Step 1: Understanding the Concept:

When a body tied to a spring moves in a horizontal circular path, the spring undergoes an extension. This extension creates a restoring force (spring force) that acts as the necessary centripetal force required for the circular motion.


Step 2: Key Formula or Approach:

1. Spring Force: \( F_s = kx \), where \( k \) is the spring constant and \( x \) is the extension.

2. Centripetal Force: \( F_c = m \omega^2 r \).

3. Radius of circular path: \( r = L + x \), where \( L \) is the natural length of the spring.


Step 3: Detailed Explanation:

At equilibrium in the rotating frame, the spring force balances the centripetal requirements:
\[ kx = m \omega^2 (L + x) \]

Given:
\( m = 200 g = 0.2 kg \)
\( k = 12.5 N/m \)
\( \omega = 5 rad/s \)

Substituting these values:
\[ 12.5x = 0.2 \times (5)^2 \times (L + x) \]
\[ 12.5x = 0.2 \times 25 \times (L + x) \]
\[ 12.5x = 5(L + x) \]
\[ 12.5x = 5L + 5x \]
\[ 12.5x - 5x = 5L \]
\[ 7.5x = 5L \]

To find the ratio of extension (\( x \)) to natural length (\( L \)):
\[ \frac{x}{L} = \frac{5}{7.5} \]
\[ \frac{x}{L} = \frac{50}{75} = \frac{2}{3} \]


Step 4: Final Answer:

The ratio of extension in the spring to its natural length is 2:3.
Quick Tip: In circular motion involving springs, always remember that the radius of the circle is the "stretched length", which is \( (Natural Length + Extension) \).

Step 1: Understanding the Concept:

Different communication systems utilize specific frequency bands of the electromagnetic spectrum as allocated by regulatory bodies for efficient transmission and to avoid interference.


Step 2: Detailed Explanation:

- AM Broadcast (Amplitude Modulation): Operates in the Medium Frequency (MF) range. The standard band is approximately 540 to 1600 kHz. Thus, A matches with II.

- FM Broadcast (Frequency Modulation): Operates in the Very High Frequency (VHF) range, commonly known to be 88 to 108 MHz. Thus, B matches with I.

- Television: Television signals use a wider range including VHF and UHF (Ultra High Frequency), spanning approximately 54 MHz to 890 MHz. Thus, C matches with IV.

- Satellite Communication: Requires very high frequencies to penetrate the ionosphere, typically in the Gigahertz (GHz) range. The 3.7 to 4.2 GHz range is common for C-band satellite communication. Thus, D matches with III.


Step 3: Final Answer:

The correct matching is A-II, B-I, C-IV, D-III.
Quick Tip: Remembering the standard FM radio band (88-108 MHz) usually allows you to eliminate most options in these matching questions immediately.

Step 1: Understanding the Concept:

According to the principle of homogeneity of dimensions, the dimensions of the physical quantities on both sides of an equation must be identical.


Step 2: Key Formula or Approach:

We determine the dimensions of each quantity:

- Frequency (\( \nu \)): \( [T^{-1}] = [M^0 L^0 T^{-1}] \)

- Radius (\( r \)): \( [L] \)

- Density (\( \rho \)): \( [Mass/Volume] = [M L^{-3}] \)

- Surface Tension (\( s \)): \( [Force/Length] = [M L T^{-2} \cdot L^{-1}] = [M T^{-2}] \)


Step 3: Detailed Explanation:

Substitute the dimensions into the given expression \( \nu = r^a \rho^b s^c \):
\[ [M^0 L^0 T^{-1}] = [L]^a [M L^{-3}]^b [M T^{-2}]^c \]
\[ [M^0 L^0 T^{-1}] = M^{b+c} L^{a-3b} T^{-2c} \]

Equating the powers of M, L, and T:

1. For \( T \): \( -2c = -1 \implies c = \frac{1}{2} \)

2. For \( M \): \( b + c = 0 \implies b = -c = -\frac{1}{2} \)

3. For \( L \): \( a - 3b = 0 \implies a = 3b = 3 \left( -\frac{1}{2} \right) = -\frac{3}{2} \)


Step 4: Final Answer:

The values are \( a = -\frac{3}{2} \), \( b = -\frac{1}{2} \), and \( c = \frac{1}{2} \).
Quick Tip: Surface tension is energy per unit area as well as force per unit length. Its dimensions \( [MT^{-2}] \) are very common in liquid properties questions.

Step 1: Understanding the Concept:

Electric potential is a scalar quantity. The total potential at a point due to multiple charge distributions is the algebraic sum of the individual potentials.


Step 2: Key Formula or Approach:

The potential \( dV \) due to a small charge element \( dq \) at a distance \( R \) is:
\[ dV = \frac{1}{4 \pi \epsilon_0} \frac{dq}{R} \]

For a continuous arc, \( V = \frac{1}{4 \pi \epsilon_0 R} \int dq = \frac{Q_{total}}{4 \pi \epsilon_0 R} \).


Step 3: Detailed Explanation:

For a half ring with radius \( R \) and linear charge density \( \lambda \), the total charge is:
\[ Q = \lambda \times (circumference of semi-circle) = \lambda (\pi R) \]

The potential at the center due to one half ring is:
\[ V = \frac{\lambda \pi R}{4 \pi \epsilon_0 R} = \frac{\lambda}{4 \epsilon_0} \]

Notice that the potential is independent of the radius \( R \) because as the radius increases, the amount of charge also increases proportionally.

Total potential at the common center:
\[ V_{total} = V_1 + V_2 \]
\[ V_{total} = \frac{\lambda}{4 \epsilon_0} + \frac{\lambda}{4 \epsilon_0} = \frac{2 \lambda}{4 \epsilon_0} = \frac{\lambda}{2 \epsilon_0} \]


Step 4: Final Answer:

The net electric potential at the center is \( \frac{\lambda}{2 \epsilon_0} \).
Quick Tip: Potential at the center of any arc with uniform \( \lambda \) is \( V = \frac{\lambda \theta}{4 \pi \epsilon_0} \) where \( \theta \) is the angle in radians. For a semi-circle, \( \theta = \pi \), so \( V = \frac{\lambda}{4 \epsilon_0} \).

Step 1: Understanding the Concept:

According to Kepler's Third Law of planetary motion, the square of the time period of revolution (\( T \)) of a planet is proportional to the cube of its mean distance (\( r \)) from the Sun.


Step 2: Key Formula or Approach:
\[ T^2 \propto r^3 \implies \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]


Step 3: Detailed Explanation:

Let \( T_1 = 1 year \) (Earth's period) and \( r_1 = 1.5 \times 10^8 km \) (Earth's distance).

Given \( T_2 = 2.83 years \). We need to find \( r_2 \).

Note that \( (2.83)^2 \approx 8 \) (since \( 2.83 \approx \sqrt{8} \approx 2\sqrt{2} \)).

Using the formula:
\[ \frac{(1)^2}{(2.83)^2} = \frac{(1.5 \times 10^8)^3}{r_2^3} \]
\[ \frac{1}{8} = \left( \frac{1.5 \times 10^8}{r_2} \right)^3 \]

Taking the cube root on both sides:
\[ \sqrt[3]{\frac{1}{8}} = \frac{1.5 \times 10^8}{r_2} \]
\[ \frac{1}{2} = \frac{1.5 \times 10^8}{r_2} \]
\[ r_2 = 2 \times 1.5 \times 10^8 = 3.0 \times 10^8 km \]


Step 4: Final Answer:

The distance of the imaginary planet from the Sun is \( 3 \times 10^8 \) km.
Quick Tip: In gravitation problems, look for numerical clues: \( 2.83 \approx 2\sqrt{2} = \sqrt{8} \). Squaring it gives 8, which is a perfect cube (\( 2^3 \)), making calculations easy.

Step 1: Understanding the Concept:

The ratio of specific heats \( \gamma \) is related to the degrees of freedom (\( f \)) of a gas molecule by the formula \( \gamma = 1 + \frac{2}{f} \).


Step 2: Key Formula or Approach:

1. For a monoatomic gas, \( f = 3 \) (3 translational).

2. For a diatomic gas (rigid rotator), \( f = 5 \) (3 translational + 2 rotational).


Step 3: Detailed Explanation:

Calculate \( \gamma_1 \) for monoatomic gas:
\[ \gamma_1 = 1 + \frac{2}{3} = \frac{5}{3} \]

Calculate \( \gamma_2 \) for diatomic rigid rotator:
\[ \gamma_2 = 1 + \frac{2}{5} = \frac{7}{5} \]

Calculate the ratio \( \frac{\gamma_1}{\gamma_2} \):
\[ \frac{\gamma_1}{\gamma_2} = \frac{5/3}{7/5} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21} \]


Step 4: Final Answer:

The ratio \( \frac{\gamma_1}{\gamma_2} \) is 25/21.
Quick Tip: Always clarify if the diatomic molecule is "rigid" or "vibrating". A non-rigid rotator has \( f=7 \) due to 2 additional vibrational degrees of freedom at high temperatures.

Step 1: Understanding the Concept:

The time period (\( T \)) of a simple pendulum is the time it takes to complete one full oscillation. If \( T \) increases, the clock takes more time for each tick, meaning it runs slow. If \( T \) decreases, it runs fast.


Step 2: Key Formula or Approach:
\[ T = 2 \pi \sqrt{\frac{L}{g}} \implies T \propto \frac{1}{\sqrt{g}} \]


Step 3: Detailed Explanation:

- Evaluating Reason R: Acceleration due to gravity (\( g \)) decreases with increase in altitude. Mount Everest is at a high altitude, so \( g_{Everest} < g_{surface} \). Reason R is correct.

- Evaluating Assertion A: Since \( g \) decreases on Mount Everest, the time period \( T \) of the pendulum clock increases (\( T \uparrow \)).

- An increased time period means the clock is ticking more slowly than a standard clock. Thus, the clock becomes slow, not fast. Assertion A is incorrect.


Step 4: Final Answer:

Assertion A is false, but Reason R is true.
Quick Tip: "Time Period Increases" = "Clock Slows Down" = "Losing Time".
"Time Period Decreases" = "Clock Speeds Up" = "Gaining Time".

Step 1: Understanding the Concept:

The value of acceleration due to gravity (\( g \)) is maximum at the Earth's surface and varies as one moves away from the surface (either into space or towards the core).


Step 2: Key Formula or Approach:

1. At height \( h \) (\( h \ll R \)): \( g_h = g \left( 1 - \frac{2h}{R} \right) \).

2. At depth \( d \): \( g_d = g \left( 1 - \frac{d}{R} \right) \).


Step 3: Detailed Explanation:

- Statement I: Moving 'up' increases the distance from the Earth's center, decreasing \( g \). Moving 'down' reduces the effective mass of the Earth attracting the object, also decreasing \( g \). Hence, \( g \) decreases in both cases. Statement I is correct.

- Statement II: For the same distance \( x \) (where \( h=d=x \)):

Decrease with height \( \Delta g_h = g \left( \frac{2x}{R} \right) \).

Decrease with depth \( \Delta g_d = g \left( \frac{x}{R} \right) \).

The decrease at height \( h \) is twice the decrease at depth \( d \) (for small distances). Therefore, the values of gravity are not the same. Statement II is incorrect.


Step 4: Final Answer:

Statement I is correct, and Statement II is incorrect.
Quick Tip: Gravity decreases twice as fast as you go up compared to as you go down (near the surface). To have equal \( g \), you would need to go to a depth \( d = 2h \).

Step 1: Understanding the Concept:

When a conductor moves across a magnetic field, the free electrons in it experience a Lorentz force, leading to a potential difference between its ends. This is called motional electromotive force (emf).


Step 2: Key Formula or Approach:

The induced emf \( d\epsilon \) for a small length element \( dr \) moving with velocity \( v \) perpendicular to magnetic field \( B \) is \( d\epsilon = B v \, dr \).


Step 3: Detailed Explanation:

For a rod of length \( L \) rotating with angular velocity \( \omega \), the linear velocity of a point at a distance \( r \) from the axis of rotation is \( v = r \omega \).

The total induced emf \( \epsilon \) is the integral over the entire length of the rod:
\[ \epsilon = \int_{0}^{L} B v \, dr \]
\[ \epsilon = \int_{0}^{L} B (r \omega) \, dr \]
\[ \epsilon = B \omega \int_{0}^{L} r \, dr \]
\[ \epsilon = B \omega \left[ \frac{r^2}{2} \right]_{0}^{L} \]
\[ \epsilon = \frac{1}{2} B L^2 \omega \]


Step 4: Final Answer:

The induced emf across the ends of the rod is \( \frac{1}{2} B L^2 \omega \).
Quick Tip: This formula can also be visualized as \( \epsilon = B \times (Area swept per second) \). The area of the circle is \( \pi L^2 \). In one period \( T \), it sweeps this area.
Area swept per sec = \( \frac{\pi L^2}{T} = \frac{\pi L^2}{2\pi/\omega} = \frac{1}{2} L^2 \omega \). Thus \( \epsilon = B \frac{1}{2} L^2 \omega \).

Step 1: Understanding the Concept:

Two vectors are said to be perpendicular (orthogonal) if their scalar (dot) product is equal to zero.


Step 2: Key Formula or Approach:

If \( \vec{A} \perp \vec{B} \), then \( \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z = 0 \).


Step 3: Detailed Explanation:

Given:
\( \vec{p} = 1 \hat{i} + 2m \hat{j} + m \hat{k} \)
\( \vec{Q} = 4 \hat{i} - 2 \hat{j} + m \hat{k} \)

Since they are perpendicular:
\[ \vec{p} \cdot \vec{Q} = (1)(4) + (2m)(-2) + (m)(m) = 0 \]
\[ 4 - 4m + m^2 = 0 \]

This is a quadratic equation which can be rewritten as:
\[ m^2 - 4m + 4 = 0 \]
\[ (m - 2)^2 = 0 \]

Taking the square root on both sides:
\[ m - 2 = 0 \implies m = 2 \]


Step 4: Final Answer:

The value of \( m \) is 2.
Quick Tip: The dot product is the fastest way to check for perpendicularity. If vectors were parallel, you would check if their components are proportional: \( \frac{p_x}{Q_x} = \frac{p_y}{Q_y} = \frac{p_z}{Q_z} \).

Step 1: Understanding the Concept:

In a plane electromagnetic wave traveling in vacuum, the amplitudes of the electric field (\( E_0 \)) and magnetic field (\( B_0 \)) are related to each other via the speed of light (\( c \)).


Step 2: Key Formula or Approach:

1. Relationship between amplitudes: \( \frac{E_0}{B_0} = c \).

2. Speed of wave: \( c = \frac{\omega}{k} \).


Step 3: Detailed Explanation:

From the standard properties of EM waves:
\[ \frac{E_0}{B_0} = c \]

Substituting the value of \( c = \frac{\omega}{k} \):
\[ \frac{E_0}{B_0} = \frac{\omega}{k} \]

Cross-multiplying gives:
\[ k E_0 = \omega B_0 \]


Step 4: Final Answer:

The correct relation is \( k E_0 = \omega B_0 \).
Quick Tip: The phase of the wave is \( (kz - \omega t) \). Dimensional check: \( [k] = [L^{-1}] \) and \( [\omega] = [T^{-1}] \). Thus \( \omega/k \) has dimensions of speed \( [LT^{-1}] \), which confirms \( E/B = \omega/k \).

Step 1: Understanding the Concept:

The given circuit is a transistor-based or simple switch-based resistor logic circuit. The output \( Y \) is taken from the collector/top node.


Step 2: Detailed Explanation:

- The output \( Y \) is connected to a +5V supply through a resistor.

- There are two parallel switches (or transistors) \( A_1 \) and \( B_1 \) connected between the output node and the ground (0V).

- If either switch \( A_1 \) or switch \( B_1 \) is closed (Input = 1), a path is created to ground. The potential at \( Y \) drops to 0V (Output = 0).

- If both switches \( A_1 \) and \( B_1 \) are open (Inputs = 0, 0), no current flows to ground. The potential at \( Y \) remains at +5V (Output = 1).


Truth Table:



The resulting logic \( Y = \overline{A + B} \) corresponds exactly to a NOR gate.


Step 3: Final Answer:

The equivalent logic gate is NOR.
Quick Tip: When switches are in \textbf{parallel} pulling the output to ground, it's a \textbf{NOR} configuration. When switches are in \textbf{series} pulling the output to ground, it's a \textbf{NAND} configuration.

Step 1: Understanding the Concept:

The de-Broglie wavelength of a particle is inversely proportional to its momentum. When particles have the same kinetic energy, their wavelengths depend on their masses.


Step 2: Key Formula or Approach:

The relation between de-Broglie wavelength (\(\lambda\)) and kinetic energy (\(K\)) is:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]


Step 3: Detailed Explanation:

Given that kinetic energy \(K\) is constant for all three particles:
\[ \lambda \propto \frac{1}{\sqrt{m}} \]

We know the relative masses of the particles are:

1. Mass of electron (\(m_e\)) \(\approx 9.1 \times 10^{-31}\) kg

2. Mass of proton (\(m_p\)) \(\approx 1.67 \times 10^{-27}\) kg

3. Mass of \(\alpha\)-particle (\(m_{\alpha}\)) \(\approx 4 \times m_p \approx 6.64 \times 10^{-27}\) kg

Comparing the masses:
\[ m_{\alpha} > m_p > m_e \]

Since wavelength is inversely proportional to the square root of mass:
\[ \lambda_{\alpha} < \lambda_p < \lambda_e \]


Step 4: Final Answer:

The correct order of wavelengths is \(\lambda_{\alpha} < \lambda_p < \lambda_e\).
Quick Tip: For the same kinetic energy, the lighter the particle, the longer its de-Broglie wavelength. Since electrons are the lightest subatomic particles among the options, they will always have the maximum wavelength.

Step 1: Understanding the Concept:

Lenses are made of dispersive materials where the refractive index varies with the wavelength of light. This causes different colors to focus at different points.


Step 2: Key Formula or Approach:

According to Lens Maker's Formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Cauchy's equation states: \(\mu \approx A + \frac{B}{\lambda^2}\).


Step 3: Detailed Explanation:

Since refractive index (\(\mu\)) depends on wavelength (\(\lambda\)), and focal length (\(f\)) depends on \(\mu\), different colors (wavelengths) will have different focal lengths.

Violet light has a higher refractive index than red light (\(\mu_V > \mu_R\)), thus violet light converges closer to the lens than red light (\(f_V < f_R\)).

The failure of a lens to focus all colors at the same point is known as chromatic aberration.


Step 4: Final Answer:

The phenomenon is called Chromatic aberration.
Quick Tip: Remember: "Chrome" means color. Aberration caused by colors is Chromatic Aberration. Aberration caused by the shape (geometry) of the lens is Spherical Aberration.

Step 1: Understanding the Concept:

A long solenoid produces a uniform magnetic field inside its core that is proportional to the number of turns per unit length and the current flowing through it.


Step 2: Key Formula or Approach:

Magnetic field inside a long solenoid is given by:
\[ B = \mu_0 n I \]

where \(n\) is the number of turns per unit meter.


Step 3: Detailed Explanation:

Given:

Number of turns per cm = 70.
\(n = 70 turns/cm = 70 \times 10^2 turns/m = 7000 turns/m\).

Current \(I = 2.0\) A.

Permeability \(\mu_0 = 4\pi \times 10^{-7} TmA^{-1}\).

Calculating \(B\):
\[ B = (4\pi \times 10^{-7}) \times 7000 \times 2 \]
\[ B = 8\pi \times 7 \times 10^{-4} \]
\[ B = 56\pi \times 10^{-4} \]

Using \(\pi \approx 3.142\):
\[ B \approx 56 \times 3.142 \times 10^{-4} \]
\[ B \approx 175.952 \times 10^{-4} \approx 176 \times 10^{-4} T \]


Step 4: Final Answer:

The magnetic field produced inside the solenoid is \(176 \times 10^{-4}\) T.
Quick Tip: Be very careful with units of '\(n\)'. Always convert 'turns per cm' to 'turns per meter' by multiplying by 100 before substituting into the standard formula.

Step 1: Understanding the Concept:

In a velocity-time (\(v-t\)) graph:

- Displacement is the algebraic sum of the areas (positive above the axis, negative below).

- Distance is the sum of the magnitudes of the areas (all treated as positive).


Step 2: Key Formula or Approach:

Displacement \(= \int v \, dt = \sum Area_i\) (with signs).

Distance \(= \int |v| \, dt = \sum |Area_i|\).


Step 3: Detailed Explanation:

From the graph, we divide the motion into segments:

1. \(t = 0\) to \(2s\): \(v = 8 m/s\). Area \(A_1 = 2 \times 8 = 16 m\).

2. \(t = 2\) to \(4s\): \(v = -4 m/s\). Area \(A_2 = 2 \times (-4) = -8 m\).

3. \(t = 4\) to \(6s\): \(v = 4 m/s\). Area \(A_3 = 2 \times 4 = 8 m\).

4. \(t = 6\) to \(8s\): \(v = 0 m/s\). Area \(A_4 = 0 m\).

5. \(t = 8\) to \(10s\): Looking at the graph, \(v = -2 m/s\). Area \(A_5 = 2 \times (-2) = -4 m\).

Total Displacement:
\[ S = 16 - 8 + 8 + 0 - 4 = 12 m \]

Total Distance:
\[ D = 16 + 8 + 8 + 0 + 4 = 36 m \]

The ratio of displacement to distance is:
\[ Ratio = \frac{12}{36} = \frac{1}{3} \]


Step 4: Final Answer:

The ratio is 1:3.
Quick Tip: Always visually double-check the height of segments on the y-axis. Segment (8-10) is half the height of segment (2-4) in the negative region, indicating a velocity of -2 m/s.

Step 1: Understanding the Concept:

When an electron transitions between energy levels in a hydrogen atom, it emits a photon with energy equal to the difference between the initial and final levels.


Step 2: Key Formula or Approach:

Energy levels in H-atom: \(E_n = -\frac{13.6}{n^2} eV\).

Energy emitted: \(\Delta E = E_4 - E_1\).

Wavelength: \(\lambda = \frac{hc}{\Delta E}\).


Step 3: Detailed Explanation:

Given \(h = 4 \times 10^{-15} eVs\) and \(c = 3 \times 10^8 m/s\).

Calculate energy difference:
\[ \Delta E = -13.6 \left( \frac{1}{4^2} - \frac{1}{1^2} \right) = 13.6 \left( 1 - \frac{1}{16} \right) \]
\[ \Delta E = 13.6 \times \frac{15}{16} = 12.75 eV \]

Calculate wavelength:
\[ \lambda = \frac{(4 \times 10^{-15} eVs) \times (3 \times 10^8 m/s)}{12.75 eV} \]
\[ \lambda = \frac{12 \times 10^{-7}}{12.75} m \]
\[ \lambda \approx 0.9411 \times 10^{-7} m = 94.11 nm \]


Step 4: Final Answer:

The corresponding wavelength is 94.1 nm.
Quick Tip: Using the simplified value \(hc \approx 1242 eV\cdotnm\) (if standard \(h\) is used) or calculating \(hc\) directly from given values helps in avoiding unit conversion errors.

Step 1: Understanding the Concept:

Elasticity in physics refers to the property of a material to return to its original shape. A "more elastic" material requires more force to produce a given strain (high Young's modulus).


Step 2: Detailed Explanation:

Assertion A: Steel is indeed the primary material for skyscrapers and bridges due to its structural integrity.

Reason R: In materials science, steel has a higher Young's modulus than materials like copper or aluminum, meaning it resists deformation under heavy loads. A high elastic limit means it can sustain significant stress without undergoing permanent plastic deformation.

The high elasticity and elastic limit are precisely why engineers choose steel for construction, as it ensures the structure remains safe and stable under load. Thus, R is the correct explanation for A.


Step 3: Final Answer:

Both A and R are correct and R is the correct explanation of A.
Quick Tip: Common misconception: People think rubber is more elastic than steel. In Physics, steel is more elastic because it resists deformation more strongly (higher Stress/Strain ratio).

Step 1: Understanding the Concept:

For an ideal gas undergoing an isothermal process, the product of pressure (\(P\)) and volume (\(V\)) is constant according to Boyle's Law.


Step 2: Key Formula or Approach:

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

For a fixed amount of gas, at a higher temperature, the \(PV\) product is larger.


Step 3: Detailed Explanation:

The graph of \(P\) vs \(V\) for an isothermal process is a rectangular hyperbola.

As temperature increases, the curve moves further away from the origin because for any given volume \(V\), the pressure \(P = \frac{nRT}{V}\) will be higher if \(T\) is higher.

Given \(T_3 > T_2 > T_1\):

- The outermost curve (highest \(P\) for same \(V\)) must correspond to \(T_3\).

- The innermost curve must correspond to \(T_1\).

Looking at the diagrams, Option 1 shows the curves correctly labeled with \(T_3\) as the outermost and \(T_1\) as the innermost curve.


Step 4: Final Answer:

The correct representation is given in Option 1.
Quick Tip: Draw a vertical line (constant Volume) across the curves. The point where the line hits the highest pressure corresponds to the highest temperature isotherm.

Step 1: Understanding the Concept:

A real voltmeter has a finite resistance. When connected in parallel to a resistor, it changes the equivalent resistance of that part of the circuit, affecting the voltage distribution.


Step 2: Key Formula or Approach:

Parallel resistance: \(R_p = \frac{R_1 R_v}{R_1 + R_v}\).

Voltage Divider Rule: \(V_{out} = V_{total} \left( \frac{R_{parallel}}{R_{parallel} + R_{other}} \right)\).


Step 3: Detailed Explanation:

Circuit components: \(V = 90\) V, \(R_1 = 100\Omega\), \(R_2 = 100\Omega\), Voltmeter resistance \(R_v = 400\Omega\).

The voltmeter is connected across one resistor (say \(R_1\)).

Equivalent resistance of the voltmeter and \(R_1\):
\[ R_p = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80\Omega \]

Total resistance of the circuit now:
\[ R_{total} = R_p + R_2 = 80 + 100 = 180\Omega \]

Current from the cell:
\[ I = \frac{V}{R_{total}} = \frac{90}{180} = 0.5 A \]

The voltmeter reading is the voltage across the parallel combination:
\[ V_{reading} = I \times R_p = 0.5 \times 80 = 40 V \]


Step 4: Final Answer:

The reading of the voltmeter will be 40 V.
Quick Tip: If the voltmeter were ideal (infinite resistance), it would read \(45\) V. Since a real voltmeter draws some current, it always reads slightly lower than the ideal value for this configuration.

Step 1: Understanding the Concept:

The magnetic force on a current-carrying straight wire segment is determined by the length of the wire, the current, the magnetic field, and the angle between the wire and the field.


Step 2: Key Formula or Approach:

Force magnitude: \(F = I L B \sin\theta\).


Step 3: Detailed Explanation:

Given a right triangle with sides \(a=5, b=12, c=13\). Since \(5^2 + 12^2 = 13^2\), the angle opposite the 13 cm side is \(90^\circ\).

The magnetic field \(\vec{B}\) is parallel to the 13 cm hypotenuse.

We need the force on the 5 cm side. Let the angle between the 5 cm side and the 13 cm side be \(\theta\).

From the triangle properties:
\[ \sin\theta = \frac{opposite side}{hypotenuse} = \frac{12}{13} \]

Now, calculate the force on the 5 cm side (\(L = 0.05\) m):
\[ F = I L B \sin\theta \]
\[ F = 2 \times 0.05 \times 0.75 \times \frac{12}{13} \]
\[ F = 0.1 \times 0.75 \times \frac{12}{13} = 0.075 \times \frac{12}{13} \]
\[ F = \frac{0.9}{13} = \frac{9}{130} N \]

Comparing with the given form \(\frac{x}{130}\), we get \(x = 9\).


Step 4: Final Answer:

The value of \(x\) is 9.
Quick Tip: For right triangles, if the field is parallel to the hypotenuse, the force on one leg involves the sine of the angle which is just the ratio of the other leg to the hypotenuse.

Step 1: Understanding the Concept:

When a body falls through a viscous fluid and reaches terminal velocity (constant velocity), the net force on it is zero. This means the viscous force plus buoyant force balances the weight.


Step 2: Key Formula or Approach:

At terminal velocity: \(F_{viscous} = W - F_b = V\rho_s g - V\rho_l g = Vg(\rho_s - \rho_l)\).


Step 3: Detailed Explanation:

Given:

Radius \(r = 1 mm = 10^{-3} m\).

Density of sphere \(\rho_s = 10.5 g/cc = 10500 kg/m^3\).

Density of liquid \(\rho_l = 1.5 g/cc = 1500 kg/m^3\).

Viscous force \(F\):
\[ F = \frac{4}{3} \pi r^3 g (\rho_s - \rho_l) \]
\[ F = \frac{4}{3} \times \frac{22}{7} \times (10^{-3})^3 \times 9.8 \times (10500 - 1500) \]
\[ F = \frac{4}{3} \times \frac{22}{7} \times 10^{-9} \times 9.8 \times 9000 \]
\[ F = \frac{4}{3} \times 22 \times 10^{-9} \times 1.4 \times 9000 \] (since \(9.8/7 = 1.4\))
\[ F = 4 \times 22 \times 1.4 \times 10^{-9} \times 3000 \]
\[ F = 88 \times 4.2 \times 10^{-6} = 369.6 \times 10^{-6} N \]

We need the form \(3696 \times 10^{-x}\):
\[ 369.6 \times 10^{-6} = 3696 \times 10^{-7} \]

Thus, \(x = 7\).


Step 4: Final Answer:

The value of \(x\) is 7.
Quick Tip: Avoid calculating terminal velocity first unless asked. Use the equilibrium condition (Weight = Buoyancy + Viscous Force) to find the force directly from the volumes and densities.

Step 1: Understanding the Concept:

The focal length of a lens depends on the relative refractive index of the lens material with respect to the surrounding medium.


Step 2: Key Formula or Approach:

Lens Maker's Formula: \(\frac{1}{f} = \left( \frac{\mu_{lens}}{\mu_{med}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).


Step 3: Detailed Explanation:

In air (\(\mu_{med} = 1\)):
\[ \frac{1}{18} = (1.5 - 1)K = 0.5K \implies K = \frac{1}{9} \]

In water (\(\mu_{med} = 4/3\)):
\[ \frac{1}{f_w} = \left( \frac{1.5}{4/3} - 1 \right)K = \left( \frac{4.5}{4} - 1 \right)K = \left( \frac{9}{8} - 1 \right)K \]
\[ \frac{1}{f_w} = \frac{1}{8} K = \frac{1}{8} \times \frac{1}{9} = \frac{1}{72} \]

So, \(f_w = 72\) cm.

The change in focal length \(\Delta f\):
\[ \Delta f = f_w - f_{air} = 72 - 18 = 54 cm \]


Step 4: Final Answer:

The change in focal length is 54 cm.
Quick Tip: A useful shortcut: For a glass lens (\(\mu=1.5\)) immersed in water (\(\mu=1.33\)), the focal length always becomes approximately 4 times its focal length in air (\(f_w \approx 4f_a\)).

Step 1: Understanding the Concept:

Power is defined as the dot product of the force acting on a body and its instantaneous velocity.


Step 2: Key Formula or Approach:
\(P = \vec{F} \cdot \vec{v}\).
\(\vec{a} = \frac{\vec{F}}{m}\).
\(\vec{v} = \int \vec{a} \, dt\).


Step 3: Detailed Explanation:

Given mass \(m = 1\) kg and \(\vec{F} = (t\hat{i} + 3t^2\hat{j})\).

Acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = (t\hat{i} + 3t^2\hat{j}) m/s^2 \]

Velocity (assuming starting from rest):
\[ \vec{v} = \int \vec{a} \, dt = \int (t\hat{i} + 3t^2\hat{j}) \, dt = \frac{t^2}{2}\hat{i} + t^3\hat{j} \]

Now, find the power at \(t = 2s\):

At \(t = 2s\), Force \(\vec{F} = 2\hat{i} + 3(2^2)\hat{j} = 2\hat{i} + 12\hat{j}\).

At \(t = 2s\), Velocity \(\vec{v} = \frac{2^2}{2}\hat{i} + 2^3\hat{j} = 2\hat{i} + 8\hat{j}\).

Power \(P = \vec{F} \cdot \vec{v}\):
\[ P = (2 \times 2) + (12 \times 8) = 4 + 96 = 100 W \]


Step 4: Final Answer:

The power developed at \(t=2s\) is 100 W.
Quick Tip: For variable force, remember that power is also \(P = \frac{d}{dt}(K.E.)\). You can find kinetic energy \(K = \frac{1}{2}mv^2\) as a function of time and then differentiate it to get power.

Step 1: Understanding the Concept:

The total energy released in a nuclear fission process is the product of the energy released per single fission event and the total number of nuclei undergoing fission.


Step 2: Key Formula or Approach:

1. Number of moles (\(n\)) = \(\frac{Given mass (m)}{Atomic mass (M)}\).

2. Number of nuclei (\(N\)) = \(n \times N_A\).

3. Total energy (\(E_{total}\)) = \(N \times E_{fission}\).


Step 3: Detailed Explanation:

Given:

Mass of \(^{240}X\), \(m = 120\) g.

Atomic mass of \(X\), \(M = 240\) g/mol.

Energy per fission, \(E_{fission} = 200\) MeV.

Avogadro number, \(N_A = 6 \times 10^{23}\) atoms/mol.


Calculate the number of moles:
\[ n = \frac{120}{240} = 0.5 mol \]

Calculate the number of nuclei:
\[ N = 0.5 \times 6 \times 10^{23} = 3 \times 10^{23} nuclei \]

Calculate the total energy released:
\[ E_{total} = (3 \times 10^{23}) \times 200 MeV \]
\[ E_{total} = 600 \times 10^{23} MeV = 6 \times 10^{25} MeV \]


Step 4: Final Answer:

Comparing this with the form \(x \times 10^{25}\) MeV, we find the value of \(x\) is 6.
Quick Tip: In nuclear physics problems, always ensure that mass is divided by atomic mass to get moles. Remember \(1 mole = 6 \times 10^{23}\) particles.

Step 1: Understanding the Concept:

The capacitance of a parallel plate capacitor depends on the area of plates, the distance between them, and the dielectric constant of the material filling the gap.


Step 2: Key Formula or Approach:

Capacitance with air: \(C_0 = \frac{\epsilon_0 A}{d}\).

Capacitance with dielectric \(K\) and new distance \(d'\): \(C = \frac{K \epsilon_0 A}{d'}\).


Step 3: Detailed Explanation:

Given:

Initial capacitance, \(C_0 = 15\) pF.

New distance, \(d' = 2d\).

Dielectric constant, \(K = 3.5\).


New capacitance \(C\):
\[ C = \frac{K \epsilon_0 A}{d'} = \frac{3.5 \epsilon_0 A}{2d} \]

Substituting \(\frac{\epsilon_0 A}{d} = C_0 = 15\):
\[ C = \frac{3.5}{2} \times 15 = 1.75 \times 15 = 26.25 pF \]

The question states the new capacitance is \(\frac{x}{4}\) pF:
\[ \frac{x}{4} = 26.25 \]
\[ x = 26.25 \times 4 = 105 \]


Step 4: Final Answer:

The value of \(x\) is 105.
Quick Tip: Capacitance increases with the dielectric constant (\(K\)) and decreases as the separation distance (\(d\)) increases. \(C \propto \frac{K}{d}\).

Step 1: Understanding the Concept:

The time period of a mass-spring system in Simple Harmonic Motion (SHM) is directly proportional to the square root of the attached mass.


Step 2: Key Formula or Approach:

Time period, \(T = 2\pi \sqrt{\frac{m}{k}}\), where \(k\) is the spring constant.


Step 3: Detailed Explanation:

Case 1: \(T_1 = 1\) s for mass \(m\).
\[ 1 = 2\pi \sqrt{\frac{m}{k}} \quad ---(i) \]

Case 2: Mass becomes \((m+3)\), and the time period increases by 1 s, so \(T_2 = 1 + 1 = 2\) s.
\[ 2 = 2\pi \sqrt{\frac{m+3}{k}} \quad ---(ii) \]

Dividing equation (ii) by equation (i):
\[ \frac{2}{1} = \frac{2\pi \sqrt{\frac{m+3}{k}}}{2\pi \sqrt{\frac{m}{k}}} = \sqrt{\frac{m+3}{m}} \]

Squaring both sides:
\[ 4 = \frac{m+3}{m} \]
\[ 4m = m+3 \]
\[ 3m = 3 \implies m = 1 kg \]


Step 4: Final Answer:

The initial mass \(m\) is 1 kg.
Quick Tip: Since \(T \propto \sqrt{m}\), doubling the time period requires the mass to be quadrupled. \(4m = m + 3\) leads directly to the answer.

Step 1: Understanding the Concept:

The moment of inertia of a solid cylinder about its geometric axis depends on its mass and the square of its radius.


Step 2: Key Formula or Approach:

Moment of Inertia of a solid cylinder, \(I = \frac{1}{2} M R^2\).

Mass, \(M = Density (\rho) \times Volume (V) = \rho \pi R^2 L\).


Step 3: Detailed Explanation:

Let \(\rho\) be the density of the cylinder material.

For the original cylinder:
\(M_1 = \rho \pi R^2 L\)
\(I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\rho \pi R^2 L) R^2 = \frac{1}{2} \rho \pi R^4 L\).


For the carved-out cylinder:

Radius, \(R' = R/2\)

Length, \(L' = L/2\)

Mass, \(M_2 = \rho \pi (R/2)^2 (L/2) = \rho \pi \frac{R^2}{4} \frac{L}{2} = \frac{1}{8} (\rho \pi R^2 L) = \frac{M_1}{8}\).

Moment of Inertia \(I_2\):
\(I_2 = \frac{1}{2} M_2 (R')^2 = \frac{1}{2} (\frac{M_1}{8}) (\frac{R}{2})^2\)
\(I_2 = \frac{1}{2} \cdot \frac{M_1}{8} \cdot \frac{R^2}{4} = \frac{1}{32} (\frac{1}{2} M_1 R^2) = \frac{I_1}{32}\).


Calculating the ratio:
\[ \frac{I_1}{I_2} = 32 \]


Step 4: Final Answer:

The ratio \(\frac{I_1}{I_2}\) is 32.
Quick Tip: For same density, \(I \propto R^4 L\). Here \(R\) is halved (\(\frac{1}{16}\) factor) and \(L\) is halved (\(\frac{1}{2}\) factor). Total factor is \(\frac{1}{16 \times 2} = \frac{1}{32}\).

Step 1: Understanding the Concept:

When a wire is stretched, its volume remains constant. An increase in length is accompanied by a decrease in cross-sectional area.


Step 2: Key Formula or Approach:

Resistance, \(R = \rho \frac{l}{A}\).

Volume, \(V = A \times l = constant\).

Therefore, \(A = \frac{V}{l}\), which gives \(R = \rho \frac{l^2}{V} \implies R \propto l^2\).


Step 3: Detailed Explanation:

Initial length = \(l_1\).

Final length, \(l_2 = l_1 + 20% of l_1 = 1.2 l_1\).

Since \(R \propto l^2\):
\[ \frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = (1.2)^2 = 1.44 \]

Percentage increase in resistance:
\[ % Increase = \frac{R_2 - R_1}{R_1} \times 100 = (1.44 - 1) \times 100 = 44% \]


Step 4: Final Answer:

The percentage increase in resistance is 44%.
Quick Tip: For a small percentage increase (\(<5%\)), use \(\Delta R/R \approx 2 \Delta l/l\). For larger changes, always use the square factor: \((1 + n/100)^2 - 1\).

Step 1: Understanding the Concept:

"Long after the switch is closed" refers to the steady-state condition. In steady state, inductors act as ideal wires (zero resistance) because the current is no longer changing (\(\frac{di}{dt} = 0\)).


Step 2: Key Formula or Approach:

1. In steady state, \(V_L = L \frac{di}{dt} = 0\). Replace inductors with short circuits.

2. Calculate the equivalent resistance (\(R_{eq}\)) of the remaining network.

3. Use Ohm's Law: \(I = \frac{V}{R_{eq}}\).


Step 3: Detailed Explanation:

In the circuit diagram:

- There are three parallel branches across the 12 V battery.

- Branch 1 contains an inductor \(L\) and a resistor \(R\). In steady state, it is just \(R = 12 \Omega\).

- Branch 2 contains a resistor \(R = 12 \Omega\).

- Branch 3 contains an inductor \(L\) and a resistor \(R\). In steady state, it is just \(R = 12 \Omega\).

The equivalent resistance of three \(12 \Omega\) resistors in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \implies R_{eq} = 4 \Omega \]

Current through the battery:
\[ I = \frac{V}{R_{eq}} = \frac{12}{4} = 3 A \]


Step 4: Final Answer:

The current is 3 A.
Quick Tip: Always remember: At \(t=0\), inductors are open circuits; at \(t \to \infty\), they are short circuits. Capacitors are exactly the opposite!

Step 1: Understanding the Concept:

The common and most stable oxidation state for lanthanoids is +3. Species in oxidation states other than +3 tend to reach the +3 state by either losing or gaining electrons.


Step 2: Detailed Explanation:

- Oxidizing Agents: Substances that undergo reduction (gain electrons).

- Lanthanoids in +4 State: \(Ce^{4+}\) and \(Tb^{4+}\) are in a higher oxidation state than the stable +3. To attain stability, they gain one electron and act as strong oxidizing agents.

- \(Ce^{4+} + e^- \to Ce^{3+}\) (Stable configuration)

- \(Tb^{4+} + e^- \to Tb^{3+}\) (Half-filled \(f^7\) stability)

- Lanthanoids in +2 State: \(Sm^{2+}\) and \(Ce^{2+}\) are in a lower state than +3. They tend to lose electrons to become +3, acting as reducing agents.


Step 3: Final Answer:

Therefore, \(Ce^{4+}\) and \(Tb^{4+}\) are good oxidizing agents.
Quick Tip: Remember: \(Ce^{4+}\) is widely used in analytical chemistry as a volumetric oxidizing agent (cerimetry) because it is a very strong oxidant.

Step 1: Understanding the Concept:

Crystal Field Theory (CFT) is an electrostatic model that considers the metal-ligand bond to be purely ionic, treating ligands as point charges.


Step 2: Detailed Explanation:

- Successes of CFT: It successfully explains the magnetic properties (pairing vs. high spin), colors (d-d transitions), and thermodynamic stability (CFSE) of coordination complexes.

- Failures of CFT: Because it treats ligands as point charges, it predicts that anionic ligands should cause greater splitting than neutral ones. However, anionic ligands like \(OH^-\) and \(I^-\) are at the low end of the spectrochemical series, while neutral ligands like \(CO\) are strong field ligands. This discrepancy is due to the covalent nature of bonding (back-bonding), which CFT ignores.


Step 3: Final Answer:

The order of the spectrochemical series cannot be explained by CFT; it requires Ligand Field Theory.
Quick Tip: CFT's biggest drawback is its complete neglect of the covalent character in metal-ligand bonds. This is why it fails to justify why \(CO\) is a stronger ligand than \(Cl^-\).

Step 1: Understanding the Concept:

Acidified potassium dichromate is a strong oxidizing agent (orange color). When it reacts with a reducing agent, it is reduced to chromium(III) ions, which are green in color.


Step 2: Detailed Explanation:

Sulphur dioxide (\(SO_2\)) is a reducing agent. When passed through acidified \(K_2Cr_2O_7\) solution or paper, it reduces the \(Cr(VI)\) in dichromate to \(Cr(III)\).

The chemical equation is:
\[ K_2Cr_2O_7 (orange) + 3SO_2 + H_2SO_4 \to Cr_2(SO_4)_3 (green) + K_2SO_4 + H_2O \]

The change from \(+6\) (orange) to \(+3\) (green) oxidation state of Chromium confirms the presence of \(SO_2\).


Step 3: Final Answer:
The gas is Sulphur dioxide.
Quick Tip: This is a characteristic test for \(SO_2\) gas. While \(H_2S\) also turns it green, it would also produce a yellow precipitate of sulphur, whereas \(SO_2\) produces a clear green solution.

Step 1: Understanding the Concept:

The reactions describe two different methods of hydration of an alkene: Hydroboration-Oxidation and Oxymercuration-Demercuration.


Step 2: Detailed Explanation:

The starting material is 2-methylbut-2-ene.

- Reaction A (Hydroboration-Oxidation): Reagents \(BH_3, THF\) followed by \(H_2O_2/OH^-\). This proceeds via Anti-Markovnikov addition of water. The OH group attaches to the less substituted carbon of the double bond.

Product A: 3-methylbutan-2-ol.

- Reaction B (Oxymercuration-Demercuration): Reagents \(Hg(OAc)_2, H_2O\) followed by \(NaBH_4\). This proceeds via Markovnikov addition of water without any carbocation rearrangement. The OH group attaches to the more substituted carbon.

Product B: 2-methylbutan-2-ol.


Step 3: Final Answer:

The products are A: 3-methylbutan-2-ol and B: 2-methylbutan-2-ol.
Quick Tip: Hydroboration = Anti-Markovnikov (OH on less substituted C).
Oxymercuration = Markovnikov (OH on more substituted C).
This rule covers almost all alkene hydration questions in exams!

Step 1: Understanding the Concept:

This question tests periodic trends and chemical properties of Group 1 (alkali) metals, including electrode potentials, solubility, and thermal stability.


Step 2: Detailed Explanation:

- Statement A: Standard reduction potentials (\(E^\circ\)) are: \(Li (-3.05 V)\), \(Rb (-2.93 V)\), \(Na (-2.71 V)\). In terms of value: \(-2.71 > -2.93 > -3.05\). So \(Na > Rb > Li\) is correct.

- Statement B: Solubility depends on lattice and hydration energies. Large cations with large anions (\(Cs^+\) and \(I^-\)) have low solubility due to size matching. \(CsI\) is poorly soluble. Incorrect.

- Statement C: \(Li_2CO_3\) is unstable because \(Li^+\) is small and polarizes the large \(CO_3^{2-}\) ion, causing it to decompose into \(Li_2O\) and \(CO_2\). Incorrect.

- Statement D: \textit{Dilute solutions are blue and paramagnetic. \textit{Concentrated solutions are bronze-colored and diamagnetic. Incorrect.

- Statement E: All alkali metal hydrides (\(LiH, NaH\), etc.) are ionic crystalline solids with high melting points. Correct.


Step 3: Final Answer:

Only statements A and E are correct.
Quick Tip: Lithium is the strongest reducing agent in aqueous solution due to its exceptionally high hydration energy, making its reduction potential the most negative.

Step 1: Understanding the Concept:

A substance acts as a reducing agent when it loses electrons (its oxidation state increases) and reduces another species. In \(H_2O_2\), oxygen is in the \(-1\) state; as a reducing agent, it is oxidized to \(O_2\) (\(0\) state).


Step 2: Detailed Explanation:

Let's analyze the oxidation states in reaction (A):

- In \(HOCl\), Chlorine is in \(+1\) state. In \(Cl^-\), it is in \(-1\) state. Chlorine is reduced.

- In \(H_2O_2\), Oxygen is in \(-1\) state. In \(O_2\), it is in \(0\) state. Oxygen is oxidized.

Since \(H_2O_2\) is oxidized while reducing \(HOCl\), it acts as a reducing agent.

In all other options (B, C, and D), \(H_2O_2\) oxidizes the other reactant (\(Mn^{2+} \to Mn^{4+}\), \(Fe^{2+} \to Fe^{3+}\), \(S^{2-} \to S^{6+}\)), thus acting as an oxidizing agent.


Step 3: Final Answer:

Reaction (A) shows \(H_2O_2\) acting as a reducing agent.
Quick Tip: Whenever you see \(O_2\) gas being evolved from \(H_2O_2\), it's a clear sign that \(H_2O_2\) is being oxidized and is thus acting as a reducing agent.

Step 1: Understanding the Concept:

This refers to the hydrometallurgical extraction process, specifically the cyanide process used for noble metals.


Step 2: Detailed Explanation:

Silver (Ag) is extracted from its ore (like Argentite, \(Ag_2S\)) using the MacArthur-Forrest cyanide process:

1. Oxidation/Leaching: The ore is treated with a dilute solution of \(NaCN\) in the presence of air (\(O_2\)). Silver is oxidized from its elemental or sulfide form to \(Ag^+\) in a complex.
\[ 4Ag + 8CN^- + O_2 + 2H_2O \to 4[Ag(CN)_2]^- + 4OH^- \]

2. Reduction/Precipitation: The complex is then treated with a more reactive metal like Zinc, which reduces \(Ag^+\) back to metallic silver.
\[ 2[Ag(CN)_2]^- + Zn \to [Zn(CN)_4]^{2-} + 2Ag \]


Step 3: Final Answer:

The metal is Silver (Ag).
Quick Tip: Noble metals like Silver and Gold are extracted via "Hydrometallurgy" using leaching. Remember that Zinc acts as the reducing agent in the final step.

Step 1: Understanding the Concept:

This question pertains to the physical properties and stability of primary aromatic amines (arylamines) like aniline.


Step 2: Detailed Explanation:

- Statement I: Pure aniline is a colorless oily liquid. Most other arylamines are also colorless when freshly prepared. This statement is correct.

- Statement II: On exposure to air and light, arylamines gradually turn brown or dark red. This is due to slow atmospheric oxidation (forming complex quinone-like structures), not reduction. Therefore, Statement II is incorrect.


Step 3: Final Answer:

Statement I is correct, but Statement II is incorrect.
Quick Tip: Aniline is very sensitive to oxidation. Even a small amount of impurity makes it look yellow or brown. Always store it in dark bottles away from air.

Step 1: Understanding the Concept:

Clemmensen and Wolff-Kishner are complementary methods for reducing aldehydes and ketones to alkanes. However, they use different reagents that may affect other sensitive functional groups in the molecule.


Step 2: Detailed Explanation:

- Statement I (Clemmensen Reduction): Uses \(Zn/Hg\) and concentrated \(HCl\). It successfully reduces the carbonyl group to a \(CH_2\) group. While amines protonate in acid, the primary reduction of the ketone still occurs. The statement depicts a valid transformation. Correct.

- Statement II (Wolff-Kishner Reduction): Uses \(NH_2NH_2\) and \(KOH\) (strong base). The ketone group should reduce to \(CH_2\). However, the molecule contains an alkyl chloride (\(Cl\) group). In the presence of a strong base like \(KOH\) at high temperatures, the alkyl halide will undergo dehydrohalogenation (elimination) to form an alkene. The product shown in the statement incorrectly retains the \(Cl\) atom. Therefore, Statement II is false.


Step 3: Final Answer:

Statement I is true, and Statement II is false.
Quick Tip: Mnemonic: Clemmensen is Acidic (HCl), avoid with acid-sensitive groups. Wolff-Kishner is Basic (KOH), avoid with base-sensitive groups like halides (\(Cl, Br\)).

Step 1: Understanding the Concept:

The standard reduction potential (\(E^{\circ}\)) of a metal is a measure of its tendency to be reduced.

For metals, it is influenced by three main energetic factors: the enthalpy of atomization (\(\Delta H_{atom}\)), the ionization enthalpy (\(IE\)), and the hydration enthalpy (\(\Delta H_{hyd}\)).


Step 2: Key Formula or Approach:

The total enthalpy change for the process \(M(s) \rightarrow M^{2+}(aq) + 2e^{-}\) is:
\[ \Delta H_{total} = \Delta H_{atom} + IE_{1} + IE_{2} + \Delta H_{hyd} \]

A less negative \(E^{\circ}\) value indicates that the oxidation process is less favorable compared to other elements in the same group.


Step 3: Detailed Explanation:

1. Assertion analysis: Beryllium (\(Be\)) has a reduction potential of \(-1.97\,V\), while other alkaline earth metals like Magnesium (\(Mg\)) and Calcium (\(Ca\)) have values around \(-2.36\,V\) and \(-2.84\,V\) respectively.

Thus, \(Be\) indeed has a less negative value. Assertion A is correct.

2. Reason analysis: Due to its exceptionally small atomic and ionic size, Beryllium has a very high hydration energy.

However, this is offset by its very high enthalpy of atomization and exceptionally high ionization enthalpies (because the electrons are close to the nucleus).

The high energy required for atomization and ionization is not fully compensated by the hydration energy, leading to a less negative reduction potential overall.

Thus, Reason R is correct and provides the physical basis for the assertion.


Step 4: Final Answer:

Both statements are correct, and R explains A.
Quick Tip: For s-block elements, reduction potential is a balance of "costly" steps (sublimation, ionization) and "profitable" steps (hydration). For Be, the cost of ionization is too high.

Step 1: Understanding the Concept:

This question pertains to human physiology and environmental exposure levels discussed in environmental chemistry.


Step 2: Detailed Explanation:

1. On average, a human being consumes about \(1.5\) to \(2.0\) kg of food per day.

2. In contrast, an average adult breathes in approximately \(10,000\) to \(20,000\) liters of air per day.

3. Given that the density of air is approximately \(1.2\,kg/m^{3}\), the mass of air consumed is roughly \(12\) to \(24\) kg per day.

4. By comparing the masses: \( Mass of air / Mass of food \approx 20/1.5 \approx 13.3 \) to \(15\).

According to standard environmental chemistry facts (NCERT), a human consumes nearly \(15\) times more air than food by weight.


Step 3: Final Answer:

The mass of air consumed daily is significantly higher, approximately 15 times that of food.
Quick Tip: This fact explains why air pollution is considered more dangerous than food contamination; we take in a much larger "dose" of air pollutants daily.

Step 1: Understanding the Concept:

Deprotonation occurs most readily from the most acidic site. For active methylene compounds, acidity depends on the resonance stabilization of the resulting carbanion by the adjacent electron-withdrawing groups (EWG).


Step 2: Detailed Explanation:

The compounds are:

(a) Pentane-2,4-dione (1,3-diketone): The central \(CH_2\) is between two ketone groups.

(b) Dimethyl malonate (1,3-diester): The central \(CH_2\) is between two ester groups.

(c) Methyl acetoacetate (keto-ester): The central \(CH_2\) is between one ketone and one ester group.


1. Ketone groups (\(-COR\)) are more electron-withdrawing than ester groups (\(-COOR\)) because the alkoxy (\(-OR\)) group in an ester donates electrons into the carbonyl via resonance, reducing the carbonyl's ability to stabilize an external negative charge.

2. Therefore, the order of acidity is: 1,3-diketone \(>\) keto-ester \(>\) 1,3-diester.

3. Compound (a) has the most stable conjugate base and thus has the lowest \(pK_a\) (highest acidity).


Step 3: Final Answer:

Compound (a) is the most acidic and will deprotonate most readily in a basic medium.
Quick Tip: Acidity Order: \(-CHO > -COR > -COOR > -CONR_2\). More ketones surrounding a \(CH_2\) mean higher acidity.

Step 1: Understanding the Concept:

This question tests the knowledge of medicinal drug classifications from Chemistry in Everyday Life.


Step 2: Detailed Explanation:

1. Antifertility drugs: These are used for birth control. Norethindrone is a synthetic progesterone used in oral contraceptives. (A \(\rightarrow\) I)

2. Tranquilizers: These are used to treat stress and anxiety. Meprobamate is a mild tranquilizer. (B \(\rightarrow\) II)

3. Antihistamines: These are used to treat allergic reactions. Seldane (Terfenadine) is a common antihistamine. (C \(\rightarrow\) III)

4. Antibiotics: These are substances used to kill or inhibit bacteria. Ampicillin is a semi-synthetic antibiotic related to penicillin. (D \(\rightarrow\) IV)


Step 3: Final Answer:

The correct match is A-I, B-II, C-III, D-IV.
Quick Tip: Commonly asked drug classes: Equanil/Meprobamate (Tranquilizers), Terfenadine/Seldane (Antihistamines), Novestrol/Norethindrone (Antifertility).

Step 1: Understanding the Concept:

Benzoic acid (\(C_6H_5COOH\)) is a weak acid, and sodium hydroxide (\(NaOH\)) is a strong base. In a conductometric titration, we monitor the change in conductivity as the titrant is added.


Step 2: Detailed Explanation:

1. Initial Phase: Benzoic acid is weakly dissociated, so the starting conductance is low.

2. Addition of NaOH: As \(NaOH\) is added, poorly conducting acid is converted into highly dissociated salt (sodium benzoate). This leads to a steady increase in the number of ions and thus an increase in conductance.

3. Equivalence Point: At this point, all acid has been converted to salt.

4. Post-Equivalence: Excess \(NaOH\) adds highly mobile \(OH^{-}\) ions to the solution. This causes the conductance to rise much more steeply than before.

5. Graph Character: The resulting graph shows an initial slow linear rise followed by a sharp linear rise after the endpoint. This matches Graph 4.


Step 3: Final Answer:

Graph 4 is the correct representation for a Weak Acid vs. Strong Base conductometric titration.
Quick Tip: Strong Acid vs. Strong Base gives a V-shaped graph. Weak Acid vs. Strong Base starts low and keeps increasing, with a steeper slope after the equivalence point.

Step 1: Understanding the Concept:

Stability of benzene is attributed to its resonance energy. Resonance involves the delocalization of electrons across several nuclei.


Step 2: Detailed Explanation:

1. Assertion analysis: Benzene has a resonance energy of \(152\,kJ/mol\), meaning it is significantly more stable than a hypothetical molecule with three localized double bonds (cyclohexatriene). Assertion A is correct.

2. Reason analysis: In localized systems, electrons are held between two nuclei. In delocalized systems like benzene, the \(\pi\) electrons are shared by six carbon nuclei. This increased nuclear attraction lowers the potential energy of the electrons, enhancing stability. Reason R is correct.

3. Connection: The fundamental reason for the "resonance stabilization" mentioned in Assertion A is the increased attraction described in Reason R.


Step 3: Final Answer:

Both statements are correct and the Reason explains the Assertion.
Quick Tip: Delocalization = Lower Energy = Higher Stability. The energy of a system decreases as the volume occupied by its electrons increases.

Step 1: Understanding the Concept:

Protons determine the atomic number (\(Z\)). \(Z = 55\) corresponds to Cesium (Cs). A "unipositive state" refers to the \(Cs^{+}\) ion.


Step 2: Detailed Explanation:

1. Neutral Cs (Z=55) Configuration:
\(1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{10}, 4s^{2} 4p^{6} 4d^{10}, 5s^{2} 5p^{6}, 6s^{1}\).

2. Formation of \(Cs^{+}\): One electron is removed from the outermost shell (6s).

Configuration of \(Cs^{+}\): \(1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{10}, 4s^{2} 4p^{6} 4d^{10}, 5s^{2} 5p^{6}\).

3. Counting s-electrons:

- \(1s \rightarrow 2\)

- \(2s \rightarrow 2\)

- \(3s \rightarrow 2\)

- \(4s \rightarrow 2\)

- \(5s \rightarrow 2\)

Total s-electrons = \(2 + 2 + 2 + 2 + 2 = 10\).


Step 3: Final Answer:

The \(Cs^{+}\) ion contains 10 s-electrons.
Quick Tip: For any alkali metal \(M\) in period \(n\), the ion \(M^{+}\) will have \(2(n-1)\) s-electrons. Since Cs is in Period 6, it has \(2 \times (6-1) = 10\) s-electrons.

Step 1: Understanding the Concept:

We apply Valence Bond Theory (VBT) or Crystal Field Theory (CFT) to determine the electronic distribution in the complex.


Step 2: Detailed Explanation:

1. Oxidation State: \(Co\) in \([Co(NH_3)_6]^{3+}\) is \(Co^{3+}\).

2. Electronic Configuration: Neutral \(Co\) is \([Ar] 3d^{7} 4s^{2}\). \(Co^{3+}\) is \([Ar] 3d^{6}\).

3. Ligand strength: \(NH_3\) acts as a strong field ligand for \(Co^{3+}\), forcing the six electrons in the \(3d\) orbitals to pair up.

4. Distribution: The 6 electrons fill the three \(t_{2g}\) orbitals: \( (t_{2g})^{6} (e_g)^{0} \). All electrons are paired.

5. Hybridization: This leaves two \(3d\) orbitals empty for bonding. Hybridization involves two \(3d\), one \(4s\), and three \(4p\) orbitals \(\rightarrow d^{2}sp^{3}\).

6. Magnetic Behaviour: Since all electrons are paired, it is diamagnetic.


Step 3: Final Answer:

The complex is \(d^{2}sp^{3}\) hybridized and diamagnetic.
Quick Tip: Cobalt (III) complexes with strong field ligands like \(NH_3, CN^{-}\), and \(NO_2^{-}\) are almost always low-spin, inner-orbital, and diamagnetic.

Step 1: Understanding the Concept:

The half-life (\(t_{1/2}\)) of a reaction is related to the initial concentration (or pressure \(p\)) by the formula \(t_{1/2} \propto 1 / p^{n-1}\), where \(n\) is the order of the reaction.


Step 2: Key Formula or Approach:
\[ \frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{p_2}{p_1} \right)^{n-1} \]


Step 3: Detailed Explanation:

Take any two points from the table:

Point 1: \(p_1 = 50, t_{1/2} = 4\)

Point 2: \(p_2 = 100, t_{1/2} = 2\)
\[ \frac{4}{2} = \left( \frac{100}{50} \right)^{n-1} \]
\[ 2 = (2)^{n-1} \]

By comparing exponents: \(n - 1 = 1 \Rightarrow n = 2\).

We can verify with other points (e.g., \(200/400\)): \(1/0.5 = 2 = (400/200)^{n-1} \Rightarrow 2 = 2^{n-1} \Rightarrow n=2\).


Step 4: Final Answer:

The order of the reaction is 2.
Quick Tip: If \(t_{1/2} \times p = constant\), the reaction is 2nd order. If \(t_{1/2} = constant\), it's 1st order. If \(t_{1/2} \propto p\), it's zero order.

Step 1: Understanding the Concept:

Molecular Orbital Theory (MOT) describes the electronic configuration and bond properties of diatomic species.


Step 2: Detailed Explanation:

1. \(N_2\) (14 \(e^{-}\)): Configuration: \(\sigma 1s^{2} \sigma^{*}1s^{2} \sigma 2s^{2} \sigma^{*}2s^{2} (\pi 2p_x^{2} = \pi 2p_y^{2}) \sigma 2p_z^{2}\). The HOMO is \(\sigma 2p_z\) with 2 electrons. Unpaired electrons = 0.

2. \(N_2^{+}\) (13 \(e^{-}\)): One electron is removed from the HOMO (\(\sigma 2p_z\)). Unpaired electrons = 1.

3. \(O_2\) (16 \(e^{-}\)): Configuration: \(\sigma 1s^{2} \sigma^{*}1s^{2} \sigma 2s^{2} \sigma^{*}2s^{2} \sigma 2p_z^{2} (\pi 2p_x^{2} = \pi 2p_y^{2}) (\pi^{*} 2p_x^{1} = \pi^{*} 2p_y^{1})\). The HOMO is \(\pi^{*}\) with 2 unpaired electrons (Hund's rule). Unpaired electrons = 2.

4. \(O_2^{+}\) (15 \(e^{-}\)): One electron is removed from the \(\pi^{*}\) orbital. Unpaired electrons = 1.


Step 3: Final Answer:

The number of unpaired electrons is 0, 1, 2, 1.
Quick Tip: Remember that \(O_2\) is one of the classic examples of a paramagnetic molecule despite having an even number of electrons because of the degenerate \(\pi^{*}\) molecular orbitals.

Step 1: Understanding the Concept:

The reaction is a diazo-coupling reaction between a benzenediazonium derivative and an aromatic amine to form an azo dye.


Step 2: Detailed Explanation:

1. The reactant on the left is diazotized sulphanilic acid.

2. It reacts with 1-naphthylamine. The coupling typically occurs at the para-position to the amine group.

3. The resulting product is an extended conjugated system containing the \(-N=N-\) (azo) group.

4. This specific product is a well-known azo dye. Azo dyes formed from naphthylamines are characterized by intense colours, typically in the red/orange spectrum. In the Griess test for nitrites, this exact reaction produces a distinctive red colour.


Step 3: Final Answer:

The product formed is a red azo dye.
Quick Tip: Benzene-based azo dyes with amines are often yellow/orange, but naphthalene-based azo dyes usually shift the absorption towards longer wavelengths (red).

Step 1: Understanding the Concept:

The number of \(\pi\)-bonds in an oxoacid of sulphur is determined by its molecular structure, specifically the number of \(S=O\) double bonds formed when sulphur is in its higher oxidation states.


Step 2: Key Formula or Approach:

Draw the chemical structures for both acids:

1. Peroxodisulphuric acid (\(H_{2}S_{2}O_{8}\)), also known as Marshall's acid.

2. Pyrosulphuric acid (\(H_{2}S_{2}O_{7}\)), also known as Oleum.


Step 3: Detailed Explanation:

1. Peroxodisulphuric acid (\(H_{2}S_{2}O_{8}\)):

Its structure is \(HO-S(=O)_{2}-O-O-S(=O)_{2}-OH\).

Each sulphur atom is bonded to two oxygen atoms via double bonds (\(S=O\)).

Since there are two sulphur atoms, total \(\pi\)-bonds = \(2 + 2 = 4\).


2. Pyrosulphuric acid (\(H_{2}S_{2}O_{7}\)):

Its structure is \(HO-S(=O)_{2}-O-S(=O)_{2}-OH\).

Similarly, each sulphur atom forms two \(S=O\) double bonds.

Total \(\pi\)-bonds = \(2 + 2 = 4\).


3. Sum of \(\pi\)-bonds:
\[ Sum = 4 (in H_{2}S_{2}O_{8}) + 4 (in H_{2}S_{2}O_{7}) = 8 \]


Step 4: Final Answer:

The sum of \(\pi\)-bonds is 8.
Quick Tip: In oxoacids of sulphur where sulphur is in the +6 oxidation state (like \(H_{2}SO_{4}\), \(H_{2}S_{2}O_{7}\), \(H_{2}S_{2}O_{8}\)), each sulphur atom typically contributes 2 \(\pi\)-bonds from its two \(S=O\) groups.

Step 1: Understanding the Concept:

Concentration is a measure of the amount of solute present in a given amount of solvent or solution. Units of concentration relate the quantity of solute to the quantity of the solution or solvent.


Step 2: Detailed Explanation:

Let's evaluate each term in the list:

1. Mass percent: Expresses the mass of solute per 100g of solution. (Concentration unit)

2. Mole: A unit of the amount of substance (SI unit). It is a quantity, not a ratio or concentration. (Not a concentration unit)

3. Mole fraction: The ratio of moles of a component to the total moles in the mixture. (Concentration unit)

4. Molarity: Moles of solute per liter of solution. (Concentration unit)

5. ppm (parts per million): Expresses very dilute concentrations as parts of solute per million parts of solution. (Concentration unit)

6. Molality: Moles of solute per kilogram of solvent. (Concentration unit)


Counting the valid units: Mass percent, Mole fraction, Molarity, ppm, and Molality. Total = 5.


Step 3: Final Answer:

The number of concentration units is 5.
Quick Tip: Always distinguish between units of quantity (like mole, gram, liter) and units of concentration (which are always ratios of solute to solution/solvent).

Step 1: Understanding the Concept:

The number of monochloro derivatives depends on the number of chemically distinct sets of hydrogen atoms in the molecule and whether substitution at those positions creates chiral centers.


Step 2: Key Formula or Approach:

Draw the structure of 2,2,5,5-tetramethylhexane:
\[ CH_{3} - C(CH_{3})_{2} - CH_{2} - CH_{2} - C(CH_{3})_{2} - CH_{3} \]

Identify equivalent groups of hydrogens based on molecular symmetry.


Step 3: Detailed Explanation:

1. Symmetry Analysis: The molecule is highly symmetric. There is a center of symmetry in the middle of the \(C3-C4\) bond.

2. Set 1 (Methyl Hydrogens): All 18 methyl hydrogens (at \(C1, C6\) and the four branch methyls) are equivalent due to rotation and symmetry. Replacing any of these with Cl gives:
\(Cl-CH_{2}-C(CH_{3})_{2}-CH_{2}-CH_{2}-C(CH_{3})_{2}-CH_{3}\) (1-chloro-2,2,5,5-tetramethylhexane). This molecule is achiral. (1 isomer)

3. Set 2 (Methylene Hydrogens): The 4 hydrogens on \(C3\) and \(C4\) are equivalent. Replacing one H at \(C3\) with Cl gives:
\(CH_{3}-C(CH_{3})_{2}-CHCl-CH_{2}-C(CH_{3})_{2}-CH_{3}\).

4. Stereochemistry: The carbon atom where Cl is attached (\(C3\)) becomes a chiral center because it is bonded to four different groups: H, Cl, a tert-butyl group, and a neopentyl-like group. This results in a pair of enantiomers (R and S). (2 isomers)

5. Total Isomers: \(1 (achiral) + 2 (enantiomers) = 3\).


Step 4: Final Answer:

The maximum number of isomeric monochloro derivatives is 3.
Quick Tip: When a question asks for "isomers" without specifying "structural isomers," you must include stereoisomers (enantiomers and diastereomers) in your count.

Step 1: Understanding the Concept:

Black body radiation curves describe the intensity of radiation emitted by a black body at different wavelengths for various temperatures. Key laws involved are Wien's Displacement Law and Planck's Law.


Step 2: Detailed Explanation:

Analyze each statement:

A. Incorrect: Looking at the graph, \(T_{1}\) has the highest peak and the peak wavelength is shifted most to the left (shortest \(\lambda\)). According to Wien's law, higher temperature means shorter peak wavelength and higher intensity. Thus, \(T_{1} > T_{2} > T_{3} > T_{4}\).

B. Incorrect: While Planck's model used "oscillators" (resonators), it is a theoretical construct for the walls of the cavity. Saying the black body "consists of particles performing SHM" is an oversimplification and not a defining characteristic of the spectrum.

C. Correct: Wien's Displacement Law (\(\lambda_{m}T = b\)) states that as temperature increases, the peak wavelength \(\lambda_{m}\) decreases.

D. Incorrect: Wien's law in terms of frequency is \(\frac{\nu_{m}}{T} = constant\). The statement says the ratio is not constant.

E. Correct: The classical Rayleigh-Jeans law failed to explain the spectrum (ultraviolet catastrophe). Planck successfully explained it by assuming energy is quantized (\(E = nh\nu\)).


Correct statements are C and E. Total = 2.


Step 3: Final Answer:

The number of correct statements is 2.
Quick Tip: Remember the two main trends in black body graphs as T increases: 1. The total area under the curve increases (\(E \propto T^{4}\)). 2. The peak shifts toward the left (shorter \(\lambda\), higher \(\nu\)).

Step 1: Understanding the Concept:

Work done in a thermodynamic cycle is the area enclosed by the \(P-V\) loop. Work for individual processes is calculated using \(W = -\int P dV\).


Step 2: Detailed Explanation:

From the graph, identify the cycle \(1 \rightarrow 2 \rightarrow 3 \rightarrow 1\):

- Process \(1 \rightarrow 2\): Isobaric expansion at \(P = 1.0\) bar.
\(W_{12} = -P \Delta V = -1.0 \times (40 - 20) = -20\) bar\(\cdot\)L.

- Process \(2 \rightarrow 3\): Isochoric cooling at \(V = 40\) L.
\(W_{23} = 0\) (as \(\Delta V = 0\)).

- Process \(3 \rightarrow 1\): Curved path. Check if it's isothermal:

At point 1: \(P \times V = 1.0 \times 20 = 20\).

At point 3: \(P \times V = 0.5 \times 40 = 20\).

Since \(PV\) is constant, it is an isothermal compression.
\(W_{31} = -nRT \ln\left(\frac{V_{final}}{V_{initial}}\right) = -P_{1}V_{1} \ln\left(\frac{V_{1}}{V_{3}}\right) = -20 \ln\left(\frac{20}{40}\right) = 20 \ln 2\).


Step 3: Calculation:

1. \(\ln 2 = 2.3 \times \log 2 = 2.3 \times 0.3 = 0.69\).

2. \(W_{31} = 20 \times 0.69 = 13.8\) bar\(\cdot\)L.

3. \(Net Work W = W_{12} + W_{23} + W_{31} = -20 + 0 + 13.8 = -6.2\) bar\(\cdot\)L.

4. \(Magnitude = 6.2\) bar\(\cdot\)L.

5. Convert to Joules (\(1 bar\cdotL = 100 J\)):
\(W = 6.2 \times 100 = 620\) J.


Step 4: Final Answer:

The magnitude of work done is 620 J.
Quick Tip: For an ideal gas, if \(P_{1}V_{1} = P_{2}V_{2}\), the process is isothermal. Use the formula \(W = 2.303 nRT \log(V_{1}/V_{2})\) carefully, keeping track of signs for expansion (-) and compression (+).

Step 1: Understanding the Concept:

A tripeptide is formed by the linkage of three amino acids. If we are "mixing" two amino acids, each of the three positions in the tripeptide chain can be occupied by either of the two available amino acids.


Step 2: Detailed Explanation:

1. Let the two amino acids be \(A\) (Valine) and \(B\) (Proline).

2. A tripeptide has 3 positions: \(Pos 1 - Pos 2 - Pos 3\).

3. Number of choices for Position 1 = 2 (either Val or Pro).

4. Number of choices for Position 2 = 2.

5. Number of choices for Position 3 = 2.

6. Total combinations = \(2 \times 2 \times 2 = 2^{3} = 8\).


The possible tripeptides are: Val-Val-Val, Val-Val-Pro, Val-Pro-Val, Pro-Val-Val, Val-Pro-Pro, Pro-Val-Pro, Pro-Pro-Val, Pro-Pro-Pro.


Step 3: Final Answer:

The total number of tripeptides is 8.
Quick Tip: If \(n\) different amino acids are available to form a peptide of length \(k\), the total number of possible peptides is \(n^{k}\). Here, \(n=2\) and \(k=3\).

Step 1: Understanding the Concept:

For an ideal solution, the total vapour pressure (\(P_{T}\)) is given by Raoult's Law: \(P_{T} = P_{A}^{\circ}X_{A} + P_{B}^{\circ}X_{B}\), where \(P^{\circ}\) are pure component pressures and \(X\) are mole fractions.


Step 2: Key Formula or Approach:

We have two unknowns (\(P_{A}^{\circ}\) and \(P_{B}^{\circ}\)) and two equations based on the two given conditions.


Step 3: Detailed Explanation:

1. Case 1: \(X_{A} = 0.7, X_{B} = 0.3, P_{T} = 350\).
\[ 0.7 P_{A}^{\circ} + 0.3 P_{B}^{\circ} = 350 \quad ---(1) \]

2. Case 2: \(X_{A} = 0.2, X_{B} = 0.8, P_{T} = 410\).
\[ 0.2 P_{A}^{\circ} + 0.8 P_{B}^{\circ} = 410 \quad ---(2) \]

3. Solving the equations:

Multiply equation (1) by 8 and equation (2) by 3 to eliminate \(P_{B}^{\circ}\):
\[ 5.6 P_{A}^{\circ} + 2.4 P_{B}^{\circ} = 2800 \]
\[ 0.6 P_{A}^{\circ} + 2.4 P_{B}^{\circ} = 1230 \]

Subtracting the two:
\[ 5.0 P_{A}^{\circ} = 1570 \Rightarrow P_{A}^{\circ} = \frac{1570}{5} = 314 mm Hg \]


Step 4: Final Answer:

The vapour pressure of pure A is 314 mm Hg.
Quick Tip: In Raoult's Law problems with two conditions, always set up a system of linear equations. If you need \(P_{A}^{\circ}\), eliminate \(P_{B}^{\circ}\) by matching its coefficients.

Step 1: Understanding the Concept:

Calcium lactate is a salt of a weak acid (lactic acid) and a strong base (calcium hydroxide). The pH of such a salt solution is calculated using the salt hydrolysis formula.


Step 2: Key Formula or Approach:

1. Salt concentration \(C = [Lactate^{-}]\). Since calcium lactate is \((CH_{3}CH(OH)COO)_{2}Ca\), one mole of salt gives 2 moles of lactate ions.

2. \(pH = 7 + \frac{1}{2}(pK_{a} + \log C)\).


Step 3: Detailed Explanation:

1. Calculate Ion Concentration:
\(Molarity of Calcium Lactate = 0.005 M\).
\([Lactate^{-}] = 2 \times 0.005 = 0.01 M = 10^{-2} M\).

2. Apply pH Formula:
\(pH = 7 + \frac{1}{2}(5 + \log 10^{-2})\)
\(pH = 7 + \frac{1}{2}(5 - 2) = 7 + \frac{3}{2} = 7 + 1.5 = 8.5\).

3. Convert to required format:
\(8.5 = 85 \times 10^{-1}\).


Step 4: Final Answer:

The value is 85.
Quick Tip: Don't forget the stoichiometry! For salts of divalent metals like \(Ca^{2+}\), the concentration of the anion (the species that hydrolyzes) is twice the molarity of the salt.

Step 1: Understanding the Concept:

Physisorption (physical adsorption) involves weak van der Waals forces between the adsorbate and adsorbent, unlike chemisorption which involves chemical bonding.


Step 2: Detailed Explanation:

Evaluate each statement:

A. Incorrect: Physisorption is not specific; any gas can be adsorbed to some extent on any solid. Chemisorption is highly specific.

B. Incorrect: Enthalpy is low (20-40 kJ/mol) because of weak forces.

C. Correct: Since adsorption is exothermic, according to Le Chatelier's principle, it decreases with increasing temperature.

D. Incorrect: It usually forms multimolecular layers. Unimolecular layers are characteristic of chemisorption.

E. Correct: Physisorption occurs almost instantaneously and does not require activation energy.


Characteristics are C and E. Total = 2.


Step 3: Final Answer:

The number of characteristic statements is 2.
Quick Tip: Think of Physisorption as "surface condensation." It's non-specific, low energy, and multilayered—just like a liquid film forming on a surface.

Step 1: Understanding the Concept:

The Andrews isotherm for \(CO_{2}\) shows the transition from gas to liquid. The horizontal portion of the curve represents the region where phase transition (liquefaction) occurs at constant pressure.


Step 2: Detailed Explanation:

Based on standard Andrews isotherm nomenclature:

A. Correct: From point (a) to point (b), the \(CO_{2}\) is being compressed as a gas. Liquefaction starts exactly at point (b).

B. Incorrect: Liquid appears for the first time at point (b), not (c). Point (c) is where the entire gas has been converted to liquid.

C. Correct: The horizontal line between (b) and (c) represents the equilibrium where both liquid and gaseous phases coexist.

D. Incorrect: As volume decreases from (b) to (c), more gas is converted to liquid, so the amount of liquid increases.


Correct statements are A and C. Total = 2.


Step 3: Final Answer:

The number of correct statements is 2.
Quick Tip: In a P-V isotherm, a horizontal line always indicates a phase change. During this change, pressure remains constant while the ratio of the two phases changes as volume is altered.

Step 1: Understanding the Concept:

We first find the value of \(\lambda\) using the position of vertex A. Then, we find the coordinates of C using properties of the orthocenter and finally compute the tangent length using the standard parabola formula.


Step 2: Key Formula or Approach:

1. A point on the y-axis has \(x = 0\).

2. The altitude from B is perpendicular to AC and passes through the orthocenter H.

3. Length of tangent from \((x_{1}, y_{1})\) to \(y^{2} = 4ax\) is \(\sqrt{y_{1}^{2} - 4ax_{1}}\).


Step 3: Detailed Explanation:

1. Find \(\lambda\): Vertex A lies on both AB and AC and has \(x = 0\).

From AB: \(\lambda y = 4 \Rightarrow y = 4/\lambda\).

From AC: \((1 - \lambda)y + \lambda = 0 \Rightarrow y = -\lambda / (1 - \lambda) = \lambda / (\lambda - 1)\).

Equating: \(4/\lambda = \lambda/(\lambda - 1) \Rightarrow 4\lambda - 4 = \lambda^{2} \Rightarrow (\lambda - 2)^{2} = 0 \Rightarrow \lambda = 2\).

Thus, \(A = (0, 2)\).

2. Equations of sides:

AB: \(3x + 2y = 4\) (Slope \(m_{AB} = -3/2\))

AC: \(2x - y + 2 = 0 \Rightarrow y = 2x + 2\) (Slope \(m_{AC} = 2\))

3. Find C: Let \(H = (1, 2)\) be the orthocenter.

The altitude from C is perpendicular to AB and passes through H.

Slope of altitude \(CH = -1/m_{AB} = 2/3\).

Equation of CH: \(y - 2 = \frac{2}{3}(x - 1) \Rightarrow 2x - 3y = -4\).

Point C is the intersection of AC and CH:
\(y = 2x + 2\) and \(2x - 3(2x + 2) = -4 \Rightarrow -4x = 2 \Rightarrow x = -1/2\).
\(y = 2(-1/2) + 2 = 1\). So \(C = (-1/2, 1)\).

4. Length of tangent: To parabola \(y^{2} - 6x = 0\) from \(C(-1/2, 1)\):
\(L = \sqrt{1^{2} - 6(-1/2)} = \sqrt{1 + 3} = \sqrt{4} = 2\).


Step 4: Final Answer:

The length of the tangent is 2.
Quick Tip: Remember that the vertex A is the intersection of the two given sides. If it lies on the y-axis, simply set \(x=0\) in both equations to solve for the parameter \(\lambda\).

Step 1: Understanding the Concept:

We use logical equivalences and De Morgan's laws to simplify the given expression.


Step 2: Detailed Explanation:

1. Start with the implication: \(p \Rightarrow \sim q \equiv \sim p \vee \sim q\).

2. Substitute into the inner term: \(p \wedge (p \Rightarrow \sim q) \equiv p \wedge (\sim p \vee \sim q)\).

3. Use the Distributive Law: \((p \wedge \sim p) \vee (p \wedge \sim q)\).

4. Since \(p \wedge \sim p\) is a contradiction (False): \(F \vee (p \wedge \sim q) \equiv p \wedge \sim q\).

5. Now apply the outer negation: \(\sim(p \wedge \sim q)\).

6. Use De Morgan's Law: \(\sim p \vee \sim(\sim q) \equiv \sim p \vee q\).


Comparing with options: Option (B) matches perfectly.


Step 3: Final Answer:

The expression is equivalent to \((\sim p) \vee q\).
Quick Tip: A quick shortcut: \(p \wedge (p \Rightarrow r)\) is simply \(p \wedge r\). Here \(r\) is \(\sim q\), so it becomes \(\sim(p \wedge \sim q)\), which is \(\sim p \vee q\).

Step 1: Understanding the Concept:

This is a homogeneous differential equation since the degree of all terms is 2. We can solve it using the substitution \(y = vx\).


Step 2: Key Formula or Approach:

Substitute \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\).


Step 3: Detailed Explanation:

1. Rearrange equation: \(3xy \frac{dy}{dx} = 3y^{2} - x^{2} \Rightarrow \frac{dy}{dx} = \frac{3y^{2} - x^{2}}{3xy}\).

2. Substitute \(y = vx\):
\(v + x \frac{dv}{dx} = \frac{3v^{2}x^{2} - x^{2}}{3vx^{2}} = \frac{3v^{2} - 1}{3v}\).
\(x \frac{dv}{dx} = \frac{3v^{2} - 1}{3v} - v = \frac{3v^{2} - 1 - 3v^{2}}{3v} = \frac{-1}{3v}\).

3. Separate variables: \(3v dv = -\frac{1}{x} dx\).

4. Integrate: \(\frac{3v^{2}}{2} = -\ln |x| + C\).

5. Back-substitute \(v = y/x\): \(\frac{3y^{2}}{2x^{2}} = -\ln |x| + C\).

6. Apply initial condition \(y(1) = 1\): \(\frac{3(1)}{2(1)} = 0 + C \Rightarrow C = \frac{3}{2}\).

Equation: \(\frac{3y^{2}}{2x^{2}} = \frac{3}{2} - \ln |x|\).

7. Find \(6y^{2}(e)\): Multiply entire equation by \(4x^{2}\):
\(6y^{2} = 6x^{2} - 4x^{2} \ln |x|\).

At \(x = e\): \(6y^{2}(e) = 6e^{2} - 4e^{2} \ln(e) = 6e^{2} - 4e^{2}(1) = 2e^{2}\).


Step 4: Final Answer:

The value of \(6y^{2}(e)\) is \(2e^{2}\).
Quick Tip: For homogeneous equations of the form \(M(x,y)dx + N(x,y)dy = 0\), if the substitution \(y=vx\) makes the integration look difficult, try \(x=vy\). In this case, \(y=vx\) led to a very simple separable form.

Step 1: Understanding the Concept:

This is a reciprocal-style quadratic equation. We can simplify it by letting \(t = x + \frac{1}{x}\).


Step 2: Key Formula or Approach:

If \(t = x + \frac{1}{x}\), then \(t^{2} = x^{2} + \frac{1}{x^{2}} + 2 \Rightarrow x^{2} + \frac{1}{x^{2}} = t^{2} - 2\).

Crucially, for real \(x\), \(|x + \frac{1}{x}| \geq 2\).


Step 3: Detailed Explanation:

1. Substitute \(t\):
\(3(t^{2} - 2) - 2t + 5 = 0\)
\(3t^{2} - 6 - 2t + 5 = 0\)
\(3t^{2} - 2t - 1 = 0\).

2. Solve for \(t\):

Factorizing: \(3t^{2} - 3t + t - 1 = 0 \Rightarrow (3t + 1)(t - 1) = 0\).

Possible values: \(t = 1\) or \(t = -1/3\).

3. Verify reality of \(x\):

For \(x\) to be real, the condition \(|t| \geq 2\) must be satisfied.

Check \(t = 1\): \(|1| < 2\). (No real \(x\))

Check \(t = -1/3\): \(|-1/3| < 2\). (No real \(x\))

Since neither value of \(t\) is valid for real \(x\), there are no real solutions.


Step 4: Final Answer:

The number of real solutions is 0.
Quick Tip: In equations involving \(x + \frac{1}{x}\), always check the range of the substituted variable. The property \(|x + \frac{1}{x}| \geq 2\) for real \(x\) (from AM-GM inequality) is a very frequent "trap" in competitive exams.

Step 1: Understanding the Concept:

The given function is of a special form where the sum of function values at symmetric points \(x\) and \(1-x\) often results in a constant. We need to check the value of \(f(x) + f(1-x)\).


Step 2: Key Formula or Approach:

Let's evaluate \(f(1-x)\):
\[ f(1-x) = \frac{2^{2(1-x)}}{2^{2(1-x)} + 2} = \frac{2^2 \cdot 2^{-2x}}{2^2 \cdot 2^{-2x} + 2} = \frac{4/2^{2x}}{4/2^{2x} + 2} \]

Multiplying the numerator and denominator by \(2^{2x}\):
\[ f(1-x) = \frac{4}{4 + 2 \cdot 2^{2x}} = \frac{2}{2 + 2^{2x}} \]

Now, compute \(f(x) + f(1-x)\):
\[ f(x) + f(1-x) = \frac{2^{2x}}{2^{2x} + 2} + \frac{2}{2^{2x} + 2} = \frac{2^{2x} + 2}{2^{2x} + 2} = 1 \]


Step 3: Detailed Explanation:

The given series is \(S = \sum_{k=1}^{2022} f\left(\frac{k}{2023}\right)\).

We can pair the terms from both ends:
\[ S = \left[ f\left(\frac{1}{2023}\right) + f\left(\frac{2022}{2023}\right) \right] + \left[ f\left(\frac{2}{2023}\right) + f\left(\frac{2021}{2023}\right) \right] + \dots \]

Each pair is of the form \(f(x) + f(1-x)\) where \(x = \frac{k}{2023}\), and thus each pair sums to 1.

There are 2022 terms in total, which forms \(\frac{2022}{2} = 1011\) such pairs.
\[ S = 1 \times 1011 = 1011 \]


Step 4: Final Answer:

The value of the sum is 1011.
Quick Tip: Whenever you see a summation of function values like this, always check if \(f(x) + f(1-x)\) or \(f(x) + f(a-x)\) is constant. It simplifies the entire series into pairs.

Step 1: Understanding the Concept:

A system of linear equations has infinitely many solutions if the rank of the augmented matrix is equal to the rank of the coefficient matrix, and both are less than the number of variables. Practically, we can use Gaussian elimination or determinants.


Step 2: Key Formula or Approach:

We can perform row operations on the augmented matrix:
\[ \begin{pmatrix} 1 & 2 & 3 & | & 3
4 & 3 & -4 & | & 4
8 & 4 & -\lambda & | & 9+\mu \end{pmatrix} \]

Perform \(R_2 \to R_2 - 4R_1\) and \(R_3 \to R_3 - 8R_1\):
\[ \begin{pmatrix} 1 & 2 & 3 & | & 3
0 & -5 & -16 & | & -8
0 & -12 & -\lambda-24 & | & \mu-15 \end{pmatrix} \]


Step 3: Detailed Explanation:

For the system to have infinitely many solutions, the third row must be a multiple of the second row:
\[ \frac{-12}{-5} = \frac{-\lambda-24}{-16} = \frac{\mu-15}{-8} \]

From the first equality:
\[ \frac{12}{5} = \frac{\lambda+24}{16} \Rightarrow 192 = 5\lambda + 120 \Rightarrow 5\lambda = 72 \Rightarrow \lambda = \frac{72}{5} \]

From the second equality:
\[ \frac{12}{5} = \frac{\mu-15}{-8} \Rightarrow -96 = 5\mu - 75 \Rightarrow 5\mu = -21 \Rightarrow \mu = -\frac{21}{5} \]


Step 4: Final Answer:

The ordered pair is \((\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}\right)\).
Quick Tip: For a \(3 \times 3\) system, "infinitely many solutions" means that the determinant \(\Delta = 0\) and all minor determinants \(\Delta_x, \Delta_y, \Delta_z\) are also 0. Row reduction is often faster for finding parameters.

Step 1: Understanding the Concept:

For a chord of a circle with radius \(R\) subtending an angle \(\theta\) at the center, the distance \(d\) of the midpoint of the chord from the center is given by \(d = R \cos\left(\frac{\theta}{2}\right)\). The locus of these midpoints is a concentric circle with radius \(r = d\).


Step 2: Key Formula or Approach:

Given circle \(C_i\) has radius \(R = \sqrt{4} = 2\).

Radius of locus \(r_i = R \cos\left(\frac{\theta_i}{2}\right) = 2 \cos\left(\frac{\theta_i}{2}\right)\).


Step 3: Detailed Explanation:

Using the given angles:

1) For \(\theta_1 = \pi/3\), \(r_1 = 2 \cos(\pi/6) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\).

2) For \(\theta_3 = 2\pi/3\), \(r_3 = 2 \cos(\pi/3) = 2 \cdot \frac{1}{2} = 1\).

Given relation: \(r_1^2 = r_2^2 + r_3^2\).
\[ (\sqrt{3})^2 = r_2^2 + 1^2 \Rightarrow 3 = r_2^2 + 1 \Rightarrow r_2^2 = 2 \Rightarrow r_2 = \sqrt{2} \]

Now find \(\theta_2\):
\[ r_2 = 2 \cos\left(\frac{\theta_2}{2}\right) \Rightarrow \sqrt{2} = 2 \cos\left(\frac{\theta_2}{2}\right) \Rightarrow \cos\left(\frac{\theta_2}{2}\right) = \frac{1}{\sqrt{2}} \]
\[ \frac{\theta_2}{2} = \frac{\pi}{4} \Rightarrow \theta_2 = \frac{\pi}{2} \]

Step 4: Final Answer:

The value of \(\theta_2\) is \(\pi/2\).
Quick Tip: Remember that the midpoint of a chord and the center of the circle form a right-angled triangle with half the chord length and the radius. The distance of the chord from the center is always \(R \cos(\theta/2)\).

Step 1: Understanding the Concept:

The equation of a plane passing through the intersection of two planes \(P_1 = 0\) and \(P_2 = 0\) is given by the family \(P_1 + k P_2 = 0\). We use the given points to find \(\lambda\) and \(k\).


Step 2: Detailed Explanation:

Let the plane be \(L : (x + (\lambda + 4)y + z - 1) + k(2x + y + z - 2) = 0\).

1) Passes through \((0, 1, 0)\):
\[ (\lambda + 4 - 1) + k(1 - 2) = 0 \Rightarrow \lambda + 3 - k = 0 \Rightarrow k = \lambda + 3 \]

2) Passes through \((1, 0, 1)\):
\[ (1 + 1 - 1) + k(2 + 1 - 2) = 0 \Rightarrow 1 + k = 0 \Rightarrow k = -1 \]

Comparing the values of \(k\):
\[ \lambda + 3 = -1 \Rightarrow \lambda = -4 \]

Now, we need to find the distance of point \(A = (2\lambda, \lambda, -\lambda) = (-8, -4, 4)\) from plane \(P_2 : 2x + y + z - 2 = 0\).


Step 3: Key Formula or Approach:

Distance \(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\).
\[ d = \frac{|2(-8) + (-4) + (4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}} = \frac{|-16 - 2|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6} \]

Step 4: Final Answer:

The distance is \(3\sqrt{6}\).
Quick Tip: When dealing with the "Family of Planes", always use the parameters \(k\) and \(\lambda\) systematically. Here, evaluating the points first made the problem straightforward.

Step 1: Understanding the Concept:

We use the properties of an Arithmetic Progression (A.P.) and the formulas for mean and variance of a data set.


Step 2: Key Formula or Approach:

Let the first term be \(a\) and common difference be \(d\).

1) \(a_1 + a_3 = 10 \Rightarrow a + (a + 2d) = 10 \Rightarrow 2a + 2d = 10 \Rightarrow a + d = 5\).

2) Mean \(\bar{x} = \frac{19}{2}\). For 6 terms in A.P., Mean \(= \frac{Sum}{6} = \frac{6/2 [2a + 5d]}{6} = \frac{2a + 5d}{2}\).
\[ \frac{2a + 5d}{2} = \frac{19}{2} \Rightarrow 2a + 5d = 19 \]


Step 3: Detailed Explanation:

Solving the system:
\(a + d = 5 \Rightarrow 2a + 2d = 10\).

Subtracting this from \(2a + 5d = 19\):
\[ 3d = 9 \Rightarrow d = 3 \]
\[ a = 5 - 3 = 2 \]

The terms are: 2, 5, 8, 11, 14, 17.

Variance for A.P. terms: \(\sigma^2 = \frac{d^2(n^2 - 1)}{12}\).
\[ \sigma^2 = \frac{3^2(6^2 - 1)}{12} = \frac{9 \times 35}{12} = \frac{3 \times 35}{4} = \frac{105}{4} \]

We need \(8\sigma^2\):
\[ 8\sigma^2 = 8 \times \frac{105}{4} = 2 \times 105 = 210 \]

Step 4: Final Answer:

The value of \(8\sigma^2\) is 210.
Quick Tip: The formula for the variance of \(n\) terms in A.P. is \(\frac{d^2(n^2 - 1)}{12}\). It is a major time-saver compared to the standard \(\frac{1}{n} \sum (x_i - \bar{x})^2\) method.

Step 1: Understanding the Concept:

For the limit of a floor function expression to be 0, the function must be constant and equal to 0 in a local neighborhood of \(a\). Any point where the floor value jumps (i.e., when the expression inside \([ \dots ]\) is an integer) requires checking left and right limits.


Step 2: Key Formula or Approach:

Let \(f(x) = [x - 5] - [2x + 2] = [x] - 5 - [2x] - 2 = [x] - [2x] - 7\).

We want \(\lim_{x \to a} ([x] - [2x]) = 7\).

Recall \([2x] = [x] + [x + 0.5]\).

Substituting this: \( [x] - ([x] + [x + 0.5]) = 7 \Rightarrow -[x + 0.5] = 7 \Rightarrow [x + 0.5] = -7\).


Step 3: Detailed Explanation:

For \([x + 0.5] = -7\), we must have:
\[ -7 \le x + 0.5 < -6 \Rightarrow -7.5 \le x < -6.5 \]

Now check the endpoints for the existence of the limit:

1) At \(x = -7.5\): L.H.L. involves \([-7.5^- + 0.5] = [-7^-] = -8\). R.H.L. involves \([-7.5^+ + 0.5] = [-7^+] = -7\). Limits are different, so it doesn't exist at \(a = -7.5\).

2) At \(x = -6.5\): L.H.L. involves \([-6.5^- + 0.5] = [-6^-] = -7\). R.H.L. involves \([-6.5^+ + 0.5] = [-6^+] = -6\). Limits are different, so it doesn't exist at \(a = -6.5\).

3) For any \(a \in (-7.5, -6.5)\), the function is locally constant because the only potential jumps for \([x]\) and \([2x]\) in this range are at integers and half-integers. In this specific range \([-7.5, -6.5)\), the values of \([x] - [2x]\) stay constant between the transition points.

Actually, at integers like \(x = -7\), \(\lim_{x \to -7} ([x] - [2x]) = 7\) because both L.H.L and R.H.L are equal to 7. Thus the limit exists for the entire open interval.


Step 4: Final Answer:

The set of values is \((-7.5, -6.5)\).
Quick Tip: For \([x] - [2x] = C\) to have a valid limit at \(a\), \(a\) cannot be a point where the value of the expression jumps between two different integers. Check integers and half-integers carefully.

Step 1: Understanding the Concept:

We are given a polynomial function where the coefficients are its own derivatives evaluated at specific points. We need to solve for these constants to find the exact function.


Step 2: Detailed Explanation:

Let \(f'(1) = A, f''(2) = B, f'''(3) = C\).

Function: \(f(x) = x^3 - Ax^2 + Bx - C\).

Differentiating:
\(f'(x) = 3x^2 - 2Ax + B\)
\(f''(x) = 6x - 2A\)
\(f'''(x) = 6\)

Now, evaluate the points:

1) \(C = f'''(3) = 6\).

2) \(B = f''(2) = 6(2) - 2A = 12 - 2A\).

3) \(A = f'(1) = 3(1)^2 - 2A(1) + B = 3 - 2A + B\).

Substitute \(B\): \(A = 3 - 2A + (12 - 2A) = 15 - 4A \Rightarrow 5A = 15 \Rightarrow A = 3\).

Then, \(B = 12 - 2(3) = 6\).

So, \(f(x) = x^3 - 3x^2 + 6x - 6\).


Step 3: Evaluating Options:

Calculate specific values:
\(f(0) = -6\)
\(f(1) = 1 - 3 + 6 - 6 = -2\)
\(f(2) = 8 - 12 + 12 - 6 = 2\)
\(f(3) = 27 - 27 + 18 - 6 = 12\)

Check option (D): \(2f(0) - f(1) + f(3) = 2(-6) - (-2) + 12 = -12 + 2 + 12 = 2\).

Since \(f(2) = 2\), option (D) is correct.


Step 4: Final Answer:

The relation \(2f(0) - f(1) + f(3) = f(2)\) holds true.
Quick Tip: For problems like \(f(x) = x^n + Ax^{n-1} + \dots\), define the derivative values as variables. It transforms a functional question into a simple system of linear equations.

Step 1: Understanding the Concept:

The functional equation \(f(x + y) = f(x) f(y)\) defines a geometric progression (or exponential function) for natural numbers.


Step 2: Key Formula or Approach:

Since \(f(1) = 3\), we have:
\(f(2) = f(1 + 1) = f(1) \cdot f(1) = 3^2\)
\(f(3) = f(2 + 1) = f(2) \cdot f(1) = 3^2 \cdot 3 = 3^3\)

In general, \(f(k) = 3^k\).


Step 3: Detailed Explanation:

The given sum is:
\[ \sum_{k=1}^n 3^k = 3 + 3^2 + \dots + 3^n = 3279 \]

This is a geometric series with \(a = 3, r = 3\). Sum formula: \(S_n = \frac{a(r^n - 1)}{r - 1}\).
\[ \frac{3(3^n - 1)}{3 - 1} = 3279 \Rightarrow \frac{3}{2}(3^n - 1) = 3279 \]
\[ 3^n - 1 = \frac{3279 \times 2}{3} = 1093 \times 2 = 2186 \]
\[ 3^n = 2187 \]

Powers of 3: \(3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187\).

Thus, \(n = 7\).


Step 4: Final Answer:

The value of \(n\) is 7.
Quick Tip: Common functional equations to remember:
1) \(f(x+y) = f(x) + f(y) \Rightarrow f(x) = cx\)
2) \(f(x+y) = f(x)f(y) \Rightarrow f(x) = a^x\)

Step 1: Understanding the Concept:

This problem involves complex numbers in polar form. The expression inside the brackets is of the form \(\frac{1 + z}{1 + \bar{z}}\).


Step 2: Key Formula or Approach:

Let \(\theta = \frac{2\pi}{9}\).

We know \(\sin\theta + i \cos\theta = \cos(\frac{\pi}{2} - \theta) + i \sin(\frac{\pi}{2} - \theta) = e^{i(\frac{\pi}{2} - \theta)}\).

Let \(z = e^{i\phi}\) where \(\phi = \frac{\pi}{2} - \theta\).

Then the expression is \(\left( \frac{1 + z}{1 + \bar{z}} \right)^3\).

Note that \(\frac{1 + z}{1 + \bar{z}} = \frac{1 + z}{1 + 1/z} = z\).


Step 3: Detailed Explanation:

The value is \(z^3 = (e^{i\phi})^3 = e^{3i\phi}\).

Calculate \(\phi\):
\[ \phi = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{9\pi - 4\pi}{18} = \frac{5\pi}{18} \]

Now find \(3\phi\):
\[ 3\phi = 3 \times \frac{5\pi}{18} = \frac{5\pi}{6} \]

The final value is \(e^{i \frac{5\pi}{6}}\):
\[ e^{i \frac{5\pi}{6}} = \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2} = -\frac{1}{2}(\sqrt{3} - i) \]

Step 4: Final Answer:

The value is \(-\frac{1}{2}(\sqrt{3} - i)\).
Quick Tip: For any complex number \(z\) on the unit circle (\(|z|=1\)), the identity \(\frac{1+z}{1+\bar{z}} = z\) is extremely useful. Also, convert \(\sin + i\cos\) to \(\cos + i\sin\) using complement angles.

Step 1: Understanding the Concept:

The numbers can be 4-digit or 5-digit. Since we have 5 distinct digits and repetitions are not allowed, 5-digit numbers will naturally be larger than any 4-digit number.


Step 2: Detailed Explanation:

Case 1: 4-digit numbers greater than 7000

To be greater than 7000, the thousands place must be either 7 or 8.

- Number of choices for thousands place = 2 (7 or 8).

- The remaining 3 places can be filled from the remaining 4 digits in \(P(4, 3)\) ways.

- Total 4-digit numbers = \(2 \times (4 \times 3 \times 2) = 2 \times 24 = 48\).


Case 2: 5-digit numbers

All 5-digit numbers formed from these digits will be greater than 7000.

- Total 5-digit numbers = \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).


Step 3: Summing the Results:

Total integers = \(48 + 120 = 168\).


Step 4: Final Answer:

There are 168 such integers.
Quick Tip: Don't forget to check numbers with more digits than the lower bound. In permutation problems with a threshold, higher-digit count cases are often overlooked.

Step 1: Understanding the Concept:

This is a standard integral of the form \(\int \frac{1}{\sqrt{a^2 - k^2x^2}} dx\), which results in an inverse sine function.


Step 2: Key Formula or Approach:

Evaluate the indefinite integral first:
\[ I = \int \frac{48}{\sqrt{9 - 4x^2}} dx = \int \frac{48}{\sqrt{4((\frac{3}{2})^2 - x^2)}} dx = \int \frac{48}{2\sqrt{(\frac{3}{2})^2 - x^2}} dx \]
\[ I = 24 \int \frac{1}{\sqrt{(\frac{3}{2})^2 - x^2}} dx = 24 \sin^{-1}\left(\frac{x}{3/2}\right) = 24 \sin^{-1}\left(\frac{2x}{3}\right) \]


Step 3: Detailed Explanation:

Apply the limits:
\[ Value = \left[ 24 \sin^{-1}\left(\frac{2x}{3}\right) \right]_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \]

Upper limit: \(24 \sin^{-1}\left(\frac{2}{3} \cdot \frac{3\sqrt{3}}{4}\right) = 24 \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 24 \cdot \frac{\pi}{3} = 8\pi\).

Lower limit: \(24 \sin^{-1}\left(\frac{2}{3} \cdot \frac{3\sqrt{2}}{4}\right) = 24 \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 24 \cdot \frac{\pi}{4} = 6\pi\).

Result \(= 8\pi - 6\pi = 2\pi\).


Step 4: Final Answer:

The integral evaluates to \(2\pi\).
Quick Tip: When integrating \(\frac{1}{\sqrt{a^2 - b^2x^2}}\), you can use \(\frac{1}{b} \sin^{-1}(\frac{bx}{a})\). It saves the step of pulling out the constant from the square root.

Step 1: Understanding the Concept:

Any vector \(\vec{\beta}\) can be decomposed into a component parallel to \(\vec{a}\) (\(\vec{\beta}_1\)) and a component perpendicular to \(\vec{a}\) (\(\vec{\beta}_2\)). \(\vec{\beta}_1\) is the vector projection of \(\vec{\beta}\) on \(\vec{a}\).


Step 2: Key Formula or Approach:
\(\vec{\beta}_1 = \left( \frac{\vec{\beta} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}\)
\(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1\)


Step 3: Detailed Explanation:

Calculate dot product and magnitude squared:
\(\vec{\beta} \cdot \vec{a} = (1)(4) + (2)(3) + (-4)(5) = 4 + 6 - 20 = -10\).
\(|\vec{a}|^2 = 4^2 + 3^2 + 5^2 = 16 + 9 + 25 = 50\).

Thus, \(\vec{\beta}_1 = \frac{-10}{50} \vec{a} = -\frac{1}{5} \vec{a}\).

Substitute \(\vec{\beta}_1\) into \(\vec{\beta}_2\):
\(\vec{\beta}_2 = \vec{\beta} + \frac{1}{5} \vec{a} \Rightarrow 5\vec{\beta}_2 = 5\vec{\beta} + \vec{a}\).
\(5\vec{\beta}_2 = 5(\hat{i} + 2\hat{j} - 4\hat{k}) + (4\hat{i} + 3\hat{j} + 5\hat{k}) = (5+4)\hat{i} + (10+3)\hat{j} + (-20+5)\hat{k}\).
\(5\vec{\beta}_2 = 9\hat{i} + 13\hat{j} - 15\hat{k}\).

Now, compute the required dot product with \(\vec{v} = \hat{i} + \hat{j} + \hat{k}\):
\(5\vec{\beta}_2 \cdot \vec{v} = (9)(1) + (13)(1) + (-15)(1) = 9 + 13 - 15 = 7\).


Step 4: Final Answer:

The value is 7.
Quick Tip: Instead of calculating \(\vec{\beta}_2\) fully as a vector, working with \(5\vec{\beta}_2 = 5\vec{\beta} + \vec{a}\) avoids fractions and makes the dot product calculation much faster.

Step 1: Understanding the Concept:

We use the determinant properties of adjoint matrices and matrix inverses. For an \(n \times n\) matrix:

1) \(|adj A| = |A|^{n-1}\)

2) \(|adj(adj A)| = |A|^{(n-1)^2}\)


Step 2: Detailed Explanation:

Given \(n = 3\) and \(|adj(adj A)| = 12^4\).

Using the formula: \(|A|^{(3-1)^2} = |A|^4 = 12^4\).

This implies \(|A| = \pm 12\).

Now, evaluate \(|A^{-1} adj A|\):
\[ |A^{-1} adj A| = |A^{-1}| \cdot |adj A| = \frac{1}{|A|} \cdot |A|^{n-1} = |A|^{n-2} \]

For \(n = 3\):
\[ |A|^{3-2} = |A|^1 = |A| \]

Since we need the value and the options are positive, we take \(|12| = 12\).


Step 3: Final Answer:

The value is 12.
Quick Tip: Memorize the generalized property \(|adj^k(A)| = |A|^{(n-1)^k}\). It appears frequently in JEE and other competitive math papers.

Step 1: Understanding the Concept:

The direction of a line parallel to two planes is perpendicular to the normal vectors of both planes. Once we find the direction, we can write the equation of the line and use the property that the vector from the given point to the foot of the perpendicular is orthogonal to the line.


Step 2: Key Formula or Approach:

1. Line direction \(\vec{d} = \vec{n}_1 \times \vec{n}_2\), where \(\vec{n}_1 = (1, 2, 1)\) and \(\vec{n}_2 = (0, 3, -1)\).

2. Foot of perpendicular \(P\) on line \(L(\vec{r}) = \vec{a} + r\vec{d}\) satisfies \(\vec{AP} \cdot \vec{d} = 0\).


Step 3: Detailed Explanation:

Direction of the line:
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 1
0 & 3 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(-1-0) + \hat{k}(3-0) = -5\hat{i} + \hat{j} + 3\hat{k} \]

Equation of the line passing through \((3, 2, 1)\) with direction \((-5, 1, 3)\):
\[ \frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3} = r \]

Any point \(P\) (foot of perpendicular) on the line is \((3-5r, 2+r, 1+3r)\).

Let \(A = (1, 9, 7)\). The vector \(\vec{AP} = (3-5r-1, 2+r-9, 1+3r-7) = (2-5r, r-7, 3r-6)\).

Since \(\vec{AP}\) is perpendicular to the line direction \((-5, 1, 3)\):
\[ -5(2-5r) + 1(r-7) + 3(3r-6) = 0 \]
\[ -10 + 25r + r - 7 + 9r - 18 = 0 \Rightarrow 35r - 35 = 0 \Rightarrow r = 1 \]

Substituting \(r=1\) into \(P\):
\((\alpha, \beta, \gamma) = (3-5(1), 2+1, 1+3(1)) = (-2, 3, 4)\).

Sum \(\alpha + \beta + \gamma = -2 + 3 + 4 = 5\).


Step 4: Final Answer:

The sum of the coordinates of the foot of the perpendicular is 5.
Quick Tip: For the foot of the perpendicular from a point \(A\) to a line \(L\), always define a general point \(P\) on the line using a parameter \(r\) and solve the linear equation generated by \(\vec{AP} \cdot \vec{d}_{line} = 0\).

Step 1: Understanding the Concept:

A matrix with entries 0 or 1 where each row and column sums to exactly 1 is known as a permutation matrix. Such a matrix has exactly one '1' in each row and each column.


Step 2: Detailed Explanation:

1. For the first row, there are 5 possible positions to place the single '1'.

2. Once the '1' is placed in the first row, the column it occupies cannot have another '1'.

3. For the second row, there are 4 remaining columns where a '1' can be placed.

4. Continuing this logic, for the third row, there are 3 options; for the fourth, 2 options; and for the fifth, only 1 option.

Total such matrices = \(5 \times 4 \times 3 \times 2 \times 1 = 5!\)
\[ 5! = 120 \]


Step 3: Final Answer:

There are 120 such matrices.
Quick Tip: The number of \(n \times n\) permutation matrices (matrices with entries \(\{0,1\}\) and row/column sum equal to 1) is always \(n!\).

Step 1: Understanding the Concept:

The series involves terms of the form \(k \cdot (^nC_k)^2\). We can use the identity \(k \cdot ^nC_k = n \cdot ^{n-1}C_{k-1}\) and properties of binomial coefficients to simplify the sum.


Step 2: Key Formula or Approach:

1. \(\sum_{k=1}^{n} k (^nC_k)^2 = n \binom{2n-1}{n-1}\).

2. Property: \(\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}\).


Step 3: Detailed Explanation:

Let \(S = \sum_{k=1}^{30} k (^nC_k)^2\), where \(n=30\).

Using the sum identity for \(n=30\):
\[ S = 30 \cdot \binom{59}{29} = 30 \cdot \frac{59!}{29! \cdot 30!} \]

We need to compare this with \(\frac{\alpha 60!}{(30!)^2}\).

Multiply the expression for \(S\) by \(\frac{60}{60}\) and \(\frac{30}{30}\):
\[ S = 30 \cdot \frac{59!}{29! \cdot 30!} \cdot \frac{30}{30} \cdot \frac{60}{60} = \frac{30 \cdot 30 \cdot 60!}{60 \cdot (30 \cdot 29!) \cdot 30!} = \frac{900 \cdot 60!}{60 \cdot 30! \cdot 30!} \]
\[ S = \frac{15 \cdot 60!}{(30!)^2} \]

Comparing with the given form, \(\alpha = 15\).


Step 4: Final Answer:

The value of \(\alpha\) is 15.
Quick Tip: A general useful result is \(\sum_{k=1}^n k \binom{n}{k}^2 = n \binom{2n-1}{n-1}\). Alternatively, the sum of \(\binom{n}{k}^2\) is \(\binom{2n}{n}\). By symmetry, the weighted sum is \(\frac{n}{2} \binom{2n}{n}\).
\(\frac{30}{2} \binom{60}{30} = 15 \binom{60}{30}\), giving \(\alpha = 15\) directly.

Step 1: Understanding the Concept:

We need to calculate the sum of the numerator and the denominator separately using standard summation formulas.


Step 2: Key Formula or Approach:

1. Sum of first \(n\) cubes: \(\sum k^3 = \frac{n^2(n+1)^2}{4}\).

2. Sum of terms in denominator: \(\sum_{k=1}^n k(2k+1) = \sum (2k^2 + k)\).


Step 3: Detailed Explanation:
Numerator: \(S_1 = \frac{n^2(n+1)^2}{4}\).

Denominator: \(S_2 = 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6} [2(2n+1) + 3] = \frac{n(n+1)(4n+5)}{6}\).

Given ratio:
\[ \frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \frac{n(n+1) \cdot 6}{4(4n+5)} = \frac{3(n^2+n)}{2(4n+5)} = \frac{9}{5} \]
\[ \frac{n^2+n}{8n+10} = \frac{3}{5} \Rightarrow 5n^2 + 5n = 24n + 30 \]
\[ 5n^2 - 19n - 30 = 0 \Rightarrow 5n^2 - 25n + 6n - 30 = 0 \]
\[ 5n(n-5) + 6(n-5) = 0 \Rightarrow n = 5 \] (since \(n \in \mathbb{N}\)).


Step 4: Final Answer:

The value of \(n\) is 5.
Quick Tip: For integer type questions in series, you can check small values of \(n\). For \(n=5\), Numerator \(= 15^2 = 225\); Denominator \(= 3 + 10 + 21 + 36 + 55 = 125\). Ratio \(225/125 = 9/5\). Verified!

Step 1: Understanding the Concept:

This is an integral equation. We can differentiate both sides with respect to \(x\) using the Leibniz Rule to obtain a differential equation.


Step 2: Detailed Explanation:

Differentiating \(f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\ln f(t))^2} dt = e\):
\[ f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0 \]
\[ \frac{df}{dx} = -f \sqrt{1 - (\ln f)^2} \Rightarrow \int \frac{df}{f \sqrt{1 - (\ln f)^2}} = -\int dx \]

Let \(\ln f = u \Rightarrow \frac{1}{f} df = du\):
\[ \int \frac{du}{\sqrt{1-u^2}} = -x + C \Rightarrow \sin^{-1}(\ln f) = -x + C \]

At \(x=0\), from original equation: \(f(0) + 0 = e \Rightarrow f(0) = e\).
\(\sin^{-1}(\ln e) = -0 + C \Rightarrow \sin^{-1}(1) = C \Rightarrow C = \frac{\pi}{2}\).

So, \(\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x \Rightarrow \ln f(x) = \sin(\frac{\pi}{2}-x) = \cos x\).

At \(x = \frac{\pi}{6}\): \(\ln f(\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\).

We need \((6 \ln f(\frac{\pi}{6}))^2 = (6 \cdot \frac{\sqrt{3}}{2})^2 = (3\sqrt{3})^2 = 27\).


Step 3: Final Answer:

The final value is 27.
Quick Tip: When a variable upper limit integral appears in an equation, use differentiation (Leibniz rule) to convert it into a solvable differential equation.

Step 1: Understanding the Concept:

The vertices of the triangle are the intersection points of the sides. The centroid formula relates the coordinates of the vertices to the coordinates of the centroid \(P(2, a)\).


Step 2: Detailed Explanation:

Let \(A\) be intersection of \(2x+y=0\) and \(x-y=3\): \(3x=3 \Rightarrow x=1, y=-2\). \(A(1, -2)\).
\(B\) lies on \(2x+y=0\) and \(x+py=21a\). \(C\) lies on \(x-y=3\) and \(x+py=21a\).

Using centroid \(x = \frac{1 + x_B + x_C}{3} = 2 \Rightarrow x_B + x_C = 5\).
\(y = \frac{-2 + y_B + y_C}{3} = a \Rightarrow y_B + y_C = 3a + 2\).

Since \(B\) is on \(2x+y=0\), \(y_B = -2x_B\). Since \(C\) is on \(y=x-3\), \(y_C = x_C - 3\).

Substitute: \(-2x_B + x_C - 3 = 3a + 2 \Rightarrow -2x_B + x_C = 3a + 5\).

From \(x_B + x_C = 5\), subtract equations: \(3x_B = -3a \Rightarrow x_B = -a\).

Then \(x_C = 5 + a\).

Coordinates: \(B(-a, 2a)\), \(C(5+a, a+2)\).

Both lie on \(x+py=21a\). For \(B\): \(-a + p(2a) = 21a \Rightarrow 2p = 22 \Rightarrow p = 11\).

For \(C\): \((5+a) + 11(a+2) = 21a \Rightarrow 5 + a + 11a + 22 = 21a \Rightarrow 9a = 27 \Rightarrow a = 3\).

Vertices: \(B(-3, 6)\), \(C(8, 5)\).
\((BC)^2 = (8 - (-3))^2 + (5 - 6)^2 = 11^2 + (-1)^2 = 121 + 1 = 122\).


Step 3: Final Answer:

The squared distance \((BC)^2\) is 122.
Quick Tip: Solve for parameters \(a\) and \(p\) by expressing vertices in terms of them and applying the centroid constraint. This systematic reduction of variables is key in coordinate geometry.

Step 1: Understanding the Concept:

Shortest distance between two skew lines \(\vec{r} = \vec{a}_1 + t\vec{d}_1\) and \(\vec{r} = \vec{a}_2 + s\vec{d}_2\) is \(d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}\).


Step 2: Detailed Explanation:

Line 1: \(\vec{a}_1 = (-\sqrt{6}, \sqrt{6}, \sqrt{6})\), \(\vec{d}_1 = (2, 3, 4)\).

Line 2: \(\vec{a}_2 = (\lambda, 2\sqrt{6}, -2\sqrt{6})\), \(\vec{d}_2 = (3, 4, 5)\).
\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
3 & 4 & 5 \end{vmatrix} = (-1, 2, -1)\). Magnitude \(= \sqrt{6}\).
\(\vec{a}_2 - \vec{a}_1 = (\lambda+\sqrt{6}, \sqrt{6}, -3\sqrt{6})\).

Shortest Distance \(d = \frac{|-1(\lambda+\sqrt{6}) + 2(\sqrt{6}) - 1(-3\sqrt{6})|}{\sqrt{6}} = 6 \Rightarrow |4\sqrt{6} - \lambda| = 6\sqrt{6}\).

Case 1: \(4\sqrt{6} - \lambda = 6\sqrt{6} \Rightarrow \lambda_1 = -2\sqrt{6}\).

Case 2: \(4\sqrt{6} - \lambda = -6\sqrt{6} \Rightarrow \lambda_2 = 10\sqrt{6}\).

Sum of values \(= 8\sqrt{6}\). Square of sum \(= (8\sqrt{6})^2 = 384\).


Step 3: Final Answer:

The result is 384.
Quick Tip: The cross product of directions gives the common perpendicular vector. The shortest distance is simply the projection of the vector joining two points on the lines onto this normal vector.

Step 1: Understanding the Concept:

The equation \(\tan A + \tan B = 0\) implies \(\tan A = \tan(-B)\), which gives \(A = n\pi - B\). This allows us to find the possible values of \(\sin \theta + \cos \theta\).


Step 2: Detailed Explanation:
\(\pi \cos \theta = n\pi - \pi \sin \theta \Rightarrow \sin \theta + \cos \theta = n\).

Since \(-\sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}\), integer \(n\) can be \(-1, 0, 1\).

1. \(n = 0\): \(\theta + \frac{\pi}{4} = \pi, 2\pi\) (as \(\sin(\theta+\pi/4)=0\)). \(\sin^2\) values are \(0, 0\).

2. \(n = 1\): \(\sin(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}\). \(\sin^2\) values are \(\frac{1}{2}, \frac{1}{2}\).

3. \(n = -1\): \(\sin(\theta + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\). \(\sin^2\) values are \(\frac{1}{2}, \frac{1}{2}\).

Total sum \(= 0 + 0 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2\).


Step 3: Final Answer:

The sum is 2.
Quick Tip: Convert sum/differences of trigonometric functions into a standard phase-shifted form \(R \sin(\theta + \phi)\) to easily solve for \(\theta\) or its functions.

Step 1: Understanding the Concept:

An equivalence relation must be reflexive, symmetric, and transitive. The set of elements connects all \(\{a, b, c, d\}\), meaning they must all belong to the same equivalence class in the smallest equivalence relation containing \(R\).


Step 2: Detailed Explanation:

1. Reflexive: Already given as \(\{(a,a), (b,b), (c,c), (d,d)\}\).

2. Symmetry: Add \((b,a), (c,b), (d,b)\). (3 elements)

3. Transitivity: Since \(a \sim b, b \sim c, b \sim d\), all pairs \((x, y)\) where \(x, y \in \{a, b, c, d\}\) must be in the relation.

Total elements in universal relation on 4 elements \(= 4^2 = 16\).

Current elements in \(R = 7\).

Elements to add \(= 16 - 7 = 9\).


Step 3: Final Answer:

Minimum elements to add is 9.
Quick Tip: If elements of a set are connected via the initial relation, the smallest equivalence relation will always be the partition where those elements form a complete equivalence class.

Step 1: Understanding the Concept:

We use Bayes' Theorem to find \(\lambda\), and then use coordinate geometry for the equilateral triangle properties within a parabola.


Step 2: Detailed Explanation:
\(P(C|R) = \frac{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}}{\frac{1}{3} [0.4 + 0.5 + \frac{\lambda}{\lambda+4}]} = 0.4 \Rightarrow \frac{\lambda}{\lambda+4} = 0.4 [0.9 + \frac{\lambda}{\lambda+4}]\).
\(0.6 \frac{\lambda}{\lambda+4} = 0.36 \Rightarrow \frac{\lambda}{\lambda+4} = 0.6 \Rightarrow \lambda = 6\).

Parabola: \(y^2 = 6x\). Equilateral triangle vertices: \((0,0), (x, y), (x, -y)\).
\(\tan 30^\circ = \frac{y}{x} = \frac{1}{\sqrt{3}} \Rightarrow y^2 = \frac{x^2}{3}\).

Substitute in parabola: \(\frac{x^2}{3} = 6x \Rightarrow x = 18\).

Side \(L = \sqrt{x^2 + y^2} = \sqrt{x^2 + x^2/3} = \sqrt{4x^2/3} = \frac{2 \times 18}{\sqrt{3}} = 12\sqrt{3}\).
\(L^2 = 144 \times 3 = 432\).


Step 3: Final Answer:

Side squared is 432.
Quick Tip: For an equilateral triangle in \(y^2=4ax\), vertex at origin, the coordinates of the other vertices satisfy \(y/x = \tan 30^\circ\).

Step 1: Understanding the Concept:

The cross product property \((\vec{a} - \vec{b}) \times \vec{c} = 0\) implies \(\vec{c}\) is parallel to \((\vec{a} - \vec{b})\). This allows us to find \(\lambda\).


Step 2: Detailed Explanation:
\(\vec{c} = k(\vec{a} - \vec{b}) = k(-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})\).
\(\vec{a} \cdot \vec{c} = k(1(-2) + 2(7) + \lambda(2\lambda)) = k(12 + 2\lambda^2) = 7\).
\(2\vec{b} \cdot \vec{c} = 2k(3(-2) - 5(7) - \lambda(2\lambda)) = 2k(-41 - 2\lambda^2) = -43 \Rightarrow k(41 + 2\lambda^2) = \frac{43}{2}\).

Ratio: \(\frac{41 + 2\lambda^2}{12 + 2\lambda^2} = \frac{43}{14} \Rightarrow 574 + 28\lambda^2 = 516 + 86\lambda^2 \Rightarrow 58\lambda^2 = 58 \Rightarrow \lambda^2 = 1\).
\(\vec{a} \cdot \vec{b} = 3 - 10 - \lambda^2 = -7 - 1 = -8\).
\(|\vec{a} \cdot \vec{b}| = 8\).


Step 3: Final Answer:

The magnitude of the dot product is 8.
Quick Tip: The relation \(\vec{u} \times \vec{w} = \vec{v} \times \vec{w}\) implies \(\vec{w}\) is collinear with \(\vec{u} - \vec{v}\).

Step 1: Understanding the Concept:

We use the binomial expansion \((a+b)^n = \sum \binom{n}{r} a^{n-r}b^r\) to find the relationship for \(n\).


Step 2: Detailed Explanation:

First three coefficients: \(1, -3n, 9 \frac{n(n-1)}{2}\).
\(1 - 3n + 4.5(n^2 - n) = 376 \Rightarrow 4.5n^2 - 7.5n - 375 = 0 \Rightarrow 3n^2 - 5n - 250 = 0\).
\((n-10)(3n+25) = 0 \Rightarrow n = 10\).

General term: \(T_{r+1} = \binom{10}{r} (x)^{10-r} (-3 x^{-2})^r = \binom{10}{r} (-3)^r x^{10-3r}\).

For \(x^4\), \(10-3r = 4 \Rightarrow r = 2\).

Coefficient \(= \binom{10}{2} (-3)^2 = 45 \times 9 = 405\).


Step 3: Final Answer:

The coefficient is 405.
Quick Tip: Coefficient sum problems are usually solved by setting the variable \(x=1\), but when restricted to "first \(k\) terms", you must explicitly calculate the initial binomial coefficients.

Step 1: Understanding the Concept:

Area between curves is found by integrating the difference \((x_{right} - x_{left})\) with respect to \(y\) between the intersection points.


Step 2: Detailed Explanation:

Curve 1: \(x = 2y - y^2\). Curve 2: \(x = -y\).

Intersections: \(2y - y^2 = -y \Rightarrow y^2 - 3y = 0 \Rightarrow y=0, 3\).

Area \(A = \int_{0}^{3} [(2y - y^2) - (-y)] dy = \int_{0}^{3} (3y - y^2) dy\).
\(A = [\frac{3y^2}{2} - \frac{y^3}{3}]_{0}^{3} = \frac{27}{2} - 9 = 4.5\).
\(8 A = 8 \times 4.5 = 36\).


Step 3: Final Answer:

The final value is 36.
Quick Tip: When a curve is given as a quadratic in \(y\), it is often simpler to integrate with respect to \(dy\) (horizontal strips) rather than \(dx\).