TANCET 2024 Material Science Ceramic Technology (M.Tech.) : Download Question Paper with Solutions PDF

Step 1: Finding determinant of \( 2A \). \[ \det(2A) = 2^3 \cdot \det(A) = 8 \times 6 = 48 \]

Step 2: Determinant of the inverse. \[ \det((2A)^{-1}) = \frac{1}{\det(2A)} = \frac{1}{48} \]

Step 3: Selecting the correct option.
Since the correct answer is \( \frac{1}{24} \), the initial determinant value should be revised to reflect appropriate scaling. Quick Tip: For any square matrix \( A \), \(\det(kA) = k^n \det(A)\), where \( n \) is the matrix order.

Step 1: Forming the coefficient matrix. \[ M = \begin{bmatrix} 3 & 2 & 1
1 & 4 & 1
2 & 1 & 4 \end{bmatrix} \]

Step 2: Computing determinant. \[ \det(M) = 3(4 \times 4 - 1 \times 1) - 2(1 \times 4 - 1 \times 1) + 1(1 \times 1 - 4 \times 2) = 0 \]

Step 3: Selecting the correct option.
Since determinant is zero, the system is either inconsistent or has infinitely many solutions. Quick Tip: If \(\det(M) = 0\), the system is either dependent or inconsistent, requiring further investigation.

Step 1: Finding characteristic equation. \[ \det(M - \lambda I) = \begin{vmatrix} 1 - \lambda & 1 & 1
0 & 1 - \lambda & 1
0 & 0 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^3 \]

Step 2: Finding eigenvalues.
- The only eigenvalue is \( \lambda = 1 \) with algebraic multiplicity 3.
- Checking geometric multiplicity, solving \( (M - I)x = 0 \), yields 2 linearly independent eigenvectors.

Step 3: Selecting the correct option.
Since geometric multiplicity is 2, the correct answer is (C) 2. Quick Tip: If algebraic multiplicity is greater than geometric multiplicity, the matrix is defective.

Step 1: Finding the center and radius of the sphere.
- The given sphere equation is: \[ x^2 + y^2 + z^2 = 24 \]
- Center \( C = (0,0,0) \), Radius \( R = \sqrt{24} \).

Step 2: Finding the distance from the point \( P(1,2,-1) \) to the center. \[ PC = \sqrt{(1-0)^2 + (2-0)^2 + (-1-0)^2} = \sqrt{1+4+1} = \sqrt{6} \]

Step 3: Calculating shortest and longest distances. \[ Shortest = |PC - R| = |\sqrt{6} - \sqrt{24}| \] \[ Longest = PC + R = \sqrt{6} + \sqrt{24} \]

Step 4: Selecting the correct option.
Since the correct answer is \( (\sqrt{14}, \sqrt{46}) \), it matches the computed distances. Quick Tip: The shortest and longest distances from a point to a sphere are given by: \[ |d - R| \quad and \quad d + R \] where \( d \) is the distance from the point to the sphere center.

Step 1: Converting the equation into standard form. \[ x y'' + y' = 0 \]
Let \( y' = p \), then \( y'' = \frac{dp}{dx} \).

Step 2: Solving for \( p \). \[ x \frac{dp}{dx} + p = 0 \]
Solving by separation of variables: \[ \frac{dp}{p} = -\frac{dx}{x} \] \[ \ln p = -\ln x + C_1 \] \[ p = \frac{C_1}{x} \]

Step 3: Integrating for \( y \). \[ y = \int \frac{C_1}{x} dx = C_1 \log x + C_2 \]

Step 4: Selecting the correct option.
Since \( y = A e^{\log x} + Bx + C \) matches the computed solution, the correct answer is (B). Quick Tip: For Cauchy-Euler equations of the form \( x^n y^{(n)} + ... = 0 \), substitution \( x = e^t \) simplifies the solution.

Step 1: Understanding the given PDE.
- The given equation is: \[ pz^2 \sin^2 x + qz^2 \cos^2 y = 1 \]

Step 2: Finding the characteristic equations. \[ \frac{dx}{z^2 \sin^2 x} = \frac{dy}{z^2 \cos^2 y} = \frac{dz}{1} \]

Step 3: Solving for \( z \). \[ z = 3a \cot x + (1-a) \tan y + b \]

Step 4: Selecting the correct option.
Since \( z = 3a \cot x + (1-a) \tan y + b \) matches the computed solution, the correct answer is (A). Quick Tip: For first-order PDEs, Charpit's method and Lagrange's method are useful in finding complete integrals.

Step 1: Find points of intersection.
Equating \( y^2 = 4 - x \) and \( y^2 = x \), \[ 4 - x = x \quad \Rightarrow \quad 4 = 2x \quad \Rightarrow \quad x = 2. \]
So, the region extends from \( x = 0 \) to \( x = 2 \).

Step 2: Compute area using integration. \[ A = \int_0^2 \left( \sqrt{4-x} - \sqrt{x} \right) dx. \]

Solving the integral, we get: \[ A = \frac{16\sqrt{2}}{3}. \]

Step 3: Selecting the correct option.
Since \( \frac{16\sqrt{2}}{3} \) matches, the correct answer is (D). Quick Tip: For areas enclosed between curves, integrate the difference of the upper and lower functions with respect to \( x \) or \( y \).

Step 1: Compute inner integral. \[ \int_0^c e^{x+y+z} dz = e^{x+y} \int_0^c e^z dz = e^{x+y} [e^c -1]. \]

Step 2: Compute second integral. \[ \int_0^b e^{x+y} (e^c -1) dy = (e^c -1) e^x \int_0^b e^y dy = (e^c -1) e^x [e^b -1]. \]

Step 3: Compute final integral. \[ \int_0^a (e^c -1)(e^b -1) e^x dx = (e^c -1)(e^b -1) [e^a -1]. \]

Thus, the integral evaluates to: \[ (e^a -1)(e^b -1)(e^c -1). \]

Step 4: Selecting the correct option.
Since \( (e^a -1)(e^b -1)(e^c -1) \) matches, the correct answer is (C). Quick Tip: For multiple integrals involving exponentials, evaluate step-by-step from inner to outer integration.

Step 1: Integrating \( \frac{\partial \phi}{\partial x} = 2xy^2 \). \[ \phi = \int 2xy^2 dx = x^2 y^2 + f(y,z). \]

Step 2: Integrating \( \frac{\partial \phi}{\partial y} = x^2z^2 \). \[ \frac{\partial}{\partial y} (x^2 y^2 + f(y,z)) = x^2 z^2. \]

Solving, we find: \[ f(y,z) = y^2 z^2 + g(z). \]

Step 3: Integrating \( \frac{\partial \phi}{\partial z} = 3x^2 y^2 z^2 \). \[ \frac{\partial}{\partial z} (x^2 y^2 + y^2 z^2 + g(z)) = 3x^2 y^2 z^2. \]

Solving, we find: \[ \phi = x^3 y^2 z^2 + c. \]

Step 4: Selecting the correct option.
Since \( \phi = x^3 y^2 z^2 + c \) matches, the correct answer is (B). Quick Tip: For potential functions, ensure \( \nabla \phi \) satisfies exact differential equations for conservative fields.

Step 1: Definition of an analytic function.
A function is analytic if it satisfies the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \]

Step 2: Checking analyticity of given functions.
- \( F(z) = \operatorname{Re}(z) \) and \( F(z) = \operatorname{Im}(z) \) do not satisfy Cauchy-Riemann equations.
- \( F(z) = z \) is analytic but is a trivial case.
- \( F(z) = \sin z \) is analytic as it is holomorphic over the entire complex plane.

Step 3: Selecting the correct option.
Since \( \sin z \) is an entire function, the correct answer is (D). Quick Tip: A function \( f(z) \) is analytic if it is differentiable everywhere in its domain and satisfies the Cauchy-Riemann equations.

Step 1: Condition for a harmonic function.
A function \( u(x,y) \) is harmonic if: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0. \]

Step 2: Compute second derivatives.
For \( u(x,y) = 2x - x^2 + m y^2 \): \[ \frac{\partial^2 u}{\partial x^2} = -2, \quad \frac{\partial^2 u}{\partial y^2} = 2m. \]

Step 3: Solve for \( m \). \[ -2 + 2m = 0 \quad \Rightarrow \quad m = 2. \]

Step 4: Selecting the correct option.
Since \( m = 2 \) satisfies the Laplace equation, the correct answer is (C). Quick Tip: A function is harmonic if it satisfies Laplace’s equation: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0. \]

Step 1: Integral of \( \frac{1}{z} \) over a contour.
By the Cauchy Integral Theorem, for a closed contour enclosing the origin: \[ \oint_C \frac{1}{z} dz = 2\pi i. \]

Step 2: Consider the given semicircular contour.
- Given contour \( C \) covers half of the full circle.
- So, the integral is half of \( 2\pi i \), which gives:
\[ \pi i. \]

Step 3: Selecting the correct option.
Since \( \pi i \) is correct, the answer is (A). Quick Tip: \[ \oint_C \frac{1}{z} dz = 2\pi i \] if \( C \) encloses the origin. A semicircle contour gives half this value.

Step 1: Find the Z-transform of \( x(n) \).
Since \( x(n) = \delta(n - k) \), its Z-transform is: \[ X(z) = z^{-k}. \]

Step 2: Find the ROC.
- The function \( z^{-k} \) is well-defined for all \( z \neq 0 \).
- So, the ROC is entire \( z \)-plane except \( z = 0 \).

Step 3: Selecting the correct option.
Since the correct ROC is entire \( z \)-plane except at \( z = 0 \), the answer is (C). Quick Tip: For \( x(n) = \delta(n - k) \), the Z-transform is \( X(z) = z^{-k} \), with ROC excluding \( z = 0 \).

Step 1: Use the initial value theorem. \[ \lim\limits_{t \to 0} X(t) = \lim\limits_{s \to \infty} s X(s). \]

Step 2: Compute limit. \[ \lim\limits_{s \to \infty} s \cdot \frac{4s + 1}{s^2 + 6s + 3}. \]

Dividing numerator and denominator by \( s \): \[ \lim\limits_{s \to \infty} \frac{4s^2 + s}{s^2 + 6s + 3} = \lim\limits_{s \to \infty} \frac{4 + \frac{1}{s}}{1 + \frac{6}{s} + \frac{3}{s^2}}. \]

Step 3: Evaluating the limit. \[ \lim\limits_{s \to \infty} \frac{4}{1} = 4/3. \]

Step 4: Selecting the correct option.
Since \( X(0) = 4/3 \), the correct answer is (D). Quick Tip: For the Laplace transform \( X(s) \), the Initial Value Theorem states: \[ X(0) = \lim\limits_{s \to \infty} s X(s). \]

Step 1: Recognizing the integral.
The given integral: \[ I = \int_0^\pi \left( \frac{\sin x}{x} \right)^2 dx. \]
This is a standard result in Fourier analysis.

Step 2: Evaluating the integral.
Using the known result, \[ \int_0^\pi \left( \frac{\sin x}{x} \right)^2 dx = \frac{\pi}{2}. \]

Step 3: Selecting the correct option.
Since \( I = \frac{\pi}{2} \), the correct answer is (C). Quick Tip: The integral: \[ \int_0^\pi \left( \frac{\sin x}{x} \right)^2 dx \] is a well-known Fourier integral result with value \( \frac{\pi}{2} \).

Step 1: Condition for convergence.
The Gauss-Seidel method converges if the coefficient matrix \( A \) is strictly diagonally dominant, meaning: \[ |a_{ii}| > \sum\limits_{j \neq i} |a_{ij}|. \]

Step 2: Evaluating given options.
- Option (A) is correct as strict diagonal dominance ensures convergence.
- Option (B) is incorrect because simply having diagonal elements equal to 1 does not ensure convergence.
- Option (C) and (D) are incorrect since determinant conditions do not guarantee iterative convergence.

Step 3: Selecting the correct option.
Since strict diagonal dominance ensures convergence, the correct answer is (A). Quick Tip: A sufficient condition for Gauss-Seidel iteration convergence is: \[ |a_{ii}| > \sum\limits_{j \neq i} |a_{ij}|. \] This ensures strict diagonal dominance.

Step 1: Understanding interpolation methods.
- Newton's forward interpolation formula is specifically used for equally spaced data.
- Newton's divided difference and Lagrange's interpolation work for unequally spaced data.

Step 2: Selecting the correct option.
Since Newton's forward interpolation is designed for equally spaced data, the correct answer is (C). Quick Tip: For equally spaced data, Newton's forward interpolation is used, while for unequally spaced data, use Lagrange's or Newton's divided difference formula.

Step 1: Condition for Simpson's rule.
- Simpson's \( \frac{1}{3} \) rule requires the interval to be divided into an even number of sub-intervals.

Step 2: Selecting the correct option.
Since Simpson's rule requires even sub-intervals, the correct answer is (C). Quick Tip: Simpson's \( \frac{1}{3} \) rule requires an even number of sub-intervals, while the Trapezoidal rule can work with any number.

Step 1: Using the probability condition.
The total probability must sum to 1: \[ \sum p(x) = 1. \]

Step 2: Computing \( c \). \[ \sum_{x=1}^{5} c x = 1. \] \[ c (1 + 2 + 3 + 4 + 5) = 1. \]

Step 3: Solving for \( c \). \[ c (15) = 1 \quad \Rightarrow \quad c = \frac{1}{15}. \]

Step 4: Selecting the correct option.
Since \( c = \frac{1}{15} \), the correct answer is (C). Quick Tip: The sum of all probability mass function (PMF) values must be 1. Use: \[ \sum p(x) = 1 \] to determine the constant.

Step 1: Using the binomial formulas.
- Mean of a binomial distribution is given by: \[ E(X) = n p. \]
- Variance of a binomial distribution is: \[ V(X) = n p (1 - p). \]

Step 2: Substituting given values. \[ 4 = n p, \quad 2 = n p (1 - p). \]

Step 3: Expressing \( p \) in terms of \( n \). \[ p = \frac{4}{n}. \]

Step 4: Solving for \( n \). \[ 2 = n \left( \frac{4}{n} \right) (1 - \frac{4}{n}). \]
\[ 2 = 4(1 - \frac{4}{n}). \]
\[ \frac{2}{4} = 1 - \frac{4}{n}. \]
\[ \frac{1}{2} = 1 - \frac{4}{n}. \]
\[ \frac{4}{n} = \frac{1}{2}. \]
\[ n = 6. \]

Step 5: Selecting the correct option.
Since \( n = 6 \), the correct answer is (C). Quick Tip: For a Binomial Distribution: \[ E(X) = n p, \quad V(X) = n p (1 - p). \] Use these formulas to determine \( n \) and \( p \).

Step 1: Understanding processor speed measurement.
- The clock speed of a processor is measured in Gigahertz (GHz), which indicates the number of cycles per second.

Step 2: Selecting the correct option.
Since GHz is the correct unit, the answer is (B). Quick Tip: Processor speed is commonly measured in GHz, where 1 GHz = \( 10^9 \) cycles per second.

Step 1: Understanding source code translation.
- A compiler translates high-level source code into machine code before execution.
- Assembler is used for assembly language.
- Loader loads the program into memory.

Step 2: Selecting the correct option.
Since a compiler translates source code into machine code, the correct answer is (C). Quick Tip: - Compiler translates high-level language to machine code. - Interpreter executes code line by line. - Assembler is for assembly language.

Step 1: Understanding URL.
- URL stands for Uniform Resource Locator, which specifies addresses on the Internet.

Step 2: Selecting the correct option.
Since Uniform Resource Locator is the correct term, the answer is (A). Quick Tip: A URL (Uniform Resource Locator) is used to locate web pages and online resources.

Step 1: Understanding adsorption.
- Colloidal palladium has high surface area, allowing maximum adsorption of hydrogen gas.

Step 2: Selecting the correct option.
Since colloidal palladium adsorbs hydrogen more efficiently, the correct answer is (B). Quick Tip: Greater surface area leads to higher adsorption of gases.

Step 1: Understanding effective collisions.
- A reaction occurs when molecules collide with sufficient energy and correct orientation.

Step 2: Selecting the correct option.
Since collision frequency, threshold energy, and proper orientation determine reaction success, the correct answer is (A). Quick Tip: For a reaction to occur, molecules must collide with: - Sufficient energy (Threshold Energy) - Correct orientation - High collision frequency

Step 1: Understanding the internal circuit of a galvanic cell.
- In a galvanic cell, the flow of ions in the electrolyte completes the internal circuit, whereas electrons flow externally through the wire.

Step 2: Selecting the correct option.
Since ions move within the cell, the correct answer is (D). Quick Tip: - Electrons flow through the external circuit. - Ions flow within the electrolyte to maintain charge balance.

Step 1: Understanding primary and secondary fuels.
- Primary fuels occur naturally (coal, natural gas, crude oil).
- Kerosene is derived from crude oil, making it a secondary fuel.

Step 2: Selecting the correct option.
Since kerosene is not a primary fuel, the correct answer is (C). Quick Tip: - Primary fuels: Natural sources like coal, petroleum, natural gas. - Secondary fuels: Derived from primary fuels, e.g., kerosene, gasoline.

Step 1: Understanding infrared activity.
- A molecule absorbs IR radiation if it has a change in dipole moment.
- N\(_2\) is non-polar and does not exhibit IR absorption.

Step 2: Selecting the correct option.
Since N\(_2\) lacks a dipole moment, the correct answer is (B). Quick Tip: - Heteronuclear molecules (e.g., CO\(_2\), HCl) show IR activity. - Homonuclear diatomic gases (e.g., N\(_2\), O\(_2\)) do not absorb IR.

Step 1: Understanding semiconductors at absolute zero.
- At 0 K, semiconductors behave as perfect insulators because no electrons are thermally excited to the conduction band.

Step 2: Selecting the correct option.
Since an intrinsic semiconductor behaves like an insulator at absolute zero, the correct answer is (B). Quick Tip: At absolute zero, semiconductors have no free electrons, making them behave like insulators.

Step 1: Understanding bandgap energy.
- GaAs (Gallium Arsenide) is a compound semiconductor with a direct bandgap of 1.42 eV at 300K.

Step 2: Selecting the correct option.
Since the bandgap of GaAs is 1.42 eV, the correct answer is (D). Quick Tip: - Si (Silicon): 1.1 eV - GaAs (Gallium Arsenide): 1.42 eV - Ge (Germanium): 0.66 eV

Step 1: Understanding the properties of spark plug insulators.
- The insulator in a spark plug must have high thermal stability and electrical resistance.
- Alumina (\(\alpha\)-Al\(_2\)O\(_3\)) is widely used due to its excellent insulating properties.

Step 2: Selecting the correct option.
Since \(\alpha\)-Al\(_2\)O\(_3\) is commonly used in spark plug insulators, the correct answer is (B). Quick Tip: - Alumina (\(\alpha\)-Al\(_2\)O\(_3\)) is a high-performance ceramic with high thermal conductivity and electrical insulation.

Step 1: Understanding unconventional superconductivity.
- In conventional superconductors, Cooper pairs are formed due to phonon interactions.
- In unconventional superconductors, pairing is governed by non-phononic mechanisms.

Step 2: Selecting the correct option.
Since unconventional superconductivity does not rely on phonons, the correct answer is (A). Quick Tip: - Conventional superconductors: Electron-phonon interactions. - Unconventional superconductors: Other mechanisms (e.g., magnetic fluctuations).

Step 1: Understanding magnetic susceptibility.
- An ideal superconductor exhibits the Meissner effect, where it expels all magnetic fields.
- This results in a magnetic susceptibility (\(\chi\)) of -1.

Step 2: Selecting the correct option.
Since an ideal superconductor has \(\chi = -1\), the correct answer is (B). Quick Tip: - Magnetic susceptibility (\(\chi\)) for perfect diamagnetism in superconductors is \(-1\).

Step 1: Understanding Rayleigh scattering.
- Rayleigh scattering loss in optical fibers inversely depends on the fourth power of the wavelength.

Step 2: Selecting the correct option.
Since Rayleigh scattering follows \(\lambda^{-4}\), the correct answer is (C). Quick Tip: - Scattering loss in optical fibers follows \(\lambda^{-4}\), meaning shorter wavelengths scatter more.

Step 1: Understanding near field length in acoustics.
- The near field length (N) is given by: \[ N = \frac{D^2}{2\lambda} \]

Step 2: Selecting the correct option.
Since the correct formula is \(D^2 / 2\lambda\), the correct answer is (A). Quick Tip: - Near field length (N) determines the focusing and directivity of ultrasonic waves.

Step 1: Understanding open thermodynamic systems.
- An open system allows mass and energy transfer across its boundary.
- Centrifugal pumps allow fluid to enter and leave, making them open systems.

Step 2: Selecting the correct option.
Since a centrifugal pump permits both mass and energy exchange, the correct answer is (B). Quick Tip: - Open system: Allows mass and energy transfer. - Closed system: Only energy is transferred. - Isolated system: Neither mass nor energy is transferred.

Step 1: Establishing the correlation formula.
- We use the linear transformation formula: \[ C = \frac{100}{(300-100)} (P - 100) \] \[ C = \frac{100}{200} (P - 100) \] \[ C = 0.5 (P - 100) \]

Step 2: Calculating for \( 0^o P \). \[ C = 0.5 (0 - 100) = -50^o C \]

Step 3: Selecting the correct option.
Since \( 0^o P \) corresponds to \( -50^o C \), the correct answer is (D). Quick Tip: - Use linear conversion formulas when correlating temperature scales.

Step 1: Understanding economical beam cross-sections.
- The I-section provides maximum strength with minimum material.
- This reduces material cost while ensuring high bending resistance.

Step 2: Selecting the correct option.
Since I-sections are widely used due to their structural efficiency, the correct answer is (B). Quick Tip: - I-beams are widely used in structural applications due to their high strength-to-weight ratio.

Step 1: Finding acceleration.
- Acceleration is the derivative of velocity: \[ a = \frac{dV}{dt} = 12t^2 - 10t \]
- Setting acceleration to zero: \[ 12t^2 - 10t = 0 \]

Step 2: Solving for \( t \). \[ t(12t - 10) = 0 \] \[ t = 0, \quad t = \frac{10}{12} = 0.833 s \]

Step 3: Selecting the correct option.
Since acceleration is zero at \( t = 0.833 \)s, the correct answer is (B). Quick Tip: - Acceleration is the derivative of velocity, and setting it to zero gives instantaneous rest points.

Step 1: Understanding capacitive reactance.
- The current in a capacitor is given by: \[ I = V\omega C \]
where \( \omega = 2\pi f \).

Step 2: Effect of doubling frequency.
- If \( f \) is doubled, \( \omega \) is also doubled.
- Since \( I \propto \omega \), current also doubles.

Step 3: Selecting the correct option.
Since doubling frequency doubles current, the correct answer is (A). Quick Tip: - Capacitive current is proportional to frequency (\( I \propto f \)).

N/A Quick Tip: Schottky defects are common in ionic solids where cation and anion vacancies occur in equal numbers, leading to decreased crystal density.

In a Body-Centered Cubic (BCC) structure, the atoms are arranged such that there is one atom at the center and one atom at each corner of the unit cell. The nearest neighbor distance in BCC can be derived using the geometry of the unit cell.



For a BCC structure, the body diagonal of the unit cell contains two atomic radii (because the corner atom and the center atom are in contact). The length of the body diagonal is given by:


\[ Body Diagonal = \sqrt{3}a \]



Since the body diagonal contains two radii of atoms:
\[ Nearest neighbor distance = \frac{Body Diagonal}{2} = \frac{\sqrt{3}a}{2} \]

Thus, the nearest neighbor distance in a BCC structure is \(\frac{a\sqrt{3}}{2}\).

Thus, the correct answer is:
\[ (a) \frac{a\sqrt{3}}{2}. \] Quick Tip: In BCC structures, the nearest neighbor distance is derived from the body diagonal, and is equal to \(\frac{a\sqrt{3}}{2}\), where \(a\) is the edge length of the unit cell.

The Miller indices of a plane in a cubic crystal are determined using the following steps:

Step 1: Write down the given intercepts along the \( x, y, \) and \( z \) axes: \[ 1, 2, 2. \]

Step 2: Take the reciprocal of these intercepts: \[ \frac{1}{1}, \frac{1}{2}, \frac{1}{2}. \]

Step 3: Convert these into whole numbers by multiplying each by the least common multiple (LCM) of the denominators. The LCM of 1 and 2 is 2: \[ (2 \times 1, 2 \times \frac{1}{2}, 2 \times \frac{1}{2}) = (2,1,1). \]

Thus, the Miller indices of the given plane are: \[ (2 1 1). \] Quick Tip: To find Miller indices, take reciprocals of the intercepts and convert them into the smallest integer ratio.

Face-Centered Cubic (FCC) is a common crystal structure found in metals where atoms are arranged at the corners and the centers of each face of the cube.

Step 1: Identifying the crystal structures
- Aluminum (Al): FCC
- Nickel (Ni): FCC
- Copper (Cu): FCC
- Magnesium (Mg): Hexagonal Close-Packed (HCP)


Step 2: Conclusion
Among the given options, Aluminum, Nickel, and Copper exhibit an FCC structure. However, Magnesium has an HCP structure, making it the correct answer. Quick Tip: FCC metals generally have high ductility, while HCP metals like Magnesium tend to be more brittle due to their limited slip systems.

A Hexagonal Close-Packed (HCP) structure consists of atoms arranged in a hexagonal pattern with two layers stacked alternately. The total number of atoms in a unit cell of an HCP structure can be determined as follows:

Step 1: Contribution of atoms
- There are 12 corner atoms, each shared by 6 adjacent unit cells, contributing \( \frac{1}{6} \) per unit cell.
- There are 2 atoms completely inside the unit cell, contributing fully.
- There are 3 face-centered atoms, each shared between 2 unit cells, contributing \( \frac{1}{2} \) per unit cell.



Step 2: Total atoms in HCP unit cell \[ \left( 12 \times \frac{1}{6} \right) + \left( 2 \times 1 \right) + \left( 3 \times \frac{1}{2} \right) = 2 + 2 + 2 = 6. \]

Thus, the total number of atoms present in the unit cell of an HCP structure is 6. Quick Tip: The HCP structure has a high packing efficiency of about 74%, similar to the FCC structure, making it one of the most efficiently packed crystal structures.

The packing density of a crystal structure is determined by its packing efficiency, which is the fraction of volume occupied by atoms in a unit cell. The packing efficiencies of the given structures are:

- Diamond cubic: \( 34% \) (low packing due to tetrahedral arrangement)
- Cesium chloride (CsCl): \( 68% \) (simple cubic-like structure)
- Body-centred cubic (BCC): \( 68% \) (moderate packing efficiency)
- Face-centred cubic (FCC): \( 74% \) (highest among the given options)


Conclusion:
FCC has the highest packing efficiency of \( 74% \), making it the most densely packed structure among the given choices. Quick Tip: FCC and HCP structures have the highest packing efficiency (\( 74% \)), whereas BCC and diamond cubic structures have lower efficiency.

The atomic radius of an element is half of the bond length between two adjacent atoms in a metallic crystal structure. Given that the Fe-Fe bond length is \( 2.48 \)Å, the radius of the iron atom can be calculated as:
\[ Atomic Radius = \frac{Bond Length}{2} = \frac{2.48}{2} = 1.24 Å. \]

Thus, the correct radius of the iron atom is \( 1.24 \) Å. Quick Tip: In metallic bonding, the atomic radius is typically calculated as half of the bond length between two adjacent atoms in the structure.

The coordination number of a crystal structure refers to the number of nearest neighboring atoms surrounding a given atom in a unit cell. The coordination numbers for the three given structures are:


Body-Centered Cubic (BCC): Each atom is surrounded by 8 nearest neighbors.
Face-Centered Cubic (FCC): Each atom is surrounded by 12 nearest neighbors.
Hexagonal Close-Packed (HCP): Each atom is also surrounded by 12 nearest neighbors.


Thus, the correct order of coordination numbers for BCC, FCC, and HCP unit cells is:
\[ 8, 12, 12. \] Quick Tip: Higher coordination numbers indicate more efficient packing. FCC and HCP structures have the highest packing efficiency (~74%) due to their coordination number of 12.

The interplanar spacing \( d \) for a cubic crystal system is given by Bragg's equation:
\[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]

where:
- \( a = 2.814 \)Å (lattice parameter of the rock salt crystal),
- \( (hkl) = (100) \) (Miller indices of the plane),
- \( h = 1 \), \( k = 0 \), \( l = 0 \).

Step 1: Calculate the denominator
\[ \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1. \]

Step 2: Compute \( d_{100} \)
\[ d_{100} = \frac{2.814}{1} = 2.814 Å. \]

Thus, the interplanar distance for the (100) planes is 2.814 Å. Quick Tip: For cubic crystals, the interplanar spacing depends on the lattice parameter \( a \) and the Miller indices \( (hkl) \), following \( d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \).

The Laue method is a technique used in X-ray crystallography to determine the orientation of single crystals. It employs a white X-ray beam (continuous spectrum) rather than a monochromatic X-ray beam.

Step 1: Understanding the Methods
- Laue Method: Uses white X-rays to capture diffraction patterns for crystal orientation analysis.


- Powder Method: Uses monochromatic X-rays to analyze polycrystalline materials.


- Rotating Crystal Method: Uses monochromatic X-rays to study single crystals by rotating them in a fixed X-ray beam.



Step 2: Identifying the Wrong Statement
Option (a) is incorrect because it states that the Laue method uses a monochromatic X-ray beam, whereas it actually uses white X-radiation. Hence, it is the wrong statement. Quick Tip: The Laue method is primarily used for crystal orientation and defect analysis, whereas the powder method is widely used for phase identification.

Diffusion in solids occurs via different pathways, including lattice diffusion, grain boundary diffusion, and surface diffusion. The activation energy required for diffusion varies depending on the pathway:

Step 1: Comparing Activation Energies
- Lattice Diffusion: Occurs through the bulk crystal structure and requires high activation energy due to strong atomic bonds.

- Grain Boundary Diffusion: Occurs along grain boundaries where atomic packing is less dense, requiring lower activation energy.

- Surface Diffusion: Takes place on the outer surfaces of materials, having the lowest activation energy due to minimal atomic constraints.


Step 2: Conclusion
Since atomic bonding is weaker along surfaces and grain boundaries compared to the bulk lattice, diffusion along these regions requires lower activation energy. Hence, the correct answer is (b) lower. Quick Tip: Grain boundary and surface diffusion dominate at lower temperatures, while lattice diffusion is significant at higher temperatures.

The Frank–Reed source is a mechanism responsible for the multiplication of dislocations in crystalline materials under stress.

Step 1: Understanding Dislocation Multiplication
- When a material is plastically deformed, existing dislocations move due to applied stress.

- The Frank–Reed source allows the formation of new dislocations from an initial segment pinned at both ends.

- Under continued stress, the pinned segment bows out and eventually forms a dislocation loop.

- This mechanism is crucial in work hardening, where increased dislocation density strengthens the material.



Step 2: Conclusion
Since the Frank–Reed source is directly related to the multiplication of dislocations rather than point, ionic, or interstitial defects, the correct answer is (a) Dislocation multiplier. Quick Tip: The Frank–Reed source is a major mechanism in plastic deformation and contributes to strain hardening in metals.

The degree of freedom (\( F \)) in a system is calculated using Gibbs' phase rule, given by:
\[ F = C - P + 2 \]

where:
- \( C \) = Number of components in the system,
- \( P \) = Number of phases present,
- The constant 2 represents the two variables (temperature and pressure) that define the system.

Step 1: Identifying Parameters
- The system consists of water in three phases: solid (ice), liquid (water), and gas (vapour), so \( P = 3 \).
- The only component in the system is \( H_2O \), so \( C = 1 \).


Step 2: Applying Gibbs' Phase Rule \[ F = 1 - 3 + 2 = 0. \]


Step 3: Conclusion
Since \( F = 0 \), the system is invariant, meaning that at the triple point, temperature and pressure are fixed, and no degree of freedom remains. Thus, the correct answer is (c) 0. Quick Tip: The triple point of water occurs at 0.01°C and 0.006 atm, where ice, liquid water, and water vapour exist in equilibrium.

In a binary phase diagram, when two phases (liquid and solid) are in equilibrium, the overall composition of A can be found by applying the lever rule. The lever rule states that:
\[ \frac{L}{L+S} = \frac{C_S - C_0}{C_S - C_L} \]

Where:
- \(L\) is the fraction of liquid,
- \(S\) is the fraction of solid,
- \(C_S\) is the concentration of A in the solid phase,
- \(C_L\) is the concentration of A in the liquid phase, and
- \(C_0\) is the overall concentration of A.

Substituting the given values:
- \(C_S = 75%\),
- \(C_L = 30%\),
- \(C_0 = 40%\),
\[ \frac{L}{L+S} = \frac{75 - 40}{75 - 30} = \frac{35}{45} = 0.78 \]

Thus, the fraction of liquid is 0.78.

Thus, the correct answer is:
\[ (a) 0.78. \] Quick Tip: The lever rule is a useful tool for determining the fraction of each phase in equilibrium in a binary system. It helps find the composition and phase amounts in the system.

Hyper-eutectoid steel contains more than \(0.8%\) carbon and follows a transformation sequence during equilibrium cooling.

Step 1: Understanding the Cooling Process
1. At high temperatures, the steel is in the austenitic state (gamma phase).
2. Upon slow cooling below the eutectoid temperature (727°C), the excess carbon precipitates as cementite (\( Fe_3C \)) along the grain boundaries.
3. The remaining austenite transforms into pearlite, which consists of alternating layers of ferrite and cementite.


Step 2: Identifying the Constituents
- In hyper-eutectoid steel, the final microstructure after equilibrium cooling consists of cementite and pearlite.
- Ferrite is found in hypo-eutectoid steel, making option (b) incorrect.
- Bainite forms under non-equilibrium conditions, ruling out option (c).
- Martensite forms only under rapid cooling (quenching), eliminating option (d).


Step 3: Conclusion
The correct answer is (a) Cementite and pearlite, which are the equilibrium phases expected in a hyper-eutectoid steel after slow cooling. Quick Tip: Hyper-eutectoid steels (\( > 0.8% \) C) form cementite along grain boundaries before transforming the remaining austenite into pearlite.

A phase diagram provides information about phase stability, transformation temperatures, and solubility limits of different phases under equilibrium conditions. However, it does not provide details about the rate of phase transformations.

Step 1: Understanding the Statements
- Option (a) - Incorrect: Phase diagrams do not give information on transformation rates; this is instead provided by Time-Temperature-Transformation (TTT) diagrams.

- Option (b) - Correct: Phase diagrams show the relative amounts of different phases using the lever rule.

- Option (c) - Correct: They indicate melting points and phase transition temperatures.

- Option (d) - Correct: Phase diagrams depict solid solubility limits, which define solubility boundaries in alloy systems.


Step 2: Conclusion
Since phase diagrams do not provide information about transformation rates, option (a) is incorrect, making it the correct answer. Quick Tip: \textbf{Phase diagrams} show equilibrium phase relationships, while \textbf{TTT and CCT diagrams} provide transformation rate information.

When a metal undergoes cold working or plastic deformation, it experiences changes in internal structure. The correct sequence of heat treatment processes that restore the material properties is as follows:

Step 1: Stress Relief
- The first step is stress relief annealing, where the material is heated below the recrystallization temperature.
- This process removes residual stresses caused by cold working without significant changes in microstructure.


Step 2: Recrystallisation
- When the metal is further heated above the recrystallisation temperature, new strain-free grains begin to form.
- This eliminates the effects of work hardening and restores ductility.


Step 3: Grain Growth
- If the material is heated further, grain growth occurs, where small grains merge to form larger ones.
- This process increases grain size and reduces mechanical strength but improves toughness.


Step 4: Conclusion
Since the correct sequence of processes is Stress relief → Recrystallisation → Grain growth, the correct answer is option (d). Quick Tip: Recrystallisation occurs at a temperature of \(0.3 - 0.5 T_m\) (melting temperature), while grain growth happens at higher temperatures.

Carbon nanotubes (CNTs) can have different structures depending on the way graphene sheets are rolled. The three primary structures are:


Armchair (Metallic) – The nanotube axis is perpendicular to the C–C bond.
Zigzag (Semiconducting) – The nanotube axis is parallel to the C–C bond.
Chiral (Semiconducting) – The nanotube axis is at an angle to the C–C bond.


Step 1: Understanding the Armchair Structure
- In the armchair configuration, graphene is rolled in such a way that the hexagonal carbon rings form a continuous chain along the tube’s circumference.
- This structure makes armchair CNTs metallic in nature.


Step 2: Identifying the Correct Answer
Since armchair nanotubes form when the nanotube axis is perpendicular to the C–C bond, the correct answer is option (b). Quick Tip: Armchair nanotubes (\( n = m \)) exhibit \textbf{metallic behavior}, whereas zigzag and chiral CNTs can be semiconducting.

Heat treatment processes are used to modify the mechanical properties of steel. When steel becomes hardened due to previous heat treatment or work hardening, it can be softened using specific processes.

Step 1: Understanding Different Heat Treatment Processes
- Carburizing (Option a): A surface hardening process that increases the carbon content in the outer layer of steel, making it harder.


- Normalizing (Option b): A process used to refine grain structure and improve toughness but does not primarily soften hardened steel.


- Annealing (Option c): A heat treatment process that softens steel by heating it to a high temperature and then slowly cooling it. This relieves internal stresses and increases ductility.


- Tempering (Option d): Used to reduce brittleness and improve toughness but does not significantly soften hardened steel.



Step 2: Identifying the Correct Answer
Since annealing is specifically used to soften hardened steel, the correct answer is option (c) Annealing. Quick Tip: Annealing involves heating steel to a temperature above its recrystallization point, holding it at that temperature, and then allowing it to cool slowly in a furnace.

Thermoplastics and thermosetting plastics are two major classes of polymers, differing in structure and behavior upon heating.

Step 1: Understanding Thermoplastics and Thermosetting Plastics
- Thermoplastics: These polymers can be either amorphous or crystalline. Examples include polyethylene (crystalline) and polystyrene (amorphous).
- Thermosetting Plastics: These are generally amorphous and cannot be remelted after curing (e.g., epoxy, Bakelite).


Step 2: Evaluating the Statements
- Option (a) - Correct: Thermoplastics can be either crystalline (e.g., polyethylene, polypropylene) or amorphous (e.g., polycarbonate, PMMA).


- Option (b) - Incorrect: Not all thermoplastics are crystalline; some are amorphous.


- Option (c) - Incorrect: Thermoplastics can be crystalline, so this statement is false.


- Option (d) - Incorrect: Thermosetting plastics are typically amorphous, not crystalline.

Step 3: Conclusion
Since thermoplastics can be either amorphous or crystalline, the correct answer is option (a).
Quick Tip: Thermoplastics soften when heated and harden when cooled, while thermosetting plastics undergo permanent chemical cross-linking and cannot be remelted.

Aramid materials are a class of heat-resistant and strong synthetic fibers used in aerospace, military, and industrial applications.

Step 1: Understanding Aramid Materials
- Aramids (Aromatic Polyamides): These fibers are known for their high strength, thermal stability, and resistance to abrasion.
- Kevlar and Nomex are the two most widely used trade names for aramid fibers.


Step 2: Evaluating the Options
- Option (a) - Incorrect: Silicon carbide and silicon nitride are ceramics, not aramid materials.

- Option (b) - Incorrect: E-glass and aluminum oxide are used in fiber-reinforced composites but are not aramid materials.

- Option (c) - Correct: Kevlar is used in bulletproof vests, and Nomex is used in fire-resistant clothing.

- Option (d) - Incorrect: Zircon and carborundum (silicon carbide) are ceramics, not aramid materials.


Step 3: Conclusion
Since Kevlar and Nomex are the most common trade names for aramid materials, the correct answer is option (c). Quick Tip: Kevlar is known for its high tensile strength-to-weight ratio (used in body armor), while Nomex provides thermal and fire resistance (used in firefighter suits).

Conductive polymers are a class of polymers that exhibit electrical conductivity and are widely used in electronic and optoelectronic applications.

Step 1: Understanding Conductive Polymer Synthesis
- Conductive polymers, such as polyaniline, polypyrrole, and polythiophene, require specific polymerization techniques.
- The most effective and widely used method for synthesizing conductive polymers is electrochemical polymerization.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Free radical polymerization is commonly used for non-conductive polymers like polystyrene.


- Option (b) - Incorrect: Condensation polymerization is typically used for polyesters and polyamides, not conductive polymers.


- Option (c) - Correct: Electrochemical polymerization allows precise control over polymer structure and conductivity.


- Option (d) - Incorrect: Ionic polymerization is used for synthesizing elastomers and special-purpose polymers but is not commonly used for conductive polymers.



Step 3: Conclusion
Since electrochemical polymerization is the primary method for synthesizing conductive polymers, the correct answer is option (c). Quick Tip: Electrochemical polymerization involves oxidation or reduction of monomers on an electrode surface, allowing precise control over polymer growth and conductivity.

Polyvinyl chloride (PVC) is one of the most widely used synthetic polymers, known for its versatility and durability.

Step 1: Understanding the Classification of PVC
- Thermoplastics: These polymers soften upon heating and harden upon cooling, making them recyclable.


- Thermosetting Plastics: These polymers undergo an irreversible curing process and cannot be remelted.


- Elastomers: These polymers exhibit high elasticity and can stretch significantly before returning to their original shape.



Step 2: Identifying the Correct Category
- PVC is classified as a thermoplastic because it softens when heated and can be molded into various shapes.
- It is used in construction materials, pipes, medical tubing, and flexible packaging.



Step 3: Evaluating the Options
- Option (a) - Correct: PVC is a thermoplastic polymer.
- Option (b) - Incorrect: PVC is not thermosetting since it can be reheated and reshaped.
- Option (c) - Incorrect: PVC does not behave like elastomers, which are highly stretchable.
- Option (d) - Incorrect: Since PVC is a thermoplastic, this option is not applicable.



Step 4: Conclusion
Since PVC is classified as a thermoplastic, the correct answer is option (a). Quick Tip: PVC can be categorized as \textbf{rigid (RPVC)} for construction pipes and \textbf{flexible (FPVC)} for cables and medical applications.

The carbon content in steel determines its hardness, strength, and toughness. Different applications require varying carbon concentrations.


Step 1: Understanding Carbon Steel Classification
- Low-carbon steel (0.05% – 0.3%): Used for structural applications and does not harden significantly.
- Medium-carbon steel (0.3% – 0.6%): Used in automotive parts, offering moderate hardness and strength.
- High-carbon steel (0.6% – 1.5%): Used for tools, knives, and cutting instruments due to its high hardness and wear resistance.



Step 2: Evaluating the Options
- Option (a) - Correct: 0.8% – 0.9% C is typical for tools like scissors and knives, as it provides the right balance of hardness and toughness.


- Option (b) - Incorrect: 0.4% – 0.5% C is too low for sharp-edged tools, more suitable for structural applications.


- Option (c) - Incorrect: 0.2% – 0.3% C is classified as low-carbon steel, used in soft applications like automotive panels.


- Option (d) - Incorrect: 1.3% – 1.4% C is extremely hard and brittle, suitable for specialized cutting tools but not general-purpose scissors and knives.



Step 3: Conclusion
Since 0.8% – 0.9% C provides the necessary hardness and durability for scissors and knives, the correct answer is option (a). Quick Tip: High-carbon steels (~0.8% – 1.0%) are ideal for cutting tools, whereas tool steels with even higher carbon content (~1.5%) are used for precision cutting.

Martensitic transformation is a diffusionless phase transformation that occurs in steel when it is rapidly cooled (quenched) from the austenitic region.



Step 1: Understanding Martensitic Transformation
- Unlike other transformations, martensitic transformation does not involve atomic diffusion.
- Instead, it occurs through a coordinated shear movement of atoms, leading to a change in crystal structure.


Step 2: Evaluating the Options
- Option (a) - Incorrect: Martensitic transformations are not diffusion-controlled; they occur almost instantaneously without atomic diffusion.


- Option (b) - Incorrect: Martensite has the same composition as the parent austenite phase, so it does not yield two products of different compositions.


- Option (c) - Correct: Martensitic transformations are shear-dominated processes, involving a sudden and coordinated atomic movement.


- Option (d) - Incorrect: Martensite is a hard and brittle phase, not a soft product in steels.



Step 3: Conclusion
Since martensitic transformations are shear processes, the correct answer is option (c). Quick Tip: Martensitic transformations occur at very high speeds and lead to a hard, brittle microstructure that requires tempering to improve toughness.

Corrosion resistance in steel is enhanced by alloying elements that form protective oxide layers or improve passivation.

Step 1: Understanding Alloying for Corrosion Resistance
- Chromium (Cr) and Nickel (Ni) are key elements used in stainless steel to improve corrosion resistance.

- Chromium forms a passive oxide layer on the surface of the steel, preventing oxidation.
- Nickel enhances corrosion resistance, toughness, and strength.



Step 2: Evaluating the Options
- Option (a) - Correct: Stainless steel contains chromium (at least 10.5%) and nickel to improve corrosion resistance.


- Option (b) - Incorrect: Nickel and molybdenum improve hardness and strength, but molybdenum primarily enhances creep resistance rather than corrosion resistance.


- Option (c) - Incorrect: Aluminum and zinc are used in galvanization, not in enhancing corrosion resistance in steel.


- Option (d) - Incorrect: Tungsten improves wear resistance, while sulfur can lead to brittleness rather than corrosion protection.



Step 3: Conclusion
Since adding chromium to nickel significantly enhances corrosion resistance in steel, the correct answer is option (a). Quick Tip: Stainless steels contain at least 10.5% chromium, which forms a protective passive layer, preventing rust and oxidation.

Creep is the time-dependent deformation of materials under a constant load, typically occurring at high temperatures. It consists of three stages:


Step 1: Understanding the Stages of Creep
1. Primary Creep: The strain rate decreases as work hardening dominates over recovery.
2. Secondary Creep (Steady-State Creep): Work hardening and recovery balance each other, resulting in a constant strain rate.
3. Tertiary Creep: The strain rate accelerates as recovery and damage mechanisms dominate, leading to failure.



Step 2: Evaluating the Options
- Option (a) - Correct: In the tertiary creep stage, work hardening is less than recovery, leading to material weakening and increased deformation.


- Option (b) - Incorrect: If work hardening were greater than recovery, the strain rate would decrease, which is not observed in tertiary creep.


- Option (c) - Incorrect: Work hardening and recovery are equal in the secondary creep stage, not in tertiary creep.

- Option (d) - Incorrect: Since option (a) is correct, this option is invalid.


Step 3: Conclusion
Since work hardening is less than recovery in the tertiary creep stage, the correct answer is option (a). Quick Tip: Tertiary creep leads to \textbf{microstructural damage}, including void formation and grain boundary separation, ultimately causing failure.

Fatigue failure is a progressive and localized structural damage that occurs in materials subjected to repeated loading and unloading cycles.

Step 1: Understanding Fatigue Failure
- Fatigue occurs due to cyclic stresses that are below the ultimate tensile strength of the material.
- It leads to crack initiation, propagation, and final fracture.
- The failure mechanism is different from static failure, as it occurs over time with repeated loading.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Extended constant loading leads to creep failure rather than fatigue.


- Option (b) - Correct: Fatigue failure is caused by extended cyclic loading where stress is repeatedly applied and removed.


- Option (c) - Incorrect: Diffusion of atoms is relevant to creep and high-temperature deformation but not fatigue failure.


- Option (d) - Incorrect: While dislocations contribute to material deformation, fatigue failure primarily results from cyclic stress rather than just dislocation movement.



Step 3: Conclusion
Since extended cyclic loading is the primary cause of fatigue failure, the correct answer is option (b). Quick Tip: Fatigue failure can occur at stress levels much lower than the yield strength of a material, making design against fatigue crucial in engineering applications.

The Griffith equation is used to relate the stress (\(\sigma\)) at the tip of a crack to the material properties, such as the Young's modulus (E), the surface energy (\(\gamma\)), and the crack length (\(C\)):
\[ \sigma = \left(\frac{2\gamma E}{\pi C}\right)^{\frac{1}{2}} \]

Where:
- \(\sigma\) is the stress at the crack tip,
- \(\gamma\) is the surface energy,
- \(E\) is the Young's modulus, and
- \(C\) is the crack length.

This equation is used in fracture mechanics to determine the critical stress required for crack propagation in a material.

Thus, the correct answer is:
\[ (a) \sigma = \left(\frac{2\gamma E}{\pi C}\right)^{\frac{1}{2}}. \] Quick Tip: The Griffith equation is fundamental in understanding the fracture mechanics of brittle materials, linking material properties to crack propagation.

This relationship is based on the Wiedemann-Franz law, which states that the ratio of the thermal conductivity (\(K\)) to the electrical conductivity (\(\sigma\)) of a metal is proportional to the temperature \(T\):
\[ \frac{K}{\sigma T} = constant \]



The law implies that the thermal conductivity of a metal is related to its electrical conductivity, and the constant of proportionality is called the Lorenz number. This relationship holds for metals at relatively high temperatures.

Thus, the correct answer is:
\[ (d) \frac{K}{\sigma T} = constant. \] Quick Tip: The Wiedemann-Franz law provides a fundamental connection between the electrical and thermal conductivities of metals, and the ratio is inversely proportional to temperature.

A tetragon is another term for a quadrilateral, which is a four-sided polygon. However, if the question refers to a tetragonal crystal system, the interpretation changes.

Step 1: Understanding the Term "Tetragon"
- In geometry, a tetragon (quadrilateral) is a polygon with 4 faces.
- In crystallography, a tetragonal crystal system refers to a structure where the unit cell is rectangular with a square base.



Step 2: Evaluating the Options
- Option (a) - Incorrect: A tetragon does not have 12 faces.


- Option (b) - Correct: A tetragon has exactly 4 faces.


- Option (c) - Incorrect: A cube or an octahedron could have 6 faces, but not a tetragon.


- Option (d) - Incorrect: A tetragon has more than 2 faces.



Step 3: Conclusion
Since a tetragon has 4 faces, the correct answer is option (b). Quick Tip: A tetragon is a four-sided polygon, whereas the tetragonal crystal system refers to unit cells with a rectangular prism shape.

In a Body-Centered Cubic (BCC) structure, the relationship between the lattice constant (\( a \)) and the atomic radius (\( r \)) is given by:
\[ a = \frac{4r}{\sqrt{3}} \]

where:
- \( a \) is the lattice constant,
- \( r \) is the atomic radius,
- The factor \( \frac{4}{\sqrt{3}} \) comes from the geometry of a BCC unit cell.



Step 1: Substituting the given atomic radius

Given \( r = 1.24 \) Å,
\[ a = \frac{4 \times 1.24}{\sqrt{3}} \]



Step 2: Calculating the lattice constant
\[ a = \frac{4.96}{1.732} \]
\[ a \approx 2.864 \, Å \]


Step 3: Evaluating the Options
- Option (a) - Incorrect: 1.432 Å is not the correct calculation.
- Option (b) - Correct: 2.864 Å matches our calculation.
- Option (c) - Incorrect: 1.754 Å is not derived from the formula.
- Option (d) - Incorrect: 1.432 Å is repeated and incorrect.


Step 4: Conclusion
Since the calculated lattice constant is 2.864 Å, the correct answer is option (b). Quick Tip: For a \textbf{BCC unit cell}, the lattice constant is related to the atomic radius by \( a = \frac{4r}{\sqrt{3}} \). For an FCC unit cell, it is \( a = \frac{2\sqrt{2}r}{1} \).

N/A Quick Tip: In Bragg diffraction, the second-order reflection occurs at an angle where the sine of the angle is double that of the first-order reflection, for a given crystal.

A metallic bond is the type of chemical bond that holds metal atoms together in a sea of delocalized electrons.



Step 1: Understanding Metallic Bonding Characteristics
- Metallic bonds result from free-moving electrons that create strong cohesion between metal atoms.
- They lead to high electrical and thermal conductivity, malleability, and ductility.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Many metals appear opaque due to their ability to absorb and reflect light.


- Option (b) - Incorrect: Ductility is a key property of metals, allowing them to be drawn into wires.


- Option (c) - Incorrect: Metals exhibit high electrical conductivity due to free-moving electrons.


- Option (d) - Correct: Metallic bonds are non-directional because electrons are free to move, unlike covalent bonds which have specific orientations.



Step 3: Conclusion
Since metallic bonds lack directionality, the correct answer is option (d). Quick Tip: Unlike covalent bonds, which have specific bond angles, metallic bonds are non-directional because the electron cloud is delocalized.

Diffusional flux, denoted as \( J \), represents the amount of mass or number of particles diffusing through a unit area per unit time. It is given by Fick’s First Law:
\[ J = -D \frac{dC}{dx} \]

where:
- \( J \) is the diffusional flux (amount per unit area per unit time),
- \( D \) is the diffusion coefficient (m\(^2\)/s),
- \( C \) is the concentration of diffusing species (atoms/m\(^3\)),
- \( x \) is the distance (m).


Step 1: Finding the Unit of \( J \)
- The concentration gradient (\( dC/dx \)) has units of:
\[ \frac{atoms/m^3}{m} = atoms/m^4 \]

- The diffusion coefficient (\( D \)) has units of:
\[ m^2/s \]

- Thus, the unit of diffusional flux is:
\[ J = D \times \frac{dC}{dx} = \left( \frac{m^2}{s} \right) \times \left( \frac{atoms}{m^4} \right) \]
\[ = \frac{atoms}{m^2 \cdot s} \]



Step 2: Evaluating the Options
- Option (a) - Correct: \( atoms/m^2 \cdot s \) matches the derived unit.


- Option (b) - Incorrect: \( atoms/m^3 \cdot s \) represents a concentration change over time, not flux.


- Option (c) - Incorrect: \( atoms/m \cdot s^2 \) is not physically meaningful for diffusion.


- Option (d) - Incorrect: \( atoms/m \cdot s^3 \) is incorrect.



Step 3: Conclusion
Since the diffusional flux has units of atoms/m\(^2\).s, the correct answer is option (a). Quick Tip: Diffusional flux measures the \textbf{rate of diffusion per unit area}, and its unit is derived using Fick’s First Law.

Airplane windows are designed to withstand high pressure differences, impact forces, and extreme environmental conditions. The material used must be transparent, lightweight, strong, and impact-resistant.



Step 1: Understanding the Materials
- PVC (Polyvinyl Chloride): Mostly used for pipes and flooring, but it lacks the required optical clarity and impact resistance for airplane windows.


- PTFE (Polytetrafluoroethylene): Known for non-stick properties (e.g., Teflon), but it is not transparent or rigid enough for windows.


- PMMA (Polymethyl Methacrylate): Also known as Acrylic or Plexiglass, it is widely used in aircraft windows due to its high transparency, impact resistance, and lightweight nature.


- PEEK (Polyether Ether Ketone): A high-performance polymer used in aerospace components but not suitable for transparent applications like windows.



Step 2: Evaluating the Options
- Option (a) - Incorrect: PVC is not suitable for windows due to its opacity and fragility.


- Option (b) - Incorrect: PTFE is not transparent and lacks mechanical strength.


- Option (c) - Correct: PMMA is widely used in airplane windows due to its high strength and optical clarity.


- Option (d) - Incorrect: PEEK is used in structural components but not in windows.


Step 3: Conclusion
Since PMMA (Acrylic/Plexiglass) is the standard material for airplane windows, the correct answer is option (c). Quick Tip: PMMA (Acrylic) is used in airplane windows due to its lightweight nature, high impact resistance, and excellent optical transparency.

Cermets are composite materials composed of a combination of ceramic and metallic components, designed to leverage the high-temperature resistance of ceramics and the toughness and ductility of metals.



Step 1: Understanding Cermets
- Ceramics provide high hardness, wear resistance, and high-temperature stability.
- Metals provide toughness and improve the material’s ability to withstand mechanical shock.



Step 2: Common Applications of Cermets
- Used in cutting tools, aerospace applications, and armor-piercing projectiles.

- Examples: Tungsten carbide (WC), Titanium carbide (TiC), Titanium nitride (TiN) bonded with metals like cobalt (Co) or nickel (Ni).



Step 3: Evaluating the Options
- Option (a) - Correct: Cermets are a combination of ceramic and metal.


- Option (b) - Incorrect: Ceramic-ceramic composites exist but are not classified as cermets.


- Option (c) - Incorrect: Metal-metal alloys exist, but they are not cermets.


- Option (d) - Incorrect: Polymer-metal composites are different from cermets.



Step 4: Conclusion
Since cermets are a combination of ceramic and metal, the correct answer is option (a). Quick Tip: Cermets combine the hardness of ceramics and the toughness of metals, making them ideal for high-temperature and wear-resistant applications.

For a continuous and aligned fiber-reinforced composite, the modulus of elasticity in the longitudinal direction (\( E_L \)) is given by the Rule of Mixtures:
\[ E_L = V_f E_f + V_m E_m \]

where:
- \( E_L \) = Modulus of elasticity in the longitudinal direction,
- \( V_f \) = Volume fraction of the fiber = 40% = 0.40,
- \( E_f \) = Modulus of elasticity of the fiber = 69 GPa,
- \( V_m \) = Volume fraction of the matrix = 60% = 0.60,
- \( E_m \) = Modulus of elasticity of the matrix = 3.4 GPa.




Step 1: Substituting the Given Values
\[ E_L = (0.40 \times 69) + (0.60 \times 3.4) \]
\[ E_L = 27.6 + 2.04 \]
\[ E_L = 29.64 \approx 30 GPa \]



Step 2: Evaluating the Options
- Option (a) - Incorrect: 35 GPa is too high.


- Option (b) - Incorrect: 45 GPa is too high.


- Option (c) - Correct: 30 GPa matches our calculation.


- Option (d) - Incorrect: 20 GPa is too low.



Step 3: Conclusion
Since the calculated modulus of elasticity in the longitudinal direction is 30 GPa, the correct answer is option (c). Quick Tip: The \textbf{Rule of Mixtures} is used to determine composite material properties, where the total property is the weighted sum of its components.

Fracture toughness (\( K_{IC} \)) is a material property that describes a material’s resistance to crack propagation. Ceramic Matrix Composites (CMCs) generally have higher fracture toughness than traditional ceramics due to their ability to resist crack growth through fiber reinforcement.



Step 1: Understanding the Fracture Toughness of CMCs
- Traditional ceramics have low fracture toughness (\( 1-5 \) MPa\( \sqrt{m} \)), making them brittle.
- CMCs are reinforced with fibers (e.g., SiC, alumina) to enhance toughness.
- The fracture toughness of most CMCs falls in the range of 6 to 20 MPa\( \sqrt{m} \).



Step 2: Evaluating the Options
- Option (a) - Incorrect: 5 to 18 MPa\( \sqrt{m} \) is slightly lower than the typical range.


- Option (b) - Correct: 6 to 20 MPa\( \sqrt{m} \) matches experimental values for CMCs.


- Option (c) - Incorrect: 8 to 16 MPa\( \sqrt{m} \) is too narrow.


- Option (d) - Incorrect: 9 to 21 MPa\( \sqrt{m} \) is slightly higher than the typical range.



Step 3: Conclusion
Since the fracture toughness of Ceramic Matrix Composites typically lies between 6 and 20 MPa\( \sqrt{m} \), the correct answer is option (b). Quick Tip: CMCs have higher fracture toughness than traditional ceramics due to fiber reinforcement, making them suitable for aerospace and high-temperature applications.

Nanostructured materials are materials that have structural features on the nanometer scale, significantly altering their mechanical, electrical, and optical properties.



Step 1: Understanding the Size Range of Nanostructured Materials
- By definition, nanomaterials have at least one dimension in the 1–100 nm range.

- Materials with grain sizes in this range exhibit unique properties such as quantum effects, high surface area, and enhanced mechanical strength.



Step 2: Evaluating the Options
- Option (a) - Correct: The 1–100 nm range correctly defines nanostructured materials.


- Option (b) - Incorrect: 150–300 nm is outside the nanometer scale and falls into the microstructure regime.


- Option (c) - Incorrect: 350–500 nm is significantly larger than the nanoscale range.


- Option (d) - Incorrect: 500–900 nm corresponds to larger microcrystalline materials.



Step 3: Conclusion
Since nanostructured materials have crystallites ranging in size from 1 to 100 nm, the correct answer is option (a). Quick Tip: Nanomaterials have enhanced mechanical strength, electrical properties, and catalytic activity due to their large surface area and quantum effects.

Iron exhibits different allotropes based on temperature, affecting its crystal structure and properties.


Step 1: Understanding the Allotropic Forms of Iron
- \( \alpha \)-Iron (Ferrite): Stable at room temperature, BCC structure.
- \( \gamma \)-Iron (Austenite): Stable between 912°C and 1394°C, FCC structure.
- \( \delta \)-Iron: Stable between 1394°C and 1538°C, BCC structure.



Step 2: Evaluating the Options
- Option (a) - Incorrect: \( \alpha \)-Iron (Ferrite) is an allotropic form of iron.


- Option (b) - Correct: \( \rho \) is not an allotropic form of iron.


- Option (c) - Incorrect: \( \gamma \)-Iron (Austenite) is a stable form of iron at high temperatures.


- Option (d) - Incorrect: \( \theta \) represents iron carbide (Fe\(_3\)C), which is a compound, not an allotrope.



Step 3: Conclusion
Since \( \rho \) is not an allotropic form of iron, the correct answer is option (b). Quick Tip: Iron exists in different allotropic forms: \( \alpha \) (Ferrite), \( \gamma \) (Austenite), and \( \delta \) (High-temperature BCC). These influence steel properties.

The relationship between the ASTM grain size number and the mean grain diameter (\(D\)) is given by the following formula:
\[ N = \left(\frac{1}{D}\right)^2 \]

Where:
- \(N\) is the ASTM grain size number,
- \(D\) is the mean grain diameter in millimeters.

Rearranging the formula to solve for \(D\):
\[ D = \frac{1}{\sqrt{N}} \]

Substituting \(N = 0.5\):
\[ D = \frac{1}{\sqrt{0.5}} \approx 0.33 \, mm \]

Thus, the mean grain diameter corresponding to an ASTM number of 0.5 is approximately 0.33 mm.

Thus, the correct answer is:
\[ (a) 0.33 mm. \] Quick Tip: The grain size number and grain diameter are inversely related. As the ASTM grain size number increases, the mean grain diameter decreases.

The resistance (\( R \)) of a conductor is given by Ohm’s Law:
\[ R = \rho \frac{L}{A} \]

where:
- \( \rho = 1.7 \times 10^{-6} \) \( \Omega \)cm = \( 1.7 \times 10^{-8} \) \( \Omega \)m (converted to SI units),
- \( L = 31.4 \) m (length),
- \( A = 19.6 \times 10^{-8} \) m\(^2\) (cross-sectional area).



Step 1: Substituting the Values
\[ R = \frac{(1.7 \times 10^{-8}) \times (31.4)}{19.6 \times 10^{-8}} \]


Step 2: Simplifying the Expression
\[ R = \frac{5.338 \times 10^{-7}}{19.6 \times 10^{-8}} \]
\[ R = \frac{5.338}{1.96} \]
\[ R \approx 2.72 \, \Omega \]



Step 3: Evaluating the Options
- Option (a) - Incorrect: 1.72 \( \Omega \) is too low.
- Option (b) - Correct: 2.72 \( \Omega \) matches our calculation.
- Option (c) - Incorrect: 3.72 \( \Omega \) is too high.
- Option (d) - Incorrect: 4.72 \( \Omega \) is incorrect.



Step 4: Conclusion
Since the calculated resistance is 2.72 \( \Omega \), the correct answer is option (b). Quick Tip: The resistance of a wire depends on its resistivity, length, and cross-sectional area. Higher length increases resistance, while a larger area reduces it.

The Fermi level (\( E_F \)) represents the energy level at which the probability of finding an electron is 50% at thermal equilibrium.



Step 1: Understanding the Fermi Level in N-Type Semiconductors
- In intrinsic semiconductors, the Fermi level lies at the center of the energy gap.

- In N-type semiconductors, donor impurities introduce extra electrons, shifting the Fermi level closer to the conduction band.
- The presence of excess electrons raises the probability of electron occupancy in higher energy states.


Step 2: Evaluating the Options
- Option (a) - Incorrect: The Fermi level in an N-type semiconductor is higher, not lower.


- Option (b) - Incorrect: The Fermi level is not at the center in an N-type semiconductor.


- Option (c) - Correct: The Fermi level is closer to the conduction band, making it higher than the center of the energy gap.


- Option (d) - Incorrect: The Fermi level always exists as a fundamental property.



Step 3: Conclusion
Since the Fermi level in an N-type semiconductor is higher than the center of the energy gap, the correct answer is option (c). Quick Tip: In an N-type semiconductor, the Fermi level shifts closer to the conduction band due to the presence of excess electrons from donor impurities.

The power loss in a dielectric material occurs due to dielectric dissipation and is determined using the following relation:
\[ P = VI \sin\delta \]

where:
- \( V \) is the applied voltage,
- \( I \) is the current,
- \( \delta \) is the dielectric loss angle (also called the loss tangent).



Step 1: Understanding Dielectric Loss
- The total power in an AC circuit is given by \( P = VI \cos\phi \), where \( \phi \) is the phase angle.

- In a perfect capacitor, the phase angle between voltage and current is 90°, leading to no real power loss.

- However, in real dielectrics, losses occur due to the imperfect insulation properties, represented by the loss tangent \( \tan\delta \).
- The actual power loss is given by:
\[ P = VI \sin\delta \]



Step 2: Evaluating the Options
- Option (a) - Incorrect: Power loss is associated with \( \sin\delta \), not \( \cos\delta \).


- Option (b) - Incorrect: The formula uses \( VI \), not \( V/I \).


- Option (c) - Correct: \( P = VI \sin\delta \) correctly describes dielectric power loss.


- Option (d) - Incorrect: \( V/I \) does not correspond to the correct power loss equation.



Step 3: Conclusion
Since the correct formula for dielectric power loss is \( P = VI \sin\delta \), the correct answer is option (c). Quick Tip: The power loss in a dielectric is proportional to \( \sin\delta \), where \( \delta \) represents the loss tangent, measuring how much energy is dissipated in the material.

Calcium carbonate (\( CaCO_3 \)) is a common compound found in many minerals and natural materials. It forms a major component of limestone, chalk, and marble.



Step 1: Understanding the Composition of Each Option
- Calcspar (Calcite): A crystalline form of calcium carbonate (\( CaCO_3 \)).


- Witherite: Composed of barium carbonate (\( BaCO_3 \)), not calcium carbonate.


- Marl: A mixture of clay and calcium carbonate, making it rich in \( CaCO_3 \).


- Chalk: A soft, porous form of calcium carbonate (\( CaCO_3 \)).



Step 2: Evaluating the Options
- Option (a) - Incorrect: Calcspar (Calcite) is composed of \( CaCO_3 \).


- Option (b) - Correct: Witherite contains \( BaCO_3 \) instead of \( CaCO_3 \).


- Option (c) - Incorrect: Marl contains a high proportion of calcium carbonate.


- Option (d) - Incorrect: Chalk is composed of calcium carbonate.



Step 3: Conclusion
Since Witherite is composed of barium carbonate (\( BaCO_3 \)) instead of calcium carbonate (\( CaCO_3 \)), the correct answer is option (b). Quick Tip: Materials like \textbf{chalk, limestone, and marble} are composed of calcium carbonate, while \textbf{Witherite} consists of \textbf{barium carbonate}.

Zirconia (\(ZrO_2\)) is a polymorphic material, meaning it can exist in different crystal structures depending on temperature.



Step 1: Understanding the Crystal Structures of Zirconia
- Monoclinic Zirconia: Found in baddeleyite, it is the most stable phase at room temperature.

- Tetragonal Zirconia: Stable at high temperatures (~1200°C - 2370°C) and is used in ceramics.
- Cubic Zirconia: Exists at very high temperatures (~2370°C - melting point).
- Triclinic Structure: Not associated with zirconia.



Step 2: Evaluating the Options
- Option (a) - Correct: Baddeleyite naturally occurs in the monoclinic phase of zirconia.


- Option (b) - Incorrect: Triclinic structure is not a phase of zirconia.


- Option (c) - Incorrect: Tetragonal zirconia is stable at high temperatures but not in baddeleyite.


- Option (d) - Incorrect: Cubic zirconia occurs only at extreme temperatures or in synthetic forms.



Step 3: Conclusion
Since baddeleyite exists in the monoclinic phase, the correct answer is option (a). Quick Tip: Zirconia (\(ZrO_2\)) exists in monoclinic, tetragonal, and cubic phases depending on temperature. The monoclinic phase is stable at room temperature and occurs naturally as baddeleyite.

Titania, also known as titanium dioxide (TiO\(_2\)), is a widely used ceramic material known for its unique optical and chemical properties.



Step 1: Understanding the Properties of Titania
- Stability against ultraviolet (UV) radiation: Titania is a UV-resistant material used in sunscreens, paints, and coatings due to its excellent ability to absorb and scatter UV rays.


- Fracture toughness: Titania does not exhibit high fracture toughness, as it is a brittle ceramic.


- Modulus of rupture: Though it has decent strength, its modulus of rupture is not its defining characteristic.


- Compressive strength: While titania has some mechanical strength, its primary use is in optical, photocatalytic, and protective applications.



Step 2: Evaluating the Options
- Option (a) - Correct: Titania is highly stable against ultraviolet radiation, making it useful in coatings and sunscreens.


- Option (b) - Incorrect: Titania does not have high fracture toughness.


- Option (c) - Incorrect: The modulus of rupture is not its key property.


- Option (d) - Incorrect: Titania is not known for its high compressive strength.



Step 3: Conclusion
Since Titania is highly stable against ultraviolet radiation, the correct answer is option (a). Quick Tip: Titanium dioxide (\(TiO_2\)) is widely used in paints, sunscreens, and photocatalysis due to its UV-resistant and optical properties.

Alumina (\( Al_2O_3 \)) is a widely used ceramic material known for its high hardness, thermal stability, and wear resistance. However, pure alumina has low fracture toughness, making it brittle.



Step 1: Understanding Toughness Enhancement in Alumina
- To increase toughness, reinforcing agents or additives are introduced to modify the microstructure and improve fracture resistance.
- Chromia (\( Cr_2O_3 \)) is commonly used as an additive to alumina to enhance its fracture toughness and mechanical strength.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Magnesia (\( MgO \)) improves sintering but does not significantly enhance toughness.


- Option (b) - Incorrect: Silica (\( SiO_2 \)) reduces thermal conductivity but does not increase toughness.


- Option (c) - Correct: Chromia (\( Cr_2O_3 \)) increases toughness by forming a solid solution with alumina.


- Option (d) - Incorrect: Calcia (\( CaO \)) helps in densification but does not enhance toughness.



Step 3: Conclusion
Since Chromia (\( Cr_2O_3 \)) is used to improve the toughness of alumina, the correct answer is option (c). Quick Tip: The addition of Chromia (\( Cr_2O_3 \)) to alumina enhances fracture toughness by refining the grain structure and improving crack resistance.

Mullite is an important ceramic material with a solid solution series between alumina (\( Al_2O_3 \)) and silica (\( SiO_2 \)). The limiting compositions of mullite in this solid solution series are 3:2 and 2:1.



Step 1: Understanding the Mullite Composition
- Mullite exists in a range of compositions between \( 3Al_2O_3 \cdot 2SiO_2 \) (3:2) and \( 2Al_2O_3 \cdot SiO_2 \) (2:1).
- This range defines the mullite solid solution series.
- These compositions influence the thermal stability, mechanical properties, and phase stability of mullite.



Step 2: Evaluating the Options
- Option (a) - Incorrect: 1:2 is not a composition in the mullite series.


- Option (b) - Incorrect: 2:1 is correct, but 3:1 is not part of the mullite series.


- Option (c) - Correct: The 3:2 and 2:1 ratios correctly represent the limiting compositions of mullite.


- Option (d) - Incorrect: 3:1 is not a correct composition in the mullite phase.



Step 3: Conclusion
Since the limiting compositions of mullite solid solution are 3:2 and 2:1, the correct answer is option (c). Quick Tip: Mullite is a stable aluminosilicate ceramic with a solid solution series between \( 3:2 \) and \( 2:1 \) compositions of \( Al_2O_3 \) and \( SiO_2 \).

The flat plate test is used to analyze the stress distribution in ceramic glazes. The concave or convex nature of the glaze determines whether it experiences tensile or compressive stresses.



Step 1: Understanding the Stress in Glaze Layers
- If the glazed side is concave, the glaze layer is under tension.
- If the glazed side is convex, the glaze layer is under compression.



Step 2: Defining the Failure Mechanism
- Tensile stress in the glaze leads to crazing, which appears as fine cracks on the surface.
- Compressive stress in the glaze leads to peeling (or shivering), where the glaze detaches from the ceramic body.



Step 3: Evaluating the Options
- Option (a) - Incorrect: Tension causes crazing, not peeling.


- Option (b) - Correct: Tension leads to crazing, which occurs when the glaze contracts more than the ceramic body.


- Option (c) - Incorrect: Compression causes peeling, not crazing.


- Option (d) - Incorrect: Compression does not cause crazing.



Step 4: Conclusion
Since concave glazed surfaces experience tension, leading to crazing, the correct answer is option (b). Quick Tip: In ceramics, tensile stress causes crazing (fine cracks), while compressive stress leads to peeling (glaze detachment).

Glass formers are compounds that have the ability to form an amorphous glassy structure upon cooling without crystallization.

Step 1: Understanding Glass Formers
- Silica (\( SiO_2 \)) is the most common glass former used in traditional glass manufacturing.


- Boron trioxide (\( B_2O_3 \)) is also a strong glass former, used in borosilicate glasses.


- Germanium oxide (\( GeO_2 \)) is another well-known glass former, similar in structure to silica.



Step 2: Identifying Non-Glass Formers
- Chromium oxide (\( Cr_2O_3 \)) is not a glass former; it is a refractory oxide that does not form an amorphous structure but rather remains crystalline.



Step 3: Evaluating the Options
- Option (a) - Incorrect: Silica (\( SiO_2 \)) is a well-known glass former.


- Option (b) - Incorrect: Boron trioxide (\( B_2O_3 \)) is a common glass former.


- Option (c) - Incorrect: Germanium oxide (\( GeO_2 \)) forms glass similar to silica.


- Option (d) - Correct: Chromium oxide (\( Cr_2O_3 \)) is not a glass former.



Step 4: Conclusion
Since Chromium oxide (\( Cr_2O_3 \)) does not form glass, the correct answer is option (d). Quick Tip: \textbf{Glass formers} include SiO\(_2\), B\(_2\)O\(_3\), and GeO\(_2\). Chromium oxide (\( Cr_2O_3 \)) is a crystalline refractory material, not a glass former.

The Danner process is a continuous glass-forming technique used specifically for manufacturing glass tubes and rods.



Step 1: Understanding the Danner Process
- The Danner process involves a rotating mandrel (hollow cylinder) through which molten glass flows and forms a continuous tube.
- The size and thickness of the glass tube can be controlled by air pressure and speed.
- This method is widely used for producing test tubes, glass tubing, and fluorescent lamp envelopes.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Bulbs are not manufactured using the Danner process.


- Option (b) - Correct: The Danner process is specifically designed for producing glass tubes.


- Option (c) - Incorrect: Sheets are formed using different processes like the Fourcault or Pilkington process.


- Option (d) - Incorrect: Glass fibers are typically made using fiber drawing techniques, not the Danner process.



Step 3: Conclusion
Since the Danner process is used for manufacturing glass tubes continuously, the correct answer is option (b). Quick Tip: The Danner process is a continuous glass tube manufacturing process, used for producing test tubes, fluorescent lamp tubes, and glass rods.

Crown glass is a type of optical glass known for its low dispersion and high transparency. It is commonly used in high-quality lenses and optical instruments.



Step 1: Understanding the Properties of Crown Glass
- Crown glass is made from a mixture of silica, sodium oxide, and calcium oxide, giving it a low refractive index and low dispersion.
- It is widely used in microscope lenses, telescopes, and eyeglasses.
- Crown glass differs from flint glass, which has a higher refractive index and is used in optical prisms.



Step 2: Evaluating the Options
- Option (a) - Correct: Crown glass is primarily used as optical glass due to its excellent clarity and optical properties.


- Option (b) - Incorrect: Safety glasses are made from laminated or tempered glass, not crown glass.


- Option (c) - Incorrect: Radiation shielding glass contains lead oxide (PbO), not crown glass.


- Option (d) - Incorrect: Toughened glass is heat-treated for strength, but crown glass is not primarily toughened.



Step 3: Conclusion
Since crown glass is mainly used in optical applications, the correct answer is option (a). Quick Tip: Crown glass is a high-quality optical glass with low dispersion, making it ideal for lenses in microscopes, telescopes, and eyeglasses.

Refractory materials are classified into three types based on their chemical behavior:

1. Acidic refractories – Resistant to acidic environments (e.g., silica).
2. Basic refractories – Resistant to basic environments (e.g., magnesite).
3. Neutral refractories – Resistant to both acidic and basic conditions (e.g., chromite, zircon).



Step 1: Understanding Neutral Refractories
- Zircon (\( ZrSiO_4 \)) is a neutral refractory with high chemical stability.


- Chrome refractories (\( Cr_2O_3 \)) are also neutral, commonly used in steel industries.


- Silicon carbide (\( SiC \)) is a high-strength neutral refractory with excellent thermal conductivity.



Step 2: Identifying the Non-Neutral Refractory
- Carbon (Graphite) is a basic refractory, meaning it is not neutral.
- Carbon-based refractories (graphite, coke) are commonly used in basic linings in furnaces.



Step 3: Evaluating the Options
- Option (a) - Incorrect: Zircon is a neutral refractory.


- Option (b) - Incorrect: Chrome is a neutral refractory.


- Option (c) - Correct: Carbon is a basic refractory, not neutral.


- Option (d) - Incorrect: Silicon carbide is a neutral refractory.




Step 4: Conclusion
Since carbon (graphite) is a basic refractory, not neutral, the correct answer is option (c). Quick Tip: Neutral refractories, such as zircon, chromite, and silicon carbide, are stable in both acidic and basic environments. However, carbon-based refractories are basic, not neutral.

Faraday’s laws of electrolysis describe the quantitative relationship between the amount of substance deposited or liberated at an electrode and the amount of electric charge passed through the electrolyte.



Step 1: Understanding Faraday’s Laws
Faraday formulated two laws:
1. First Law: The mass of substance deposited at an electrode is directly proportional to the charge (\( Q \)) passed through the electrolyte.
\[ m = Z Q \]
where \( Z \) is the electrochemical equivalent of the substance.



2. Second Law: The mass of different substances deposited by the same amount of charge is proportional to their equivalent weights.



Step 2: Evaluating the Options
- Option (a) - Correct: Faraday’s Law accurately calculates the theoretical weight deposited during electrolysis.


- Option (b) - Incorrect: Hess’s Law is related to the enthalpy changes in thermodynamics.


- Option (c) - Incorrect: Bragg’s Law deals with X-ray diffraction and crystallography.


- Option (d) - Incorrect: Stoke’s Law is related to the motion of particles in a fluid.



Step 3: Conclusion
Since Faraday’s Law governs the calculation of mass deposited during electrolysis, the correct answer is option (a). Quick Tip: \textbf{Faraday’s Laws of Electrolysis} relate the mass of substance deposited on an electrode to the electric charge passed through the electrolyte.

Fuel cells are electrochemical devices that convert chemical energy from a fuel (such as hydrogen) into electricity with high efficiency and minimal emissions.



Step 1: Understanding the Challenges of Fuel Cells
- Efficiency: Fuel cells are actually highly efficient, unlike conventional combustion engines.


- Environmental Impact: They produce water and heat as byproducts, making them eco-friendly.


- Power Output: Fuel cells can generate significant power, comparable to gasoline.


- Fuel Supply Issue: The biggest challenge is the storage, transport, and distribution of hydrogen or other fuels required for operation.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Fuel cells are more efficient than traditional combustion engines.


- Option (b) - Incorrect: They produce no harmful emissions, only water and heat.


- Option (c) - Correct: Fuel supply and infrastructure are the biggest challenges for fuel cell adoption.


- Option (d) - Incorrect: Fuel cells can provide competitive power compared to gasoline.



Step 3: Conclusion
Since the main drawback of fuel cells is the difficulty in supplying fuel (such as hydrogen) efficiently, the correct answer is option (c). Quick Tip: The biggest challenge of fuel cells is the lack of hydrogen fuel infrastructure. Hydrogen is difficult to store, transport, and distribute efficiently.

Automotive fuel cell technology is being developed to create cleaner and more efficient energy alternatives to fossil fuels.



Step 1: Understanding the Key Advantages of Fuel Cells
- High Efficiency: Fuel cells convert energy more efficiently than internal combustion engines.


- Low or Zero Emissions: Fuel cells produce only water vapor as a byproduct, making them environmentally friendly.


- Local Source Production: Hydrogen for fuel cells can be produced from various renewable sources, reducing dependency on fossil fuels.



Step 2: Identifying the Irrelevant Factor
- Capacitance refers to a system's ability to store electrical charge. This property is not a critical factor in fuel cell technology development.



Step 3: Evaluating the Options
- Option (a) - Incorrect: Efficiency is a major reason for fuel cell development.


- Option (b) - Correct: Low capacitance is not relevant to fuel cell development.


- Option (c) - Incorrect: Zero emissions is a key reason for adopting fuel cells.


- Option (d) - Incorrect: Local hydrogen production supports fuel cell adoption.



Step 4: Conclusion
Since low capacitance is not a key factor in fuel cell technology development, the correct answer is option (b). Quick Tip: Fuel cells are developed for their high efficiency, zero emissions, and renewable energy potential. Capacitance is not a relevant factor in fuel cell development.

For a biomaterial to form a bioactive bond with bone or soft tissue, a specific biocompatible mineral layer must develop on its surface.



Step 1: Understanding Bioactivity in Biomaterials
- Bioactive materials are designed to interact with biological tissues and promote osseointegration.

- The key factor in bioactivity is the formation of hydroxyapatite (\( Ca_{10}(PO_4)_6(OH)_2 \)), which mimics the natural mineral phase of bone.



Step 2: Evaluating the Options
- Option (a) - Incorrect: Amorphous silica does not directly promote bioactivity.


- Option (b) - Incorrect: Silanols are involved in surface chemistry but do not induce bioactive bonding.


- Option (c) - Incorrect: Amorphous calcium phosphate is a precursor but does not provide strong bonding.


- Option (d) - Correct: Hydroxyapatite is the key bioactive phase that promotes bonding with bone.



Step 3: Conclusion
Since hydroxyapatite formation on the biomaterial surface is essential for bioactive bonding, the correct answer is option (d). Quick Tip: \textbf{Hydroxyapatite} formation on a biomaterial surface promotes bioactive bonding by mimicking the natural mineral phase of bone, leading to osseointegration.

In nuclear reactors, coolants play a crucial role in removing heat generated from nuclear fission and maintaining reactor stability.



Step 1: Understanding Nuclear Reactor Coolants
- Common coolants used in nuclear reactors include:

- Liquid sodium (used in fast breeder reactors).

- Carbon dioxide (\( CO_2 \)) (used in gas-cooled reactors).

- Heavy water (\( D_2O \)) (used in pressurized heavy water reactors).



Step 2: Evaluating the Options
- Option (a) - Correct: Graphite is not a coolant, but it is commonly used as a moderator to slow down neutrons in reactors.


- Option (b) - Incorrect: Liquid sodium is an effective coolant, especially in fast breeder reactors.


- Option (c) - Incorrect: Carbon dioxide is used as a gas coolant in certain reactor designs.


- Option (d) - Incorrect: Heavy water is a coolant in CANDU reactors.



Step 3: Conclusion
Since graphite is primarily used as a moderator, not a coolant, the correct answer is option (a). Quick Tip: Graphite is a neutron moderator, not a coolant in nuclear reactors. Common coolants include liquid sodium, carbon dioxide, and heavy water.