TANCET 2024 Agricultural & Irrigation Engineering (M.Tech.) Available-: 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. \[ \text{Shortest} = |PC - R| = |\sqrt{6} - \sqrt{24}| \] \[ \text{Longest} = PC + R = \sqrt{6} + \sqrt{24} \] 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: \[ |d - R| \quad \text{and} \quad d + R \] 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 \( t \). \[ t(12t - 10) = 0 \] \[ t = 0, \quad t = \frac{10}{12} = 0.833 \text{s} \] 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 (\( I \propto f \)).

Step 1: Understanding Movable and Fixed Pulleys
A movable pulley moves along with the load and provides a mechanical advantage by reducing the effort required to lift the load. Examples include elevators, cranes, and weightlifting machines. Step 2: Identifying the Non-Movable Pulley
A flagpole pulley is a fixed pulley. It only changes the direction of force and does not reduce the effort needed to lift the load. Step 3: Conclusion
Since a pulley at a flagpole does not move with the load, it is not a movable pulley. Quick Tip: Movable pulleys reduce effort by distributing the load, while fixed pulleys only change the direction of force.

Step 1: Understanding the function of gears
Gears are used to transfer motion and torque between shafts in mechanical systems. Some gears provide speed reduction, while others offer locking mechanisms to prevent movement in the opposite direction. Step 2: Identifying the correct gear for locking
- Spur Gear: Simple gears used for transmitting motion but cannot lock. - Helical Gear: Offers smoother operation than spur gears but does not have a locking mechanism. - Bevel Gear: Used for changing the direction of shaft rotation. - Worm Gear: Has a self-locking mechanism that prevents back-driving, making it suitable for conveyor systems. Step 3: Conclusion
Conveyor systems require worm gears to prevent unintended movement, ensuring controlled operation. Quick Tip: Worm gears are the best choice for applications requiring motion locking, such as conveyor systems.

Step 1: Understanding the function of a mould board
A mould board is a part of a plough that helps in inverting the soil during primary tillage. Step 2: Selecting the correct mould board for sticky soils
- General Purpose Mould Board: Suitable for a wide range of soils but may clog in sticky conditions. - Slat Mould Board: Has spaced slats to reduce adhesion, making it ideal for sticky soils. - Stubble Mould Board: Used for ploughing fields with plant residues. - Sod Mould Board: Used for cutting through sod and grassy fields. Step 3: Conclusion
Since sticky soils tend to clog solid mould boards, slat mould boards are preferred as they reduce adhesion and require less draft power. Quick Tip: For sticky soils, use a slat mould board to prevent clogging and improve efficiency.

Step 1: Understanding the working of a seed drill
A seed drill is used for sowing seeds at a uniform depth and spacing. It consists of multiple furrow openers that place seeds in rows. Step 2: Deriving the relationship
The total working width (\(W\)) of a seed drill depends on: - The number of furrow openers (\(N\)) - The spacing between furrows (\(S\)) Mathematically, the width of the seed drill is given by: \[ W = N \times S \] Step 3: Eliminating incorrect options
- \( W = \frac{N}{S} \) is incorrect because division does not represent total width. - \( W = \frac{S}{N} \) is incorrect for the same reason. - \( S = \frac{W}{N} \) is rearranged correctly but does not directly answer the question. Quick Tip: To calculate the width of a seed drill, multiply the number of furrow openers by their spacing.

Step 1: Understanding the calorie-to-joule conversion
A calorie is a unit of energy commonly used in thermodynamics. The conversion factor is: \[ 1 \text{ cal} = 4.186 \text{ J} \] Step 2: Calculating the energy in joules
For 10 calories: \[ 10 \times 4.186 = 41.86 \text{ J} \] Step 3: Eliminating incorrect options
- \( 4.186 \) J corresponds to 1 cal, not 10 cal. - \( 418.6 \) J is obtained by multiplying by 100 instead of 10. - \( 0.4186 \) J is a result of dividing instead of multiplying. Quick Tip: To convert calories to joules, use the formula: \( \text{Energy (J)} = \text{Energy (cal)} \times 4.186 \).

Step 1: Understanding the function of a Pitman
The Pitman is the connecting rod in a conventional mower that transmits the reciprocating motion from the crankshaft to the cutter bar. Step 2: Understanding why other options are incorrect
- Ledger Plate: A part of the sickle bar used for cutting, not for transmitting motion. - Wearing Plate: A protective plate that reduces friction between moving parts. - Grass Board: Guides cut grass away from uncut areas, does not transmit motion. Step 3: Conclusion
Since the Pitman is responsible for transmitting reciprocating motion, it is the correct answer. Quick Tip: A Pitman is a key component in a reciprocating mower, converting rotary motion into linear motion for cutting grass efficiently.

Step 1: Understanding Wind Power Formula
The power available from wind energy is given by: \[ P = \frac{1}{2} \rho A v^3 \] where: - \( P \) = Power (W) - \( \rho \) = Air density (kg/m³) - \( A \) = Swept area of turbine blades (m²) - \( v \) = Wind velocity (m/s) Step 2: Eliminating Incorrect Options
- \( P = \frac{1}{2} mv^2 \) is the kinetic energy formula, not for wind power. - \( P = \frac{1}{2} \rho v^3 \) does not account for blade area \( A \), making it incomplete. - \( P = \frac{1}{2} \rho A v^2 \) is incorrect because wind power is proportional to the cube of velocity. Step 3: Conclusion
The correct expression for wind power density includes air density, swept area, and wind velocity cubed. Quick Tip: Wind power is proportional to the cube of wind velocity. Small increases in wind speed significantly increase power generation.

Step 1: Comparing Battery Efficiencies
Rechargeable battery efficiency is measured by energy density, cycle life, and charge efficiency. Step 2: Evaluating Each Option
- Lead Acid: Low energy density and short lifespan. Used in automotive applications. - Nickel-Cadmium: Better than lead-acid but suffers from memory effect, reducing efficiency over time. - Zinc-Air: High energy density but mainly used in non-rechargeable applications. - Lithium-Ion: Highest energy density, long lifespan, and high efficiency. Step 3: Conclusion
Lithium-Ion batteries are the most efficient, making them the best choice for energy storage. Quick Tip: Lithium-Ion batteries offer the highest energy density, faster charging, and longer cycle life, making them ideal for modern energy storage.

Step 1: Understanding Displacement Volume Formula
The displacement volume of a single cylinder in a four-stroke engine is given by: \[ V = \frac{\pi}{4} D^2 L \] where: - \( D \) = Bore diameter (cm) - \( L \) = Stroke length (cm) Step 2: Substituting Given Values
Given \( D = 10 \) cm and \( L = 8 \) cm, we calculate: \[ V = \frac{\pi}{4} \times (10)^2 \times 8 \] \[ V = \frac{3.1416}{4} \times 100 \times 8 \] \[ V = 3.1416 \times 200 \] \[ V = 251.2 \text{ cm}^3 \] Step 3: Eliminating Incorrect Options
- \( 2512 \) cm\(^3\) is 10 times the correct answer. - \( 25.12 \) cm\(^3\) is 1/10th of the correct answer. - \( 25120 \) cm\(^3\) is an incorrect magnitude. Quick Tip: Use the formula \( V = \frac{\pi}{4} D^2 L \) to find displacement volume in a single-cylinder engine.

Step 1: Understanding the Torque-Speed Relationship
Torque (\( T \)) and power (\( P \)) in an engine are related by: \[ P = T \times \omega \] where: - \( P \) = Power (constant in this case) - \( T \) = Torque (Nm) - \( \omega \) = Angular speed (rad/s) Rearranging the equation: \[ T = \frac{P}{\omega} \] Step 2: Effect of Lower Speed on Torque
Since power remains constant, reducing speed (\( \omega \)) will result in an increase in torque (\( T \)). This means: - At lower speed, torque is higher. - At higher speed, torque decreases. Step 3: Eliminating Incorrect Options
- "Very High" and "High" speeds decrease torque, making these incorrect. - "Constant" is incorrect as torque varies with speed. Step 4: Conclusion
For maximum torque at the rear wheels, the engine must run at a lower speed. Quick Tip: To increase torque in a tractor with constant power, reduce the speed. Torque is inversely proportional to speed when power is constant.

Step 1: Purpose of Antifreeze Solutions
Antifreeze solutions lower the freezing point of water, preventing the coolant from freezing in extremely cold conditions. Step 2: Evaluating the Options
- Sodium chloride (Salt): Corrosive and not suitable for engine cooling systems. - Glycerine: Used as an antifreeze agent because it prevents freezing and has low toxicity. - Caustic soda (NaOH): Highly reactive and not used as an antifreeze. - Sodium hydroxide: A strong base that damages cooling systems. Step 3: Conclusion
Glycerine is the best antifreeze agent for tractor cooling systems in cold conditions. Quick Tip: Use glycerine or ethylene glycol as antifreeze in cold conditions to prevent radiator freezing.

Step 1: Understanding Power Requirements in Farm Operations
Power-intensive farm operations require high energy consumption, usually from tractors or tillers. Step 2: Comparing Power Needs
- Plant Protection: Uses low power, mainly for spraying chemicals. - Tilling: Requires maximum power to break soil and prepare the seedbed. - Transplanting: Requires moderate power for planting crops. - Weeding: Needs less power as it mainly involves uprooting unwanted plants. Step 3: Conclusion
Tilling is the most power-intensive farm operation because it requires deep soil penetration. Quick Tip: Tilling consumes the highest power as it involves breaking, cutting, and turning the soil.

Step 1: Understanding the Dashboard Components
The dashboard of a tractor includes essential controls and monitoring devices. Step 2: Identifying the Incorrect Option
- Main Switch: Controls the tractor’s ignition. - Decompression Lever: Helps in starting the engine. - Water Temperature Gauge: Monitors engine temperature. - Steering Wheel: Not part of the dashboard, but an external control component. Step 3: Conclusion
The steering wheel is separate from the dashboard and does not belong in this category. Quick Tip: The dashboard includes gauges and switches, while the steering wheel is part of the control mechanism.

Step 1: Defining Each Term
- Ground Clearance: The height between the lowest point of the tractor and the ground. - Track: Distance between the two wheels on the same axle. - Wheelbase: Distance between the front and rear wheels, measured at ground contact. - Turning Space: The area required for the tractor to make a turn. Step 2: Conclusion
The wheelbase is the correct term for the horizontal distance between the front and rear wheels. Quick Tip: Wheelbase is the key dimension affecting stability, turning radius, and ride quality of a tractor.

Step 1: Understanding Power Transmission in Power Tillers
Power tillers use a transmission system where the main clutch engages the transmission gears. Step 2: Evaluating Power Flow
- Steering Clutch: Engages only for directional control. - Tilling Attachment: Receives power after the centre drive. - Brakes: Not directly connected to the power flow. - Centre Drive: Distributes power to the wheels and tilling mechanism. Step 3: Conclusion
The centre drive is responsible for power distribution in a power tiller. Quick Tip: In a power tiller, the centre drive is the main link between the transmission and working components.

Step 1: Understanding Fluid Behavior
Fluids are classified based on their response to shear stress. Step 2: Identifying the Correct Type
- Ideal Plastic: Does not flow until stress exceeds yield value. - Newtonian Fluid: Follows \(\tau = \mu \frac{du}{dy}\), meaning shear stress is proportional to shear strain rate. - Non-Newtonian Fluid: Shear stress is not proportional to shear rate. - Ideal Fluid: Has no viscosity. Step 3: Conclusion
A Newtonian fluid exhibits a direct proportionality between shear stress and shear rate. Quick Tip: Newtonian fluids have constant viscosity, meaning their flow behavior is predictable.

Step 1: Reynolds Number Formula
Reynolds number (\(Re\)) is given by: \[ Re = \frac{\rho V D}{\mu} \] where: - \( \rho \) = Density of water (\( 1000 \) kg/m³) - \( V \) = Velocity of flow (\( 2 \) m/s) - \( D \) = Diameter of pipe (\( 0.1 \) m) - \( \mu \) = Dynamic viscosity (\( 1 \) centipoise = \( 0.001 \) Pa.s) Step 2: Substituting Values
\[ Re = \frac{1000 \times 2 \times 0.1}{0.001} \] \[ Re = 2.00 \times 10^4 \] Step 3: Conclusion
The calculated Reynolds number matches option (B). Quick Tip: Reynolds number determines flow type: - \( Re < 2000 \) (Laminar), - \( 2000 < Re < 4000 \) (Transition), - \( Re > 4000 \) (Turbulent).

Step 1: Understanding Hydraulic Jump
A hydraulic jump occurs when high-velocity water suddenly slows down, converting kinetic energy into turbulence and heat. Step 2: Purpose of Hydraulic Jump
- Prevents downstream erosion. - Reduces energy to protect structures. - Prevents excessive scour. Step 3: Conclusion
The primary purpose is to dissipate excess energy. Quick Tip: Hydraulic jumps prevent erosion by converting kinetic energy into turbulence.

Step 1: Understanding Soil Shear Strength
Soil transitions through limits based on moisture content: - Liquid limit: Soil behaves like a liquid. - Plastic limit: Soil is moldable but not liquid. - Shrinkage limit: Moisture removal causes minimal volume reduction. Step 2: Shear Strength at Limits
At the liquid limit, shear strength is low (~1.7 kN/m²), making it the correct choice. Quick Tip: At liquid limit, soil behaves like a thick slurry with minimal shear strength.

Step 1: Understanding Flow Nets
Flow nets consist of two types of lines: - Streamlines: Show direction of flow. - Equipotential lines: Represent equal hydraulic potential. Step 2: Importance of Equipotential Lines
Equipotential lines indicate locations of equal pressure head, with uniform head loss between successive lines. Step 3: Conclusion
Since the question asks about uniform head loss, the correct answer is equipotential lines. Quick Tip: In flow nets, streamlines show direction, while equipotential lines indicate pressure distribution.

Step 1: Understanding Tile Drainage Systems
Tile drainage involves subsurface pipes to remove excess water. Common layouts: - Herringbone: V-shaped branches converging into a main drain. - Grid Iron: Parallel drains leading to a collector. - Intercepting Drain: Blocks lateral water movement. Step 2: Why French Drain is Incorrect?
French drains use gravel-filled trenches, not tile pipes. Step 3: Conclusion
French drains are not classified as tile drain layouts. Quick Tip: French drains use gravel, while tile drains use perforated pipes.

Step 1: Understanding Zones in Groundwater
- Water Table: Upper boundary of saturated zone. - Capillary Fringe: Zone where water rises due to capillary action. - Phreatic Zone: Fully saturated region below the water table. - Aquifer: A water-bearing geological formation. Step 2: Identifying the Correct Answer
Since the question asks about water rising above the water table, the capillary fringe is the correct answer. Quick Tip: The capillary fringe occurs due to surface tension, pulling water above the saturated zone.

Step 1: Understanding Darcy’s Law
Darcy’s law describes the flow of a fluid through a porous medium: \[ Q = K A \frac{\Delta H}{L} \] where: - \( K \) = Hydraulic conductivity (m/s) - \( A \) = Cross-sectional area - \( \Delta H \) = Hydraulic head difference - \( L \) = Length of flow path Step 2: Identifying the Correct Option
- Density: Refers to mass per unit volume, not permeability. - Dynamic Viscosity: Relates to fluid resistance, not media property. - Kinematic Viscosity: Ratio of viscosity to density, irrelevant to Darcy’s equation. - Intrinsic Permeability: Represents the ability of a porous medium to transmit fluids, making it the correct answer. Quick Tip: Intrinsic permeability is a property of the medium, while hydraulic conductivity depends on both the medium and fluid properties.

Step 1: Understanding Groundwater Flow
Groundwater moves through different zones, influenced by topography and geology. Step 2: Identifying the Correct Option
- Discharge Area: Where groundwater flows out to the surface (e.g., rivers, springs). - Recharge Area: Where water infiltrates into the groundwater. - Midline: Not a standard groundwater term. - Pathline: Describes the trajectory of a single water particle, not the overall flow. Step 3: Conclusion
Water moving away from the water table flows towards the discharge area. Quick Tip: A discharge area is where groundwater naturally emerges, such as springs, rivers, or wetlands.

Step 1: Understanding Spring Formation
Springs form when groundwater naturally flows to the surface due to geological conditions. Step 2: Identifying the Correct Option
- Fault Spring: Occurs along fault lines. - Sinkhole Spring: Forms in collapsed limestone areas. - Fracture Spring: Occurs due to rock fractures. - Depression Spring: Forms when land dips below the water table, making it the correct answer. Quick Tip: Depression springs occur naturally in low-lying areas where the land surface meets the water table.

Step 1: Understanding Infiltration
Infiltration is the process by which water enters the soil. Step 2: Evaluating the Given Conditions
- If \( i < f_c \), the soil absorbs all rainwater. - If \( i > f_c \), infiltration is limited to \( f_c \), meaning \( f \leq i \). Quick Tip: When rainfall exceeds infiltration capacity, runoff occurs.

Step 1: Understanding Flow Routing Methods
- Lumped routing considers only time variations. - Distributed routing accounts for spatial variations. - Hydraulic routing includes fluid mechanics principles. - Dynamic wave routing considers unsteady flow effects. Step 2: Conclusion
Lumped routing is correct because it calculates flow as a function of time only. Quick Tip: Lumped routing is simple and used when spatial variations are negligible.

Step 1: Understanding Recording vs. Non-Recording Gauges
- Recording Gauges: Automatically record rainfall over time. - Non-Recording Gauges: Require manual measurement. Step 2: Identifying the Incorrect Option
- Tipping Bucket: Records rainfall electronically. - Weighing Bucket: Measures weight over time. - Natural Siphon: Uses a siphon mechanism to record rainfall. - Symon’s Gauge: A non-recording gauge, requiring manual reading. Quick Tip: Recording rain gauges automatically track rainfall, while non-recording gauges require manual measurement.

Step 1: Understanding Well Construction
Wells are classified based on the geological formation in which they are constructed. Step 2: Evaluating Options
- Alluvial Formation: Requires full lining due to unstable loose sediments. - Rocky Formation: Stable, requiring lining only at the top, making it the correct answer. - Fracture Zones: Require complete reinforcement. - Fault Zones: Prone to collapse, requiring full lining. Quick Tip: Wells in rocky formations require minimal lining as the rock structure itself supports the excavation.

Step 1: Understanding Aquifer Testing
Aquifer tests determine hydraulic conductivity, transmissivity, and storage coefficient. Step 2: Evaluating Options
- Time Drawdown Test: Measures water level changes over time, determining aquifer properties. - Step–Drawdown Test: Determines well performance, not aquifer properties. - Step–Injection Test: Used in artificial recharge studies. - Time–Groundwater Level Rise Test: Used in monitoring recharge, not hydraulic parameters. Quick Tip: Time drawdown tests measure aquifer characteristics by analyzing water level decrease over time.

Step 1: Water Power Formula
Water power (\( P \)) is given by: \[ P = \frac{\rho g Q H}{3600} \] where: - \( \rho \) = Density of water (1000 kg/m³) - \( g \) = Acceleration due to gravity (9.81 m/s²) - \( Q \) = Discharge in m³/s - \( H \) = Head in meters Step 2: Converting Given Data
\[ Q = \frac{100000}{3600} = 27.78 \text{ l/s} = 0.02778 \text{ m³/s} \] Step 3: Substituting Values
\[ P = 1000 \times 9.81 \times 0.02778 \times 20 \] \[ P = 5493.6 \text{ W} \] Quick Tip: Convert discharge to m³/s and use \( P = \rho g Q H \) to calculate water power.

Step 1: Understanding Surveying Types
- Plane Surveying: Assumes the earth is flat, used for small areas. - Compass Surveying: Measures bearings using a compass. - Topographic Surveying: Maps land features. - Geodetic Surveying: Accounts for the curvature of the earth, making it correct. Quick Tip: Geodetic surveying considers the earth’s curvature and is used for large-scale mapping.

Step 1: Understanding Survey Methods
- Traverse Surveying: Establishes perimeter lines first. - Plane Table Surveying: Uses direct plotting. - Theodolite Surveying: Measures angles. - Triangulation Surveying: Uses interconnected triangles. Step 2: Conclusion
Traverse surveying is the correct method for plotting details from established perimeter lines. Quick Tip: Traverse surveying is useful for defining boundaries before plotting internal details.

Step 1: Understanding Compass Surveying
- Levelling: Ensures the compass is horizontal. - Focusing: Adjusts telescope clarity. - Centering: Places the compass directly over the station point. - Bearing: Measures direction. Step 2: Conclusion
Centering ensures accurate measurement by aligning the compass with the station. Quick Tip: Centering ensures the compass is positioned exactly over the survey station for accurate readings.

Step 1: Understanding Soil Structure and Water Movement
- Granular: Allows good infiltration. - Columnar: Somewhat restrictive but allows vertical movement. - Platy: Forms horizontal layers, restricting downward water movement. - Prismoidal: Allows moderate water movement. Step 2: Conclusion
Platy structure restricts vertical movement due to its horizontal arrangement. Quick Tip: Platy soil structure impedes water infiltration, leading to waterlogging issues.

Step 1: Bulk Density Formula
\[ \text{Bulk Density} = \frac{\text{Mass of Soil}}{\text{Volume of Soil}} \] Step 2: Substituting Values
\[ \text{Bulk Density} = \frac{20}{15} = 1.2 \text{ g/cm}^3 \] Quick Tip: Bulk density helps assess soil compaction, affecting root penetration and water movement.

Step 1: Understanding Matric Potential
Matric potential measures soil moisture retention. Step 2: Evaluating Soil Water Retention
- Clay soil: Retains maximum water due to fine particles. - Loam soil: Holds moderate water. - Silty soil: Holds more than sand but less than clay. - Sandy soil: Holds the least due to large pores. Step 3: Conclusion
Clay has the highest water retention due to small particle size. Quick Tip: Fine-textured soils like clay retain the most water due to their high surface area.

Step 1: Understanding Erosion Control
Slopes cause excessive runoff and soil erosion. Step 2: Evaluating Options
- Conservation tillage: Reduces soil disturbance but does not shape slopes. - Vegetative hedge: Slows runoff but does not stop erosion completely. - Check dam: Used in streams, not for slope management. - Terraces: Modify slope geometry to reduce erosion. Step 3: Conclusion
Terraces are the best method for managing slopes and reducing erosion. Quick Tip: Terracing slows down water movement, reducing soil erosion on slopes.

Step 1: Understanding Windbreak Effectiveness
Windbreaks reduce wind speed and prevent soil erosion. Step 2: Evaluating Distance Protected
Windbreaks protect an area about 5-10 times their height. Step 3: Conclusion
Thus, the correct answer is 5-10 times tree height. Quick Tip: Windbreaks reduce wind speed over an area 5-10 times their height.

Step 1: Understanding Grassed Waterways
Grassed waterways prevent soil erosion by slowing down runoff. Step 2: Evaluating Cover Conditions
- Good grass cover: Supports moderate flow. - Sod of excellent cover: Supports high-velocity flow (2.0 m/s). - Sparse grass cover: Unsuitable for high velocity. - No vegetation: Leads to erosion. Step 3: Conclusion
For 2.0 m/s flow, sod of excellent cover is needed. Quick Tip: Dense vegetation stabilizes grassed waterways, preventing erosion at high velocities.

Step 1: Understanding Delta
Delta (\(\Delta\)) is the total depth of water required by a crop during its entire growing period: \[ \Delta = \text{Water applied per irrigation} \times \frac{\text{Total Crop Period}}{\text{Irrigation Interval}} \] Step 2: Substituting Values
\[ \Delta = 10 \times \frac{120}{10} = 120 \text{ cm} \] Quick Tip: Delta (\(\Delta\)) is calculated by multiplying the irrigation depth by the number of irrigations.

Step 1: Understanding Furrow Irrigation
Furrow irrigation is effective for row crops but unsuitable for crops that require standing water. Step 2: Evaluating Crops
- Cotton: Suitable for furrow irrigation. - Rice: Requires flooded conditions, making furrow irrigation ineffective. - Sugarcane: Suitable due to its row planting. - Potato: Grown in ridges, making furrow irrigation suitable. Quick Tip: Furrow irrigation is not suitable for rice because it requires continuous standing water.

Step 1: Understanding Fertigation
Fertigation requires water-soluble fertilizers to be delivered efficiently through a drip system. Step 2: Evaluating Options
- Urea: Highly soluble, making it suitable. - Super Phosphate: Poor solubility, leading to clogging, making it unsuitable. - Gypsum: Moderately soluble but usable in small quantities. - Muriate of Potash: Highly soluble, making it suitable. Quick Tip: Use only highly water-soluble fertilizers in drip irrigation to avoid clogging.

Step 1: Understanding Soil Salinity
Soil salinity is measured in terms of electrical conductivity (\(EC\)). Step 2: Evaluating Salinity Levels
- Slightly saline: \(EC < 4\) mmhos/cm - Medium saline: \(4 < EC < 8\) mmhos/cm - Highly saline: \(EC > 18\) mmhos/cm Step 3: Conclusion
Since the given soil has \(EC > 18\) mmhos/cm, it is classified as highly saline. Quick Tip: Soils with \( EC > 18 \) mmhos/cm are considered highly saline and require special irrigation management.

Step 1: Understanding Drainage Coefficient
\[ \text{Drainage Coefficient} = \frac{\text{Discharge (m}^3\text{/s) } \times 86400}{\text{Area (ha) } \times 10000} \] Step 2: Substituting Values
\[ = \frac{2.5 \times 86400}{1500 \times 10000} = 1.44 \text{ cm/day} \] Quick Tip: Drainage coefficient is the volume of water drained per unit area per day, expressed in cm/day.

Step 1: Understanding Drainable Porosity Formula
\[ n_d = \frac{\text{Total Water Discharge} \times \text{Time}}{\text{Total Area} \times \text{Water Table Drop}} \] Step 2: Substituting Values
\[ n_d = \frac{10 \times 86400 \times 3}{10 \times 40 \times 150 \times 0.4 \times 10000} = 7.8% \] Quick Tip: Drainable porosity represents the percentage of soil volume that can be drained under gravity.

Step 1: Understanding Magnetic Separators
Magnetic separators are used to remove ferrous metal contaminants from grains, seeds, and food products. Step 2: Evaluating Options
- Sorting: Separates materials based on size or weight. - Cleaning: Removes unwanted particles, making it the correct choice. - Grading: Classifies based on quality. - Milling: Involves grinding or processing grains. Step 3: Conclusion
Magnetic separators remove metal contaminants, making them useful for cleaning. Quick Tip: Magnetic separators ensure food safety by removing ferrous metal impurities during processing.

Step 1: Understanding Rice Components
- Husk: Outer protective layer, removed during milling. - Bran: Nutrient-rich layer containing fiber, fatty acids, proteins, and vitamins. - Whole kernel: Retains bran but not as nutrient-dense. - Oil: Extracted from bran but lacks fiber. Step 2: Conclusion
Bran contains the highest concentration of nutrients. Quick Tip: Rice bran is the most nutritious part, containing fiber, essential fats, and vitamins.

Step 1: Purpose of Blanching
Blanching involves briefly boiling vegetables before freezing to deactivate spoilage-causing enzymes. Step 2: Evaluating Other Options
- Softening cellulose: Happens but is not the main reason. - Increasing color: Retains color but not primary purpose. - Preventing vitamin C loss: May slow degradation but is not its main function. Step 3: Conclusion
Blanching prevents spoilage by deactivating enzymes. Quick Tip: Blanching stops enzymatic activity, preserving flavor, texture, and color in frozen vegetables.

Step 1: Understanding Pantnagar Process
A specialized milling method developed for pigeon pea (\textit{Cajanus cajan) to enhance dal recovery. Step 2: Evaluating Options
- Black gram, green gram, and green pea are processed differently. - The Pantnagar process was developed to improve milling efficiency of pigeon pea. Quick Tip: The Pantnagar process enhances pigeon pea milling efficiency, reducing processing losses.

Step 1: Understanding Size Reduction Methods
- Hammer mill: Uses impact force. - Disc attrition mill: Uses shear force. - Roller mill: Uses compression force, making it the correct answer. - Knife cutter: Uses cutting force. Step 2: Conclusion
Compression force is applied in roller mills. Quick Tip: Roller mills reduce particle size by applying compression between rollers.

Step 1: Understanding Belt Conveyor Speed
Too slow a speed leads to inefficiency, while too high a speed may cause grain damage. Step 2: Conclusion
2.5-2.8 m/s provides an optimal balance. Quick Tip: For grain transport, 2.5-2.8 m/s prevents damage while maintaining efficiency.

Step 1: Understanding mKisan Portal
mKisan provides SMS-based advisory services to farmers. Step 2: Evaluating Options
- Pusa Krishi: Provides research updates. - mKisan: Correct answer, used for SMS advisories. - Farm Opera: Not a known government initiative. - Kisan Suvidha: Provides weather and market information. Step 3: Conclusion
mKisan is the correct choice for SMS-based advisories. Quick Tip: mKisan enables farmers to receive agricultural advisories via SMS in multiple languages.

Step 1: Understanding FAO's Role
The Food and Agriculture Organization (FAO) collaborates with Google Earth Engine to monitor environmental changes affecting agriculture. Step 2: Identifying the Correct Option
- FAO uses satellite imagery to track locust movements, predicting outbreaks. - H5 Ni Virus and Avian Influenza relate to poultry health, not environmental tracking. - Corona Virus is unrelated to FAO’s work in environmental monitoring. Step 3: Conclusion
Locust control requires large-scale forecasting, making it the correct answer. Quick Tip: FAO uses satellite technology to predict and control locust outbreaks to protect crops.

Step 1: Understanding Crop Growth Monitoring
Remote sensing involves satellite and aerial imagery to assess crop health, moisture levels, and growth patterns. Step 2: Evaluating Other Methods
- GIS: Useful for mapping but does not directly monitor crop growth. - UAVs: Useful but limited in scale compared to satellites. - Modelling: Used for prediction but requires input from remote sensing. Step 3: Conclusion
Remote sensing provides real-time data for large-scale crop monitoring. Quick Tip: Remote sensing allows large-scale, real-time monitoring of crop growth and stress conditions.

Step 1: Understanding Linear Programming
Linear programming is an optimization technique used in resource allocation. Step 2: Identifying the Correct Answer
- George B. Dantzig formulated the simplex method, making him the founder of linear programming. - Richard Bellman is known for dynamic programming. - Narendra Karmarkar developed Karmarkar’s algorithm for optimization. - Bolton is not related to linear programming. Step 3: Conclusion
George B. Dantzig is the correct answer. Quick Tip: Linear programming, developed by Dantzig, optimizes resource allocation in agriculture and industries.

Step 1: Understanding OFD
On-Farm Development (OFD) works include land leveling, drainage, and irrigation infrastructure. Step 2: Identifying the Correct Answer
- CADP (Command Area Development Program) supports OFD work. - HADP, DPAP, and RVP focus on hilly and arid regions. Quick Tip: OFD works under CADP improve irrigation efficiency and agricultural productivity.

Step 1: Understanding Neerkatti
Neerkatti is a traditional water management system where a village official oversees water distribution. Step 2: Identifying the Correct Answer
Since it is a locally managed practice passed through generations, it is classified as traditional. Quick Tip: Neerkatti is a community-managed water distribution system in Tamil Nadu.

Step 1: Understanding Kudimaramathu
Kudimaramathu is a traditional system where local communities contribute labor and resources for water body maintenance. Step 2: Conclusion
Kudimaramathu is the correct answer. Quick Tip: Kudimaramathu is a traditional system of community participation in tank rehabilitation.

Step 1: Understanding Shejpali System
The Shejpali system is a rotational irrigation method used in Maharashtra. Step 2: Evaluating Other States
- Bihar, Karnataka, and Andhra Pradesh have different irrigation systems. Quick Tip: The Shejpali system in Maharashtra regulates water supply through rotational irrigation.