CUET PG Physics 2025 Question Paper (Available): Download Question Paper with Answer Key And Solutions PDF

Step 1: Recall the formula for interplanar spacing in a cubic lattice.
The interplanar spacing \(d_{hkl}\) for a cubic lattice with lattice constant \(a\) is given by: \[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]
where \((h, k, l)\) are the Miller indices of the plane.

Step 2: Identify the given values.
We are given:
- Interplanar spacing, \(d = 3 \, A°\)
- Miller indices for the plane, \((hkl) = (002)\)

Step 3: Substitute the values into the formula and solve for \(a\). \[ 3 \, A° = \frac{a}{\sqrt{0^2 + 0^2 + 2^2}} \] \[ 3 \, A° = \frac{a}{\sqrt{4}} \] \[ 3 \, A° = \frac{a}{2} \] \[ a = 2 \times 3 \, A° = 6 \, A° \]
Thus, the lattice constant is \( 6.0 \, A° \). Quick Tip: For any cubic crystal system (Simple, BCC, or FCC), the formula for interplanar spacing \( d = a / \sqrt{h^2 + k^2 + l^2} \) remains the same. Memorizing this single formula is sufficient for all cubic lattice problems involving Miller indices and lattice parameters.

Step 1: Recall the formula for intrinsic carrier concentration (\(n_i\)).
The intrinsic carrier concentration in a semiconductor is a function of temperature \(T\) and the energy bandgap \(E_g\). The standard formula is: \[ n_i = A T^{3/2} \exp\left(-\frac{E_g}{2k_B T}\right) \]
where \(A\) is a material-specific constant and \(k_B\) is the Boltzmann constant.

Step 2: Analyze the temperature dependence.
The formula shows that \(n_i\) depends on temperature in two ways: through the \(T^{3/2}\) term and the exponential term \(\exp(-E_g / 2k_B T)\). The exponential term describes the dominant temperature dependence, but the question asks how the concentration *varies with* T, and the pre-exponential factor \(T^{3/2}\) is a key part of this variation.

Step 3: Compare with the given options.
The options provided are power-law dependencies on \(T\). The pre-exponential factor in the expression for \(n_i\) is \(T^{3/2}\), which matches option (3). Quick Tip: While the exponential term \(\exp(-E_g / 2k_B T)\) causes the most significant change in carrier concentration with temperature, the pre-exponential \(T^{3/2}\) term is also a fundamental part of the relationship derived from the density of states. Always check if this term is among the options.

Step 1: Define the First Brillouin Zone.
The First Brillouin Zone is a fundamental concept in solid-state physics used to describe electron wave propagation in a crystal lattice.

Step 2: Evaluate each statement.
- A. Wigner-Seitz cell of reciprocal lattice: This is the standard definition of the First Brillouin Zone. The Wigner-Seitz cell is the region of space that is closer to one particular lattice point than to any other. When applied to the reciprocal lattice, this defines the Brillouin zone. This statement is correct.
- B. Primitive unit cell: The First Brillouin Zone is, by its construction as a Wigner-Seitz cell, a primitive unit cell of the reciprocal lattice. It has the volume of one primitive cell. This statement is correct.
- C. The locus of all k-values...which are Bragg reflected: The boundaries of the Brillouin Zone are defined by planes that are the perpendicular bisectors of the vectors connecting the origin of the reciprocal lattice to the nearest lattice points. These planes represent the wave vectors (\(k\)-values) that satisfy the Bragg condition for diffraction, \(2\mathbf{k} \cdot \mathbf{G} = |\mathbf{G}|^2\). Therefore, the zone represents the set of k-vectors that are not Bragg reflected, and its boundaries are where Bragg reflection begins. The statement is conceptually correct in linking the zone to Bragg reflection.
- D. Wigner-Seitz cell of direct lattice: This is incorrect. The Wigner-Seitz cell of the *direct* lattice is a primitive cell in real space, not reciprocal space. The Brillouin zone is exclusively a concept of the reciprocal lattice.

Step 3: Conclude the correct statements.
Statements A, B, and C are correct descriptions of the Brillouin zone. Statement D is incorrect. Therefore, the correct option includes A, B, and C only. Quick Tip: The key to understanding the Brillouin zone is to remember that it lives in "reciprocal space" or "k-space," not real space. It is the Wigner-Seitz cell of the reciprocal lattice, and its boundaries are directly related to the condition for Bragg diffraction.

Step 1: Determine the coordination number for each crystal structure.
The coordination number is the number of nearest neighbors of an atom in the crystal lattice.
- A. Diamond: In the diamond cubic structure, each carbon atom is tetrahedrally bonded to four other carbon atoms. The coordination number is 4.
- B. Sodium Chloride (NaCl): In the rock salt structure, each ion (e.g., Na\(^+\)) is surrounded by six ions of the opposite charge (e.g., Cl\(^-\)) in an octahedral arrangement. The coordination number is 6.
- C. Cesium Chloride (CsCl): In the CsCl structure, each ion is at the center of a cube of eight ions of the opposite charge. The coordination number is 8.
- D. Zinc (HCP): Zinc has a Hexagonal Close-Packed (HCP) structure. In any close-packed structure (HCP or FCC), each atom is in contact with 12 other atoms (6 in its own plane, 3 above, and 3 below). The coordination number is 12.

Step 2: Arrange the structures in ascending order of their coordination number.
The coordination numbers are:
- Diamond (A): 4
- Sodium Chloride (B): 6
- Cesium Chloride (C): 8
- Zinc (HCP) (D): 12

Arranging these in ascending order gives the sequence: A (4) < B (6) < C (8) < D (12).
The correct order is A, B, C, D. Quick Tip: Memorize the coordination numbers for common crystal structures: Diamond (4), NaCl (6), BCC/CsCl (8), and HCP/FCC (12). This is a frequent topic in solid-state physics questions.

Step 1: Set up the hexadecimal subtraction problem.
We need to calculate:

Step 2: Subtract the fractional parts.
The rightmost column is B - A. In decimal, this is \(11 - 10 = 1\). The result is \(1_{16}\).

Step 3: Subtract the integer parts column by column, from right to left.
First, subtract the units column: F - 9. In decimal, this is \(15 - 9 = 6\). The result is \(6_{16}\).

Next, subtract the 16's column: 4 - 2. In decimal, this is \(4 - 2 = 2\). The result is \(2_{16}\).

Step 4: Combine the results.
The final result of the subtraction is (26.1)\(_{16}\). Quick Tip: When performing hexadecimal arithmetic, remember the decimal equivalents: A=10, B=11, C=12, D=13, E=14, F=15. For subtraction, if you need to borrow, you borrow 16 from the column to the left. In this case, no borrowing was needed.

Step 1: Analyze an ideal Zener regulator circuit.
A simple Zener regulator consists of a series resistor (\(R_S\)) connected to an input voltage (\(V_{in}\)), and a Zener diode in parallel with the load resistor (\(R_L\)). The purpose of the Zener diode is to maintain a constant voltage (\(V_Z\)) across the load.

Step 2: Determine the series current (\(I_S\)).
The series current flows through the series resistor \(R_S\). The voltage across \(R_S\) is the difference between the input voltage and the Zener voltage: \(V_{RS} = V_{in} - V_Z\).
According to Ohm's law, the series current is: \[ I_S = \frac{V_{in} - V_Z}{R_S} \]

Step 3: Evaluate the effect of a change in load resistance.
Assuming the regulator is operating correctly, the Zener diode maintains \(V_Z\) at a constant value. The input voltage \(V_{in}\) and the series resistance \(R_S\) are also constant. Since all the terms on the right side of the equation for \(I_S\) are constant, the series current \(I_S\) remains constant.

Step 4: Consider the current distribution.
The series current splits into the Zener current (\(I_Z\)) and the load current (\(I_L\)): \(I_S = I_Z + I_L\). If the load resistance \(R_L\) decreases, the load current \(I_L = V_Z / R_L\) increases. Since \(I_S\) is constant, the Zener current \(I_Z\) must decrease to compensate. The regulator works as long as \(I_Z\) remains positive. Quick Tip: In a working Zener regulator, think of the series resistor and the constant Zener voltage as setting a constant total current supply (\(I_S\)). The Zener diode then "absorbs" whatever current is not drawn by the load to keep the voltage constant.

Step 1: Understand the function of a controlled current source.
A controlled current source is an electronic circuit that delivers a constant current to a load, where the value of this current is set by an input signal.

Step 2: Analyze the role of the Op-Amp in this application.
In a typical Op-Amp based controlled current source (like a Howland current pump or a simple transistor-based design), an input voltage (\(V_{in}\)) is applied to one of the Op-Amp's inputs. The Op-Amp uses its high gain and feedback to adjust its output in such a way that the current flowing through the load (\(I_{out}\)) becomes directly proportional to the input voltage (\(I_{out} \propto V_{in}\)).

Step 3: Classify the circuit based on its input and output signals.
The circuit takes a voltage as its input signal and produces a proportional current as its output signal. This is the definition of a voltage-to-current converter. Another name for this type of circuit is a transconductance amplifier. Quick Tip: Remember the four basic types of amplifiers based on their input and output signals: - Voltage In, Voltage Out \(\rightarrow\) Voltage Amplifier - Current In, Voltage Out \(\rightarrow\) Transresistance Amplifier (Current-to-Voltage Converter) - Voltage In, Current Out \(\rightarrow\) Transconductance Amplifier (Voltage-to-Current Converter) - Current In, Current Out \(\rightarrow\) Current Amplifier

Step 1: Analyze each logic gate and its corresponding Boolean expression.
- A. EX-OR: The Exclusive-OR gate outputs true only when the inputs are different. Its expression is \( A \oplus B = A\bar{B} + \bar{A}B \). This matches expression I. So, A \(\rightarrow\) I.
- C. OR: The OR gate outputs true if at least one input is true. Its expression is \( A + B \). This matches expression II. So, C \(\rightarrow\) II.
- D. EX-NOR: The Exclusive-NOR gate outputs true only when the inputs are the same. Its expression is \( \overline{A \oplus B} = AB + \bar{A}\bar{B} \). This matches expression IV. So, D \(\rightarrow\) IV.
- B. NAND: The NAND gate is the negation of the AND gate. Its expression is \( \overline{AB} \). Expression III is \(AB\), which is the expression for an AND gate.

Step 2: Evaluate the options based on the correct matches.
We have confirmed: A\(\rightarrow\)I, C\(\rightarrow\)II, D\(\rightarrow\)IV.
Let's check the given options:
- Option (1) has C\(\rightarrow\)III, which is incorrect.
- Option (2) has A\(\rightarrow\)I, C\(\rightarrow\)II, and D\(\rightarrow\)IV. This aligns with our findings. It matches B (NAND) with III (AND). This implies a likely error in the question, where gate B should have been AND, or expression III should have been \(\overline{AB}\). However, given the other three perfect matches, this option is the most plausible intended answer.
- Option (3) has C\(\rightarrow\)IV, which is incorrect.
- Option (4) has A\(\rightarrow\)III, which is incorrect.

Therefore, option (2) is the correct choice, assuming a typo in the question's list. Quick Tip: When facing matching questions with a potential error, first identify all the pairs you are certain about. Then, use the process of elimination to find the option that correctly matches all the unambiguous pairs. The remaining pair in that option is likely the intended, albeit flawed, answer.

Step 1: Convert all numbers to a common base (decimal is easiest).
- A. (10110.011)\(_2\):
\[ (1 \cdot 2^4) + (0 \cdot 2^3) + (1 \cdot 2^2) + (1 \cdot 2^1) + (0 \cdot 2^0) + (0 \cdot 2^{-1}) + (1 \cdot 2^{-2}) + (1 \cdot 2^{-3}) \]
\[ = 16 + 0 + 4 + 2 + 0 + 0 + 0.25 + 0.125 = 22.375_{10} \]
- B. (32)\(_{10}\):
This is already in decimal. Value is \(32_{10}\).
- C. (5F.8)\(_{16}\):
\[ (5 \cdot 16^1) + (F \cdot 16^0) + (8 \cdot 16^{-1}) \]
\[ = (5 \cdot 16) + (15 \cdot 1) + (8 / 16) = 80 + 15 + 0.5 = 95.5_{10} \]
- D. F\(_{16}\):
\[ F_{16} = 15_{10} \]

Step 2: Compare the decimal values.
The decimal values are:
- A = 22.375
- B = 32
- C = 95.5
- D = 15

Step 3: Arrange them in ascending order.
The smallest value is D (15), followed by A (22.375), then B (32), and the largest is C (95.5).
The ascending order is D, A, B, C. Quick Tip: To compare numbers in different bases (binary, decimal, hexadecimal), the most reliable method is to convert all of them to a single base, usually decimal. Remember the positional values are powers of the base (2, 10, or 16).

Step 1: Analyze the characteristics of each BJT configuration.
- A. Common Base (CB): This configuration has a low input impedance and high output impedance. Its current gain (\(\alpha\)) is slightly less than 1. However, it can have a very high voltage gain. Therefore, it provides "Voltage Gain but no Current Gain" (since the gain is less than unity). This matches II.
- B. Common Emitter (CE): This is the most widely used configuration. It has a medium input and output impedance. It provides significant gain for both current (\(\beta\)) and voltage. Therefore, it provides "Both Current and Voltage Gain". This matches III.
- C. Common Collector (CC): Also known as an emitter follower, this configuration has a high input impedance and low output impedance. Its voltage gain is approximately 1 (unity). It has a high current gain (\(\beta+1\)). Therefore, it provides "Current Gain but no Voltage Gain". This matches I.

Step 2: Formulate the correct matching sequence.
Based on the analysis:
- A \(\rightarrow\) II
- B \(\rightarrow\) III
- C \(\rightarrow\) I

This sequence corresponds to option (2). Quick Tip: A simple way to remember BJT configurations: - **Common Emitter (CE):** The all-rounder. Good voltage and current gain. Inverts the signal. - **Common Collector (CC):** The "buffer". High current gain, unity voltage gain. Used for impedance matching. - **Common Base (CB):** The "current follower". Unity current gain, high voltage gain. Used for high-frequency applications.

Step 1: Recall Bragg's Law for X-ray diffraction.
Bragg's Law is given by the formula: \[ n\lambda = 2d\sin\theta \]
where \(n\) is the order of the maximum, \(\lambda\) is the wavelength of the X-rays, \(d\) is the distance between atomic planes, and \(\theta\) is the angle of incidence.

Step 2: Identify the given values from the problem.
- Order of maximum, \(n = 1\) (since it's the "first maxima").
- Wavelength, \(\lambda = 0.32 \, nm\).
- Angle of diffraction, \(\theta = 30^{\circ}\).

Step 3: Substitute the values into Bragg's Law and solve for \(d\). \[ % Option (1)(0.32 \, nm) = 2 \cdot d \cdot \sin(30^{\circ}) \]
We know that \(\sin(30^{\circ}) = 0.5\). \[ 0.32 \, nm = 2 \cdot d \cdot (0.5) \] \[ 0.32 \, nm = d \]
Thus, the distance between the atomic planes is \(0.32 \, nm\). Quick Tip: Bragg's Law is a fundamental equation in solid-state physics. Remember that \(n\) must be an integer (1, 2, 3, ...) representing the order of the reflection. For "first maxima" or "first-order diffraction," always use \(n=1\).

Step 1: Recall the photoelectric effect equation.
Einstein's photoelectric equation relates the maximum kinetic energy (\(K_{max}\)) of a photoelectron to the frequency of the incident photon (\(f\)) and the work function of the metal (\(\phi\)): \[ K_{max} = hf - \phi \]
where \(h\) is Planck's constant.

Step 2: Relate stopping potential to kinetic energy.
The stopping potential (\(V_s\)) is the potential required to stop the most energetic photoelectrons. It is related to the maximum kinetic energy by: \[ e V_s = K_{max} \]
where \(e\) is the elementary charge. Combining the two equations gives: \[ V_s = \frac{h}{e}f - \frac{\phi}{e} \]

Step 3: Analyze the dependencies of the stopping potential.
From the equation, we can see that \(V_s\) depends on:
- The frequency (\(f\)) of the incident photon.
- The wavelength (\(\lambda\)), since \(f = c/\lambda\).
- The work function (\(\phi\)), which is a property of the specific metal (type of metals).

The intensity of the incident light determines the *number* of photons arriving per unit time, which in turn determines the number of photoelectrons emitted (the photoelectric current). However, it does not affect the energy of individual photons, and thus does not affect the maximum kinetic energy or the stopping potential of the photoelectrons. Quick Tip: For the photoelectric effect, remember this key distinction: - **Frequency/Wavelength** determines the **Energy** of photoelectrons (\(K_{max}\), \(V_s\)). - **Intensity** determines the **Number** of photoelectrons (photocurrent).

Step 1: Identify the physical principle.
Since the source of the signals (Swati) is moving away from the receiver (Ravi) at a relativistic speed, we must use the relativistic Doppler effect to find the time interval between the reception of signals.

Step 2: Recall the formula for the relativistic Doppler effect (time interval).
The time interval between received signals (\(T_{received}\)) is related to the time interval between sent signals (\(T_{sent}\)) by the formula: \[ T_{received} = T_{sent} \sqrt{\frac{1 + v/c}{1 - v/c}} \]
where \(v\) is the relative velocity of separation.

Step 3: Substitute the given values.
- Time interval between sent signals, \(T_{sent} = 1\) year.
- Relative velocity, \(v = 0.80c\). \[ T_{received} = (1 \, year) \sqrt{\frac{1 + 0.80}{1 - 0.80}} = \sqrt{\frac{1.8}{0.2}} = \sqrt{9} = 3 \, years \]
This means Ravi receives a signal from Swati every 3 years.

Step 4: Calculate the total number of signals received.
The trip lasts for 15 years from Ravi's perspective. Since he receives one signal every 3 years, the total number of signals received is: \[ Number of signals = \frac{Total trip time}{T_{received}} = \frac{15 \, years}{3 \, years/signal} = 5 \, signals \] Quick Tip: This is a classic twin paradox-style problem. The key is recognizing that the rate of receiving signals is altered by the relative motion. For objects moving apart, the time between received signals is longer than the time between sent signals.

Step 1: Write down the general form of the time-dependent Schrodinger equation. \[ i\hbar \frac{\partial \Psi(\mathbf{r}, t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}, t) \right] \Psi(\mathbf{r}, t) \]
where \(\Psi\) is the wave function.

Step 2: Analyze the properties of the equation.
- **Linearity:** The equation is linear because the wave function \(\Psi\) and its derivatives appear only to the first power. There are no terms like \(\Psi^2\) or \(\Psi \frac{\partial \Psi}{\partial t}\). This means that if \(\Psi_1\) and \(\Psi_2\) are solutions, then any linear combination \(c_1\Psi_1 + c_2\Psi_2\) is also a solution. This is the principle of superposition.
- **Order in time:** The equation contains a first derivative with respect to time, \(\frac{\partial}{\partial t}\). It is first-order in time.
- **Order in space:** The equation contains the Laplacian operator, \(\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\), which involves second derivatives with respect to spatial coordinates. It is second-order in space.

Step 3: Evaluate the given options.
1. non-linear differential equation: Incorrect.
2. linear differential equation: Correct.
3. second order equation in time: Incorrect. It is first-order in time.
4. first order equation in space: Incorrect. It is second-order in space. Quick Tip: The linearity of the Schrodinger equation is one of its most important features. It is the mathematical basis for the principle of superposition in quantum mechanics, which allows for phenomena like interference of wave functions.

Step 1: Recall the formula for the Compton shift.
The Compton shift, \(\Delta\lambda\), which is the change in wavelength of a photon after scattering off an electron, is given by: \[ \Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) \]
where \(\theta\) is the scattering angle.

Step 2: Define the Compton wavelength.
The Compton wavelength, \(\lambda_c\), is a constant defined as: \[ \lambda_c = \frac{h}{m_e c} \]

Step 3: Set the Compton shift equal to the Compton wavelength and solve for \(\theta\).
We are given the condition \(\Delta\lambda = \lambda_c\). \[ \frac{h}{m_e c}(1 - \cos\theta) = \frac{h}{m_e c} \]
Dividing both sides by \(\frac{h}{m_e c}\) gives: \[ 1 - \cos\theta = 1 \] \[ -\cos\theta = 0 \] \[ \cos\theta = 0 \]
The angle \(\theta\) for which \(\cos\theta = 0\) (in the range \(0 \le \theta \le \pi\)) is \(\theta = \pi/2\). Quick Tip: Remember the physical meaning of the limits for Compton scattering: - \(\theta = 0\): No scattering, \(\Delta\lambda = 0\). - \(\theta = \pi/2\) (90 degrees): Shift equals the Compton wavelength, \(\Delta\lambda = \lambda_c\). - \(\theta = \pi\) (180 degrees, backscattering): Maximum shift, \(\Delta\lambda = 2\lambda_c\).

Step 1: Relate penetrating power to the energy of a photon.
The penetrating power of electromagnetic radiation, like X-rays, is directly proportional to the energy of its photons. Higher energy photons are more penetrating.

Step 2: Recall the energy-wavelength relationship for a photon.
The energy (\(E\)) of a photon is inversely proportional to its wavelength (\(\lambda\)): \[ E = hf = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light.

Step 3: Determine which wavelength corresponds to the highest energy.
To have the largest penetrating power, the X-ray must have the highest energy. According to the formula, the highest energy corresponds to the shortest (smallest) wavelength.

Step 4: Compare the given wavelengths.
The given options are \(1.2 \, A°\), \(6 \, A°\), \(9 \, A°\), and \(12 \, A°\). The smallest value among these is \(1.2 \, A°\). This wavelength will have the highest energy and therefore the largest penetrating power. Quick Tip: This is a fundamental concept in electromagnetic radiation: Short Wavelength \(\leftrightarrow\) High Frequency \(\leftrightarrow\) High Energy \(\leftrightarrow\) High Penetrating Power. This applies to the entire EM spectrum, from radio waves to gamma rays.

Step 1: Analyze each type of radioactive decay.
- A. Alpha decay: The emission of an alpha particle (\(^{4}_{2}He\)) reduces the mass number by 4 and the atomic number by 2. This is the most common decay mode for very heavy nuclei (typically with Z > 82, i.e., heavier than Lead) to become more stable. This matches III.
- B. Beta negative decay: A neutron transforms into a proton, emitting an electron (\(e^-\)) and an antineutrino. This process (\(n \to p + e^- + \bar{\nu}_e\)) increases the proton number and decreases the neutron number. It occurs in nuclei that have an excess of neutrons compared to protons (a high N/Z ratio). This matches IV.
- C. Gamma decay: An excited nucleus transitions to a lower energy state by emitting a high-energy photon (gamma ray). The number of protons and neutrons does not change. This occurs when a nucleus has excess energy, often after a previous alpha or beta decay. This matches I.
- D. Positron Emission (\(\beta^+\) decay): A proton transforms into a neutron, emitting a positron (\(e^+\)) and a neutrino. This process (\(p \to n + e^+ + \nu_e\)) decreases the proton number and increases the neutron number. It occurs in nuclei that have an excess of protons compared to neutrons (a low N/Z ratio). This matches II.

Step 2: Combine the matches to find the correct sequence.
- A \(\rightarrow\) III
- B \(\rightarrow\) IV
- C \(\rightarrow\) I
- D \(\rightarrow\) II
This sequence is A - III, B - IV, C - I, D - II, which corresponds to option (4). Quick Tip: Think of nuclear decay as a nucleus's way of adjusting its proton-to-neutron ratio to reach the "valley of stability". - Too heavy? \(\rightarrow\) Alpha decay. - Too many neutrons? \(\rightarrow\) Beta-minus decay. - Too many protons? \(\rightarrow\) Positron emission or electron capture. - Too much energy? \(\rightarrow\) Gamma decay.

Step 1: Recall the de-Broglie wavelength formula.
The de-Broglie wavelength (\(\lambda\)) of a particle is given by: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity.

Step 2: List the required constants and given values.
- Planck's constant, \(h \approx 6.626 \times 10^{-34} \, J\cdots\).
- Mass of an electron, \(m_e \approx 9.11 \times 10^{-31} \, kg\).
- Velocity of the electron, \(v = 10^7 \, m/s\).

Step 3: Substitute the values and calculate the wavelength. \[ \lambda = \frac{6.626 \times 10^{-34} \, J\cdots}{(9.11 \times 10^{-31} \, kg) \times (10^7 \, m/s)} \] \[ \lambda = \frac{6.626}{9.11} \times 10^{-34 - (-31) - 7} \, m \] \[ \lambda \approx 0.727 \times 10^{-10} \, m \] \[ \lambda \approx 7.27 \times 10^{-11} \, m \]
This value is approximately \(7.3 \times 10^{-11}\) m. Quick Tip: For calculations involving fundamental constants, it's often useful to know approximate ratios. For instance, \(h/m_e \approx 7.27 \times 10^{-4}\). This can sometimes speed up calculations. Also, always double-check the powers of ten.

Step 1: Analyze the photoelectric effect.
The photoelectric effect involves the emission of electrons from a material when light shines on it. Key observations, such as the existence of a threshold frequency and the instantaneous emission of electrons, could not be explained by the wave theory of light. Albert Einstein explained it by postulating that light consists of discrete packets of energy called photons (particles). Therefore, the photoelectric effect is a primary evidence for the particle nature of light. Statement A is incorrect, and statement C is correct.

Step 2: Analyze the Compton effect.
The Compton effect describes the scattering of X-rays or gamma rays by electrons. The observed change in wavelength of the scattered radiation is explained by treating the interaction as a collision between a photon and an electron, conserving both energy and momentum. This billiard-ball-like interaction is a clear demonstration of the particle nature of light. Therefore, the Compton effect exhibits the particle nature of light. Statement B is incorrect, and statement D is correct.

Step 3: Conclude which statements are correct.
Both statements C and D are correct. They both point to phenomena that support the particle (photon) model of light. Quick Tip: Remember the key experiments for wave-particle duality: - **Wave Nature:** Interference, Diffraction, Polarization. - **Particle Nature:** Photoelectric Effect, Compton Scattering, Blackbody Radiation.

Step 1: Recall the energy formula for a particle in a 3D cubic box.
The energy levels for a particle of mass \(m\) in a cubic box of side length \(L\) are given by: \[ E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2) \]
where \(n_x, n_y, n_z\) are positive integers (1, 2, 3, ...). Degeneracy is the number of different combinations of \((n_x, n_y, n_z)\) that yield the same energy value.

Step 2: Calculate the degeneracy for each energy level.
- A. Energy \(14h^2/(8mL^2)\):
We need to find the number of integer sets \((n_x, n_y, n_z)\) such that \(n_x^2 + n_y^2 + n_z^2 = 14\).
By inspection, \(1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\). The distinct sets are permutations of (1, 2, 3). The number of permutations of three distinct numbers is \(3! = 6\). The states are (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). The degeneracy is 6. So, A \(\rightarrow\) III.

- B. Energy \(11h^2/(8mL^2)\):
We need \(n_x^2 + n_y^2 + n_z^2 = 11\).
By inspection, \(1^2 + 1^2 + 3^2 = 1 + 1 + 9 = 11\). The distinct sets are permutations of (1, 1, 3). The number of permutations is \(\frac{3!}{2!} = 3\). The states are (1,1,3), (1,3,1), (3,1,1). The degeneracy is 3. So, B \(\rightarrow\) II.

- C. Energy \(3h^2/(8mL^2)\):
We need \(n_x^2 + n_y^2 + n_z^2 = 3\).
The only possible combination is \(1^2 + 1^2 + 1^2 = 3\). The state is (1,1,1). There is only one such state. The degeneracy is 1. So, C \(\rightarrow\) I.

Step 3: Formulate the correct matching sequence.
The correct matches are: A \(\rightarrow\) III, B \(\rightarrow\) II, and C \(\rightarrow\) I. This corresponds to option (3). Quick Tip: To find the degeneracy for a 3D particle in a box, you are looking for the number of ways you can sum the squares of three positive integers to get a specific number. Systematically test combinations of small integers (1, 2, 3, 4, ...) to find the sets that work.

Step 1: Analyze the given wave function.
The state is \( \Psi = c_1\Psi_1 + c_2\Psi_2 \) with \(c_1 = \sqrt{1/3}\) and \(c_2 = \sqrt{2/3}\). The state is normalized because \(|c_1|^2 + |c_2|^2 = 1/3 + 2/3 = 1\). The energy eigenvalues are \( E_n = n^2h^2 / (8mL^2) \).

Step 2: Evaluate each statement.
- A. Expectation value of momentum \(< p >\): For any energy eigenstate in an infinite square well, \(< p >=0\). For a superposition of such states, the expectation value of momentum is also 0. So, statement A is correct.
- C. Expectation value of energy \(< E >\):
\[ < E > = |c_1|^2 E_1 + |c_2|^2 E_2 = \left(\frac{1}{3}\right)\left(\frac{1^2 h^2}{8mL^2}\right) + \left(\frac{2}{3}\right)\left(\frac{2^2 h^2}{8mL^2}\right) \]
\[ < E > = \frac{h^2}{8mL^2} \left( \frac{1}{3} + \frac{2 \cdot 4}{3} \right) = \frac{h^2}{8mL^2} \left( \frac{1+8}{3} \right) = \frac{3h^2}{8mL^2} \]
Statement C is correct.
- B. Uncertainty in momentum \(\Delta p\):
\( (\Delta p)^2 = < p^2 > - < p >^2 \). Since \(< p >=0\), \( (\Delta p)^2 = < p^2 > \).
We use the relation \( \hat{H} = \hat{p}^2 / (2m) \), so \( < p^2 > = 2m< E > \).
\[ < p^2 > = 2m \left( \frac{3h^2}{8mL^2} \right) = \frac{3h^2}{4L^2} \]
\[ \Delta p = \sqrt{< p^2 >} = \sqrt{\frac{3h^2}{4L^2}} = \frac{\sqrt{3}h}{2L} \]
Statement B is correct.
- D. Uncertainty in position \(\Delta x\): \(\Delta x\) is the standard deviation in position. It cannot be zero for a particle in a box, as this would violate the Heisenberg Uncertainty Principle. Statement D is incorrect.

Step 3: Conclude the correct set of statements.
Statements A, B, and C are correct. This corresponds to option (2). Quick Tip: For a superposition state \( \Psi = \sum c_n \Psi_n \), the expectation value of an operator \(\hat{A}\) with eigenvalues \(a_n\) is \( < A > = \sum |c_n|^2 a_n \), provided the \(\Psi_n\) are eigenstates of \(\hat{A}\). This works for energy, but for operators like momentum, you need to be more careful. The relation \(< p^2 > = 2m < E >\) is a useful shortcut for the infinite square well.

This is a fundamental property of determinants. If a matrix has two identical rows or two identical columns, its determinant is zero. This can be understood by considering that swapping two rows negates the determinant. If the two rows are identical, swapping them changes nothing about the matrix, but the determinant must be negated. The only number that is equal to its own negative is zero.
Therefore, the value of the determinant is 0. Quick Tip: Memorizing the basic properties of determinants is essential for linear algebra. The key properties include: det(A\(^T\))=det(A), det(AB)=det(A)det(B), and the effect of row/column operations (swapping, scaling, adding a multiple of another row/column).

Step 1: Recall the property of eigenvalues for a polynomial of a matrix.
If \(\lambda\) is an eigenvalue of a matrix \(A\), then for any polynomial \(P(x)\), the corresponding eigenvalue of the matrix \(P(A)\) is \(P(\lambda)\).

Step 2: Define the polynomial and list the eigenvalues of A.
The matrix polynomial is \(P(A) = 3I - 2A + A^2\).
The corresponding scalar polynomial is \(P(\lambda) = 3 - 2\lambda + \lambda^2\).
The eigenvalues of A are \(\lambda_1 = 1\), \(\lambda_2 = -2\), and \(\lambda_3 = 3\).

Step 3: Calculate the new eigenvalues by applying the polynomial to each eigenvalue of A.
- For \(\lambda_1 = 1\): \(P(1) = 3 - 2(1) + 1^2 = 3 - 2 + 1 = 2\).
- For \(\lambda_2 = -2\): \(P(-2) = 3 - 2(-2) + (-2)^2 = 3 + 4 + 4 = 11\).
- For \(\lambda_3 = 3\): \(P(3) = 3 - 2(3) + 3^2 = 3 - 6 + 9 = 6\).

Step 4: Match the calculated eigenvalues with the given statements.
The eigenvalues of \(3I - 2A + A^2\) are 2, 6, and 11. These correspond to statements A, B, and D. Statement C (8) is not an eigenvalue. Quick Tip: This property is very powerful. It means you don't need to find the matrix \(A\) itself. You can find the eigenvalues of any function of \(A\) (like \(A^2\), \(A^{-1}\), or \(e^A\)) directly from the eigenvalues of \(A\).

Step 1: Define an even function and the Fourier coefficients.
An even function satisfies \(f(-x) = f(x)\). The Fourier series is \(f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))\).
- \(a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx\)
- \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) dx\)
- \(b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) dx\)

Step 2: Analyze the coefficients for an even function.
- \(a_0\) and \(a_n\): The cosine function, \(\cos(nx)\), is also an even function. The product of two even functions (\(f(x)\) and \(\cos(nx)\)) is even. The integral of a non-zero even function over a symmetric interval (like \(-\pi\) to \(\pi\)) is generally non-zero. Thus, for a general even function, we expect \(a_0 \neq 0\) and \(a_n \neq 0\). Statements A and B are generally correct. Statement C is incorrect.
- \(b_n\): The sine function, \(\sin(nx)\), is an odd function. The product of an even function (\(f(x)\)) and an odd function (\(\sin(nx)\)) is an odd function. The integral of any odd function over a symmetric interval is always zero. Thus, for any even function, \(b_n = 0\). Statement D is correct.

Step 3: Conclude the correct statements.
The correct statements describing the Fourier coefficients for an even function are that \(a_0\) and \(a_n\) can be non-zero, while \(b_n\) must be zero. This corresponds to statements A, B, and D. Quick Tip: Remember the symmetry properties for Fourier series: - **Even function:** Contains only cosine terms (\(b_n = 0\)). - **Odd function:** Contains only sine terms (\(a_0 = 0\) and \(a_n = 0\)). This saves a lot of calculation time.

Step 1: Recall the cyclic property of powers of \(i\). \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\). The pattern repeats every 4 powers. To find \(i^n\), we can evaluate \(i^k\) where \(k\) is the remainder of \(n\) divided by 4.

Step 2: Evaluate each expression.
- A. \(i^{49}\): The remainder of 49 divided by 4 is 1. So, \(i^{49} = i^1 = i\). This matches III.
- B. \(i^{38}\): The remainder of 38 divided by 4 is 2. So, \(i^{38} = i^2 = -1\). This matches IV.
- C. \(i^{103}\): The remainder of 103 divided by 4 is 3. So, \(i^{103} = i^3 = -i\). This matches II.
- D. \(i^{92}\): 92 is perfectly divisible by 4 (remainder is 0). We can treat this as \(i^4\). So, \(i^{92} = (i^4)^{23} = 1^{23} = 1\). This matches I.

Step 3: Formulate the correct sequence.
The matches are: A\(\rightarrow\)III, B\(\rightarrow\)IV, C\(\rightarrow\)II, D\(\rightarrow\)I. This corresponds to option (3). Quick Tip: To quickly find the remainder when dividing a number by 4, you only need to look at its last two digits. For example, for \(i^{103}\), you just need the remainder of 03 divided by 4, which is 3.

Step 1: Express the complex number \(z = x+iy\) in polar form.
In polar coordinates, \(z = r e^{i\theta}\), where the modulus is \(r = |z| = \sqrt{x^2 + y^2}\) and the argument is \(\theta = \arg(z) = \tan^{-1}(y/x)\).

Step 2: Apply the complex logarithm.
The natural logarithm of \(z\) is: \[ \log(z) = \log(r e^{i\theta}) \]
Using the property \(\log(ab) = \log(a) + \log(b)\), we get: \[ \log(z) = \log(r) + \log(e^{i\theta}) \]
Using the property \(\log(e^w) = w\), we get: \[ \log(z) = \log(r) + i\theta \]

Step 3: Identify the real and imaginary parts.
- The real part is \(Re(\log(z)) = \log(r) = \log(\sqrt{x^2+y^2})\).
- The imaginary part is \(Im(\log(z)) = \theta = \tan^{-1}(y/x)\).

This matches the expressions in option (4). Note that \(\log(\sqrt{x^2+y^2})\) is also equal to \(\frac{1}{2}\log(x^2+y^2)\). Quick Tip: To find the logarithm of a complex number, the first step is always to convert it from Cartesian form (\(x+iy\)) to polar form (\(re^{i\theta}\)). The logarithm then separates neatly into its real and imaginary parts.

Step 1: Establish the relationships between Cartesian and polar coordinates.
We are given \(x = r\cos\theta\) and \(y = r\sin\theta\). The inverse relationships are \(r^2 = x^2 + y^2\) and \(\theta = \tan^{-1}(y/x)\).

Step 2: Calculate the partial derivatives.
- A. \(\frac{\partial r}{\partial x}\): Differentiate \(r^2 = x^2 + y^2\) with respect to \(x\):
\(2r \frac{\partial r}{\partial x} = 2x \implies \frac{\partial r}{\partial x} = \frac{x}{r}\). This matches III.
- B. \(\frac{\partial r}{\partial y}\): Differentiate \(r^2 = x^2 + y^2\) with respect to \(y\):
\(2r \frac{\partial r}{\partial y} = 2y \implies \frac{\partial r}{\partial y} = \frac{y}{r}\). This matches II.

Step 3: Calculate the Jacobians.
- C. \(\frac{\partial(x,y)}{\partial(r,\theta)}\): This is the determinant of the Jacobian matrix for the transformation from polar to Cartesian coordinates.

This matches IV.
- D. \(\frac{\partial(r,\theta)}{\partial(x,y)}\): This is the inverse of the Jacobian calculated in C.
\[ \frac{\partial(r,\theta)}{\partial(x,y)} = \left(\frac{\partial(x,y)}{\partial(r,\theta)}\right)^{-1} = \frac{1}{r} \]
This matches I.

Step 4: Formulate the correct matching sequence.
The matches are: A\(\rightarrow\)III, B\(\rightarrow\)II, C\(\rightarrow\)IV, D\(\rightarrow\)I. This corresponds to option (2). Quick Tip: The Jacobian \(\frac{\partial(x,y)}{\partial(r,\theta)} = r\) is a crucial result used for changing variables in double integrals from Cartesian to polar coordinates: \(dx dy = r dr d\theta\).

Step 1: Determine the order of the ODE.
The order is the highest derivative present in the equation. Here, the highest derivative is \(\frac{d^2y}{dx^2}\). Thus, the order is 2.

Step 2: Determine the degree of the ODE.
The degree is the highest power of the highest-order derivative after the equation has been cleared of radicals and fractions in its derivatives. First, square both sides to remove the \(3/2\) power: \[ \left[ \left(1+\frac{d^2y}{dx^2}\right)^{3/2} \right]^2 = \left(y\frac{d^2y}{dx^2}\right)^2 \] \[ \left(1+\frac{d^2y}{dx^2}\right)^{3} = y^2\left(\frac{d^2y}{dx^2}\right)^2 \]
Now, expand the left side using the binomial theorem \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\): \[ 1 + 3\left(\frac{d^2y}{dx^2}\right) + 3\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^3 = y^2\left(\frac{d^2y}{dx^2}\right)^2 \]
The highest power of the highest derivative (\(\frac{d^2y}{dx^2}\)) is 3. Thus, the degree is 3.

Step 3: Determine the linearity.
An equation is linear if the dependent variable \(y\) and all its derivatives appear only to the first power and are not multiplied together. This equation contains terms like \(\left(\frac{d^2y}{dx^2}\right)^3\) and \(y^2\left(\frac{d^2y}{dx^2}\right)^2\). Because of these higher powers and products, the equation is non-linear.

The correct description is order 2, degree 3, and non-linear. Quick Tip: To find the degree, always make sure the equation is a polynomial in its derivatives first. This means eliminating all fractional powers and denominators involving derivatives.

Step 1: Solve for the particular integral of each expression.
- B. \(\frac{1}{D^2+D+1}\cos x\): For cosine functions, we substitute \(D^2 \to -a^2\), so \(D^2 \to -1^2 = -1\).
\[ P.I. = \frac{1}{-1+D+1}\cos x = \frac{1}{D}\cos x = \int \cos x \,dx = \sin x \]
This matches II.
- C. \(\frac{1}{(D-1)^2}e^x\): This is a case of failure, as substituting \(D=1\) makes the denominator zero. We use the formula \(\frac{1}{f(D)}e^{ax}V = e^{ax}\frac{1}{f(D+a)}V\).
\[ P.I. = e^x \frac{1}{((D+1)-1)^2}(1) = e^x \frac{1}{D^2}(1) = e^x \iint 1 \,dx\,dx = e^x \frac{x^2}{2} \]
This matches III.
- D. \(\frac{1}{D^3-3D^2+4D-2}e^x\): Substitute \(D=1\): \(1-3+4-2=0\). It's a failure case. We use the rule: If \(f(a)=0\), P.I. is \(x \frac{1}{f'(a)}e^{ax}\).
Let \(f(D) = D^3-3D^2+4D-2\). Then \(f'(D) = 3D^2-6D+4\).
\(f'(1) = 3(1)^2 - 6(1) + 4 = 3-6+4=1\).
\[ P.I. = x \frac{1}{1}e^x = xe^x \]
This matches I.
- A. \(\frac{1}{(D-1)}x^2\): We use binomial expansion: \(\frac{1}{D-1} = -(1-D)^{-1} = -(1+D+D^2+\dots)\).
\[ P.I. = -(1+D+D^2)(x^2) = -(x^2 + D(x^2) + D^2(x^2)) = -(x^2 + 2x + 2) \]
This matches IV.

Step 2: Formulate the correct matching sequence.
The matches are: A\(\rightarrow\)IV, B\(\rightarrow\)II, C\(\rightarrow\)III, D\(\rightarrow\)I. This corresponds to option (3). Quick Tip: Master the different methods for finding particular integrals based on the form of the function on the right side: exponential (\(e^{ax}\)), trigonometric (\(\sin(ax), \cos(ax)\)), polynomial (\(x^n\)), and their products. Special attention is needed for "cases of failure" where the simple substitution method fails.

Step 1: Use vector calculus identities.
There are two fundamental vector identities that can solve this problem without direct calculation:
1. The curl of the gradient of any scalar field is always zero: \(\vec{\nabla} \times (\vec{\nabla}\phi) = \vec{0}\).
2. The divergence of the gradient is the Laplacian: \(\vec{\nabla} \cdot (\vec{\nabla}\phi) = \nabla^2\phi\).

Step 2: Apply the identities to the given problem.
- Curl of \(\vec{A}\): We are given \(\vec{A} = \vec{\nabla}\phi\). Using the first identity, we immediately know that \(\vec{\nabla} \times \vec{A} = \vec{\nabla} \times (\vec{\nabla}\phi) = \vec{0}\). Therefore, statement C is true and D is false. A vector field that is the gradient of a scalar potential is called a conservative or irrotational field.

- Divergence of \(\vec{A}\): Using the second identity, we need to calculate the Laplacian of \(\phi\).
\[ \vec{\nabla} \cdot \vec{A} = \nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} \]
First derivatives are: \(\frac{\partial\phi}{\partial x} = y+z\), \(\frac{\partial\phi}{\partial y} = x+z\), \(\frac{\partial\phi}{\partial z} = y+x\).
Second derivatives are:
\[ \frac{\partial^2\phi}{\partial x^2} = 0, \quad \frac{\partial^2\phi}{\partial y^2} = 0, \quad \frac{\partial^2\phi}{\partial z^2} = 0 \]
Therefore, \(\vec{\nabla} \cdot \vec{A} = 0+0+0=0\). Statement A is true and B is false.

Step 3: Conclude the correct statements.
Both statements A and C are true. Quick Tip: Knowing the identities \(\nabla \times (\nabla \phi) = 0\) and \(\nabla \cdot (\nabla \times \vec{F}) = 0\) can save you a lot of time. If a vector field is given as a gradient of a scalar (\(\vec{A} = \vec{\nabla}\phi\)), its curl must be zero. If it's given as the curl of another vector (\(\vec{B} = \vec{\nabla} \times \vec{F}\)), its divergence must be zero.

Step 1: Recall the formula for the scalar projection of vector \(\vec{A}\) onto vector \(\vec{B}\).
The projection is given by: \[ proj_{\vec{B}}\vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \]

Step 2: Calculate the dot product \(\vec{A} \cdot \vec{B}\). \[ \vec{A} \cdot \vec{B} = (1)(4) + (-2)(-4) + (1)(7) \] \[ \vec{A} \cdot \vec{B} = 4 + 8 + 7 = 19 \]

Step 3: Calculate the magnitude of vector \(\vec{B}\). \[ |\vec{B}| = \sqrt{4^2 + (-4)^2 + 7^2} \] \[ |\vec{B}| = \sqrt{16 + 16 + 49} = \sqrt{81} = 9 \]

Step 4: Compute the projection. \[ proj_{\vec{B}}\vec{A} = \frac{19}{9} \] Quick Tip: Be careful to distinguish between scalar projection (which is a number, as asked for here) and vector projection (which would be a vector: \( \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\vec{B} \)). The formula divides by the magnitude \(|\vec{B}|\), not the magnitude squared.

Step 1: Apply the Divergence Theorem (Gauss's Theorem).
The Divergence Theorem relates a surface integral over a closed surface S to a volume integral over the volume V enclosed by that surface: \[ \iint_S \vec{A} \cdot \hat{n} dS = \iiint_V (\vec{\nabla} \cdot \vec{A}) dV \]

Step 2: Substitute the given expression for \(\vec{A}\).
We are given that \(\vec{A} = \vec{\nabla} \times \vec{F}\). Substituting this into the volume integral: \[ \iiint_V \vec{\nabla} \cdot (\vec{\nabla} \times \vec{F}) dV \]

Step 3: Use the vector calculus identity for the divergence of a curl.
A fundamental identity in vector calculus is that the divergence of the curl of any vector field is always identically zero: \[ \vec{\nabla} \cdot (\vec{\nabla} \times \vec{F}) = 0 \]

Step 4: Evaluate the integral.
Since the integrand is zero, the volume integral is zero. \[ \iiint_V (0) dV = 0 \]
Therefore, the surface integral is also zero. Quick Tip: The identity \(div(curl \, \vec{F}) = 0\) is extremely useful. It implies that a vector field which is the curl of another field (like the magnetic field \(\vec{B} = \vec{\nabla} \times \vec{A}\)) must be solenoidal (divergence-free). This means it has no sources or sinks, and its field lines must form closed loops.

Step 1: Square both sides of the given equation.
The magnitude of a vector is always non-negative, so we can square both sides without loss of information. \[ |\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2 \]

Step 2: Express the square of the magnitude using the dot product.
Recall that for any vector \(\vec{V}\), \(|\vec{V}|^2 = \vec{V} \cdot \vec{V}\). \[ (\vec{A}+\vec{B}) \cdot (\vec{A}+\vec{B}) = (\vec{A}-\vec{B}) \cdot (\vec{A}-\vec{B}) \]

Step 3: Expand the dot products. \[ \vec{A}\cdot\vec{A} + \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - \vec{A}\cdot\vec{B} - \vec{B}\cdot\vec{A} + \vec{B}\cdot\vec{B} \]
Since \(\vec{A}\cdot\vec{A} = |\vec{A}|^2\) and the dot product is commutative (\(\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\)): \[ |\vec{A}|^2 + 2(\vec{A}\cdot\vec{B}) + |\vec{B}|^2 = |\vec{A}|^2 - 2(\vec{A}\cdot\vec{B}) + |\vec{B}|^2 \]

Step 4: Simplify the equation to find the condition on the dot product.
Canceling the \(|\vec{A}|^2\) and \(|\vec{B}|^2\) terms from both sides: \[ 2(\vec{A}\cdot\vec{B}) = -2(\vec{A}\cdot\vec{B}) \] \[ 4(\vec{A}\cdot\vec{B}) = 0 \] \[ \vec{A}\cdot\vec{B} = 0 \]

Step 5: Interpret the result.
The dot product of two non-zero vectors is zero if and only if they are orthogonal (perpendicular). The angle between them is 90 degrees, or \(\pi/2\) radians. Quick Tip: Geometrically, the vectors \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\) represent the diagonals of a parallelogram formed by vectors \(\vec{A}\) and \(\vec{B}\). The condition that the diagonals have equal length means the parallelogram must be a rectangle, which implies that \(\vec{A}\) and \(\vec{B}\) are perpendicular.

Step 1: Recall the formula for the rate of precession of a Foucault pendulum.
The angular speed \(\omega_P\) of the precession of the plane of oscillation is given by: \[ \omega_P = \Omega \sin\phi \]
where \(\Omega\) is the angular speed of the Earth's rotation, and \(\phi\) is the latitude of the pendulum's location.

Step 2: Find the condition for no rotation.
For the plane of vibration to not rotate at all, the rate of precession \(\omega_P\) must be zero. \[ \Omega \sin\phi = 0 \]

Step 3: Solve for the latitude \(\phi\).
Since the Earth is rotating, \(\Omega \neq 0\). Therefore, we must have: \[ \sin\phi = 0 \]
This condition is met when the latitude \(\phi = 0^{\circ}\).

Step 4: Identify the geographical location corresponding to this latitude.
A latitude of \(0^{\circ}\) corresponds to the Earth's Equator. At the North or South Pole, \(\phi = \pm 90^{\circ}\), \(\sin\phi = \pm 1\), and the rotation is maximum. Quick Tip: Visualize the Earth's rotation. At the poles, an observer is simply spinning in place, so the pendulum's plane appears to rotate a full circle in 24 hours. At the equator, an observer is carried along without any local twisting motion, so the pendulum's plane remains fixed relative to the ground.

Step 1: Define a coordinate system and calculate the initial momentum components.
Let east be the positive x-direction and north be the positive y-direction. The collision is perfectly inelastic, so momentum is conserved.
- Momentum of the car (x-direction): \(p_x = m_{car} v_{car} = (1500 \, kg)(25 \, m/s) = 37500 \, kg\cdot m/s\).
- Momentum of the van (y-direction): \(p_y = m_{van} v_{van} = (2500 \, kg)(20 \, m/s) = 50000 \, kg\cdot m/s\).
The initial total momentum is \(\vec{P}_{initial} = 37500\hat{i} + 50000\hat{j}\).

Step 2: Apply the law of conservation of momentum.
For a perfectly inelastic collision, the vehicles stick together. The final momentum \(\vec{P}_{final}\) must be equal to the initial momentum. \[ \vec{P}_{final} = 37500\hat{i} + 50000\hat{j} \]

Step 3: Determine the direction of the final momentum vector.
The direction of the wreckage is the direction of the final momentum vector. The angle \(\theta\) this vector makes with the positive x-axis (East) is given by: \[ \tan\theta = \frac{P_y}{P_x} = \frac{50000}{37500} \]

Step 4: Calculate the angle. \[ \tan\theta = \frac{500}{375} = \frac{20}{15} = \frac{4}{3} \approx 1.333 \] \[ \theta = \arctan\left(\frac{4}{3}\right) \approx 53.13^{\circ} \]
The direction is approximately \(53.1^{\circ}\) (north of east). Quick Tip: In 2D collision problems, always break the momentum into x and y components. Momentum is conserved independently in each direction. For a perfectly inelastic collision, the final velocity vector points in the same direction as the total initial momentum vector.

This is a standard result from classical mechanics. The moment of inertia \(I\) of a solid cone of mass \(M\) and base radius \(R\) about its central axis of symmetry (the vertical axis passing through the apex and the center of the base) is given by the formula: \[ I = \frac{3}{10}MR^2 \]
This can be derived by integrating the moment of inertia of infinitesimal circular disks that make up the cone. The moment of inertia of a disk of mass \(dm\) and radius \(r\) is \(\frac{1}{2}r^2 dm\). Integrating this from the apex to the base yields the final result. Quick Tip: It's helpful to memorize the moments of inertia for common shapes: - Thin Rod (center): \(ML^2/12\) - Hoop (central axis): \(MR^2\) - Solid Cylinder/Disk (central axis): \(MR^2/2\) - Solid Sphere (center): \(2MR^2/5\) - Solid Cone (central axis): \(3MR^2/10\)

Step 1: Recall the definition of work done by a constant force in linear motion.
Work \(W\) is defined as the product of the force \(F\) and the displacement \(d\) in the direction of the force: \(W = F \cdot d\).

Step 2: Use the rotational analogues for force and displacement.
In rotational motion, the analogue of force is torque (\(\tau\)), and the analogue of linear displacement is angular displacement (\(\theta\)).

Step 3: Formulate the expression for rotational work.
By analogy, the work done by a constant torque is the product of the torque and the angular displacement. \[ W = \tau \theta \]
This corresponds to option (3). If the torque is not constant, the work done is found by integrating the torque over the angular displacement: \(W = \int \tau \, d\theta\). Quick Tip: Many concepts in rotational mechanics have direct analogues in linear mechanics. - Position \(x\) \(\leftrightarrow\) Angle \(\theta\) - Velocity \(v\) \(\leftrightarrow\) Angular Velocity \(\omega\) - Mass \(m\) \(\leftrightarrow\) Moment of Inertia \(I\) - Force \(F\) \(\leftrightarrow\) Torque \(\tau\) - Momentum \(p=mv\) \(\leftrightarrow\) Angular Momentum \(L=I\omega\) - Work \(W=Fd\) \(\leftrightarrow\) Work \(W=\tau\theta\)

Step 1: Recall the definition of a conservative force.
A force field \(\vec{F}\) is conservative if its curl is equal to the zero vector. \[ \vec{\nabla} \times \vec{F} = \vec{0} \]

Step 2: Write out the components of the curl.
The curl of \(\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}\) is given by the determinant:

Step 3: Set the curl to zero and analyze the components.
For the curl to be zero, each of its vector components must be zero.
- \(\hat{i}\) component: \(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0\). This is statement B.
- \(\hat{j}\) component: \(-\left(\frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z}\right) = 0 \implies \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0\). This is statement C.
- \(\hat{k}\) component: \(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 0\). This is statement A.

Therefore, for a force to be conservative, all three relations A, B, and C must be satisfied. Statement D describes a non-conservative force. Quick Tip: A force is conservative if it can be written as the gradient of a scalar potential, \(\vec{F} = -\vec{\nabla}V\). The condition \(\vec{\nabla} \times \vec{F} = 0\) is equivalent to this, due to the identity \(\vec{\nabla} \times (\vec{\nabla}V) = 0\).

Step 1: Analyze the motion of the water.
As a stream of water falls from a faucet, it is accelerated downwards by gravity. This means its speed increases as it descends.

Step 2: Apply the continuity equation for an incompressible fluid.
The continuity equation is an expression of the conservation of mass for a fluid. For a steady flow of an incompressible fluid (like water), it takes the form: \[ A_1 v_1 = A_2 v_2 = Constant \]
where \(A\) is the cross-sectional area of the stream and \(v\) is the fluid velocity.

Step 3: Relate the change in velocity to the change in area.
Let point 1 be at the faucet and point 2 be some distance below. Due to gravity, \(v_2 > v_1\). For the product \(Av\) to remain constant, if the velocity \(v\) increases, the cross-sectional area \(A\) must decrease. A smaller cross-sectional area means the stream becomes narrower. This directly explains the observation. Bernoulli's equation relates pressure, velocity, and height, but the continuity equation is the primary principle explaining the change in shape. Quick Tip: The continuity equation \(Av = constant\) is a very intuitive principle of fluid flow. It simply means that "what goes in must come out." If you squeeze a hose to make the opening smaller (decrease A), the water must speed up (increase v).

Step 1: Analyze the initial state (anchor in the boat).
The boat and the anchor together are floating. According to Archimedes' principle, a floating object displaces a volume of water with a weight equal to the total weight of the object.
Let \(W_{boat}\) be the weight of the boat and \(W_{anchor}\) be the weight of the anchor. The total weight is \(W_{total} = W_{boat} + W_{anchor}\).
The weight of the displaced water is \(W_{disp,1} = W_{total}\). The volume of displaced water is \(V_{disp,1} = \frac{W_{disp,1}}{\rho_{water}g} = \frac{W_{boat} + W_{anchor}}{\rho_{water}g}\).

Step 2: Analyze the final state (anchor in the water).
The boat is now floating by itself, and the anchor is fully submerged at the bottom of the pond.
- The boat displaces a volume of water corresponding to its own weight: \(V_{boat,disp} = \frac{W_{boat}}{\rho_{water}g}\).
- The anchor is submerged and displaces a volume of water equal to its own volume. The volume of the anchor is \(V_{anchor} = \frac{W_{anchor}}{\rho_{anchor}g}\).
The total volume of water displaced in the final state is the sum of these two volumes: \[ V_{disp,2} = V_{boat,disp} + V_{anchor} = \frac{W_{boat}}{\rho_{water}g} + \frac{W_{anchor}}{\rho_{anchor}g} \]

Step 3: Compare the initial and final displaced volumes.
We compare \(V_{disp,1}\) and \(V_{disp,2}\). \[ V_{disp,1} = \frac{W_{boat}}{\rho_{water}g} + \frac{W_{anchor}}{\rho_{water}g} \]
The difference between the two states is the volume displaced by the anchor's weight. We are comparing \(\frac{W_{anchor}}{\rho_{water}g}\) (when it's part of the floating system) with \(\frac{W_{anchor}}{\rho_{anchor}g}\) (when it's submerged).
Since an anchor is made of a dense material (e.g., iron), its density \(\rho_{anchor}\) is much greater than the density of water \(\rho_{water}\).
Therefore, \(\frac{1}{\rho_{water}} > \frac{1}{\rho_{anchor}}\), which implies \(\frac{W_{anchor}}{\rho_{water}g} > \frac{W_{anchor}}{\rho_{anchor}g}\).
This means \(V_{disp,1} > V_{disp,2}\).

Step 4: Conclude the effect on the water level.
Since the total volume of water displaced by the system decreases when the anchor is thrown overboard, the overall water level in the pond must go down. Quick Tip: The key insight is that when the anchor is in the boat, it contributes its full *weight* to water displacement. When it's at the bottom, it only displaces its own *volume*. Since the anchor is denser than water, the volume of water equivalent to its weight is much larger than its actual volume.

Step 1: Define impulse.
Impulse is the change in momentum of an object (\(\Delta p\)). By the impulse-momentum theorem, the impulse imparted to the skater is equal to the change in her momentum. By Newton's third law, this is equal in magnitude to the impulse imparted to the Frisbee. We need to find the case with the largest change in the Frisbee's momentum. Let the initial momentum of the Frisbee be \(p_i = mv\).

Step 2: Analyze the change in momentum for each case.
- Case 1 (Catches and holds): The Frisbee's final momentum, \(p_f\), becomes part of the skater's system. The change in momentum is \(\Delta p = p_f - p_i\). The impulse to stop the Frisbee relative to the skater is \(0 - mv = -mv\). The magnitude is \(|mv|\).
- Case 2 (Catches and drops): The horizontal impulse is the same as in case 1, as the skater must first absorb the Frisbee's momentum to catch it. The magnitude is \(|mv|\).
- Case 3 (Catches and throws back): The skater first absorbs the initial momentum \(mv\) and then imparts an additional momentum to throw it back, giving it a final momentum \(p_f = -mv'\) (in the opposite direction). The total change in the Frisbee's momentum is \(\Delta p = p_f - p_i = (-mv') - (mv) = -m(v+v')\). The magnitude of the impulse is \(m(v+v')\), which is larger than \(|mv|\).
- Case 4 (Can't catch): No interaction means zero impulse.

Step 3: Compare the impulses.
The greatest change in momentum occurs when the Frisbee's velocity is reversed. This action imparts the largest impulse to the skater. Quick Tip: Impulse is maximized when there is a reversal of momentum. Think of a bouncy ball versus a lump of clay hitting a wall; the bouncy ball imparts a greater impulse because its momentum changes from \(+p\) to \(-p\), a total change of \(2p\).

Step 1: Recall the rocket equation for thrust.
The thrust force \(F\) on a rocket is given by the equation: \[ F = v_{rel} \left| \frac{dm}{dt} \right| \]
where \(v_{rel}\) is the exhaust velocity relative to the rocket, and \(\left| \frac{dm}{dt} \right|\) is the rate of mass ejection.

Step 2: Relate thrust to acceleration using Newton's second law.
The force on the rocket produces an acceleration \(a\) according to \(F = M a\), where \(M\) is the instantaneous mass of the rocket. \[ M a = v_{rel} \left| \frac{dm}{dt} \right| \implies a = \frac{v_{rel}}{M} \left| \frac{dm}{dt} \right| \]

Step 3: Determine the values for the initial acceleration.
- We need the initial acceleration, so we use the initial mass of the rocket, \(M = M_0\).
- The relative speed is given as \(v_{rel} = 2000 \, m/s\).
- The rocket ejects \(M_0/100\) of its mass in the first second, so the rate of mass ejection is \(\left| \frac{dm}{dt} \right| = \frac{M_0/100}{1 \, s} = \frac{M_0}{100}\).

Step 4: Calculate the initial acceleration. \[ a_{initial} = \frac{2000 \, m/s}{M_0} \left( \frac{M_0}{100} \right) = \frac{2000}{100} \, m/s^2 = 20 \, m/s^2 \]
The acceleration is positive, indicating it is in the forward direction. Quick Tip: The rocket thrust equation is a direct application of the conservation of momentum. The force on the rocket is the reaction force to the force required to eject the fuel mass.

Step 1: Analyze the superposition of two perpendicular SHMs of the same frequency. The resulting figure depends on the phase difference \(\delta\).
- D. Linearly polarized vibrations: The resulting motion is along a straight line. This occurs when the phase difference is \(\delta = 0\) or \(\pi\). This matches III (No phase difference).
- C. Circularly polarized vibrations: The resulting motion is a circle. This occurs for the special case where the amplitudes are equal and the phase difference is \(\delta = \pi/2\) or \(3\pi/2\). This matches IV.
- A. & B. Elliptically polarized vibrations: For any other phase difference, the resulting motion is an ellipse.
- For phase differences between \(0\) and \(\pi\), such as \(\pi/4\) and \(3\pi/4\), the polarization is elliptical. By convention, phase differences in this range can define right and left-handedness based on the tracing direction. \(\delta = \pi/4\) corresponds to a right-handed ellipse (A matches I), and \(\delta = 3\pi/4\) to another ellipse, defined here as left-handed (B matches II).

Step 2: Formulate the correct matching sequence based on the analysis.
- A \(\rightarrow\) I
- B \(\rightarrow\) II
- C \(\rightarrow\) IV
- D \(\rightarrow\) III
This sequence is A - I, B - II, C - IV, D - III, which corresponds to option (3). Quick Tip: For Lissajous figures with two SHMs of the same frequency: - Phase diff \(0\) or \(\pi\) \(\rightarrow\) Straight Line - Phase diff \(\pi/2\) or \(3\pi/2\) (and equal amplitudes) \(\rightarrow\) Circle - Any other phase diff \(\rightarrow\) Ellipse

Step 1: Identify the standard equation for simple harmonic motion (SHM).
The displacement \(y\) in SHM is given by \(y = A \sin(\omega t + \phi)\), where \(A\) is amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase angle.

Step 2: Extract the angular frequency \(\omega\) from the given equation.
The given equation is \(y = 5 \sin(100\pi t + \phi)\). Comparing this to the standard form, we can see that the angular frequency \(\omega = 100\pi\) rad/s.

Step 3: Convert angular frequency (\(\omega\)) to linear frequency (\(f\)).
The relationship between them is \(\omega = 2\pi f\). \[ f = \frac{\omega}{2\pi} \]

Step 4: Calculate the frequency. \[ f = \frac{100\pi}{2\pi} = 50 \, Hz \]
The frequency of oscillation is 50 Hz. Quick Tip: Always be careful to distinguish between angular frequency \(\omega\) (in rad/s) and frequency \(f\) (in Hz). The term multiplying \(t\) inside the sine/cosine function is always \(\omega\).

Step 1: Define dispersion.
Dispersion is the phenomenon where the phase velocity of a wave depends on its frequency. The relationship is characterized by group velocity (\(v_g = \frac{d\omega}{dk}\)) and phase velocity (\(v_p = \frac{\omega}{k}\)).

Step 2: Characterize the types of dispersion.
- **No Dispersion:** The phase velocity is constant for all frequencies. In this case, \(v_g = v_p\).
- **Normal Dispersion:** The phase velocity decreases as frequency increases (e.g., light in glass). This corresponds to \(v_g < v_p\).
- **Anomalous Dispersion:** The phase velocity increases as frequency increases. This phenomenon occurs in specific frequency bands, typically near a medium's absorption frequency. This condition corresponds to \(v_g > v_p\).

Therefore, anomalous dispersion occurs when the group velocity is greater than the phase velocity. Quick Tip: A simple way to remember is: - Normal: \(v_g < v_p\) (The "normal" situation for light through a prism). - Anomalous: \(v_g > v_p\) (The "anomalous" or unusual case). - Non-dispersive: \(v_g = v_p\) (e.g., light in a vacuum).

Step 1: Determine the possible initial frequencies of the unknown fork.
Let the unknown frequency be \(f_1\). The reference frequency is \(f_2 = 256\) Hz. The beat frequency is \(f_{beat} = |f_1 - f_2| = 4\) Hz.
This gives two possibilities for \(f_1\):
- \(f_1 = 256 + 4 = 260\) Hz
- \(f_1 = 256 - 4 = 252\) Hz

Step 2: Analyze the effect of adding wax.
Adding wax to a tuning fork increases its mass, which causes its frequency to decrease. Let the new frequency of the first fork be \(f_1'\), where \(f_1' < f_1\).

Step 3: Test the two possibilities with the new information.
The new beat frequency is 3 Hz.
- Case A: Assume initial \(f_1 = 260\) Hz. When wax is added, its frequency \(f_1'\) will be slightly less than 260 Hz. The new beat frequency is \(|f_1' - 256|\). As \(f_1'\) moves from 260 towards 256, the difference decreases. It is possible for the beat frequency to become 3 Hz. This is consistent.
- Case B: Assume initial \(f_1 = 252\) Hz. When wax is added, its frequency \(f_1'\) will be slightly less than 252 Hz. The new beat frequency is \(|f_1' - 256| = 256 - f_1'\). As \(f_1'\) decreases from 252, it moves further away from 256, so the difference will increase. The beat frequency would become greater than 4 Hz, not 3 Hz. This contradicts the observation.

Step 4: Conclude the initial frequency.
Only the first case is consistent with the experimental result. Therefore, the original frequency of the first fork was 260 Hz. Quick Tip: Remember the effects of modifying a tuning fork: - **Adding mass (waxing):** Decreases frequency. - **Removing mass (filing):** Increases frequency.

The guiding principle for the path of light, which covers reflection and refraction, is Fermat's Principle of Least Time. This principle states that the path a light ray takes between two points is the path that can be traveled in the least amount of time.
When light travels in a uniform medium, its speed is constant, and the path of minimum time is a straight line (the minimum distance). However, when light refracts, its speed changes as it enters the new medium. To minimize the total travel time, the light bends, following a path that is not the shortest distance but is the quickest. Snell's law can be derived from this principle. Quick Tip: Fermat's Principle is a powerful concept in optics. Think of it like a lifeguard saving a swimmer: the lifeguard doesn't run in a straight line to the swimmer, but runs a longer distance on the sand (where they are fast) and a shorter distance in the water (where they are slow) to minimize the total time.

Step 1: Define the system.
Let the first lens (\(L_1\)) be at the origin (\(x=0\)) and the second lens (\(L_2\)) be at \(x=4\) cm. We have \(f_1 = 2\) cm, \(f_2 = 6\) cm, and \(d = 4\) cm.

Step 2: Calculate the equivalent focal length (\(F\)). \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = \frac{1}{2} + \frac{1}{6} - \frac{4}{12} = \frac{6+2-4}{12} = \frac{4}{12} \implies F = 3 \, cm \]

Step 3: Calculate the positions of the Principal Points (\(P_1, P_2\)).
- The position of \(P_1\) relative to \(L_1\) is \(\alpha_1 = \frac{dF}{f_2} = \frac{4 \times 3}{6} = 2\) cm. So, the absolute position of \(P_1\) is at \(x=2\).
- The position of \(P_2\) relative to \(L_2\) is \(\alpha_2 = -\frac{dF}{f_1} = -\frac{4 \times 3}{2} = -6\) cm. So, the absolute position of \(P_2\) is at \(x = 4 - 6 = -2\).

Step 4: Calculate the positions of the Focal Points (\(F_1, F_2\)).
- The position of \(F_1\) is at a distance \(-F\) from \(P_1\). Absolute position of \(F_1\) is \(x = 2 - 3 = -1\).
- The position of \(F_2\) is at a distance \(+F\) from \(P_2\). Absolute position of \(F_2\) is \(x = -2 + 3 = 1\).

Step 5: Determine the positions of the Nodal Points (\(N_1, N_2\)).
Since the medium is air on both sides, the nodal points coincide with the principal points. So, \(N_1\) is at \(x=2\) and \(N_2\) is at \(x=-2\).

Step 6: Calculate the distance of each point from the second lens (\(L_2\) at \(x=4\)) and order them.
- A (First Principal Point at \(x=2\)): Distance = \(|2-4| = 2\) cm.
- B (First Focal Point at \(x=-1\)): Distance = \(|-1-4| = 5\) cm.
- C (Second Focal Point at \(x=1\)): Distance = \(|1-4| = 3\) cm.
- D (Second Nodal Point at \(x=-2\)): Distance = \(|-2-4| = 6\) cm.

Arranging these distances in ascending order: A (2 cm) < C (3 cm) < B (5 cm) < D (6 cm). The correct order is A, C, B, D. Quick Tip: For a two-lens system, always establish a coordinate system first (e.g., first lens at the origin). Calculate the positions of the principal points relative to the physical lenses, then use these principal points as the reference for locating the focal points.

Step 1: Analyze each phenomenon in List-I and match it with the underlying principle in List-II.
- A. Compton Effect: This involves the change in wavelength of an X-ray photon when it collides with an electron. It is a classic example of particle-like Scattering. This matches IV.
- B. Colors in thin film: The vibrant colors seen in soap bubbles or oil slicks are caused by the constructive and destructive Interference of light waves reflecting off the top and bottom surfaces of the thin film. This matches II.
- C. Double Refraction: Certain crystals, like calcite, split an unpolarized light beam into two separate, plane-polarized beams. This phenomenon is a direct consequence of the Polarization of light. This matches III.
- D. Bragg's Equation: This equation (\(n\lambda = 2d\sin\theta\)) describes the condition for constructive interference of waves scattered by crystal planes. It is the fundamental principle of X-ray Diffraction. This matches I.

Step 2: Assemble the correct pairings.
- A \(\rightarrow\) IV
- B \(\rightarrow\) II
- C \(\rightarrow\) III
- D \(\rightarrow\) I
This corresponds to option (1). Quick Tip: Associate these key pairs: - Thin Films \(\leftrightarrow\) Interference - Crystals/Slits \(\leftrightarrow\) Diffraction - Calcite/Polaroids \(\leftrightarrow\) Polarization - Photon-Electron Collision \(\leftrightarrow\) Scattering

Step 1: Recall the formula for the resolving power of a diffraction grating.
The resolving power \(R\) is defined as \(R = \frac{\lambda}{\Delta\lambda}\), where \(\Delta\lambda\) is the smallest wavelength difference that can be distinguished at wavelength \(\lambda\). The formula for a grating is: \[ R = mN \]
where \(m\) is the spectral order (an integer) and \(N\) is the total number of lines (or rulings) on the grating.

Step 2: Analyze the given statements based on the formula.
- A. increases with increase in total number of lines (N): From \(R=mN\), it is clear that if \(N\) increases, \(R\) increases. This statement is correct.
- B. increases with increase in total width of grating (W): The total width \(W\) is related to \(N\) and the grating element \(d\) by \(W = Nd\). The grating equation is \(d\sin\theta = m\lambda\). Substituting \(d = W/N\) gives \(\frac{W}{N}\sin\theta = m\lambda \implies mN = \frac{W\sin\theta}{\lambda}\). So, \(R = \frac{W\sin\theta}{\lambda}\). Thus, resolving power is directly proportional to the total width \(W\). This statement is correct.
- C. increases with increasing the order of spectrum (m): From \(R=mN\), if \(m\) increases, \(R\) increases. An Echelon grating is specifically designed to work at very high orders (\(m\)), giving it very high resolving power. This statement is correct.
- D. increases with decreasing the order of spectrum: This is the opposite of C and is incorrect.

Step 3: Conclude the correct statements.
Statements A, B, and C are correct descriptions of the factors affecting the resolving power of a grating. Quick Tip: To get high resolution with a grating, you want to use a grating with many total lines (\(N\)) and observe the spectrum in a high order (\(m\)). Both factors directly improve the ability to separate close spectral lines.

Step 1: Recall the formula for specific rotation.
Specific rotation \([\alpha]\) is defined as: \[ [\alpha] = \frac{\theta}{l \times c} \]
where \(\theta\) is the observed optical rotation in degrees, \(l\) is the path length in decimeters (dm), and \(c\) is the concentration of the solution in g/cc (or g/mL).

Step 2: Convert the given units to the required units.
- Observed rotation: \(\theta = 53^{\circ}30' = 53.5^{\circ}\).
- Path length: \(l = 20 \, cm = 2 \, dm\) (since 1 dm = 10 cm).
- Concentration: Mass = 20 g, Volume = 50 cc.
\[ c = \frac{mass}{volume} = \frac{20 \, g}{50 \, cc} = 0.4 \, g/cc \]

Step 3: Calculate the specific rotation. \[ [\alpha] = \frac{53.5}{2 \times 0.4} = \frac{53.5}{0.8} = 66.875 \]

Step 4: Match the result with the options.
The calculated value is approximately 66.9 degree (decimeter)\(^{-1}\) (g/cc)\(^{-1}\). Quick Tip: Pay close attention to units when calculating specific rotation. The path length must be in decimeters and the concentration in g/cc or g/mL. A common mistake is forgetting to convert from centimeters to decimeters.

Step 1: Analyze each condition for sustained interference.
- A. Coherence: The sources must be coherent, meaning they must maintain a constant phase relationship. This is the most fundamental condition for interference to be observable. If the phase relationship changes randomly, the interference pattern will wash out. So, A is true.
- B. Propagation along the same line: For interference to occur, the waves must overlap in space. While they don't have to be perfectly collinear, they must be propagating in nearly the same direction to produce a stable interference pattern. So, B is considered a necessary condition.
- C. Equal amplitude: For maximum contrast between bright and dark fringes (i.e., perfect destructive interference where intensity is zero), the amplitudes of the interfering waves should be equal. While interference can be observed with unequal amplitudes, the condition for ideal, sustained interference often includes equal amplitudes for best visibility. So, C is a condition for ideal interference.
- D. Same state of polarization: Two light waves with perpendicular polarization cannot interfere. For interference to occur, the electric field vectors of the waves must have components along the same direction. Therefore, they must be in the same state of polarization. So, D is true.

Step 2: Conclude the set of true conditions.
All four statements (A, B, C, and D) represent the ideal conditions required to observe a clear, stable, and high-contrast interference pattern. Quick Tip: The three "Cs" of interference are Coherence, Collinearity (or near-collinearity), and Comparable amplitudes. A fourth condition, related to polarization, is also crucial for light waves.

Step 1: Recall the formula for the magnetic field of a long, straight wire.
Ampere's Law gives the magnetic field \(B\) at a perpendicular distance \(r\) from a long, straight wire carrying a current \(I\): \[ B = \frac{\mu_0 I}{2\pi r} \]
where \(\mu_0\) is the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T\(\cdot\)m/A.

Step 2: Identify the given values and convert them to SI units.
- Current \(I = 10\) A.
- Distance \(r = 5 \, cm = 0.05 \, m\).

Step 3: Substitute the values into the formula and calculate \(B\). \[ B = \frac{(4\pi \times 10^{-7} \, T\cdot m/A) \times (10 \, A)}{2\pi \times (0.05 \, m)} \] \[ B = \frac{2 \times 10^{-7} \times 10}{0.05} \, T = \frac{2 \times 10^{-6}}{5 \times 10^{-2}} \, T \] \[ B = \frac{2}{5} \times 10^{-4} \, T = 0.4 \times 10^{-4} \, T = 4 \times 10^{-5} \, T \] Quick Tip: The expression \(\frac{\mu_0}{2\pi}\) can be simplified to \(2 \times 10^{-7}\) T\(\cdot\)m/A, which makes calculations for the magnetic field of a straight wire quicker.

Step 1: Recall the formula for the electric potential of a point charge.
The electric potential \(V\) at a distance \(r\) from a point charge \(Q\) is given by: \[ V = \frac{kQ}{r} \]
where \(k = \frac{1}{4\pi\epsilon_0}\) is Coulomb's constant.

Step 2: Calculate the distance \(r\) from the origin to the given point.
The charge is at the origin (0,0,0) and the point is at (3a,4a,0). The distance \(r\) is the magnitude of the position vector, calculated using the distance formula in 3D: \[ r = \sqrt{(3a-0)^2 + (4a-0)^2 + (0-0)^2} \] \[ r = \sqrt{(3a)^2 + (4a)^2} = \sqrt{9a^2 + 16a^2} = \sqrt{25a^2} = 5a \]

Step 3: Substitute the distance \(r\) into the potential formula. \[ V = \frac{kQ}{5a} \] Quick Tip: Electric potential is a scalar quantity, so you only need the magnitude of the distance between the charge and the point of interest. Remember the 3-4-5 right triangle, as it frequently appears in physics problems to simplify distance calculations.

Step 1: Apply Gauss's Law for a point outside the sphere.
For a spherically symmetric charge distribution, the electric field outside the distribution (at \(r > R\)) is the same as if all the charge were concentrated at a point charge at the center.

Step 2: State the formula for the electric field of a point charge.
The electric field \(E\) at a distance \(r\) from a point charge \(Q\) is given by Coulomb's Law: \[ E = \frac{kQ}{r^2} \]
where \(k = \frac{1}{4\pi\epsilon_0}\).

Step 3: Conclude the electric field for the sphere.
Since the point is outside the conducting sphere, we can treat the sphere's charge \(Q\) as a point charge at its center. Therefore, the electric field at a distance \(r > R\) is: \[ E = \frac{kQ}{r^2} \] Quick Tip: This is a key result from Gauss's Law. For a spherical shell or solid conducting sphere:
- \(\textbf{Outside (r > R):}\) \(E = kQ/r^2\) (acts like a point charge).
- \(\textbf{Inside (r < R):}\) \(E = 0\) (for a conductor or hollow shell).

Step 1: Recall the relationship between intensity and distance for an isotropic source.
For a source radiating uniformly in all directions, the intensity \(I\) decreases with the square of the distance \(r\) from the source. This is the inverse square law: \[ I \propto \frac{1}{r^2} \]
This means that the product \(I \cdot r^2\) is constant. Therefore, we can write \(I_1 r_1^2 = I_2 r_2^2\).

Step 2: Identify the given values.
- Initial intensity \(I_1 = 0.250\) W/m\(^2\).
- Initial distance \(r_1 = 15\) m.
- Final distance \(r_2 = 75\) m.
- We need to find the final intensity \(I_2\).

Step 3: Solve for \(I_2\). \[ I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2 \] \[ I_2 = 0.250 \, W/m^2 \times \left( \frac{15 \, m}{75 \, m} \right)^2 \] \[ I_2 = 0.250 \times \left( \frac{1}{5} \right)^2 = 0.250 \times \frac{1}{25} \] \[ I_2 = \frac{0.250}{25} = 0.010 \, W/m^2 \] Quick Tip: The inverse square law is fundamental for any quantity that spreads out uniformly from a point source, including sound intensity, light intensity, and gravitational and electrostatic fields. If the distance increases by a factor of \(n\), the intensity decreases by a factor of \(n^2\).

Step 1: Apply Gauss's Law for a point inside the shell.
Consider a spherical Gaussian surface of radius \(r\) (where \(r\) is less than the radius of the shell) concentric with the charged shell.
Gauss's Law states that the net electric flux through this surface is equal to the enclosed charge divided by \(\epsilon_0\): \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} \]

Step 2: Determine the enclosed charge.
For a spherical shell, all the charge resides on the surface of the shell. The Gaussian surface at radius \(r\) inside the shell encloses no charge. \[ Q_{enclosed} = 0 \]

Step 3: Solve for the electric field.
Since \(Q_{enclosed} = 0\), the net flux must be zero. \[ \oint \vec{E} \cdot d\vec{A} = 0 \]
Due to the spherical symmetry, if the electric field \(\vec{E}\) were non-zero, it would have a constant magnitude on the Gaussian surface. The integral would be \(E \times (4\pi r^2)\). For this to be zero, the electric field \(E\) must be zero. \[ E = 0 \]
Thus, the electric field is zero everywhere inside a uniformly charged spherical shell. Quick Tip: This is a classic result of Gauss's Law. It also applies to the inside of any hollow conductor in electrostatic equilibrium, a principle used in Faraday cages for electrostatic shielding.

Step 1: Relate the Poynting vector to momentum.
The average Poynting vector \(S_{av}\) represents the average energy flux (energy per unit area per unit time) of an electromagnetic wave. The momentum flux (momentum per unit area per unit time) is given by \(\frac{S_{av}}{c}\). This momentum flux is equal to the radiation pressure exerted by the wave when it is completely absorbed.

Step 2: Consider the case of total reflection.
When the wave is totally reflected, its momentum is reversed.
- The initial momentum arriving per unit area per unit time is \(\frac{S_{av}}{c}\).
- The final momentum leaving per unit area per unit time is \(-\frac{S_{av}}{c}\).
The change in momentum per unit area per unit time is the difference between the final and initial momentum fluxes: \[ \Delta (momentum flux) = \left(-\frac{S_{av}}{c}\right) - \left(\frac{S_{av}}{c}\right) = -\frac{2S_{av}}{c} \]

Step 3: Determine the pressure on the surface.
The radiation pressure on the surface is the force per unit area, which is equal to the magnitude of the change in momentum per unit area per unit time transferred to the surface. \[ P_{rad} = \left| \Delta (momentum flux) \right| = \frac{2S_{av}}{c} \] Quick Tip: Remember the two cases for radiation pressure: - **Perfect Absorption:** \(P_{rad} = S_{av}/c\) - **Perfect Reflection:** \(P_{rad} = 2S_{av}/c\) This is analogous to the impulse imparted by a particle: a particle that bounces back imparts twice the impulse of a particle that sticks.

Step 1: Relate electric potential and electric field.
The electric field \(\vec{E}\) is related to the electric potential \(V\) by \(\vec{E} = -\vec{\nabla}V\). In one dimension, this is \(E = -dV/dr\).

Step 2: Apply the given condition.
We are given that the potential \(V\) is constant inside the sphere. If \(V\) is constant, its derivative with respect to position must be zero. \[ \frac{dV}{dr} = 0 \implies E = 0 \]
So, the electric field inside the conductor is zero.

Step 3: Relate electric field to charge distribution using Gauss's Law.
Gauss's Law in differential form (Poisson's equation) is \(\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0\), where \(\rho\) is the volume charge density.
Since \(\vec{E} = 0\) everywhere inside the sphere, its divergence \(\vec{\nabla} \cdot \vec{E}\) must also be zero. \[ \vec{\nabla} \cdot (0) = 0 \implies \frac{\rho}{\epsilon_0} = 0 \implies \rho = 0 \]
This means there is no net charge at any point inside the volume of the conductor. Any net charge on a conductor must reside on its surface. Quick Tip: For a conductor in electrostatic equilibrium: 1. The electric field inside is zero. 2. The electric potential inside is constant and equal to the potential on the surface. 3. Any net charge resides entirely on the surface.

Step 1: Recall the properties of a conductor in electrostatic equilibrium.
One of the key properties is that the electric potential is constant throughout the entire conductor, including its surface. This means the surface of a conductor is an equipotential surface.

Step 2: Relate electric field lines to equipotential surfaces.
Electric field lines are always perpendicular to equipotential surfaces. If there were a component of the electric field parallel (tangential) to the surface, it would exert a force on the charges on the surface and cause them to move.

Step 3: Apply the condition of electrostatic equilibrium.
"Electrostatic equilibrium" means that the charges are no longer moving. For the charges to be stationary, there must be no net force on them parallel to the surface. This implies that the tangential component of the electric field must be zero.

Step 4: Conclude the relationship.
If the tangential component of the electric field is zero, the electric field vector at the surface must be entirely perpendicular to the surface. Note that the field is not zero on the surface unless there is no charge on the conductor at all. Quick Tip: Remember that electric field lines point from higher potential to lower potential and are always perpendicular to equipotential lines/surfaces. Since a conductor's surface is an equipotential, the E-field lines must emerge from it at a right angle.

A fundamental property of a conductor in electrostatic equilibrium is that the net electric field inside the conductor must be zero. If there were an electric field inside, it would exert a force on the free charges (electrons) within the conductor, causing them to move. This movement of charge would mean the conductor is not in equilibrium. Therefore, the charges rearrange themselves on the surface of the conductor in such a way that they cancel out any external electric field within the bulk of the conductor. Regardless of the strength of the electric field just outside the conductor, the field just inside is always zero. Quick Tip: This principle is the basis for electrostatic shielding. A conducting box, known as a Faraday cage, allows the electric field inside to remain zero even when the box is placed in a strong external electric field.

Step 1: Recall the formula for the capacitance of a parallel-plate capacitor.
The capacitance \(C_0\) of a parallel-plate capacitor with plate area \(A\) and separation \(d\), filled with vacuum, is: \[ C_0 = \frac{\epsilon_0 A}{d} \]
where \(\epsilon_0\) is the permittivity of free space.

Step 2: Recall the effect of inserting a dielectric material.
When a dielectric material with dielectric constant \(K\) is inserted to completely fill the space between the plates, the permittivity of the space becomes \(\epsilon = K \epsilon_0\).

Step 3: Write the formula for the new capacitance.
The new capacitance \(C\) with the dielectric slab is: \[ C = \frac{K \epsilon_0 A}{d} \]

Step 4: Compare the new capacitance with the original capacitance. \[ C = K \left( \frac{\epsilon_0 A}{d} \right) = K C_0 \]
Thus, the capacitance increases by a factor of K. Since the dielectric constant \(K\) is always greater than 1 for materials, the capacitance always increases. Quick Tip: A dielectric material reduces the electric field between the capacitor plates for a given charge. This allows more charge to be stored at the same potential difference, thereby increasing the capacitance (\(C=Q/V\)).

The gravitational field \(\vec{g}\) at a point in space is defined as the gravitational force \(\vec{F}_g\) experienced by a small test mass \(m\) placed at that point, divided by the magnitude of the test mass. \[ \vec{g} = \frac{\vec{F}_g}{m} \]
Therefore, the gravitational field is the force per unit mass. Its units are Newtons per kilogram (N/kg), which is equivalent to meters per second squared (m/s\(^2\)), the unit of acceleration. This is why it is also called the acceleration due to gravity. Quick Tip: This definition is directly analogous to the definition of the electric field, which is the electric force per unit charge (\(\vec{E} = \vec{F}_e/q\)).

Step 1: Recall Newton's second law for a system of particles.
The net external force \(\vec{F}_{net, ext}\) acting on a system is equal to the total mass \(M\) of the system times the acceleration of its center of mass \(\vec{a}_{CM}\). \[ \vec{F}_{net, ext} = M \vec{a}_{CM} \]

Step 2: Apply the given condition.
We are given that the net external force is zero, \(\vec{F}_{net, ext} = 0\). \[ 0 = M \vec{a}_{CM} \]

Step 3: Solve for the acceleration of the center of mass.
Since the total mass \(M\) is not zero, the acceleration of the center of mass must be zero. \[ \vec{a}_{CM} = 0 \]

Step 4: Interpret the result.
Zero acceleration means that the velocity of the center of mass, \(\vec{v}_{CM}\), does not change. This means the center of mass moves at a constant velocity. Being "at rest" is a special case of moving at a constant velocity where the constant velocity is zero. Option (2) is the more general and correct statement. Quick Tip: This is a statement of the conservation of momentum for a system. If the net external force is zero, the total momentum of the system (\(M\vec{v}_{CM}\)) is conserved.

Step 1: Analyze each statement.
- A. Specific heat of saturated water vapour at 100\(^{\circ}\)C is negative: Saturated vapor is vapor in equilibrium with its liquid phase. If you add heat to saturated steam, it will superheat, and to keep it saturated, you must compress it. This compression work increases the temperature more than the added heat, so to maintain a constant temperature, you would actually have to remove heat. Alternatively, if you expand saturated steam adiabatically (no heat exchange), it cools and condenses. To prevent condensation and keep it saturated during expansion, you must add heat. The specific heat is defined per unit temperature change, and under certain processes for saturated steam, it can indeed be negative. So, A is correct.
- B. There is only one triple point of a substance: The triple point is the unique combination of temperature and pressure at which the solid, liquid, and gas phases of a substance can all coexist in thermodynamic equilibrium. For a pure substance, this is a unique, fixed point. So, B is correct.
- C. Boiling point of every liquid rises with increase in pressure: The boiling point is the temperature at which the vapor pressure of a liquid equals the surrounding pressure. For all liquids, vapor pressure increases with temperature. Therefore, to make the liquid boil at a higher temperature, you must increase the external pressure. So, C is correct.
- D. Latent heat can not become zero: The latent heat of vaporization is the energy required to change a substance from liquid to gas at constant temperature and pressure. As you increase the temperature and pressure along the boiling curve, you approach the critical point. At the critical point, the distinction between liquid and gas phases disappears, and the latent heat of vaporization becomes zero. So, D is incorrect.

Step 2: Conclude the set of correct statements.
Statements A, B, and C are correct. Quick Tip: Phase diagrams are essential for understanding these concepts. The triple point is a single point, the boiling point is a line (the liquid-vapor coexistence curve), and this line terminates at the critical point, where the latent heat of vaporization vanishes.

Step 1: Recall Fourier's Law of Heat Conduction.
In the steady state, the rate of heat flow (\(H\) or \(\frac{dQ}{dt}\)) through a body is given by: \[ H = -kA \frac{dT}{dx} \]
For a slab of material with cross-sectional area A, thickness L, and a temperature difference \(\Delta T\) across its faces, the law is often written as: \[ H = \frac{kA\Delta T}{L} \]
where \(k\) is the thermal conductivity.

Step 2: Analyze the factors in the equation.
The flow of heat \(H\) depends on:
- k (thermal conductivity): Yes. This is statement B.
- \(\Delta T\) (temperature difference): Yes. This is statement C.
- The geometry (A and L).
Thermal resistivity (\(\rho_T\)) is the reciprocal of thermal conductivity (\(\rho_T = 1/k\)). Since the heat flow depends on \(k\), it also depends on \(\rho_T\). So, statement D is also correct.
Thermal capacity (\(C = mc\), where c is specific heat) relates to the amount of heat a body can store to change its temperature. It does not determine the rate of heat flow in the steady state, where by definition, the temperature at any point in the body is constant. So, statement A is incorrect.

Step 3: Conclude the correct factors.
The steady-state flow of heat depends on thermal conductivity (B), temperature difference (C), and thermal resistivity (D). Quick Tip: Think of the analogy with electrical circuits (Ohm's Law, \(I = V/R\)): - Heat Flow \(H \leftrightarrow\) Current \(I\) - Temperature Difference \(\Delta T \leftrightarrow\) Voltage \(V\) - Thermal Resistance \(R_T = L/(kA) \leftrightarrow\) Electrical Resistance \(R\) Thermal capacity is analogous to electrical capacitance, which is irrelevant for steady DC current.

A quantum gas is said to be "degenerate" when quantum effects (due to the Pauli exclusion principle for fermions or Bose-Einstein condensation for bosons) become important. This happens when the thermal de Broglie wavelength of the particles becomes comparable to or larger than the average inter-particle separation.

The thermal de Broglie wavelength is \(\lambda_{th} = \frac{h}{\sqrt{2\pi m k_B T}}\). It is large when the temperature \(T\) is low.
The average inter-particle separation is related to the particle density \(n = N/V\). It is small when the particle density \(n\) is large.

Therefore, the system becomes degenerate (the degree of degeneracy is large) when the temperature is low and the particle density is high. This forces particles to occupy higher energy states (for fermions) or crowd into the ground state (for bosons), which are distinctly quantum behaviors. Quick Tip: Degeneracy in this context means "quantum-ness." Quantum effects dominate when particles are cold (slow-moving, large de Broglie wavelength) and crowded (small inter-particle spacing).

Step 1: Understand the terms.
- \(n_i\) is the number of particles in energy level \(i\).
- \(g_i\) is the degeneracy of energy level \(i\), which is the number of available states at that energy level.
- The ratio \(\frac{n_i}{g_i}\) is the "occupation index," representing the average number of particles per available quantum state.

Step 2: Characterize quantum vs. classical statistics.
- **Quantum Statistics (Fermi-Dirac and Bose-Einstein):** These are necessary when the occupation index is not small. The distinguishability of particles (or lack thereof) and limits on occupation (Pauli exclusion principle for fermions) are critical. This corresponds to a high density of particles in the available quantum states.
- **Classical Statistics (Maxwell-Boltzmann):** This is the limit where particles are treated as distinguishable and there are no restrictions on how many can occupy a state. This limit is valid when the occupation index is very small, meaning the particles are sparsely distributed among a large number of available states.

Step 3: Formulate the condition for the classical limit.
The classical limit is reached when the occupation index is much less than 1: \[ \frac{n_i}{g_i} \ll 1 \]
This is equivalent to saying that the number of available states \(g_i\) is much larger than the number of particles \(n_i\) that need to occupy them. \[ g_i \gg n_i \implies \frac{g_i}{n_i} \gg 1 \]
This condition corresponds to a low-density, high-temperature gas where quantum effects are negligible. Quick Tip: The classical limit is the "non-crowded" limit. If there are many more available quantum "seats" (\(g_i\)) than there are particles (\(n_i\)) to sit in them, the particles are unlikely to interact in a way that requires quantum rules, so classical statistics work fine.

In quantum mechanics, particles are classified into two fundamental types based on their intrinsic angular momentum, or spin.
- Fermions: These are particles that have half-integer spin values (\(\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \dots\)) in units of \(\hbar\). They obey the Pauli exclusion principle, which states that no two identical fermions can occupy the same quantum state simultaneously. Examples include electrons, protons, and neutrons (all have spin 1/2).
- Bosons: These are particles that have integer spin values (\(0, 1, 2, \dots\)) in units of \(\hbar\). They do not obey the Pauli exclusion principle. Examples include photons (spin 1) and Higgs bosons (spin 0).

The option provided that is a half-integer spin is \(1/2\). Quick Tip: A simple mnemonic: **F**ermions have **F**ractional spin (half-integer), while **B**osons have... not fractional spin (integer).

The Law of Dulong and Petit is a classical thermodynamic law that describes the molar specific heat capacity of a solid element. It is based on the classical equipartition theorem. The law states that, at sufficiently high temperatures, the molar specific heat at constant volume (\(C_V\)) for all solid elements is approximately constant and equal to \(3R\), where \(R\) is the universal gas constant. \[ C_V \approx 3R \approx 3 \times 8.314 \, J/(mol\cdot K) \approx 25 \, J/(mol\cdot K) \]
"Atomic heat" is another term for molar specific heat. The law predicts this value to be constant, independent of the substance and temperature (at high temperatures). While modern quantum theories (like the Debye model) show that the specific heat does decrease and approach zero at absolute zero, the classical law of Dulong and Petit itself predicts a constant value. The question asks what the law states, not what is observed experimentally at low temperatures. Quick Tip: Dulong-Petit is a classical high-temperature limit. It fails at low temperatures where quantum effects become important. Einstein's and Debye's models were developed to explain this low-temperature deviation.

Step 1: Define compressibility.
Compressibility (\(\kappa\)) is a measure of the relative volume change of a fluid or solid as a response to a pressure change. It is defined as the reciprocal of the bulk modulus (\(B\)). \[ \kappa = \frac{1}{B} \]

Step 2: Determine the S.I. unit of bulk modulus.
Bulk modulus is defined as the ratio of pressure stress to volumetric strain: \[ B = -\frac{\Delta P}{\Delta V / V_0} \]
Since volumetric strain (\(\Delta V / V_0\)) is dimensionless, the unit of bulk modulus is the same as the unit of pressure. The S.I. unit of pressure is the Pascal (Pa), which is equivalent to Newtons per square meter (N/m\(^2\)).

Step 3: Determine the S.I. unit of compressibility.
Since compressibility is the reciprocal of bulk modulus, its unit is the reciprocal of the unit of pressure. \[ Unit of \kappa = \frac{1}{Unit of B} = \frac{1}{N/m^2} = \frac{m^2}{N} \] Quick Tip: Remembering that compressibility is the inverse of stiffness (bulk modulus) is key. Stiff materials have a high bulk modulus and low compressibility, while easily compressed materials have a low bulk modulus and high compressibility.

The specific heat of saturated water vapor is a non-intuitive concept. Saturated vapor is vapor that is in equilibrium with its liquid phase (i.e., it's on the point of condensing).
Consider a process where we add heat to saturated water vapor. This added heat will tend to increase its temperature (superheat it). However, to keep the vapor in a saturated state at this new, higher temperature, the pressure must also increase significantly along the vaporization curve. To achieve this higher pressure, the vapor must be compressed.
The work done on the vapor during this compression can increase its internal energy (and thus temperature) by more than the amount of heat added. In order to only reach the new saturation temperature without overshooting it, heat must actually be removed from the system.
Since specific heat is defined as the heat added per unit mass per unit temperature change (\(c = \frac{dQ}{m dT}\)), and in this process \(dQ\) can be negative while \(dT\) is positive, the specific heat of saturated water vapor is negative. Quick Tip: This is a famous paradox in thermodynamics. Adding heat to saturated steam can make it condense if the pressure isn't increased enough, and expanding it adiabatically causes condensation. This behavior leads to the negative specific heat capacity along the saturation curve.

Step 1: Recall the two forms of the ideal gas law.
- The molar form is \(PV = nRT\), where \(n\) is the number of moles and \(R\) is the universal gas constant.
- The molecular form is \(PV = Nk_BT\), where \(N\) is the number of molecules and \(k_B\) is the Boltzmann constant.

Step 2: Relate the number of moles (\(n\)) to the number of molecules (\(N\)).
The number of molecules is equal to the number of moles multiplied by Avogadro's number (\(N_A\)), which is the number of molecules per mole. \[ N = n N_A \]

Step 3: Equate the two forms of the ideal gas law and solve for R. \[ nRT = Nk_BT \]
Substitute \(N = n N_A\): \[ nRT = (n N_A) k_B T \]
Cancel \(n\) and \(T\) from both sides: \[ R = N_A k_B \]
Thus, the universal gas constant \(R\) is the product of Avogadro's number and the Boltzmann constant. Quick Tip: Think of the constants this way: The universal gas constant \(R\) is for macroscopic amounts (per mole), while the Boltzmann constant \(k_B\) is for microscopic amounts (per molecule). Avogadro's number is the conversion factor between them.

This question requires knowledge of the approximate refractive indices of common materials. We can match them by considering their typical values.
- D. Diamond: Diamond is famous for its very high refractive index, which gives it its brilliance. The highest value in the list is 2.417. So, D matches IV.
- A. Ice: Ice is frozen water, and its refractive index is close to that of water (which is about 1.33). The lowest value in the list is 1.309, which is a very reasonable value for ice. So, A matches I.
- B. Rock salt (NaCl): This is a crystalline solid. Its refractive index is known to be in the mid-range, typically around 1.54. So, B matches III.
- C. CCl4 (Carbon Tetrachloride): This is a dense organic liquid. Its refractive index is typically in the mid-range, lower than that of rock salt. The value 1.460 is correct for CCl4. So, C matches II.

The correct set of matches is A - I, B - III, C - II, D - IV. Quick Tip: It's useful to have a mental scale of refractive indices: - Air: \(\approx\) 1.00 - Water/Ice: \(\approx\) 1.3 - Common Glass/Plastics/Liquids: \(\approx\) 1.4 - 1.6 - Diamond/High-index materials: > 2.0

Step 1: Understand the concept of mean free path.
The mean free path (\(\lambda\)) is the average distance a particle (like a molecule) travels between successive collisions with other particles.

Step 2: Consider a simplified model.
If we imagine one molecule moving through a gas of stationary molecules, it will collide with any molecule whose center is within a cylinder of radius \(d\) (or cross-sectional area \(\sigma = \pi d^2\)). The volume of this collision cylinder is \(Area \times length = \sigma v t\). The number of collisions is the number of particles in this volume, which is \(n(\sigma v t)\). The rate of collision is \(n\sigma v\). The mean free path would be the average distance traveled per collision, \(\lambda = \frac{v}{n\sigma v} = \frac{1}{n\sigma} = \frac{1}{\pi n d^2}\).

Step 3: Apply the correct model accounting for the motion of all particles.
The simplified model is incorrect because it assumes the target molecules are stationary. When the relative motion of all molecules is properly accounted for using the Maxwell-Boltzmann distribution of speeds, an additional factor of \(\sqrt{2}\) appears in the denominator. The correct formula for the mean free path is: \[ \lambda = \frac{1}{\sqrt{2} n \sigma} = \frac{1}{\sqrt{2}\pi n d^2} \]
where \(n\) is the number density of molecules and \(d\) is the molecular diameter. Quick Tip: The mean free path is inversely proportional to the density of the gas (\(n\)) and the collision cross-section (\(\pi d^2\)). A denser gas or larger molecules will lead to a shorter mean free path. The \(\sqrt{2}\) factor is a crucial correction that arises from considering the relative speeds of the colliding molecules.