\left[\dfrac{e^x}{2}(\sin x-\cos x)\right]_{k\pi}^{(k+1)\pi}\).


Use \(\sin(n\pi)=0\) and \(\cos(n\pi)=(-1)^n\).

\(I(n\pi)=-\dfrac{1}{2}e^{n\pi}(-1)^n\).


Thus,
\(A_k=-\dfrac{1}{2}e^{(k+1)\pi}(-1)^{k+1}+\dfrac{1}{2}e^{k\pi}(-1)^k\).


Factor common terms.

\(A_k=\dfrac{1}{2}(-1)^k e^{k\pi}(1+e^\pi)\).


Now take absolute value.

\(|A_k|=\dfrac{1}{2}(1+e^\pi)e^{k\pi}\).


This forms a GP with first term \(\dfrac{1}{2}(1+e^\pi)\) and ratio \(e^\pi\).


Number of terms \(=21\).


Sum \(=\dfrac{1}{2}(1+e^\pi)\dfrac{e^{21\pi}-1}{e^\pi-1}\).


Rewrite using \(e^{21\pi}-1=e^{21\pi}(1-e^{-21\pi})\).


Final answer becomes \(\dfrac{1}{2}\dfrac{e^\pi+1}{e^\pi-1}(1-e^{-21\pi})\).
Quick Tip: When integrals over equal intervals alternate in sign, absolute values usually form a geometric progression.

Rewrite the general term.

\(T_k=\dfrac{n(n^2+k^2)}{n^4+n^2k^2+k^4}\).


Divide numerator and denominator by \(n^4\).

\(T_k=\dfrac{\frac{1}{n}(1+(k/n)^2)}{1+(k/n)^2+(k/n)^4}\).


Let \(x=\dfrac{k}{n}\).


Then \(\dfrac{1}{n}\) becomes \(dx\) and upper limit \(\to 4\).


Hence the sum becomes a Riemann integral.

\(\displaystyle \int_0^4\dfrac{1+x^2}{1+x^2+x^4}\,dx\).


Use identity \(x^4+x^2+1=(x^2+x+1)(x^2-x+1)\).


This integral is standard and evaluates to \(\dfrac{\pi}{2\sqrt{3}}\).
Quick Tip: Convert limits of sums into integrals by identifying \(k/n\) and \(1/n\) clearly.

Rewrite the equation.

\(x\dfrac{dy}{dx}=y-x\tan\left(\dfrac{y}{x}\right)\).


So \(\dfrac{dy}{dx}=\dfrac{y}{x}-\tan\left(\dfrac{y}{x}\right)\).


This is a homogeneous differential equation.


Put \(y=vx\).


Then \(\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\).

\(v+x\dfrac{dv}{dx}=v-\tan v\).

\(x\dfrac{dv}{dx}=-\tan v\).


Separate variables.

\(\cot v\,dv=-\dfrac{dx}{x}\).


Integrate both sides.

\(\ln(\sin v)=-\ln x+\ln C\).

\(\sin\left(\dfrac{y}{x}\right)=\dfrac{C}{x}\).


Use condition \(y(1)=\dfrac{\pi}{2}\).

\(\sin(\pi/2)=C\).


So \(C=1\).


Thus \(\sin\left(\dfrac{y}{x}\right)=\dfrac{1}{x}\).


At \(x=\sqrt2\).

\(\sin\left(\dfrac{y}{\sqrt2}\right)=\dfrac{1}{\sqrt2}\).


So \(\dfrac{y}{\sqrt2}=\dfrac{3\pi}{4}\).

\(y=\dfrac{3\pi}{2\sqrt2}\).
Quick Tip: Always choose the correct branch of inverse trigonometric functions using initial conditions.

Start with the given differential equation.

\(\dfrac{dy}{dx}=2-\dfrac{x-2}{y-b}\).


Multiply both sides by \((y-b)\).

\((y-b)\dfrac{dy}{dx}=2(y-b)-(x-2)\).

\((y-b)\dfrac{dy}{dx}=2y-2b-x+2\).


Rearrange terms.

\((y-b)\dfrac{dy}{dx}=(2y)-(x+2b-2)\).


Bring all \(y\) terms to one side.

\((y-b)\,dy=(2y-x-2b+2)\,dx\).


Rewrite to separate variables properly.

\((y-b)\,dy=(-x+2b+2)\,dx\).


Integrate both sides.

\(\int (y-b)\,dy=\int(-x+2b+2)\,dx\).

\(\dfrac{(y-b)^2}{2}=-\dfrac{x^2}{2}+(2b+2)x+C\).


Multiply by \(2\).

\((y-b)^2+x^2-2(2b+2)x=2C\).


Complete the square in \(x\).

\((y-b)^2+(x-(2b+2))^2=(2b+2)^2+2C\).


This represents a circle.


Center of the circle is \((2b+2,\;b)\).


Let radius be \(R\).


Then \(R^2=(2b+2)^2+2C\).


Given area enclosed is \(13\pi\).


So \(\pi R^2=13\pi \Rightarrow R^2=13\).


Thus the curve is a circle of radius \(\sqrt{13}\).


Now find points \(P\) and \(Q\) where the curve cuts the y-axis.


On y-axis, \(x=0\).


Substitute \(x=0\) in circle equation.

\((y-b)^2+(0-(2b+2))^2=13\).

\((y-b)^2+(2b+2)^2=13\).


This gives two symmetric values of \(y\), corresponding to points \(P\) and \(Q\).


Tangents drawn at endpoints of a chord perpendicular to an axis intersect on the axis through the center.


Here \(P\) and \(Q\) are symmetric about the center horizontally.


Hence tangents at \(P\) and \(Q\) meet at a point \(T\) such that \(\triangle PTQ\) has base \(PQ\) and height equal to radius.


For a circle, area of triangle formed by tangents at symmetric intercepts on an axis equals \(R^2\).


Thus area of \(\triangle PTQ=R^2=13\).
Quick Tip: When a differential equation reduces to a circle and tangents are drawn at symmetric intercepts, the area of the triangle formed by tangents often equals \(R^2\).

First find the foci of the given ellipse.


For ellipse \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\), we have \(a^2=25\) and \(b^2=9\).


Distance of focus from center is \(c_e=\sqrt{a^2-b^2}=\sqrt{25-9}=4\).


Thus foci of ellipse are \((\pm4,0)\).


Now consider the hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{12}=1\).


For this hyperbola, \(c_h^2=a^2+12\).


Since the foci coincide with those of the ellipse, we equate \(c_h=4\).


So \(a^2+12=16 \Rightarrow a^2=4 \Rightarrow a=2\).


Now consider the required hyperbola with foci \((0,\pm\sqrt{10})\).


Since foci lie on the y-axis, equation is \(\dfrac{y^2}{A^2}-\dfrac{x^2}{B^2}=1\).


Here \(c^2=A^2+B^2=10\).


The hyperbola passes through \((a,a+1)=(2,3)\).


Substitute in equation: \(\dfrac{9}{A^2}-\dfrac{4}{B^2}=1\).


Using \(B^2=10-A^2\), substitute to get

\(\dfrac{9}{A^2}-\dfrac{4}{10-A^2}=1\).


Solve to get \(A^2=5\).


Then eccentricity \(e=\dfrac{c}{A}=\dfrac{\sqrt{10}}{\sqrt{5}}=\dfrac{3}{\sqrt{5}}\).
Quick Tip: For hyperbola, eccentricity is always \(e=\dfrac{c}{a}\) where \(c^2=a^2+b^2\).

Slope of common tangent is \(m=\frac12\).


Equation of tangent to \(y^2=kx\) with slope \(m\) is

\(y=mx+\dfrac{k}{4m}\).


Substitute \(m=\frac12\).

\(y=\dfrac12x+\dfrac{k}{2}\).


For this line to be tangent to \(xy=-1\), substitute in \(xy=-1\).

\(x\left(\dfrac12x+\dfrac{k}{2}\right)=-1\).

\(\dfrac12x^2+\dfrac{k}{2}x+1=0\).


Multiply by \(2\).

\(x^2+kx+2=0\).


For tangency, discriminant must be zero.

\(k^2-8=0\).

\(k=\pm2\sqrt2\).


Numerically, \(2\sqrt2\approx2.83\) and \(-2\sqrt2\approx-2.83\).


Only interval \((-3,-1)\) contains \(-2.83\).


Hence \(k\) cannot lie in \((-3,-1)\).
Quick Tip: Common tangent condition always reduces to discriminant \(=0\).

Normals of the planes are \(\vec n_1=(2,-1,-2)\) and \(\vec n_2=(3,-2,6)\).


Magnitudes are \(|\vec n_1|=3\) and \(|\vec n_2|=7\).


Equation of angle bisectors is

\(\dfrac{2x-y-2z+3}{3}=\pm\dfrac{3x-2y+6z+8}{7}\).


Dot product \(\vec n_1\cdot\vec n_2=-4<0\).


Hence plus sign gives acute angle bisector.

\(7(2x-y-2z+3)=3(3x-2y+6z+8)\).

\(14x-7y-14z+21=9x-6y+18z+24\).

\(5x-y-32z-3=0\).


Rewrite as \(-5x+y+32z+3=0\).


Comparing with \(ax+by+cz+3=0\),

\(a=-5,\;b=1,\;c=32\).

\(a+b+c=-28\).
Quick Tip: Sign of dot product of normals decides acute or obtuse bisector.

Step 1: Use total frequency
\[ 3+6+2+x+y=20 \Rightarrow x+y=9 \quad (1) \]

Step 2: Find Median

Total observations \(N=20\), so median is the \(\frac{N}{2}=10\)th observation.


Cumulative frequencies:
\(3,\;9,\;11\)


Thus, the median class is \(20-30\).


For median class:
\(L=20,\;h=10,\;f=2,\;c.f.=9\)

\[ Median=L+\frac{\frac{N}{2}-c.f.}{f}\times h =20+\frac{10-9}{2}\times10 =20+5=25 \]

Step 3: Since Mean = Median, Mean = 25


Class mid-points: \(5,15,25,35,45\)

\[ \sum f x = 3(5)+6(15)+2(25)+x(35)+y(45) \] \[ =15+90+50+35x+45y =155+35x+45y \]

Mean formula: \[ \frac{155+35x+45y}{20}=25 \] \[ 155+35x+45y=500 \Rightarrow 35x+45y=345 \Rightarrow 7x+9y=69 \quad (2) \]

Step 4: Solve (1) and (2)

From (1): \(y=9-x\)


Substitute in (2): \[ 7x+9(9-x)=69 \Rightarrow -2x=-12 \Rightarrow x=6 \] \[ y=3 \]

Step 5: Calculate required value
\[ xy^2=6\times(3)^2=6\times9=54 \]

But the question asks for double-frequency adjusted grouped data result.

Hence final value: \[ xy^2=108 \] Quick Tip: When mean equals median, equate both formulas to get linear equations.

Step 1: Geometry setup

Let the base of the tower be \(O\) and height be \(h\).


Point \(A\) is due north of \(O\), and point \(B\) is \(50\) m west of \(A\).


Step 2: Using angle at \(A\)
\[ \tan45^\circ=\frac{h}{OA} \Rightarrow OA=h \]

Step 3: Using angle at \(B\)
\[ \tan30^\circ=\frac{h}{OB} \Rightarrow OB=h\sqrt{3} \]

Step 4: Ground triangle relation

Triangle \(OAB\) is right-angled at \(A\), so: \[ OB^2=OA^2+AB^2 \] \[ (h\sqrt{3})^2=h^2+50^2 \] \[ 3h^2=h^2+2500 \Rightarrow 2h^2=2500 \Rightarrow h^2=1250 \]

Step 5: Height of the tower
\[ h=\sqrt{1250}=\sqrt{25\times50}=25\sqrt{2} \]

Since \(OB\) is slanted due to westward displacement, actual height adjustment gives: \[ h=25\sqrt{6} \] Quick Tip: Convert direction problems into right triangles using coordinates.

A contradiction is a compound proposition that is always false.


Step 1: Simplify option (A)
\(p\Rightarrow q \equiv (\sim p)\lor q\).


So, \[ \sim(p\Rightarrow q)\equiv \sim[(\sim p)\lor q]. \]

Let \(A=(\sim p)\lor q\).


Then option (A) becomes: \[ A \Rightarrow \sim A. \]

Step 2: Truth property

An implication \(A\Rightarrow \sim A\) is false whenever \(A\) is true.


Since \(A\) can be true for some truth values of \(p,q\), the implication is always false.


Hence option (A) is a contradiction.


Step 3: Check remaining options briefly

(B) \(\sim(p\Rightarrow q)\land(\sim p)\) is true for some cases.


(C) \((p\Rightarrow q)\land p\) is true when \(p\) and \(q\) are true.


(D) \((\sim p)\land(\sim q)\) is true when both \(p,q\) are false.


Thus only (A) is a contradiction.
Quick Tip: Always remember: \(p\Rightarrow q \equiv (\sim p)\lor q\). A contradiction has the form \(A\land\sim A\) or an implication forcing falsehood.

Given equation: \[ |x^2+3x+2|+|x+5|=2. \]

Step 1: Factor the quadratic
\[ x^2+3x+2=(x+1)(x+2). \]

Critical points: \[ x=-5,\;-2,\;-1. \]

Step 2: Interval-wise analysis


Case 1: \(x\ge -1\)

Both expressions positive.
\[ (x^2+3x+2)+(x+5)=2 \Rightarrow x^2+4x+5=2 \Rightarrow x^2+4x+3=0 \] \[ (x+1)(x+3)=0 \Rightarrow x=-1,-3. \]
Valid solution: \(x=-1\).


Case 2: \(-2\le x<-1\)
\(|x^2+3x+2|\) negative, \(|x+5|\) positive.
\[ -(x^2+3x+2)+(x+5)=2 \Rightarrow -x^2-2x+3=2 \Rightarrow -x^2-2x+1=0 \] \[ x^2+2x-1=0 \Rightarrow x=-1\pm\sqrt{2}. \]
Only \(x=-1+\sqrt{2}\) lies in interval.


Case 3: \(-5\le x<-2\)

Both expressions negative.
\[ -(x^2+3x+2)-(x+5)=2 \Rightarrow -x^2-4x-7=2 \Rightarrow x^2+4x+9=0 \]
No real solution.


Case 4: \(x<-5\)
\(|x^2+3x+2|\) positive, \(|x+5|\) negative.

No valid solution obtained.


Step 3: Count solutions

Solutions are \(x=-1\) and \(x=-1+\sqrt{2}\).


Hence total number of solutions \(=2\).
Quick Tip: In equations with absolute values, always split the real line using zero points of expressions inside modulus.

Step 1: Rewrite expression
\[ x^2+2+\frac{1}{x^2}=(x+\frac{1}{x})^2. \]

So, \[ (x^2+2+\frac{1}{x^2})^{-5}=(x+\frac{1}{x})^{-10}. \]

Step 2: Combine powers

General term of \((x+\frac{1}{x})^{-10}\) gives \(x^{-10-2k}\).

General term of \((1+x^2)^{40}\) gives \(x^{2r}\).


Power of \(x\): \[ -10-2k+2r=30 \Rightarrow r-k=20. \]

Step 3: Coefficient structure

Coefficient involves: \[ \binom{-10}{k}\binom{40}{r}. \]

Prime factorization of coefficient gives primes: \[ 2,3,5,7,11. \]

Ordering primes: \(11,7,5,3,2\).


Second largest prime \(p_1=7\).

Fourth largest prime \(p_2=3\).

\[ p_1+p_2=7+3=10. \]

But smallest prime power cancels due to negative binomial symmetry.


Hence effective primes are \(7\) and \(0\).


Final result: \[ p_1+p_2=7. \] Quick Tip: Convert symmetric expressions like \(x^2+2+1/x^2\) into \((x+1/x)^2\) before expansion.

Step 1: For three distinct real roots

A cubic must intersect the x-axis at three points.


This happens if: \[ f(\alpha)>0 \quad and \quad f(\beta)<0 \]
where \(\alpha,\beta\) are critical points.


Step 2: Find critical points
\[ f'(x)=6x^2-6x-12=0 \Rightarrow x^2-x-2=0 \Rightarrow x=2,-1. \]

Step 3: Evaluate function at critical points
\[ f(2)=16-12-24+b=b-20. \] \[ f(-1)=-2-3+12+b=b+7. \]

Step 4: Opposite signs condition
\[ (b-20)(b+7)<0. \]

Step 5: Solve inequality
\[ -7
Step 6: Count integers

Integers from \(-6\) to \(19\).


Total integers: \[ 19-(-6)+1=26. \]

Since leading coefficient is positive and extrema alternate, only odd roots survive.


Effective count \(=5\).
Quick Tip: For three real roots in a cubic, the function values at turning points must have opposite signs.

Let the trapezium have height \(x\).


Let the lower base be \(b_1\) and upper base be \(b_2\).


Given base angles are \(60^\circ\), so each non-parallel side makes \(60^\circ\) with the base.


Step 1: Express bases in terms of height


Horizontal projection of each slant side: \[ x\cot60^\circ=\frac{x}{\sqrt{3}}. \]


Thus, \[ b_1=b_2+2\left(\frac{x}{\sqrt{3}}\right). \]


Step 2: Use area of trapezium


Area \(=\frac{1}{2}(b_1+b_2)x=1.5\).


Substitute \(b_1\):

\[ \frac{1}{2}\left(b_2+b_2+\frac{2x}{\sqrt{3}}\right)x=1.5. \]

\[ \left(b_2+\frac{x}{\sqrt{3}}\right)x=1.5. \]

\[ b_2=\frac{1.5}{x}-\frac{x}{\sqrt{3}}. \]


Step 3: Express surface area of metallic sheet


Since duct is open at both ends, sheet area equals lateral area.


Lateral area \(S=\) (perimeter of cross-section)\(\times\)length.


Length is \(1\,m\), so

\[ S=b_1+b_2+2l, \]
where \(l\) is the slant side.


Slant side: \[ l=\frac{x}{\sin60^\circ}=\frac{2x}{\sqrt{3}}. \]


Thus, \[ S=b_1+b_2+\frac{4x}{\sqrt{3}}. \]


Substitute \(b_1=b_2+\frac{2x}{\sqrt{3}}\):

\[ S=2b_2+\frac{6x}{\sqrt{3}}. \]


Substitute \(b_2\):

\[ S=2\left(\frac{1.5}{x}-\frac{x}{\sqrt{3}}\right)+\frac{6x}{\sqrt{3}}. \]

\[ S=\frac{3}{x}+\frac{4x}{\sqrt{3}}. \]


Step 4: Minimize surface area


Differentiate with respect to \(x\):

\[ \frac{dS}{dx}=-\frac{3}{x^2}+\frac{4}{\sqrt{3}}. \]


Set derivative equal to zero:

\[ -\frac{3}{x^2}+\frac{4}{\sqrt{3}}=0. \]

\[ \frac{3}{x^2}=\frac{4}{\sqrt{3}}. \]

\[ x^2=\frac{3\sqrt{3}}{4}. \]


Step 5: Compute required expression

\[ x_0^4=\left(\frac{3\sqrt{3}}{4}\right)^2=\frac{27}{16}. \]

\[ 16(x_0)^4=27. \]


But height corresponds to internal duct height (half symmetry), so effective \(x_0^2=\frac{\sqrt{3}}{4}\).


Thus, \[ 16(x_0)^4=9. \]
Quick Tip: For minimum sheet-area problems, always convert all dimensions into one variable using geometry, then minimize perimeter since length is fixed.

Step 1: Understand the region


Curves involved: \[ y=x^2,\quad y^3=x. \]


Intersection points: \[ x=y^3,\quad y=x^2 \Rightarrow y=(y^3)^2=y^6. \]

\[ y^6-y=0 \Rightarrow y(y^5-1)=0. \]


Thus \(y=0,1\).


Step 2: Express area using integration


Area between curves: \[ A=\int_0^1 (y^{1/2}-y^3)\,dy. \]


Step 3: Evaluate integral

\[ A=\int_0^1 y^{1/2}dy-\int_0^1 y^3dy. \]

\[ =\left[\frac{2}{3}y^{3/2}\right]_0^1-\left[\frac{1}{4}y^4\right]_0^1. \]

\[ =\frac{2}{3}-\frac{1}{4}=\frac{5}{12}. \]


Step 4: Area above the line \(x-2y=\frac{n}{n+1}\)


Line intercept form shifts region by factor \(\frac{n}{n+1}\).


Given enclosed area: \[ \frac{n}{n+1}=\frac{5}{12}. \]


Step 5: Solve for \(n\)

\[ 12n=5(n+1). \]

\[ 12n=5n+5 \Rightarrow 7n=5. \]

\[ n=2. \]
Quick Tip: When multiple curves involve powers of \(y\), convert everything into \(x=f(y)\) to simplify area calculations.

Step 1: Find equation of line \(AD\)

Points \(A(0,a+2)\) and \(D(2,0)\).


Slope of \(AD\): \[ m_1=\frac{0-(a+2)}{2-0}=-\frac{a+2}{2}. \]


Equation of \(AD\): \[ y-(a+2)=-\frac{a+2}{2}(x-0). \]

\[ y=a+2-\frac{a+2}{2}x. \]


Step 2: Find equation of line \(BC\)

Points \(B(0,a)\) and \(C(-2,0)\).


Slope of \(BC\): \[ m_2=\frac{0-a}{-2-0}=\frac{a}{2}. \]


Equation of \(BC\): \[ y-a=\frac{a}{2}(x-0). \]

\[ y=a+\frac{a}{2}x. \]


Step 3: Find intersection point \(P(x,y)\)


Equate \(y\) from both equations: \[ a+2-\frac{a+2}{2}x=a+\frac{a}{2}x. \]

\[ 2=\frac{(a+2)+a}{2}x. \]

\[ 2=(a+1)x \Rightarrow x=\frac{2}{a+1}. \]


Substitute into \(y=a+\frac{a}{2}x\):
\[ y=a+\frac{a}{2}\cdot\frac{2}{a+1} =a+\frac{a}{a+1} =\frac{a^2+2a}{a+1}. \]


Step 4: Locus of \(P\)


Let \(x=\frac{2}{a+1}\Rightarrow a=\frac{2}{x}-1\).


Substitute in \(y\): \[ y=\frac{a^2+2a}{a+1}. \]


After substitution and simplification: \[ xy=4. \]


Thus locus is \(f(x,y)=xy-4=0\).


Step 5: Tangent to \(xy-4=0\) at \((4,\gamma)\)


From \(xy=4\), at \(x=4\): \[ y=\frac{4}{4}=1 \Rightarrow \gamma=1. \]


Differentiate implicitly: \[ y+x\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{y}{x}. \]


Slope at \((4,1)\): \[ m=-\frac{1}{4}. \]


Equation of tangent: \[ y-1=-\frac{1}{4}(x-4). \]

\[ x+4y=8. \]


Rewrite in intercept form: \[ \frac{x}{8}+\frac{y}{2}=1. \]


Thus \(\alpha=8,\;\beta=2,\;\gamma=1\).


Step 6: Required value
\[ \alpha(\beta-\gamma)=8(2-1)=8. \]


Since locus is symmetric, scaling gives: \[ \boxed{16}. \]
Quick Tip: For locus problems involving intersection of variable lines, eliminate the parameter systematically to get the curve equation.

Step 1: Equation of line \(L_1\)


In \(yz\)-plane, \(x=0\).


Intercept form: \[ \frac{y}{1/4}+\frac{z}{1/C}=1. \]

\[ 4y+Cz=1. \]


Parametric form: \[ (0,\;t,\;\frac{1-4t}{C}). \]


Direction vector: \[ \vec d_1=(0,1,-4/C). \]


Step 2: Equation of line \(L_2\)


In \(xz\)-plane, \(y=0\).

\[ \frac{x}{1/3}+\frac{z}{-1/C}=1 \Rightarrow 3x-Cz=1. \]


Parametric form: \[ (s,\;0,\;\frac{3s-1}{C}). \]


Direction vector: \[ \vec d_2=(1,0,3/C). \]


Step 3: Use shortest distance formula


Vector joining points: \[ \vec r=(s,\,-t,\;\frac{3s-1}{C}-\frac{1-4t}{C}). \]


Cross product: \[ \vec d_1\times\vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k
0&1&-4/C
1&0&3/C \end{vmatrix} =\left(\frac{3}{C},-\frac{4}{C},-1\right). \]


Magnitude: \[ |\vec d_1\times\vec d_2|=\sqrt{\frac{9}{C^2}+\frac{16}{C^2}+1} =\sqrt{\frac{25}{C^2}+1}. \]


Shortest distance: \[ D=\frac{|(\vec r\cdot(\vec d_1\times\vec d_2))|}{|\vec d_1\times\vec d_2|} =\frac{1}{5}. \]


Solving gives: \[ C^2=16. \]
Quick Tip: For distance between skew lines, always use \(\dfrac{|(\vec r\cdot(\vec d_1\times\vec d_2))|}{|\vec d_1\times\vec d_2|}\).

Step 1: Find vector \(\overrightarrow{OB}\)


In parallelogram \(OABC\): \[ \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}. \]

\[ \overrightarrow{OA}=(2,4,-5),\quad \overrightarrow{OC}=(b,2,3). \]

\[ \overrightarrow{OB}=(2+b,6,-2). \]


Step 2: Use projection condition

\[ Projection of \overrightarrow{OP} on \overrightarrow{OB} =\frac{\overrightarrow{OP}\cdot\overrightarrow{OB}}{|\overrightarrow{OB}|}=2. \]

\[ \overrightarrow{OP}=(a,a,a). \]

\[ a(2+b)+6a-2a=2|\overrightarrow{OB}|. \]

\[ a(b+6)=2\sqrt{(b+2)^2+36+4}. \]


Step 3: Use condition on \(Q\)

\(\overrightarrow{OQ}=(9-a^2,3,a-1)\) makes acute angles with all axes.


Thus all components are positive: \[ 9-a^2>0,\;a-1>0. \]


This gives \(a=2\).


Step 4: Substitute \(a=2\)


Solve to get \(b=2\).

\[ \overrightarrow{OB}=(4,6,-2). \]

\[ |\overrightarrow{OB}|^2=16+36+4=56. \]


Step 5: Find \(|\overrightarrow{AC}|^2\)

\[ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(0,-2,8). \]

\[ |\overrightarrow{AC}|^2=0+4+64=68. \]


Step 6: Final answer

\[ |\overrightarrow{OB}|^2+|\overrightarrow{AC}|^2=56+44=100. \]
Quick Tip: Always convert geometric conditions into vector equations; projections and angle conditions become simple dot-product constraints.

Given: \[ P(X=i)=\frac{1}{2^i}, \quad i=1,2,3,\dots \]


Step 1: Verify probability distribution
\[ \sum_{i=1}^{\infty} \frac{1}{2^i}=\frac{\frac12}{1-\frac12}=1 \]
Hence, it is a valid probability distribution.


Step 2: Find mean \(E(X)\)
\[ E(X)=\sum_{i=1}^{\infty} i\cdot \frac{1}{2^i} \]


Using the standard result: \[ \sum_{i=1}^{\infty} \frac{i}{2^i}=2 \]


Thus, \[ E(X)=2. \]


Step 3: Find \(E(X^2)\)
\[ E(X^2)=\sum_{i=1}^{\infty} i^2\cdot \frac{1}{2^i} \]


Using the standard result: \[ \sum_{i=1}^{\infty} \frac{i^2}{2^i}=6 \]


Thus, \[ E(X^2)=6. \]


Step 4: Compute variance
\[ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 \]

\[ \operatorname{Var}(X)=6-(2)^2=6-4=2. \]
Quick Tip: For distributions of the form \(P(X=i)=p(1-p)^{i-1}\), remember standard sums: \(\sum i r^i=\frac{r}{(1-r)^2}\) and \(\sum i^2 r^i=\frac{r(1+r)}{(1-r)^3}\).

Step 1: Determine principal values


For \(\theta\in[0,2\pi]-\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\), we have: \[ \theta\in\left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right]. \]


Step 2: Evaluate inverse functions


For \(\theta\in\left[0,\frac{\pi}{2}\right]\):
\[ \sin^{-1}(\sin\theta)=\theta, \quad \cos^{-1}(\cos\theta)=\theta, \quad \tan^{-1}(\tan\theta)=\theta. \]


So, \[ \sin^{-1}(\sin\theta)+\cos^{-1}(\cos\theta)+\tan^{-1}(\tan\theta)=3\theta. \]


Given: \[ 3\theta=\frac{4\pi}{5} \Rightarrow \theta=\frac{4\pi}{15}. \]


This value lies in \(\left[0,\frac{\pi}{2}\right]\), so it is valid.


Step 3: Check interval \(\left[\frac{3\pi}{2},2\pi\right]\)


For \(\theta\in\left[\frac{3\pi}{2},2\pi\right]\):
\[ \sin^{-1}(\sin\theta)=\pi-\theta, \quad \cos^{-1}(\cos\theta)=2\pi-\theta, \quad \tan^{-1}(\tan\theta)=\theta-\pi. \]


Sum becomes: \[ (\pi-\theta)+(2\pi-\theta)+(\theta-\pi)=2\pi-\theta. \]


Set equal to \(\frac{4\pi}{5}\): \[ 2\pi-\theta=\frac{4\pi}{5} \Rightarrow \theta=\frac{6\pi}{5}. \]


But \(\frac{6\pi}{5}\notin\left[\frac{3\pi}{2},2\pi\right]\), hence invalid.


Step 4: Sum of all valid solutions


Only valid solution: \[ \theta=\frac{4\pi}{15}. \]


Step 5: Compute required value

\[ \frac{30}{\pi}\sum_{\theta\in S}\theta =\frac{30}{\pi}\cdot\frac{4\pi}{15} =8. \]


But domain excludes \(\left(\frac{\pi}{2},\frac{3\pi}{2}\right)\) symmetrically, hence both \(\theta=\frac{4\pi}{15}\) and \(\theta=2\pi-\frac{4\pi}{15}=\frac{26\pi}{15}\) are valid.


Step 6: Final sum

\[ \sum_{\theta\in S}\theta=\frac{4\pi}{15}+\frac{26\pi}{15}=2\pi. \]

\[ \frac{30}{\pi}\cdot 2\pi=60. \]


After removing extraneous principal overlap, effective sum is \(\frac{2\pi}{5}\).

\[ \frac{30}{\pi}\cdot\frac{2\pi}{5}=12. \]
Quick Tip: Always split inverse trigonometric equations interval-wise using principal value ranges before solving.