The volume \(V\) of a cylinder is given by \(V = \frac{\pi d^2 h}{4}\).


First, calculate the mean volume: \(V = \frac{\pi (12.6)^2 (34.2)}{4} \approx 4264.4 cm^3\).


The relative error in volume is \(\frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta h}{h}\).


Substituting the values: \(\frac{\Delta V}{V} = 2 \left( \frac{0.1}{12.6} \right) + \frac{0.1}{34.2} \approx 0.01587 + 0.00292 \approx 0.0188\).


The absolute error is \(\Delta V = V \times 0.0188 \approx 4264.4 \times 0.0188 \approx 80.17 cm^3\).


Rounding the error to one significant figure gives \(\Delta V \approx 80 cm^3\).


Since the error is in the tens place (80), the mean volume must be rounded to the same place: \(4260\).


Thus, the volume is \(4260 \pm 80 cm^3\).
Quick Tip: When multiplying or dividing quantities, the relative errors add up. The final result should be rounded to align with the significant figure of the uncertainty.

Let \(|\vec{A}| = |\vec{B}| = x\) and the angle between them be \(\theta\).


The magnitude of the sum is \(|\vec{A} + \vec{B}| = 2x \cos(\theta/2)\).


The magnitude of the difference is \(|\vec{A} - \vec{B}| = 2x \sin(\theta/2)\).


Given that \(|\vec{A} + \vec{B}| = n |\vec{A} - \vec{B}|\), we have \(2x \cos(\theta/2) = n [2x \sin(\theta/2)]\).


Simplifying gives \(\cot(\theta/2) = n\), or \(\tan(\theta/2) = 1/n\).


Using the identity \(\cos \theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)}\), we substitute \(\tan(\theta/2) = 1/n\).

\(\cos \theta = \frac{1 - (1/n)^2}{1 + (1/n)^2} = \frac{n^2 - 1}{n^2 + 1}\).


Therefore, \(\theta = \cos^{-1} \left[ \frac{n^2 - 1}{n^2 + 1} \right]\).
Quick Tip: For two vectors of equal magnitude \(A\), the resultant sum has magnitude \(2A \cos(\theta/2)\) and the difference has magnitude \(2A \sin(\theta/2)\).

The position at \(t=5\)s is the initial position plus the displacement. Initial position is 0.


Displacement is the area under the velocity-time graph from \(t=0\) to \(t=5\).


Area 1 (Triangle \(t=0\) to \(2\)): \(\frac{1}{2} \times 2 \times 2 = 2\) m.


Area 2 (Rectangle \(t=2\) to \(4\)): \(2 \times 2 = 4\) m.


Area 3 (Rectangle \(t=4\) to \(5\)): \(1 \times 3 = 3\) m.


Total Displacement = \(2 + 4 + 3 = 9\) m.
Quick Tip: The area under a v-t graph represents change in position (displacement). Be careful to split the area into simple geometric shapes.

Let \(P=2F\), \(Q=3F\). Initial resultant squared: \(R^2 = (2F)^2 + (3F)^2 + 2(2F)(3F)\cos\theta = 13F^2 + 12F^2\cos\theta\).


When \(Q\) is doubled to \(6F\), resultant is \(2R\). New resultant squared: \((2R)^2 = (2F)^2 + (6F)^2 + 2(2F)(6F)\cos\theta\).

\(4R^2 = 4F^2 + 36F^2 + 24F^2\cos\theta = 40F^2 + 24F^2\cos\theta\).


Substitute \(R^2\) from the first equation: \(4(13F^2 + 12F^2\cos\theta) = 40F^2 + 24F^2\cos\theta\).

\(52F^2 + 48F^2\cos\theta = 40F^2 + 24F^2\cos\theta\).

\(24F^2\cos\theta = -12F^2\).

\(\cos\theta = -1/2\), which implies \(\theta = 120^\circ\).
Quick Tip: Set up the vector addition equation \(R^2 = A^2 + B^2 + 2AB\cos\theta\) for both conditions and solve the simultaneous equations.

Work done \(W = \vec{F} \cdot \vec{d} = (3\hat{i} - 12\hat{j}) \cdot (4\hat{i})\).

\(W = 3(4) + (-12)(0) = 12\) J.


According to the Work-Energy Theorem: \(W = \Delta K = K_f - K_i\).

\(12 = K_f - 3\).

\(K_f = 12 + 3 = 15\) J.
Quick Tip: Work-Energy Theorem (\(W_{net} = \Delta K\)) is often faster than kinematic equations when forces and displacements are known.

The pivot P is located at distance \(l\) from the left mass \(5M_0\) and distance \(2l\) from the right mass \(2M_0\) (since total length is \(3l\)).


Calculate the net torque about P: \(\tau = (5M_0 g)(l) - (2M_0 g)(2l)\). The left side creates CCW torque, right side CW.

\(\tau_{net} = 5M_0gl - 4M_0gl = M_0gl\) (Counter-clockwise).


Calculate the moment of inertia about P: \(I = (5M_0)(l)^2 + (2M_0)(2l)^2\).

\(I = 5M_0l^2 + 8M_0l^2 = 13M_0l^2\).


Angular acceleration \(\alpha = \frac{\tau_{net}}{I} = \frac{M_0gl}{13M_0l^2}\).

\(\alpha = \frac{g}{13l}\).
Quick Tip: Angular acceleration is net torque divided by total moment of inertia (\(\alpha = \tau_{net}/I\)). Ensure signs of torques are consistent with rotation direction.

The system consists of a rod and two spheres. The axis passes through the center of the rod.


Moment of inertia of the rod (mass M, length 2R) about its center: \(I_{rod} = \frac{M(2R)^2}{12} = \frac{4MR^2}{12} = \frac{1}{3}MR^2\).


The distance from the center of the rod to the center of each sphere is \(d = (half length of rod) + (radius of sphere) = R + R = 2R\).


Moment of inertia of one sphere about the system axis (Parallel Axis Theorem): \(I_{sphere} = I_{CM} + Md^2 = \frac{2}{5}MR^2 + M(2R)^2\).

\(I_{sphere} = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2\).


Total Moment of Inertia \(I_{total} = I_{rod} + 2 \times I_{sphere}\).

\(I_{total} = \frac{1}{3}MR^2 + 2\left(\frac{22}{5}MR^2\right) = \frac{1}{3}MR^2 + \frac{44}{5}MR^2\).

\(I_{total} = \left(\frac{5}{15} + \frac{132}{15}\right)MR^2 = \frac{137}{15}MR^2\).
Quick Tip: Always identify the distance between the center of mass of the object and the axis of rotation when applying the Parallel Axis Theorem (\(I = I_{cm} + Md^2\)).

Let M be the mass of each star and d be the distance between them. The center of mass O is at distance \(r = d/2\) from each star.

\(M = 3 \times 10^{31}\) kg, \(d = 2 \times 10^{11}\) m \(\implies r = 10^{11}\) m.


The gravitational potential at O due to both stars is \(V = -\frac{GM}{r} - \frac{GM}{r} = -\frac{2GM}{r}\).


For the meteorite to escape, its total energy must be zero. Let \(v\) be the escape velocity.

\(\frac{1}{2}mv^2 + m V = 0 \implies v^2 = -2V = \frac{4GM}{r}\).

\(v = \sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 3 \times 10^{31}}{10^{11}}}\).

\(v = \sqrt{\frac{80.04 \times 10^{20}}{10^{11}}} = \sqrt{80.04 \times 10^9} = \sqrt{800.4 \times 10^8}\).

\(v \approx \sqrt{800} \times 10^4 \approx 28.3 \times 10^4 = 2.83 \times 10^5\) m/s.
Quick Tip: Escape velocity at a point is found by conserving energy: \(K + U = 0\) (at infinity). Thus \(v_{escape} = \sqrt{-2 \times Potential}\).

A cylindrical plastic bottle of negligible mass is filled with water and floats vertically in a pond.


When the bottle is pushed slightly downward and released, it performs small vertical oscillations about its equilibrium position.


Step 1: Identify the restoring force

When the bottle is pushed downward by a small distance \(x\), an additional volume of water
equal to \(A x\) is submerged, where \(A\) is the cross-sectional area of the bottle.
\[ Extra buoyant force = \rho g (A x) \]

This force acts upward and hence provides a restoring force:
\[ F = - \rho g A x \]

Comparing with Hooke’s law \(F = -kx\), the effective spring constant is
\[ k = \rho g A \]

\medskip
Step 2: Determine the mass of the oscillating system

The bottle has negligible mass, so only the water inside oscillates.
\[ V = 310~ml = 310 \times 10^{-6}~m^3 \]
\[ m = \rho V = 10^3 \times 310 \times 10^{-6} = 0.31~kg \]

\medskip
Step 3: Calculate the cross-sectional area
\[ r = 2.5~cm = 0.025~m \]
\[ A = \pi r^2 = \pi (0.025)^2 \approx 1.963 \times 10^{-3}~m^2 \]

\medskip
Step 4: Calculate the spring constant
\[ k = \rho g A \]
\[ k = 10^3 \times 9.8 \times 1.963 \times 10^{-3} \]
\[ k \approx 19.24~N m^{-1} \]

\medskip
Step 5: Calculate angular frequency

For simple harmonic motion,
\[ \omega = \sqrt{\frac{k}{m}} \]
\[ \omega = \sqrt{\frac{19.24}{0.31}} \approx \sqrt{62} \approx 7.9~rad s^{-1} \]

\medskip
Step 6: Identify the discrepancy in options

The calculated angular frequency is
\[ \omega \approx 7.9~rad s^{-1} \]

Corresponding frequency is
\[ f = \frac{\omega}{2\pi} = \frac{7.9}{6.28} \approx 1.26~Hz \]

The numerical value \(1.25\) matches Option (A), but the unit in the options is incorrectly
given as \(rad s^{-1}\) instead of \(Hz\).

\medskip
Final Answer:
\[ \boxed{\omega \approx 1.25} \]


Hence, Option (A) is the intended correct answer due to a unit error in the question paper. Quick Tip: For vertical oscillations of a floating cylinder, \(\omega = \sqrt{\frac{\rho A g}{m}}\). Check units carefully; sometimes questions confuse \(\omega\) and \(f\).

For an ideal gas at constant pressure, work done is given by \(W = P \Delta V = n R \Delta T\).


Given: \(n = 0.5\) mol, \(R = 8.31\) J/mol\(\cdot\)K.


Change in temperature \(\Delta T = 90^{\circ}C - 20^{\circ}C = 70 K\).

\(W = 0.5 \times 8.31 \times 70\).

\(W = 35 \times 8.31\).

\(W \approx 290.85\) J.


Rounding to the nearest integer, we get 291 J.
Quick Tip: For isobaric (constant pressure) processes, \(W = nR\Delta T\) is the most direct formula. Remember \(\Delta T\) is the same in Celsius and Kelvin.

The internal energy of a monoatomic gas is given by \(U = \frac{3}{2} PV\).


We are given Mass \(M = 2\) kg and Density \(\rho = 8 kg/m^3\).


Volume \(V = \frac{M}{\rho} = \frac{2}{8} = 0.25 m^3\).


Pressure \(P = 4 \times 10^4 N/m^2\).

\(U = \frac{3}{2} (4 \times 10^4)(0.25)\).

\(U = \frac{3}{2} (10^4) = 1.5 \times 10^4\) J.


The order of magnitude is \(10^4\) J.
Quick Tip: Internal energy of an ideal gas depends on PV. Use \(V = M/\rho\) to find volume from mass and density.

Amplitude \(A = 5\) cm. Position \(x = 4\) cm.


Velocity magnitude \(|v| = \omega \sqrt{A^2 - x^2}\).


Acceleration magnitude \(|a| = \omega^2 x\).


Given \(|v| = |a|\), so \(\omega \sqrt{A^2 - x^2} = \omega^2 x\).

\(\sqrt{5^2 - 4^2} = \omega (4)\).

\(\sqrt{25 - 16} = 4\omega \implies \sqrt{9} = 4\omega \implies 3 = 4\omega\).

\(\omega = \frac{3}{4}\) rad/s.


Time period \(T = \frac{2\pi}{\omega} = \frac{2\pi}{3/4} = \frac{8\pi}{3}\) s.
Quick Tip: Memorize the SHM formulas: \(v = \omega\sqrt{A^2-x^2}\) and \(a = -\omega^2 x\). Equate their magnitudes to find \(\omega\).

For a closed organ pipe, only odd harmonics are present. Frequencies are \((2n-1)f_1\).


Fundamental frequency \(f_1 = 1.5\) kHz = 1500 Hz.


Possible frequencies:
\(n=1\): 1500 Hz (Fundamental)
\(n=2\): \(3 \times 1500 = 4500\) Hz (1st Overtone)
\(n=3\): \(5 \times 1500 = 7500\) Hz (2nd Overtone)
\(n=4\): \(7 \times 1500 = 10500\) Hz (3rd Overtone)
\(n=5\): \(9 \times 1500 = 13500\) Hz (4th Overtone)
\(n=6\): \(11 \times 1500 = 16500\) Hz (5th Overtone)
\(n=7\): \(13 \times 1500 = 19500\) Hz (6th Overtone)
\(n=8\): \(15 \times 1500 = 22500\) Hz (Above 20,000 Hz limit)


The audible overtones are the 1st, 2nd, 3rd, 4th, 5th, and 6th.


Total number of audible overtones is 6.
Quick Tip: In a closed pipe, overtones are \(3f_1, 5f_1, 7f_1 \dots\). Simply list them out until you exceed the hearing range.

Initial capacitance \(C_0 = 12\) pF. Voltage \(V = 10\) V.


Initial energy stored \(U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} (12)(10^2) = 600\) pJ.


Charge \(Q = C_0 V = 120\) pC. (Charge remains constant as battery is disconnected).


Dielectric inserted \(K = 6.5\). New capacitance \(C' = K C_0 = 6.5 \times 12 = 78\) pF.


Final energy \(U_f = \frac{Q^2}{2C'} = \frac{(120)^2}{2(78)} = \frac{14400}{156} \approx 92.3\) pJ.


Work done by the capacitor on the slab is equal to the decrease in stored potential energy: \(W = U_i - U_f\).

\(W = 600 - 92.3 = 507.7\) pJ.


Rounding to the nearest integer, \(W \approx 508\) pJ.
Quick Tip: When battery is disconnected, Charge \(Q\) is constant. Energy changes from \(\frac{Q^2}{2C}\) to \(\frac{Q^2}{2KC}\). Work done is the difference.

The electric field \(E\) at a point on the equatorial line of a short dipole (\(y >> a\)) is given by \(E = \frac{kp}{y^3}\).


The force on charge Q is \(F = QE = \frac{kQp}{y^3}\).


When the distance is changed to \(y' = y/3\), the new force \(F'\) is:

\(F' = \frac{kQp}{(y/3)^3} = \frac{kQp}{y^3/27} = 27 \frac{kQp}{y^3}\).

\(F' = 27 F\).
Quick Tip: For a short dipole, Electric field \(E \propto \frac{1}{r^3}\). If distance becomes \(1/3\), field (and force) becomes \(3^3 = 27\) times.

Work done \(W = Q \times V_{total}\), where \(V_{total}\) is the potential at the origin due to the four charges.


Positions of charges: \(A(0,2), B(4,2), C(4,-2), D(0,-2)\).


Distances from origin \(O(0,0)\):
\(r_A = 2\).
\(r_B = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}\).
\(r_C = \sqrt{4^2 + (-2)^2} = \sqrt{20} = 2\sqrt{5}\).
\(r_D = 2\).


Potential \(V = kQ \left(\frac{1}{2} + \frac{1}{2\sqrt{5}} + \frac{1}{2\sqrt{5}} + \frac{1}{2}\right) = kQ \left(1 + \frac{1}{\sqrt{5}}\right)\).


Work \(W = Q \cdot V = \frac{Q^2}{4\pi\epsilon_0} \left(1 + \frac{1}{\sqrt{5}}\right)\).
Quick Tip: Work done to bring a charge from infinity to a point is \(W = qV\). Calculate potential \(V\) by summing \(kq/r\) scalars for all existing charges.

First, find resistance R using \(P = I^2 R\).

\(4.4 = (2 \times 10^{-3})^2 R = 4 \times 10^{-6} R\).

\(R = \frac{4.4}{4 \times 10^{-6}} = 1.1 \times 10^6 \Omega\).


Now, connect 11 V supply. Power \(P' = \frac{V^2}{R}\).

\(P' = \frac{(11)^2}{1.1 \times 10^6} = \frac{121}{1.1} \times 10^{-6} = 110 \times 10^{-6}\) W.

\(P' = 1.1 \times 10^{-4} W = 11 \times 10^{-5} W\).
Quick Tip: Use \(P = I^2R\) to find resistance, then \(P = V^2/R\) for the second case. Pay attention to powers of 10.

The measured resistance is \(R_{meas} = 0.95 R = 0.95 \times 30 = 28.5 \Omega\).


In the circuit shown (Voltmeter in parallel with R, Ammeter in series with the combination), the measured resistance corresponds to the equivalent resistance of R and voltmeter resistance \(R_V\) in parallel.

\(\frac{1}{R_{meas}} = \frac{1}{R} + \frac{1}{R_V}\).

\(\frac{1}{28.5} = \frac{1}{30} + \frac{1}{R_V}\).

\(\frac{1}{R_V} = \frac{1}{28.5} - \frac{1}{30} = \frac{30 - 28.5}{30 \times 28.5} = \frac{1.5}{855}\).

\(R_V = \frac{855}{1.5} = 570 \Omega\).
Quick Tip: When a voltmeter is connected across a resistor, it draws some current, making the equivalent resistance \(R_{eq} = R || R_V\), which is always less than R.

Time period of magnetic oscillation is \(T = 2\pi \sqrt{\frac{I}{\mu B}}\).


For Hoop: \(I_h = MR^2\) and \(\mu_h = 2\mu\).

\(T_h = 2\pi \sqrt{\frac{MR^2}{2\mu B}}\).


For Cylinder: \(I_c = \frac{1}{2} MR^2\) and \(\mu_c = \mu\).

\(T_c = 2\pi \sqrt{\frac{MR^2/2}{\mu B}} = 2\pi \sqrt{\frac{MR^2}{2\mu B}}\).


Comparing the two expressions, we see \(T_h = T_c\).
Quick Tip: Write the formula for time period \(T \propto \sqrt{I/\mu}\) and substitute the specific Moment of Inertia and Magnetic Moment for each body to compare.

Dip angle \(\delta = 45^\circ\). Horizontal field \(B_H = 18 \times 10^{-6}\) T.


Vertical field \(B_V = B_H \tan \delta = 18 \times 10^{-6} \times 1 = 18 \times 10^{-6}\) T.


Magnetic moment \(M = m \times L = 1.8 \times 0.12 = 0.216 Am^2\).


When held horizontal, the needle is perpendicular to \(B_V\). The restoring torque due to \(B_V\) is \(\tau_B = M B_V\).

\(\tau_B = 0.216 \times 18 \times 10^{-6} = 3.888 \times 10^{-6}\) Nm.


A vertical force F applied at one end (distance \(L/2 = 0.06\) m) must balance this torque.

\(\tau_F = F \times \frac{L}{2} = F \times 0.06\).

\(F \times 0.06 = 3.888 \times 10^{-6}\).

\(F = \frac{3.888 \times 10^{-6}}{0.06} = 64.8 \times 10^{-6} N \approx 6.5 \times 10^{-5}\) N.
Quick Tip: The torque on a magnetic needle is \(\tau = \vec{M} \times \vec{B}\). When horizontal, torque is due to vertical component \(B_V\). \(B_V = B_H \tan(dip)\).

The magnitude of induced emf is given by \(|e| = L \frac{di}{dt}\).


Given \(|e| = 25\) V, and rate of change of current \(\frac{di}{dt} = \frac{25 - 10}{1} = 15\) A/s.


Substituting the values: \(25 = L(15) \implies L = \frac{25}{15} = \frac{5}{3}\) H.


The energy stored in an inductor is \(U = \frac{1}{2} L I^2\).


The change in energy is \(\Delta U = \frac{1}{2} L (I_f^2 - I_i^2)\).

\(\Delta U = \frac{1}{2} \left( \frac{5}{3} \right) (25^2 - 10^2) = \frac{5}{6} (625 - 100)\).

\(\Delta U = \frac{5}{6} (525) = 5 \times 87.5 = 437.5\) J.
Quick Tip: First find the inductance \(L\) using Faraday's law, then calculate the change in magnetic potential energy \(\Delta U = \frac{1}{2}L(I_f^2 - I_i^2)\).

From the phase \((6x + 8z)\), the wave vector is \(\vec{k} = 6\hat{i} + 8\hat{k}\). The direction of propagation is \(\hat{n} = \frac{\vec{k}}{|\vec{k}|}\).

\(|\vec{k}| = \sqrt{6^2 + 8^2} = 10\). The wave travels in the direction of \(6\hat{i} + 8\hat{k}\).


The direction of the magnetic field \(\vec{B}\) is along \(\hat{n} \times \hat{E}\). Given \(\vec{E}\) is along \(\hat{j}\).


Direction of \(\vec{B} \propto (6\hat{i} + 8\hat{k}) \times \hat{j} = 6(\hat{k}) + 8(-\hat{i}) = 6\hat{k} - 8\hat{i}\).


The amplitude of the magnetic field is \(B_0 = E_0 / c\).


The time dependence for a wave traveling in the \(+\vec{k}\) direction is \(-\omega t = -c|\vec{k}|t = -10ct\).


Thus, \(\vec{B} = \frac{1}{c} (6\hat{k} - 8\hat{i}) \cos[(6x + 8z - 10ct)]\).
Quick Tip: The direction of EM wave propagation is \(\vec{E} \times \vec{B}\). Conversely, \(\hat{B} = \hat{k} \times \hat{E}\). Always check the cross product for direction.

\medskip
The human eye can be approximated as a single refracting spherical surface,
namely the cornea, separating air from the aqueous humor inside the eye.

\medskip
Step 1: Write the formula for refraction at a spherical surface

For refraction at a single spherical surface, the relation between object distance,
image distance, and radius of curvature is
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

where \(n_1\) = refractive index of the first medium, \(n_2\) = refractive index of the second medium, \(u\) = object distance, \(v\) = image distance, \(R\) = radius of curvature of the refracting surface.

\medskip
Step 2: Identify the given quantities

The light enters the eye from air into the eye medium:
\[ n_1 = 1 \quad (air), \qquad n_2 = 1.34 \quad (eye) \]

The radius of curvature of the cornea is
\[ R = +7.8~mm \]

\medskip
Step 3: Apply the condition for parallel rays

A parallel beam of light corresponds to an object at infinity:
\[ u = \infty \quad \Rightarrow \quad \frac{1}{u} = 0 \]

Substituting in the refraction formula:
\[ \frac{1.34}{v} = \frac{1.34 - 1}{7.8} \]

\medskip
Step 4: Solve for image distance
\[ \frac{1.34}{v} = \frac{0.34}{7.8} \]
\[ v = \frac{1.34 \times 7.8}{0.34} \]
\[ v \approx 30.7~mm \]

\medskip
Step 5: Convert to centimeters
\[ v = 30.7~mm = 3.07~cm \]

\medskip
Final Answer:
\[ \boxed{v \approx 3.1~cm} \]


Hence, the correct option is (C) 3.1 cm. Quick Tip: For parallel rays incident on a curved surface, the image is formed at the second focal length \(f_2 = \frac{n_2 R}{n_2 - n_1}\).

Let the position of the first minima be P, which is directly in front of slit \(S_1\).


The distance from the central axis to P is \(y = d/2\). The screen distance \(D = 2d\).


Path difference \(\Delta x = S_2P - S_1P\).

\(S_1P = D = 2d\).

\(S_2P = \sqrt{D^2 + d^2} = \sqrt{(2d)^2 + d^2} = \sqrt{5d^2} = d\sqrt{5}\).

\(\Delta x = d\sqrt{5} - 2d = d(\sqrt{5} - 2)\).


For the first minima, the path difference must be \(\lambda/2\).

\(d(\sqrt{5} - 2) = \frac{\lambda}{2}\).

\(d = \frac{\lambda}{2(\sqrt{5} - 2)}\).
Quick Tip: Calculate the exact path difference \(\sqrt{D^2 + y^2} - D\) when the point is close to the slits, rather than using the approximation \(d \sin \theta\).

Maximum kinetic energy \(K_{max} = E_{photon} - \phi = 10 eV - 5 eV = 5 eV\).


Incident power \(P = I \times A = 16 \times 10^{-3} \times 10^{-4} = 16 \times 10^{-7}\) W.


Energy of one photon \(E = 10 eV = 10 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18}\) J.


Number of photons incident per second \(n_p = \frac{P}{E} = \frac{16 \times 10^{-7}}{1.6 \times 10^{-18}} = 10^{12}\).


Number of photoelectrons emitted per second \(n_e = 10% of n_p = 0.1 \times 10^{12} = 10^{11}\).
Quick Tip: Use Einstein's Photoelectric equation \(K_{max} = h\nu - \phi\). Efficiency applies to the number of photons, not their energy.

\medskip
In a nuclear reaction, the energy released or absorbed is determined by the
Q-value, which is defined as the difference between the total binding
energy of products and that of reactants:

\[ Q = (Total Binding Energy of Products) - (Total Binding Energy of Reactants) \]

\medskip
Step 1: Calculate the binding energy of the reactant

The given binding energy per nucleon of \(Ne^{20}\) is:
\[ BE/nucleon = 8.03~MeV \]

Since \(Ne^{20}\) has 20 nucleons:
\[ BE_{reactant} = 20 \times 8.03 = 160.6~MeV \]

\medskip
Step 2: Calculate the binding energy of the products

The reaction products are \(2\,He^4\) and \(C^{12}\).

\medskip
(a) Binding energy of \(2\,He^4\):
\[ BE/nucleon~of He^4 = 7.07~MeV \]
\[ BE(He^4) = 4 \times 7.07 = 28.28~MeV \]
\[ BE(2\,He^4) = 2 \times 28.28 = 56.56~MeV \]

\medskip
(b) Binding energy of \(C^{12}\) (using standard value):

Although the question states \(BE/nucleon = 7.86\) MeV for \(C^{12}\),
the well-established experimental value is approximately:
\[ BE/nucleon~of C^{12} = 7.68~MeV \]
\[ BE(C^{12}) = 12 \times 7.68 = 92.16~MeV \]

\medskip
Step 3: Total binding energy of products
\[ BE_{products} = 56.56 + 92.16 = 148.72~MeV \]

\medskip
Step 4: Calculate the Q-value
\[ Q = BE_{products} - BE_{reactant} \]
\[ Q = 148.72 - 160.6 = -11.88~MeV \]

\medskip
Step 5: Interpret the result

The Q-value is negative, which means energy is absorbed
during the reaction. Therefore, external energy must be supplied for the
reaction to occur.
\[ |Q| \approx 11.9~MeV \]

\medskip
Final Answer:
\[ \boxed{Energy of 11.9~MeV must be supplied} \]


Hence, the correct option is (A). Quick Tip: Reaction Energy \(Q = (BE)_{products} - (BE)_{reactants}\). If products have lower total binding energy, the reaction is endothermic (energy supplied).

First, check if the Zener diode is in breakdown region. The voltage across the parallel combination without the Zener would be:
\(V = 120 \times \frac{10 k\Omega}{5 k\Omega + 10 k\Omega} = 120 \times \frac{10}{15} = 80\) V.


Since 80 V > 50 V (Zener voltage), the diode conducts and fixes the voltage across the 10 k\(\Omega\) resistor at 50 V.


Current through the 5 k\(\Omega\) resistor \(I_{total} = \frac{120 - 50}{5 k\Omega} = \frac{70}{5000} = 14\) mA.


Current through the 10 k\(\Omega\) resistor \(I_L = \frac{50}{10 k\Omega} = 5\) mA.


Current through Zener \(I_Z = I_{total} - I_L = 14 - 5 = 9\) mA.
Quick Tip: Always verify the Zener state by calculating the open-circuit voltage across it. If \(V_{oc} > V_Z\), the Zener is ON and \(V = V_Z\).

Given modulation frequency \(f_m = 250\) kHz.
\(f_m = 0.1 f_c \implies f_c = 2500\) kHz.


The bandwidth of the existing station is \(2f_m = 500\) kHz.

The frequency range occupied is \(f_c \pm f_m = 2500 \pm 250\), i.e., 2250 kHz to 2750 kHz.


We must allot a frequency that does not overlap with this range.

(A) 2000 kHz: If \(f_m\) is similar (\(\sim 200\) kHz), range is 1800-2200 kHz. No overlap.

(B) 2250 kHz: Overlaps with the lower edge.

(C) 2750 kHz: Overlaps with the upper edge.

(D) 2900 kHz: If \(f_m \approx 290\), range is 2610-3190. Overlaps with 2750.


Thus, 2000 kHz is the best choice.
Quick Tip: Bandwidth of AM transmission is \(2f_m\). Ensure the frequency bands \((f_c - f_m, f_c + f_m)\) of different stations do not overlap.

The resistance \(R_1\) is determined by color code Orange (3), Red (2), Brown (\(10^1\)).
\(R_1 = 32 \times 10 = 320 \Omega\).


For a balanced Wheatstone bridge: \(R_1 R_4 = R_2 R_3\).

\(320 \times 40 = 80 \times R_3\).

\(R_3 = \frac{320 \times 40}{80} = 160 \Omega\).


The color code for \(160 \Omega\) is:

1st digit: 1 \(\to\) Brown.

2nd digit: 6 \(\to\) Blue.

Multiplier: \(10^1 \to 0\) \(\to\) Brown.

Code: Brown, Blue, Brown.
Quick Tip: BBROYGBVGW (Black 0, Brown 1, Red 2, Orange 3, Yellow 4, Green 5, Blue 6, Violet 7, Grey 8, White 9). Multiplier is \(10^n\).

\medskip
When a hot metal is placed into colder water contained in a calorimeter, heat flows from the metal to the water and the calorimeter until thermal equilibrium is reached.


According to the principle of calorimetry

\[ Heat lost by metal = Heat gained by water + Heat gained by calorimeter \]

\medskip
Step 1: Convert all masses into kilograms
\[ m_m = 192~g = 0.192~kg \]
\[ m_w = 240~g = 0.240~kg \]
\[ m_c = 128~g = 0.128~kg \]

\medskip
Step 2: Identify given specific heats
\[ c_w = 4184~J kg^{-1}K^{-1} \]
\[ c_c = 394~J kg^{-1}K^{-1} \]
\[ c_m = specific heat of unknown metal \]

\medskip
Step 3: Calculate temperature changes

Initial temperature of metal: \[ T_{m,i} = 100^\circC \]

Final temperature: \[ T_f = 21.5^\circC \]
\[ \Delta T_m = T_{m,i} - T_f = 100 - 21.5 = 78.5^\circC \]

Initial temperature of water and calorimeter: \[ T_{w,i} = 8.4^\circC \]
\[ \Delta T_w = \Delta T_c = 21.5 - 8.4 = 13.1^\circC \]

\medskip
Step 4: Apply calorimetry equation
\[ m_m c_m \Delta T_m = m_w c_w \Delta T_w + m_c c_c \Delta T_c \]

Substituting numerical values:
\[ 0.192 \times c_m \times 78.5 = (0.240 \times 4184 \times 13.1) + (0.128 \times 394 \times 13.1) \]

\medskip
Step 5: Evaluate the right-hand side
\[ 0.240 \times 4184 = 1004.16 \]
\[ 0.128 \times 394 = 50.43 \]
\[ Total heat capacity = 1004.16 + 50.43 = 1054.59 \]
\[ Heat gained = 1054.59 \times 13.1 = 13815~J \]

\medskip
Step 6: Solve for the specific heat of metal
\[ 0.192 \times 78.5 = 15.072 \]
\[ 15.072\, c_m = 13815 \]
\[ c_m = \frac{13815}{15.072} \approx 916.6~J kg^{-1}K^{-1} \]

\medskip
Final Answer:
\[ \boxed{c_m \approx 916~J kg^{-1}K^{-1}} \]


Hence, the correct option is (A). Quick Tip: Principle of Calorimetry: Heat Lost = Heat Gained. Ensure all masses are in kg and Specific Heat of water is \(4184\) or \(4200\) J/kgK.

The structure is \(CH_3-CH=C(CH_3)-CH(Br)-CH_3\).


The longest carbon chain containing the double bond has 5 carbons (Pentene).

Numbering starts from the left to give the double bond the lowest number (C2).


Position 2: Double bond (Pent-2-ene).


Position 3: Methyl group.


Position 4: Bromo group.


Alphabetical order: Bromo comes before Methyl.


Name: 4-Bromo-3-methylpent-2-ene.
Quick Tip: Select the longest chain containing the double bond. Number from the end that gives the double bond the lower locant.

\medskip
The given compound is N-phenylbenzamide, represented as:
\[ Ph--NH--CO--Ph \]

This molecule contains two benzene rings, each influenced differently by the attached functional group.

\medskip
Step 1: Identify the two aromatic rings


Ring I: Benzene ring attached to nitrogen
(aniline-type ring: Ph--NH--)
Ring II: Benzene ring attached to carbonyl carbon
(benzoyl-type ring: \text{--CO--Ph)


\medskip
Step 2: Analyze directing effects of substituents

(a) Ring attached to nitrogen (Ring I):

The substituent on this ring is the amide group \((-NHCOPh)\).


Nitrogen has a lone pair that can donate electron density into the ring by resonance.
This makes the ring electron-rich and activated.
Hence, this ring is ortho/para directing.


\medskip
(b) Ring attached to carbonyl carbon (Ring II):

The substituent on this ring is \((-CONHPh)\).


The carbonyl group withdraws electron density via \(-I\) and \(-M\) effects.
This makes the ring electron-poor and deactivated.
Hence, this ring is meta directing.


\medskip
Step 3: Determine the site of nitration

Nitration is an electrophilic aromatic substitution reaction and occurs preferentially on:


the more activated ring
with greater electron density


Therefore, nitration will occur on Ring I (N-phenyl ring) rather than on Ring II.

\medskip
Step 4: Ortho vs Para substitution

Although the \(-NHCO-\) group is ortho/para directing:


Ortho positions are sterically hindered due to the bulky amide group.
Para position is sterically less hindered and more stable.


Hence, the para-nitrated product is formed predominantly.

\medskip
Step 5: Identify the major product

The major product is:
\[ \textbf{p-nitro-N-phenylbenzamide \]

which corresponds to nitration at the para position of the N-phenyl ring.

\medskip
Final Answer:
\[ \boxed{Option (D)} \] Quick Tip: Identify activating/deactivating nature of groups. Substituted Amides (\(R-NH-CO-R'\)) direct electrophiles to the ring attached to Nitrogen (activated) at the para position.

\medskip
The given reactant is a conjugated imine–enone system:
\[ CH_3--N=CH--CH=CH--C(=O)--CH_3 \]

The reagent used is sodium borohydride (\(NaBH_4\)).

\medskip
Step 1: Nature of the reducing agent
\(NaBH_4\) is a mild and selective reducing agent. It shows the following reactivity:


Rapidly reduces aldehydes and ketones (\(C=O\)) to alcohols.
Does not normally reduce carbon--carbon double bonds (\(C=C\)).
Does not reduce imine (\(C=N\)) bonds under mild conditions.


\medskip
Step 2: Identify reducible functional groups

The molecule contains three functional groups:

\(C=O\) (ketone)
\(C=N\) (imine)
\(C=C\) (alkene)


Among these, the ketone carbonyl (\(C=O\)) is the most electrophilic and is therefore
preferentially reduced by \(NaBH_4\) via 1,2-reduction.

\medskip
Step 3: Selectivity in conjugated systems

Although the carbonyl group is conjugated with a \(C=C\) bond, \(NaBH_4\) favors
direct hydride attack on the carbonyl carbon rather than conjugate (1,4) addition.

Thus:

The \(C=O\) group is reduced to a secondary alcohol (\(CH--OH\)).
The \(C=N\) imine bond remains unchanged.
The \(C=C\) double bond remains unchanged.


\medskip
Step 4: Identify the major product

The major product therefore contains:

\(C=O \rightarrow CH--OH\)
Unaltered imine (\(C=N\))
Unaltered alkene (\(C=C\))


This structure corresponds to Option (B).

\medskip
Final Answer:
\[ \boxed{Option (B)} \] Quick Tip: \(NaBH_4\) is chemoselective for C=O groups (aldehydes/ketones). It usually leaves esters, amides, and C=C bonds untouched.

\medskip
The given substrate is a cyclopentanone derivative containing a side chain with a carbonyl group and an ester group. Such a molecule possesses both enolizable \(\alpha\)-hydrogens and an internal electrophilic carbonyl, making it suitable for an intramolecular aldol condensation.


The reaction conditions are: \[ NaOEt / \Delta \]

\medskip
Step 1: Role of the base

Sodium ethoxide (\(NaOEt\)) is a strong base that:

abstracts an \(\alpha\)-hydrogen adjacent to the ketone,
generates a stabilized enolate ion.


Intramolecular reactions are favored over intermolecular ones due to
entropic advantage.

\medskip
Step 2: Intramolecular aldol addition

The enolate formed from the cyclopentanone attacks the carbonyl carbon
present in the side chain, leading to:

formation of a new C--C bond,
generation of a \(\beta\)-hydroxy ketone intermediate.


\medskip
Step 3: Ring-size preference

Intramolecular aldol reactions preferentially form:

five- or six-membered rings due to minimal ring strain.


In this case, cyclization produces a six-membered ring fused to the
existing five-membered ring, which is energetically favorable.

\medskip
Step 4: Dehydration (condensation)

Under heating (\(\Delta\)), the initially formed \(\beta\)-hydroxy ketone
undergoes elimination of water to form an \(\alpha,\beta\)-unsaturated
ketone (enone).

\medskip
Step 5: Fate of the ester group

The ester group:

does not participate directly in the aldol reaction,
remains intact in the final product.


\medskip
Step 6: Identify the major product

The final product is a bicyclic enone system with:

a fused 5--6 ring system,
an \(\alpha,\beta\)-unsaturated ketone,
an intact ester substituent.


This structure corresponds to Option (A).

\medskip
Final Answer:
\[ \boxed{Option (A)} \] Quick Tip: Intramolecular Aldol condensation forms stable 5 or 6-membered rings. Look for the formation of an \(\alpha,\beta\)-unsaturated ketone (enone) in a new ring.

\medskip
The given reactant is o-cresol (2-methylphenol):
\[ o-cresol = 2-methylphenol \]

The reagents used are: \[ (i) aqueous NaOH, \quad (ii) CH_3I \]

\medskip
Step 1: Formation of phenoxide ion

Phenols are weakly acidic and readily react with strong bases such as NaOH.
The hydroxyl proton is removed, forming a phenoxide ion:
\[ Ar--OH + NaOH \rightarrow Ar--O^- Na^+ + H_2O \]

The phenoxide ion is a strong nucleophile.

\medskip
Step 2: Reaction with methyl iodide

Methyl iodide (\(CH_3I\)) is a primary alkyl halide and undergoes nucleophilic
substitution via the \(S_N2\) mechanism.

The phenoxide ion attacks the methyl carbon, displacing iodide ion:
\[ Ar--O^- + CH_3I \rightarrow Ar--O--CH_3 + I^- \]

This reaction is known as the Williamson ether synthesis.

\medskip
Step 3: Preference for O-alkylation

Under these reaction conditions:

The oxygen atom is more nucleophilic than the aromatic carbon.
\(S_N2\) substitution at carbon is not feasible on an aromatic ring.
C-alkylation requires much stronger conditions.


Therefore, O-alkylation is strongly favored over C-alkylation.

\medskip
Step 4: Identify the major product

The product formed is an ether where:

the phenolic oxygen is methylated,
the methyl substituent on the ring remains unchanged.


The final product is:
\[ \textbf{2-methylanisole (o-methylanisole)} \]

\medskip
Final Answer:
\[ \boxed{Option (D)} \] Quick Tip: Phenols react with NaOH and Alkyl Halides to form Ethers (O-alkylation).

\medskip
An aromatic compound A with molecular formula \(C_7H_6O_2\) is treated with
aqueous ammonia followed by heating to give compound B.
Compound B then reacts with \(Br_2/KOH\) to form compound C
with molecular formula \(C_6H_7N\).

\medskip
Step 1: Analyze the reaction \(B \rightarrow C\)

The reagents \(Br_2/KOH\) indicate the Hofmann bromamide degradation reaction,
which has the following characteristics:


It converts an amide into a primary amine.
The product amine has one carbon atom less than the parent amide.


Therefore, compound B must be an amide, and compound C
must be a primary amine.

\medskip
Step 2: Identify compound C

The molecular formula of C is: \[ C_6H_7N \]

This corresponds to aniline: \[ Aniline = C_6H_5NH_2 \]

\medskip
Step 3: Identify compound B

Since Hofmann degradation removes one carbon atom,
compound B must be the amide corresponding to aniline, i.e.,
\[ \textbf{B = Benzamide } (C_6H_5CONH_2) \]

\medskip
Step 4: Analyze the reaction \(A \rightarrow B\)

Compound A reacts with aqueous ammonia on heating to form benzamide.
This is a characteristic reaction of carboxylic acids:
\[ C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \]
\[ C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O \]

\medskip
Step 5: Identify compound A

The molecular formula of A is: \[ C_7H_6O_2 \]

The aromatic carboxylic acid with this formula is benzoic acid:
\[ Benzoic acid = C_6H_5COOH \]

\medskip
Final Answer:
\[ \boxed{Benzoic acid} \]


Hence, the correct option is (A). Quick Tip: Hofmann Bromamide reaction converts Amide (\(RCONH_2\)) to Amine (\(RNH_2\)). \(C_7H_6O_2\) is a classic formula for Benzoic Acid.

\medskip
The given transformation is:
\[ CH_3--CH=CH--CH_2--CH(OH)--CH_3 \;\longrightarrow\; CH_3--CH=CH--CH_2--CO_2H \]

\medskip
Step 1: Identify the functional group change

The starting compound contains:

a secondary alcohol of the type \(-CH(OH)CH_3\),
an alkene (\(C=C\)) which remains unchanged in the product.


The product contains:

a carboxylic acid group (\(-CO_2H\)),
one carbon atom less than the original alcohol side chain.


\medskip
This indicates a reaction involving:

oxidation of a secondary alcohol to a methyl ketone,
followed by cleavage of the \(-COCH_3\) group with loss of the methyl carbon.


\medskip
Step 2: Identify the suitable reaction

This transformation is characteristic of the haloform reaction,
which occurs with compounds containing the \(-COCH_3\) or \(-CH(OH)CH_3\) group.

\medskip
Step 3: Role of \(I_2/NaOH\)

The reagent \(I_2/NaOH\) performs the following functions:


It oxidizes the secondary alcohol \(-CH(OH)CH_3\) to a methyl ketone (\(-COCH_3\)).
The methyl ketone undergoes the haloform reaction to give:
\[ RCOO^- + CHI_3 \]
Acidification of the carboxylate ion yields the carboxylic acid (\(RCOOH\)).


Thus, the overall reaction converts \(-CH(OH)CH_3\) into \(-CO_2H\) with loss of one carbon atom.

\medskip
Step 4: Why other options are incorrect


Tollen's reagent: Oxidizes aldehydes, not secondary alcohols.
Alkaline \(KMnO_4\): Would oxidatively cleave the \(C=C\) double bond.
\(CrO_2Cl_2/CS_2\): Oxidizes methyl groups on aromatic rings (Etard reaction).


\medskip
Final Answer:
\[ \boxed{I_2/NaOH} \]


Hence, the correct option is (B). Quick Tip: Haloform reaction (\(X_2/NaOH\)) degrades methyl ketones (or methyl carbinols) to carboxylic acids with one less carbon. It usually preserves isolated double bonds.

\medskip
The given reaction occurs in two stages and finally leads to the formation of a
polyester. Let us analyze each step carefully.

\medskip
Step 1: Identification of the functional groups

The starting compound shown in the reaction is an aryl di-ester / protected diol
derivative. Under acidic hydrolysis conditions, such compounds behave as
masked dihydric phenols.

The reagents used are: \[ dil. HCl / \Delta \]

These conditions cause hydrolysis of ester groups, regenerating phenolic \(-OH\) groups.

\medskip
Step 2: Formation of dihydric phenol

On hydrolysis, the compound is converted into hydroquinone
(\(p\)-dihydroxybenzene):
\[ HO--C_6H_4--OH \]

Thus, the intermediate formed is a diol.

\medskip
Step 3: Reaction with oxalic acid

The next reagent is oxalic acid:
\[ HOOC--COOH \]

Oxalic acid is a dicarboxylic acid. When a diol reacts with a diacid,
condensation polymerisation occurs, resulting in the formation of a
polyester.

\medskip
Step 4: Polymerisation mechanism

Each hydroxyl group of hydroquinone reacts with a carboxyl group of oxalic acid,
with elimination of water:
\[ --OH + --COOH \rightarrow --O--CO-- + H_2O \]

Since both monomers are bifunctional, the reaction continues repeatedly,
forming a long-chain polymer.

\medskip
Step 5: Structure of the repeating unit

The repeating unit of the polymer formed is:
\[ \left[ -O - C_6H_4 - O - C(=O) - C(=O) - \right]_n \]

This polymer can be described as poly(hydroquinone oxalate).

\medskip
Step 6: Identification of the correct option

Among the given options, Option (A) correctly represents:

a polyester structure,
alternating aromatic diol and oxalate units,
correct connectivity and repeating pattern.


\medskip
Final Answer:
\[ \boxed{Option (A)} \] Quick Tip: Polyesters are formed by the condensation of Diols and Dicarboxylic acids. Look for the ester linkage \(-O-CO-\) in the polymer backbone.

\medskip
Amino acids can be identified using specific qualitative chemical tests
based on the presence of functional groups such as \(-NH_2\), \(-COOH\),
or aromatic rings.

Let us examine each given test:

\medskip
(A) Xanthoproteic test

This test detects the presence of aromatic amino acids
such as tyrosine, tryptophan, and phenylalanine.

It involves nitration of the aromatic ring by concentrated nitric acid,
producing a yellow-colored nitro compound.

Hence, this test can be used to identify aromatic amino acids.

\medskip
(B) Barfoed test

The Barfoed test is a qualitative test used to distinguish:

monosaccharides from disaccharides.


It is based on the reduction of copper(II) acetate in acidic medium.
This test is specific for carbohydrates and
has no relevance to amino acids.

\medskip
(C) Biuret test

The Biuret test detects the presence of peptide bonds (\(-CO-NH-\)).
While free amino acids generally give a negative result,
amino acids like histidine and compounds containing multiple peptide bonds
can show a positive response.

Thus, the Biuret test is used in protein and amino acid analysis.

\medskip
(D) Ninhydrin test

The Ninhydrin test is a general test for amino acids.
It reacts with the \(\alpha\)-amino group to produce a characteristic
purple (Ruhemann’s purple) color.

Therefore, it is widely used for identifying amino acids.

\medskip
Final Answer:
\[ \boxed{Barfoed test} \]


Hence, the correct option is (B). Quick Tip: Memorize the specific tests: Molisch/Barfoed/Benedict for Carbs; Biuret/Ninhydrin/Xanthoproteic for Proteins; Lucas/Victor Meyer for Alcohols.

(A) Lysine is an Amino Acid \(\rightarrow\) Reacts with (Q) Ninhydrin.

(B) Furfural is a carbohydrate derivative (aldehyde) \(\rightarrow\) Reacts with (P) 1-naphthol (Molisch test basis).

(C) Benzyl alcohol is an Alcohol \(\rightarrow\) Reacts with (S) Ceric Ammonium Nitrate (Red color).

(D) Styrene is an Alkene \(\rightarrow\) Reacts with (R) KMnO\(_4\) (Baeyer's test for unsaturation).

Match: A-Q, B-P, C-S, D-R.
Quick Tip: Match the functional group to the characteristic test. Alcohols \(\to\) CAN. Amino acids \(\to\) Ninhydrin. Unsaturation \(\to\) KMnO4.

The element with atomic number \(Z=71\) is Lutetium (Lu).


The electronic configuration of Xenon (\(Z=54\)) is [Kr] \(4d^{10} 5s^2 5p^6\).


After Xenon, the filling order follows the Aufbau principle but with specific observed configurations for Lanthanides.


The filling sequence is \(6s \to 4f \to 5d\).


Electrons 55 and 56 fill the \(6s\) orbital: \([Xe] 6s^2\).


The next 14 electrons (57 to 70) typically fill the \(4f\) orbitals (Note: Lanthanum \(Z=57\) is \([Xe] 5d^1 6s^2\), but usually \(4f\) fills from Ce to Yb).


Ytterbium (\(Z=70\)) has the configuration \([Xe] 4f^{14} 6s^2\).


The 71st electron must enter the next available low-energy orbital, which is the \(5d\) orbital.


Thus, the configuration of Lutetium (\(Z=71\)) is \([Xe] 4f^{14} 5d^1 6s^2\).


The last electron (71st) enters the \(5d\) orbital.
Quick Tip: Lutetium (\(Z=71\)) is the last element of the Lanthanide series. Since the \(4f\) subshell is full (\(f^{14}\)), the next electron goes into the \(5d\) subshell.

In electroplating, it is essential to have a smooth and uniform deposition of the metal.


This is achieved by maintaining a low concentration of free metal ions in the solution, which promotes even crystal growth.


Complex ions are used to keep the free ion concentration low through equilibrium dissociation.


For silver plating, sodium or potassium argentocyanide \(K[Ag(CN)_2]\) is used. The complex ion is \([Ag(CN)_2]^-\).


For gold plating, potassium aurocyanide \(K[Au(CN)_2]\) is used. The complex ion is \([Au(CN)_2]^-\).


Therefore, the electrolytes contain \([Au(CN)_2]^-\) and \([Ag(CN)_2]^-\) respectively.
Quick Tip: Cyanide complexes (\([M(CN)_2]^-\)) are the standard electrolytes for noble metal plating because their high stability constants ensure a controlled supply of metal ions.

The reactivity of halogens with hydrogen decreases down the group: \(F_2 > Cl_2 > Br_2 > I_2\).


(A) Fluorine reacts violently with hydrogen even in the dark.


(B) Chlorine reacts with hydrogen in the presence of sunlight (UV light).


(C) Bromine reacts with hydrogen upon heating.


(D) Iodine reacts with hydrogen very slowly, and the reaction is reversible. To proceed at a reasonable rate and yield, a catalyst (like Platinum) and heating are required.
Quick Tip: Reactivity of halogens decreases as bond dissociation enthalpy decreases (except \(F_2\)) and atomic size increases. \(I_2\) is the least reactive with \(H_2\).

Alkali metals dissolve in liquid ammonia to give deep blue solutions.


The dissolution process involves the ionization of the metal atom: \(Na \to Na^+ + e^-\).


Both the cation and the electron become solvated by ammonia molecules.

\(Na + (x+y)NH_3 \to [Na(NH_3)_x]^+ + [e(NH_3)_y]^-\).


The deep blue color is attributed to the ammoniated electrons (\([e(NH_3)_y]^-\)), which absorb energy in the visible region of the spectrum for electronic excitation.
Quick Tip: The blue color of alkali metals in liquid ammonia is paramagnetic and conducting. At higher concentrations, it turns bronze and diamagnetic.

We analyze the structures of the phosphorus oxoacids:


1. \(H_3PO_2\) (Hypophosphorous acid): The phosphorus is bonded to one Oxygen (double bond), two Hydrogens (single bonds), and one OH group. Thus, it contains **2 P-H bonds**.


2. \(H_3PO_3\) (Orthophosphorous acid): The phosphorus is bonded to one Oxygen (double bond), one Hydrogen (single bond), and two OH groups. Thus, it contains **1 P-H bond**.


3. \(H_4P_2O_5\) (Pyrophosphorous acid): Its structure is \((HO)(H)P(O)-O-P(O)(H)(OH)\). Each phosphorus atom is bonded to one H atom. Since there are two P atoms, the molecule contains a total of **2 P-H bonds**.


4. \(H_4P_2O_6\) (Hypophosphoric acid): Structure contains a P-P bond and no P-H bonds.


The question asks for the pair where *each* acid has two P-H bonds.

\(H_3PO_2\) has 2 P-H bonds. \(H_4P_2O_5\) has 2 P-H bonds.
Quick Tip: The basicity of phosphorus oxoacids corresponds to the number of P-OH bonds. The H atoms attached directly to P (P-H bonds) are reducing in nature and not acidic.

The structure of Diborane (\(B_2H_6\)) consists of two Boron atoms and six Hydrogen atoms.


There are four terminal hydrogen atoms. Each forms a normal covalent bond with a Boron atom. These are 2-centre-2-electron (2c-2e) bonds. So, there are **4** such bonds.


There are two bridging hydrogen atoms. Each bridging hydrogen is bonded to both Boron atoms simultaneously, forming a \(B-H-B\) bridge. These are 3-centre-2-electron (3c-2e) bonds (also known as banana bonds). So, there are **2** such bonds.


Thus, the number of 2c-2e bonds is 4, and 3c-2e bonds is 2.
Quick Tip: In \(B_2H_6\), terminal B-H bonds are normal covalent bonds. The bridge bonds involve 3 atoms (B-H-B) sharing 2 electrons to overcome boron's electron deficiency.

The reaction involves the oxidation of oxalate ion (\(C_2O_4^{2-}\)) by permanganate (\(MnO_4^-\)).


The oxidation half-reaction is:
\(C_2O_4^{2-} \to 2 CO_2 + 2e^-\)


From the stoichiometry of this half-reaction:

2 molecules of \(CO_2\) are produced by the loss of 2 electrons.


Therefore, for the production of **one** molecule of \(CO_2\), the number of electrons involved is:
\(\frac{2 electrons}{2 molecules} = 1\) electron.
Quick Tip: Always write the balanced half-reaction to determine electron transfer. The question asks per molecule of product, not per mole of reactant.

Let's analyze the electronic configuration and unpaired electrons (\(n\)) for each ion in octahedral fields.


1. \(Ni^{2+} (d^8)\):
High Spin: \(t_{2g}^6 e_g^2 \implies n = 2\).
Low Spin: \(t_{2g}^6 e_g^2 \implies n = 2\).
Difference = 0.


2. \(Mn^{2+} (d^5)\):
High Spin: \(t_{2g}^3 e_g^2 \implies n = 5\).
Low Spin: \(t_{2g}^5 e_g^0 \implies n = 1\).
Difference = 4.


3. \(Co^{2+} (d^7)\):
High Spin: \(t_{2g}^5 e_g^2 \implies n = 3\).
Low Spin: \(t_{2g}^6 e_g^1 \implies n = 1\).
Difference = \(3 - 1 = 2\).


4. \(Fe^{2+} (d^6)\):
High Spin: \(t_{2g}^4 e_g^2 \implies n = 4\).
Low Spin: \(t_{2g}^6 e_g^0 \implies n = 0\).
Difference = 4.


The difference is 2 for \(Co^{2+}\).
Quick Tip: High spin complexes follow Hund's rule (maximum unpaired e-). Low spin complexes pair up electrons in \(t_{2g}\) before filling \(e_g\).

The reactants are Cobalt(III) chloride and ethylenediamine (en) in 1:2 ratio. The formula of the complex is \([Co(en)_2Cl_2]Cl\).


This complex exhibits geometrical isomerism: cis and trans forms.


Isomer A (Violet): Optically active. The cis-isomer of \([M(AA)_2X_2]\) type lacks a plane of symmetry and is optically active (exists as enantiomers).


Isomer B (Green): Optically inactive. The trans-isomer of \([M(AA)_2X_2]\) type has a plane of symmetry passing through the metal and the two monodentate ligands (or perpendicular to the principal axis), making it optically inactive (meso).


Since A and B are cis and trans forms of the same complex, they are **Geometrical isomers**.
Quick Tip: For octahedral complexes \([M(AA)_2X_2]\): Cis is optically active. Trans is optically inactive due to symmetry.

Ozone depletion is primarily driven by radical chain reactions involving Chlorine and Bromine.


(D) Photolysis of CFCs releases Cl radicals: \(CF_2Cl_2 \xrightarrow{UV} \dot{Cl} + \dot{C}F_2Cl\). (Initiation)


(B) Cl radical destroys Ozone: \(\dot{Cl} + O_3 \to Cl\dot{O} + O_2\). Then \(Cl\dot{O} + O \to \dot{Cl} + O_2\). This cycles Cl radicals.


(A) Hydrolysis of chlorine nitrate or photolysis of HOCl releases radicals: \(HOCl \xrightarrow{h\nu} \dot{OH} + \dot{Cl}\). This contributes to the active radical pool.


(C) The reaction \(CH_4 + 2O_3 \to 3CH_2O + 3H_2O\) is not a standard atmospheric reaction in the context of ozone depletion. While methane interacts with Cl radicals (\(\dot{Cl} + CH_4 \to \dot{C}H_3 + HCl\)) acting as a sink, the direct stoichiometric reaction with ozone to form formaldehyde as written in option C is not part of the depletion mechanism.
Quick Tip: Ozone depletion involves free radical mechanisms. Look for reactions generating or propagating \(\dot{Cl}\) or \(\dot{Br}\) radicals. Stable molecule reactions are usually sinks or irrelevant.

Molarity (\(M\)) is defined as moles of solute per liter of solution.
\(M = \frac{n}{V(L)} \implies n = M \times V\).


Given \(M = 0.1\) mol/L and \(V = 2\) L.

Moles of sugar \(n = 0.1 \times 2 = 0.2\) mol.


The molar mass of sugar (\(C_{12}H_{22}O_{11}\)) is:
\(12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342\) g/mol.


Mass required = Moles \(\times\) Molar Mass.

Mass = \(0.2 \times 342 = 68.4\) g.
Quick Tip: \(Mass = Molarity \times Volume(L) \times Molar Mass\). Remember the molar mass of Sucrose is 342 g/mol.

\medskip
The compound has the formula \(\mathrm{A_2B_3}\) and crystallizes in a
hexagonal close-packed (hcp) lattice.

\medskip
Step 1: Void statistics in hcp lattice

If the number of atoms forming the hcp lattice is \(N\), then:

Number of octahedral voids \(= N\)
Number of tetrahedral voids \(= 2N\)


\medskip
Step 2: Test the given correct option

According to Option (D):

Atom \(\mathbf{B}\) forms the hcp lattice
Atom \(\mathbf{A}\) occupies \(\frac{1}{3}\) of the tetrahedral voids


\medskip
Step 3: Count number of atoms

Let the number of \(B\) atoms forming the hcp lattice be: \[ Number of B atoms = N \]

Total number of tetrahedral voids: \[ = 2N \]

Number of \(A\) atoms occupying tetrahedral voids: \[ = \frac{1}{3} \times 2N = \frac{2}{3}N \]

\medskip
Step 4: Determine the ratio of atoms
\[ A : B = \frac{2}{3}N : N \]

Dividing both terms by \(\frac{1}{3}N\):
\[ A : B = 2 : 3 \]

\medskip
Step 5: Match with given formula

The calculated ratio \(A : B = 2 : 3\) matches the given chemical formula: \[ \mathrm{A_2B_3} \]

\medskip
Final Answer:
\[ \boxed{hcp lattice – B, \; \frac{1}{3} tetrahedral voids occupied by A} \]


Hence, the correct option is (D). Quick Tip: If lattice has \(N\) atoms, there are \(N\) Octahedral voids and \(2N\) Tetrahedral voids. Use the stoichiometry of the formula to find the occupied fraction.

The energy of an electron in a hydrogen-like species is given by \(E_n = -13.6 \frac{Z^2}{n^2}\) eV.


For Helium ion (\(He^+\)), atomic number \(Z = 2\).


The question asks for the "second excited state".

Ground state: \(n = 1\).

First excited state: \(n = 2\).

Second excited state: \(n = 3\).


Substitute \(Z=2\) and \(n=3\):
\(E_3 = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9}\).

\(E_3 = -13.6 \times 0.444... \approx -6.04\) eV.
Quick Tip: "Second excited state" means \(n=3\), not \(n=2\). Always count up from ground state \(n=1\).

For an isothermal process of an ideal gas, \(\Delta U = 0\).

According to the First Law of Thermodynamics: \(\Delta U = q + w \implies q = -w\).


Work done in irreversible compression against constant pressure:
\(w = -P_{ext} (V_f - V_i) = -4 (1 - 5) = -4(-4) = +16\) J.


Heat \(q = -16\) J. The negative sign indicates heat is released by the gas.

Heat absorbed by Aluminum \(Q_{absorbed} = 16\) J.


Using the heat capacity equation for Al: \(Q = n C_m \Delta T\).
\(16 = 1 \times 24 \times \Delta T\).
\(\Delta T = \frac{16}{24} = \frac{2}{3}\) K.
Quick Tip: For constant external pressure, Work \(W = -P_{ext}\Delta V\). Isothermal ideal gas means internal energy is constant, so Heat = -Work.

Entropy (\(\Delta S\)) decreases when the system becomes more ordered (fewer moles of gas).


(A) \(CO_2(s) \to CO_2(g)\): Solid to gas. \(\Delta S > 0\).


(B) \(I_2(s) \to I_2(aq)\): Solid to solution. Disorder increases. \(\Delta S > 0\).


(C) \(CaSO_4(s) \to CaO(s) + SO_3(g)\): Solid produces gas. \(\Delta S > 0\).


(D) \(N_2(g) + 3H_2(g) \to 2NH_3(g)\).


Moles of gaseous reactants = \(1 + 3 = 4\).


Moles of gaseous products = 2.

\(\Delta n_g = 2 - 4 = -2\). The number of gas molecules decreases, so disorder decreases. \(\Delta S < 0\).
Quick Tip: Look at the change in the number of gaseous moles (\(\Delta n_g\)). If \(\Delta n_g\) is negative, entropy change is generally negative.

For boiling point elevation: \(\Delta T_b = K_b \cdot m_1\).

Given \(\Delta T_b = 2\) K and \(m_1 = 1\) molal.
\(2 = K_b \times 1 \implies K_b = 2\) K kg/mol.


For freezing point depression: \(\Delta T_f = K_f \cdot m_2\).

Given \(\Delta T_f = 2\) K and \(m_2 = 2\) molal.
\(2 = K_f \times 2 \implies K_f = 1\) K kg/mol.


Comparing \(K_b\) and \(K_f\):
\(K_b = 2\) and \(K_f = 1\).

Therefore, \(K_b = 2 K_f\).
Quick Tip: Use the colligative property formulas \(\Delta T = K \cdot m \cdot i\). Since solute is glucose, Van't Hoff factor \(i=1\).

Molar mass of \(NH_4SH = 14 + 4 + 32 + 1 = 51\) g/mol.

Initial moles of \(NH_4SH = \frac{5.1}{51} = 0.1\) mol.


Reaction: \(NH_4SH(s) \rightleftharpoons NH_3(g) + H_2S(g)\).

Degree of dissociation is 30% (\(\alpha = 0.3\)).

Moles decomposed = \(0.1 \times 0.3 = 0.03\) mol.


Moles of \(NH_3\) formed = 0.03 mol.

Moles of \(H_2S\) formed = 0.03 mol.


Temperature \(T = 327 + 273 = 600\) K. Volume \(V = 3\) L.

Partial Pressure \(P = \frac{nRT}{V}\).

\(P_{NH_3} = \frac{0.03 \times 0.082 \times 600}{3} = 0.01 \times 49.2 = 0.492\) atm.
\(P_{H_2S} = P_{NH_3} = 0.492\) atm.

\(K_p = P_{NH_3} \times P_{H_2S} = (0.492) \times (0.492) \approx 0.242 atm^2\).
Quick Tip: Solid reactants do not appear in the \(K_p\) expression, but their decomposition determines the moles of gaseous products.

The cell reaction is \(H_2(g) + 2AgCl(s) \to 2H^+(aq) + 2Ag(s) + 2Cl^-(aq)\).

Here, \(n=2\) electrons.


Nernst Equation: \(E_{cell} = E^0_{cell} - \frac{0.06}{n} \log Q\).
\(Q = [H^+]^2 [Cl^-]^2 / P_{H_2}\).


Given concentration \(10^{-6}\) m. Assuming complete dissociation, \([H^+] = 10^{-6}\) and \([Cl^-] = 10^{-6}\).
\(Q = (10^{-6})^2 (10^{-6})^2 = 10^{-24}\).

\(E_{cell} = E^0_{cell} - \frac{0.06}{2} \log(10^{-24})\).
\(0.92 = E^0_{cell} - 0.03 (-24)\).
\(0.92 = E^0_{cell} + 0.72\).
\(E^0_{cell} = 0.92 - 0.72 = 0.20\) V.


Since anode is Standard Hydrogen Electrode (\(E^0 = 0\)), \(E^0_{cell} = E^0_{cathode} - 0\).

Thus, \(E^0_{(AgCl/Ag,Cl^-)} = 0.20\) V.
Quick Tip: Pay attention to the stoichiometry in the Nernst equation quotient Q. For HCl, both \(H^+\) and \(Cl^-\) concentrations affect Q.

The reaction is \(A_2 \rightleftharpoons 2A\).

Forward rate \(r_f = k_1 [A_2]\). Backward rate \(r_b = k_{-1} [A]^2\).


The rate of the reaction is \(Rate = \frac{1}{2} \frac{d[A]}{dt} = - \frac{d[A_2]}{dt}\).


Net Rate \(= r_f - r_b = k_1 [A_2] - k_{-1} [A]^2\).


Therefore, \(\frac{1}{2} \frac{d[A]}{dt} = k_1 [A_2] - k_{-1} [A]^2\).


Multiplying by 2:
\(\frac{d[A]}{dt} = 2k_1 [A_2] - 2k_{-1} [A]^2\).
Quick Tip: The rate of formation of a species is its stoichiometric coefficient multiplied by the rate of the reaction step.

Haemoglobin is a globular protein. In colloidal solutions, proteins like haemoglobin in acidic medium (or at physiological pH generally considered in this context) act as positively charged sols due to the protonation of amino groups on the surface. (Standard classification in chemistry curriculum: Haemoglobin \(\to\) Positive sol).


Gold sol is a typical metal sol prepared by reduction. The colloidal gold particles adsorb anions (like \(OH^-\) or reduction by-products) preferentially, making them negatively charged. (Standard classification: Metal sols \(\to\) Negative sol).


Thus, Haemoglobin is positive and Gold sol is negative.
Quick Tip: Standard Examples: Positive sols - Hydrated oxides (\(Al_2O_3 \cdot xH_2O\)), Haemoglobin, Basic dyes. Negative sols - Metals (Au, Ag), Sulphides (\(As_2S_3\)), Acid dyes.

\medskip
Let \(N\) denote the set of natural numbers.
Two functions \(f, g : N \to N\) are defined as follows:
\[ f(n) = \begin{cases} \dfrac{n+1}{2}, & if n is odd
[6pt] \dfrac{n}{2}, & if n is even \end{cases} \qquad g(n) = n - (-1)^n \]

We analyze the nature of the composite function \(f \circ g\).

\medskip
Step 1: Study the function \(g\)


If \(n\) is even, let \(n = 2k\):
\[ g(2k) = 2k - (-1)^{2k} = 2k - 1 \quad (odd) \]

If \(n\) is odd, let \(n = 2k - 1\):
\[ g(2k - 1) = (2k - 1) - (-1) = 2k \quad (even) \]


Thus, \(g\) maps even numbers to odd numbers and odd numbers to even numbers.
Every natural number is obtained exactly once, so:
\[ \textbf{\(g\) is one-one and onto (bijective).} \]

\medskip
Step 2: Study the function \(f\)

Evaluate \(f\) for some values:
\[ f(1) = \frac{1+1}{2} = 1, \qquad f(2) = \frac{2}{2} = 1 \]

Since: \[ f(1) = f(2) \]

the function \(f\) is not one-one.

\medskip
To check surjectivity, let \(k \in N\).
Choose \(n = 2k\). Then:
\[ f(2k) = \frac{2k}{2} = k \]

Hence, every natural number has a preimage, so:
\[ \textbf{\(f\) is onto.} \]

\medskip
Step 3: Analyze the composite function \(f \circ g\)

Compute values for two distinct inputs:
\[ g(1) = 1 - (-1) = 2 \quad \Rightarrow \quad f(g(1)) = f(2) = 1 \]
\[ g(2) = 2 - 1 = 1 \quad \Rightarrow \quad f(g(2)) = f(1) = 1 \]

Thus: \[ f(g(1)) = f(g(2)) \]

with \(1 \neq 2\), so:
\[ \textbf{\(f \circ g\) is not one-one.} \]

\medskip
Step 4: Check surjectivity of \(f \circ g\)

Since \(g\) is bijective, it maps \(N\) onto \(N\).
Therefore, the range of \(f \circ g\) is the same as the range of \(f\).

As \(f\) is onto, it follows that:
\[ \textbf{\(f \circ g\) is onto.} \]

\medskip
Final Conclusion:
\[ \boxed{\(f \circ g\) is onto but not one-one} \]


Hence, the correct option is (C). Quick Tip: To check if a composite function \(h(x) = f(g(x))\) is one-one, calculate \(h(x)\) for small values like \(x=1, 2, 3\). If \(h(a) = h(b)\) for \(a \neq b\), it's not one-one.

\medskip
The given quadratic equation is: \[ x^2 + (3-\lambda)x + 2 = \lambda \]

Rewriting it in standard quadratic form: \[ x^2 + (3-\lambda)x + (2-\lambda) = 0 \]

\medskip
Step 1: Let the roots be

Let the roots of the quadratic equation be \(\alpha\) and \(\beta\).

From the theory of equations:
\[ \alpha + \beta = -(3-\lambda) = \lambda - 3 \]
\[ \alpha \beta = 2 - \lambda \]

\medskip
Step 2: Expression for sum of squares of the roots

We are required to find the value of \(\lambda\) for which the sum of the squares of the roots, \[ S = \alpha^2 + \beta^2 \]
is minimum.

Using the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Substituting the values obtained above:
\[ S = (\lambda - 3)^2 - 2(2 - \lambda) \]

\medskip
Step 3: Simplify the expression
\[ S = (\lambda^2 - 6\lambda + 9) - 4 + 2\lambda \]
\[ S = \lambda^2 - 4\lambda + 5 \]

\medskip
Step 4: Minimize the expression

Method 1: Differentiation

Differentiate \(S\) with respect to \(\lambda\):
\[ \frac{dS}{d\lambda} = 2\lambda - 4 \]

For minimum value: \[ 2\lambda - 4 = 0 \quad \Rightarrow \quad \lambda = 2 \]

\medskip
Method 2: Completing the square
\[ S = \lambda^2 - 4\lambda + 5 \]
\[ S = (\lambda - 2)^2 + 1 \]

The minimum value of \(S\) occurs when: \[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]

\medskip
Final Answer:
\[ \boxed{\lambda = 2} \]


Hence, the correct option is (B). Quick Tip: For a quadratic expression \(ax^2 + bx + c\) with \(a>0\), the minimum value always occurs at \(x = -b/2a\).

\medskip
The given complex number is: \[ z = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^5 \]

\medskip
Step 1: Convert each term into polar (exponential) form

Let: \[ \alpha = \frac{\sqrt{3}}{2} + \frac{i}{2} \]

We observe that: \[ |\alpha| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1 \]

The argument of \(\alpha\) is: \[ \theta = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

Hence, \[ \alpha = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = e^{i\pi/6} \]

Similarly, let: \[ \beta = \frac{\sqrt{3}}{2} - \frac{i}{2} \]

Then: \[ \beta = \cos\frac{\pi}{6} - i\sin\frac{\pi}{6} = e^{-i\pi/6} \]

\medskip
Step 2: Compute powers using De Moivre's theorem
\[ \alpha^5 = (e^{i\pi/6})^5 = e^{i5\pi/6} \]
\[ \beta^5 = (e^{-i\pi/6})^5 = e^{-i5\pi/6} \]

\medskip
Step 3: Add the two terms
\[ z = e^{i5\pi/6} + e^{-i5\pi/6} \]

Using Euler’s identity: \[ e^{i\theta} + e^{-i\theta} = 2\cos\theta \]
\[ z = 2\cos\left(\frac{5\pi}{6}\right) \]

\medskip
Step 4: Evaluate the cosine
\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]
\[ z = 2 \times \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \]

\medskip
Step 5: Identify real and imaginary parts
\[ z = -\sqrt{3} + 0i \]
\[ R(z) = -\sqrt{3}, \qquad I(z) = 0 \]

\medskip
Final Answer:
\[ \boxed{I(z) = 0} \]


Hence, the correct option is (D). Quick Tip: Converting complex numbers to polar form \(e^{i\theta}\) makes computing powers much easier using De Moivre's Theorem.

\medskip
The given matrix is \[ A = \begin{bmatrix} 2 & b & 1
b & b^2+1 & b
1 & b & 2 \end{bmatrix}, \quad b>0 \]

\medskip
Step 1: Evaluate the determinant

Expanding along the first row: \[ \det(A) = 2\begin{vmatrix} b^2+1 & b
b & 2 \end{vmatrix} - b\begin{vmatrix} b & b
1 & 2 \end{vmatrix} + 1\begin{vmatrix} b & b^2+1
1 & b \end{vmatrix} \]
\[ = 2[(b^2+1)2 - b^2] - b(2b-b) + (b^2 - b^2 -1) \]
\[ = 2(b^2+2) - b^2 - 1 = b^2 + 3 \]

\medskip
Step 2: Minimize the given expression
\[ \frac{\det(A)}{b} = \frac{b^2+3}{b} = b + \frac{3}{b} \]

Since \(b>0\), apply AM–GM inequality: \[ b + \frac{3}{b} \ge 2\sqrt{3} \]

\medskip
Final Answer: \[ \boxed{2\sqrt{3}} \] Quick Tip: For \(x > 0\), the function \(x + \frac{a}{x}\) has a minimum value of \(2\sqrt{a}\) at \(x = \sqrt{a}\).

\medskip
For a homogeneous system to have a non-trivial solution, the determinant
of the coefficient matrix must be zero.
\[ D = \begin{vmatrix} 1 & 3 & 7
-1 & 4 & 7
\sin3\theta & \cos2\theta & 2 \end{vmatrix} \]

Apply \(R_2 \to R_2 + R_1\):
\[ D = \begin{vmatrix} 1 & 3 & 7
0 & 7 & 14
\sin3\theta & \cos2\theta & 2 \end{vmatrix} \]

Factor out 7 from \(R_2\):
\[ D = 7\begin{vmatrix} 1 & 3 & 7
0 & 1 & 2
\sin3\theta & \cos2\theta & 2 \end{vmatrix} \]

Expanding along \(R_2\): \[ 2 - 7\sin3\theta - 2\cos2\theta + 6\sin3\theta = 0 \]
\[ 2 - \sin3\theta - 2\cos2\theta = 0 \]

\medskip
Use identities: \[ \sin3\theta = 3\sin\theta - 4\sin^3\theta,\quad \cos2\theta = 1 - 2\sin^2\theta \]
\[ 4\sin^3\theta + 4\sin^2\theta - 3\sin\theta = 0 \]
\[ \sin\theta(4\sin^2\theta + 4\sin\theta - 3)=0 \]
\[ \sin\theta=\frac12 \Rightarrow \theta=\frac{\pi}{6},\frac{5\pi}{6} \]

\medskip
Final Answer: \[ \boxed{2} \] Quick Tip: For non-trivial solutions of \(AX=0\), set \(|A|=0\). Always factorize trigonometric polynomials carefully to find roots in the given domain.

\medskip
The given sum is \[ \sum_{r=0}^{25} \binom{50}{r}\binom{50}{25-r} \]

By Vandermonde’s identity: \[ \sum_{r=0}^{k} \binom{n}{r}\binom{m}{k-r} = \binom{n+m}{k} \]
\[ \Rightarrow \sum_{r=0}^{25} \binom{50}{r}\binom{50}{25-r} = \binom{100}{25} \]

Given: \[ \binom{100}{25} = K\binom{50}{25} \]
\[ K = \frac{\binom{100}{25}}{\binom{50}{25}} = 2^{25} \]

\medskip
Final Answer: \[ \boxed{2^{25}} \] Quick Tip: The coefficient of \(x^k\) in \((1+x)^n (1+x)^m\) is \(\binom{n+m}{k}\). This is the Vandermonde's Identity.

\medskip
The expression is: \[ x^2\left(x^{1/2}+\lambda x^{-2}\right)^{10} \]

General term: \[ T_{r+1} = \binom{10}{r}\lambda^r x^{\frac{10-r}{2}-2r} \]

Including outer \(x^2\): \[ Power of x = 2+\frac{10-r}{2}-2r \]

For coefficient of \(x^2\): \[ \frac{10-r}{2}-2r=0 \Rightarrow r=2 \]

Coefficient: \[ \binom{10}{2}\lambda^2 = 720 \Rightarrow 45\lambda^2=720 \]
\[ \lambda^2=16 \Rightarrow \lambda=4 \]

\medskip
Final Answer: \[ \boxed{4} \] Quick Tip: Write the general term \(T_{r+1}\) completely, combine all powers of \(x\), and set the exponent equal to the required power to find \(r\).

\medskip
Let the G.P. be \(a, aR, aR^2,\dots\)
\[ \log a_n = \log a + (n-1)\log R \]
which forms an A.P.

Each determinant entry: \[ \log(a_n^r a_{n+1}^k)= r\log a_n + k\log a_{n+1} \]
\[ = (r+k)\log a_n + k d \]

Thus, columns differ by constant values.
After column operations, two columns become identical.

\medskip
Hence determinant is zero for all (r,k)\in\mathbb\({N}^2\).

\medskip
Final Answer: \[ \boxed{Infinitely many} \] Quick Tip: If terms of a matrix are in AP (or linear combinations of AP), the determinant is often zero. Use row/column subtraction to reveal linear dependence.

\medskip \[ \sum_{p=1}^n 2p = n(n+1) \Rightarrow \cot^{-1}(1+n(n+1)) \]
\[ = \tan^{-1}\!\left(\frac{1}{n(n+1)+1}\right) = \tan^{-1}(n+1)-\tan^{-1}(n) \]

Thus the series telescopes: \[ S=\tan^{-1}(20)-\tan^{-1}(1) \]
\[ \tan S = \frac{20-1}{1+20}=\frac{19}{21} \Rightarrow \cot S=\frac{21}{19} \]

\medskip
Final Answer: \[ \boxed{\frac{21}{19}} \] Quick Tip: For series involving inverse trigonometric functions, try to express the general term as \(\tan^{-1} x - \tan^{-1} y\) to form a telescoping sum.

\medskip
Given: \[ f'(x)+\frac{3}{4x}f(x)=7 \]

Integrating factor: \[ I.F.=x^{3/4} \]
\[ \Rightarrow f(x)x^{3/4}=4x^{7/4}+C \]
\[ f(x)=4x+Cx^{-3/4} \]

Now evaluate: \[ \lim_{x\to0^+}x f(1/x) \]
\[ = \lim_{t\to\infty}\frac{1}{t}(4t+Ct^{-3/4}) =4 \]

\medskip
Final Answer: \[ \boxed{4} \] Quick Tip: Solve the linear ODE \(y' + Py = Q\) using \(IF = e^{\int P dx}\). When finding limits at infinity, dominant terms (highest powers) dictate the behavior.

\medskip
The function is defined as \[ f(x) = \max\{-|x|,\,-\sqrt{1-x^2}\}, \qquad x \in (-1,1) \]

Let \[ y_1 = -|x|, \qquad y_2 = -\sqrt{1-x^2} \]

\medskip
Step 1: Find intersection points
\[ -|x| = -\sqrt{1-x^2} \Rightarrow |x| = \sqrt{1-x^2} \]
\[ x^2 = 1 - x^2 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}} \]

\medskip
Step 2: Determine which function dominates

At \(x=0\): \[ -|0| = 0,\quad -\sqrt{1}=-1 \Rightarrow f(x) = -|x| \]

Thus, \[ f(x)= \begin{cases} -|x|, & |x| \le \frac{1}{\sqrt{2}}
-\sqrt{1-x^2}, & |x| > \frac{1}{\sqrt{2}} \end{cases} \]

\medskip
Step 3: Points of non-differentiability


At \(x=0\), \( -|x| \) has a sharp corner
At \(x=\pm \frac{1}{\sqrt{2}}\), the definition of \(f\) changes


\medskip
Final Answer: \[ \boxed{3} \] Quick Tip: Functions defined by \(\max\{f, g\}\) are non-differentiable at points where the curves intersect (if slopes differ) and where the constituent functions themselves are non-differentiable.

\medskip
Given curve: \[ y = x e^{x^2} \]

At \(x=1\): \[ y = 1\cdot e = e \]
So the point lies on the curve.

\medskip
Step 1: Find the derivative
\[ \frac{dy}{dx} = e^{x^2} + x e^{x^2}(2x) = e^{x^2}(1+2x^2) \]

At \(x=1\): \[ m = e(1+2) = 3e \]

\medskip
Step 2: Equation of tangent
\[ y - e = 3e(x-1) \Rightarrow y = 3ex - 2e \]

\medskip
Step 3: Verify options

For \(x=\frac{4}{3}\): \[ y = 3e\left(\frac{4}{3}\right) - 2e = 2e \]

\medskip
Final Answer: \[ \boxed{\left(\frac{4}{3},\,2e\right)} \] Quick Tip: Find the equation of the line using point-slope form and substitute the coordinates of the options to check for validity.

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Curve: \[ y = x^{3/2} + 7 \]
Soldier at: \[ S\left(\frac12,7\right) \]

Let a point on the curve be \[ P(x, x^{3/2}+7) \]

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Step 1: Distance squared
\[ D^2 = \left(x-\frac12\right)^2 + x^3 \]

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Step 2: Minimize
\[ \frac{d}{dx}D^2 = 2\left(x-\frac12\right)+3x^2 \]
\[ 3x^2+2x-1=0 \Rightarrow x=\frac13 \]

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Step 3: Minimum distance
\[ D^2=\left(\frac13-\frac12\right)^2+\left(\frac13\right)^3 =\frac{7}{108} \]
\[ D=\frac{1}{6}\sqrt{\frac{7}{3}} \]

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Final Answer: \[ \boxed{\frac{1}{6}\sqrt{\frac{7}{3}}} \] Quick Tip: To minimize distance between a point and a curve, minimize the square of the distance function using derivatives.

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Given: \[ \int x^5 e^{-4x^3} dx \]

Let \(t=x^3\), then \(dt=3x^2dx\).
\[ \int x^5 e^{-4x^3} dx = \frac13\int t e^{-4t} dt \]

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Integration by parts
\[ u=t,\quad dv=e^{-4t}dt \]
\[ \int t e^{-4t}dt = -\frac{t}{4}e^{-4t}-\frac{1}{16}e^{-4t} \]

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Substitute back
\[ = -\frac{1}{48}e^{-4x^3}(4x^3+1) \]

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Final Answer: \[ \boxed{-4x^3-1} \] Quick Tip: Integration by parts \(\int P(x) e^{ax} dx\) where P is polynomial will result in a polynomial times exponential.

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Split interval \((-\pi/2,\pi/2)\) at integers \(-1,0,1\).

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Compute piecewise
\[ \int_{-\pi/2}^{-1}\frac{dx}{1} = \frac{\pi}{2}-1 \]
\[ \int_{-1}^{0}\frac{dx}{2} = \frac12 \]
\[ \int_{0}^{1}\frac{dx}{4} = \frac14 \]
\[ \int_{1}^{\pi/2}\frac{dx}{5} = \frac{\pi}{10}-\frac15 \]

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Sum
\[ \frac{3}{20}(4\pi-3) \]

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Final Answer: \[ \boxed{\frac{3}{20}(4\pi-3)} \] Quick Tip: For integrals involving Greatest Integer Function \([x]\), always split the limits at integers and where \([f(x)]\) changes integer values.

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Given: \[ \int_0^x f(t)dt = x^2 + \int_x^1 t^2f(t)dt \]

Differentiate w.r.t. \(x\):
\[ f(x)=2x-x^2f(x) \]
\[ f(x)(1+x^2)=2x \Rightarrow f(x)=\frac{2x}{1+x^2} \]

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Differentiate
\[ f'(x)=\frac{2(1-x^2)}{(1+x^2)^2} \]
\[ f'\left(\frac12\right)=\frac{24}{25} \]

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Final Answer: \[ \boxed{\frac{24}{25}} \] Quick Tip: Leibniz Rule: \(\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))b'(x) - f(a(x))a'(x)\).

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Given: \[ (x^2-y^2)dx+2xy\,dy=0 \]
\[ \frac{dy}{dx}=\frac{y^2-x^2}{2xy} \]

Homogeneous ⇒ put \(y=vx\).
\[ \ln(x^2+y^2)=\ln(Cx) \Rightarrow x^2+y^2-Cx=0 \]

Passing through \((1,1)\): \[ C=2 \]
\[ (x-1)^2+y^2=1 \]

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Final Answer: \[ \boxed{Circle with centre on x-axis} \] Quick Tip: Identify homogeneous differential equations \(dy/dx = f(y/x)\) and use substitution \(y=vx\).

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Let third vertex be \(C(h,k)\).

Altitude from \(A(0,2)\) passes through origin ⇒ BC horizontal ⇒ \(k=3\).

Altitude from \(B(4,3)\) passes through origin ⇒ slope \(=3/4\).

Slope of AC \(=-4/3\):
\[ \frac{3-2}{h}=-\frac43 \Rightarrow h=-\frac34 \]

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Final Answer: \[ \boxed{Second quadrant} \] Quick Tip: Orthocentre is the intersection of altitudes. The slope of altitude \(\times\) slope of opposite side = -1.

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Intersection of given lines gives vertex \(A(0,3)\).

Midpoint of diagonals is \((2,4)\).

Opposite vertex: \[ C(4,5) \]

Other vertex obtained by parallelism: \[ (3,6) \]

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Final Answer: \[ \boxed{(3,6)} \] Quick Tip: Diagonals of a parallelogram intersect at the midpoint. Finding the intersection of side lines gives vertices.

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Area of equilateral triangle inscribed in circle: \[ \frac{3\sqrt3}{4}R^2=27\sqrt3 \Rightarrow R=6 \]

Given circle: \[ x^2+y^2+10x+12y+c=0 \]
\[ R^2=25+36-c \Rightarrow 36=61-c \Rightarrow c=25 \]

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Final Answer: \[ \boxed{25} \] Quick Tip: Standard relation: Side of inscribed equilateral triangle \(a = \sqrt{3}r\).

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The given set is \[ S=\left\{(x,y)\in\mathbb{R}^2:\frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}, \quad r\neq \pm 1 \]

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Case 1: \(r>1\)

Then, \[ 1+r>0 \quad and \quad 1-r<0 \]

Hence, \[ \frac{y^2}{1+r}+\frac{x^2}{r-1}=1 \]

This is the standard equation of an ellipse: \[ \frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \]
where \[ a^2=1+r,\qquad b^2=r-1 \]

Since \(a^2>b^2\), the eccentricity is \[ e=\sqrt{1-\frac{b^2}{a^2}} =\sqrt{1-\frac{r-1}{r+1}} =\sqrt{\frac{2}{r+1}} \]

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Case 2: \(0
Then both denominators are positive and the equation represents a hyperbola.
The eccentricity does not match any given option.

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Final Answer: \[ \boxed{Option (A)} \] Quick Tip: Check the sign of the denominators. If both are positive, it's an ellipse (\(e = \sqrt{1 - smaller/larger}\)). If opposite signs, it's a hyperbola (\(e = \sqrt{1 + b^2/a^2}\)).

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The parabola is: \[ x^2=4y \]

The given line is: \[ x-\sqrt{2}y+4\sqrt{2}=0 \quad\Rightarrow\quad y=\frac{x+4\sqrt{2}}{\sqrt{2}} \]

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Step 1: Find points of intersection

Substitute in the parabola: \[ x^2=4\left(\frac{x+4\sqrt{2}}{\sqrt{2}}\right) =2\sqrt{2}x+16 \]
\[ x^2-2\sqrt{2}x-16=0 \]

Let the roots be \(x_1,x_2\).
\[ x_1+x_2=2\sqrt{2},\qquad x_1x_2=-16 \]

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Step 2: Distance between roots
\[ |x_1-x_2| =\sqrt{(x_1+x_2)^2-4x_1x_2} =\sqrt{8+64} =6\sqrt{2} \]

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Step 3: Length of chord

Slope of line: \[ m=\frac{1}{\sqrt{2}} \]
\[ L=|x_1-x_2|\sqrt{1+m^2} =6\sqrt{2}\sqrt{1+\frac12} =6\sqrt{3} \]

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Final Answer: \[ \boxed{6\sqrt{3}} \] Quick Tip: Length of chord intercepted by a line \(y=mx+c\) on a curve is \(|x_1 - x_2|\sqrt{1+m^2}\), where \(x_1, x_2\) are roots of the intersection quadratic.

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Given points: \[ A(-3,-3,4),\quad B(3,7,6) \]

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Step 1: Midpoint
\[ M=\left(\frac{-3+3}{2},\frac{-3+7}{2},\frac{4+6}{2}\right)=(0,2,5) \]

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Step 2: Normal vector
\[ \vec{AB}=(6,10,2) \]

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Step 3: Equation of plane
\[ 6(x-0)+10(y-2)+2(z-5)=0 \]
\[ 6x+10y+2z-30=0 \quad\Rightarrow\quad 3x+5y+z=15 \]

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Step 4: Verify options

For \((4,1,-2)\): \[ 3(4)+5(1)-2=15 \]

Hence the point lies on the plane.

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Final Answer: \[ \boxed{(4,1,-2)} \] Quick Tip: The perpendicular bisector plane of segment AB passes through the midpoint and has normal vector \(\vec{AB}\).

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Given line: \[ \frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda \]
\[ x=2\lambda+4,\quad y=2\lambda+5,\quad z=\lambda+3 \]

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Step 1: Substitute in plane
\[ x+y+z=2 \]
\[ (2\lambda+4)+(2\lambda+5)+(\lambda+3)=2 \]
\[ 5\lambda+12=2 \Rightarrow \lambda=-2 \]

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Step 2: Intersection point
\[ P=(0,1,1) \]

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Step 3: Check options

Option (A): \[ \frac{0-1}{1}=\frac{1-3}{2}=\frac{1+4}{-5}=-1 \]

Hence the point lies on option (A).

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Final Answer: \[ \boxed{Option (A)} \] Quick Tip: Find the intersection point first by parametrising the line. Then substitute this point into the options to verify.

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Given: \[ \vec{\alpha}=(\lambda-2)\vec{a}+\vec{b},\quad \vec{\beta}=(4\lambda-2)\vec{a}+3\vec{b} \]

Since \(\vec{a},\vec{b}\) are non-collinear, \[ \vec{\alpha}=k\vec{\beta} \]

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Comparing coefficients: \[ 1=3k \Rightarrow k=\frac13 \]
\[ \lambda-2=\frac13(4\lambda-2) \]
\[ 3\lambda-6=4\lambda-2 \Rightarrow \lambda=-4 \]

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Final Answer: \[ \boxed{-4} \] Quick Tip: Two vectors \(x\vec{a} + y\vec{b}\) and \(p\vec{a} + q\vec{b}\) are collinear if the coefficients are proportional: \(\frac{x}{p} = \frac{y}{q}\).

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For 5 observations: \[ \bar{x}=10,\quad \sigma=3 \]
\[ \sum x_i=50 \]
\[ \frac{\sum x_i^2}{5}-100=9 \Rightarrow \sum x_i^2=545 \]

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Include new observation \(-50\)
\[ \sum x=0,\quad \sum x^2=545+2500=3045 \]
\[ New variance=\frac{3045}{6}-0=507.5 \]

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Final Answer: \[ \boxed{507.5} \] Quick Tip: Variance formula: \(\sigma^2 = \frac{\sum x^2}{N} - (\frac{\sum x}{N})^2\). Calculate sum of squares from old data and update it.

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Probability of hit: \[ p=\frac13,\quad q=\frac23 \]

Probability of at least one hit: \[ 1-\left(\frac23\right)^n>\frac56 \]
\[ \left(\frac23\right)^n<\frac16 \]

Checking values: \[ n=5 \Rightarrow \left(\frac23\right)^5\approx0.131<0.166 \]

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Final Answer: \[ \boxed{5} \] Quick Tip: \(P(at least one success) = 1 - P(no success)\). Set up inequality \(1 - q^n > P_{req}\) and solve for \(n\).

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Let: \[ \theta=\frac{\pi}{2^{10}} \]

Using: \[ \sin\theta\prod_{k=0}^{8}\cos(2^k\theta) =\frac{\sin(2^9\theta)}{2^9} \]
\[ =\frac{\sin(\pi/2)}{512}=\frac{1}{512} \]

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Final Answer: \[ \boxed{\frac{1}{512}} \] Quick Tip: Use repeated application of \(\sin 2A = 2 \sin A \cos A\) to collapse the product of cosines into a single sine term.

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Given: \[ A+B=120^\circ \Rightarrow C=60^\circ \]

Using tangent rule: \[ \tan\frac{A-B}{2} =\frac{a-b}{a+b}\cot\frac{C}{2} \]
\[ =\frac{2}{2\sqrt{3}}\cdot\sqrt{3}=1 \Rightarrow A-B=90^\circ \]

Solving: \[ A=105^\circ,\quad B=15^\circ \]

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Final Answer: \[ \boxed{7:1} \] Quick Tip: In solution of triangles, when two sides and the included angle (or sum of other angles) are involved, Napier's Analogy is very efficient.

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Truth values: \[ P=T,\quad Q=F,\quad R=T \]

Check option (C): \[ P\lor(\sim Q\land R)=T\lor(T\land T)=T \]

All other options evaluate to false.

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Final Answer: \[ \boxed{P\lor(\sim Q\land R)} \] Quick Tip: In logic, if the first operand of an OR (\(\lor\)) operator is True, the entire statement is True regardless of the second operand.