AP EAPCET 2026 May 14 Shift 1 Engineering Question Paper- Download Solution PDF

Concept:
For a system of linear equations to have infinitely many solutions, the equations must be linearly dependent. This typically means that one equation can be expressed as a linear combination of the others, or the determinant of the coefficient matrix \( (\Delta) \) and the determinants \( \Delta_x, \Delta_y, \Delta_z \) must all be zero.


Step 1: Comparing Equation 1 and Equation 3

Equation 1: \( 2x + py + 6z = 8 \)

Equation 3: \( x + y + 3z = 4 \)

Notice that if we multiply Equation 3 by 2, we get: \[ 2(x + y + 3z) = 2(4) \quad \Rightarrow \quad 2x + 2y + 6z = 8 \]


Step 2: Equating the coefficients to find \( p \).

For the system to have infinitely many solutions, Equation 1 and the modified Equation 3 must represent the same plane (or Equation 1 must be a multiple of Equation 3).
Comparing \( 2x + py + 6z = 8 \) with \( 2x + 2y + 6z = 8 \):
Matching the coefficients of \( y \): \[ p = 2 \]


Step 3: Verification with Equation 2 (Optional for \( p \)).

Although not required to find \( p \), the value of \( q \) can be found by ensuring Equation 2 is also consistent. Subtracting Eq(3) from Eq(2): \[ (x + 2y + qz) - (x + y + 3z) = 5 - 4 \] \[ y + (q - 3)z = 1 \]
Since we found from Eq(1) and Eq(3) that \( y = 1 - 3z + const \), the system remains consistent. Specifically, \( p=2 \) satisfies the primary dependency. Quick Tip: Always look for proportional equations first! If the coefficients of \( x, z \) and the constant term in one equation are a direct multiple of another, the remaining coefficient must follow that same multiple for infinite solutions to be possible.

Concept:
A system of non-homogeneous linear equations is inconsistent if the determinant of the coefficient matrix (\( \Delta \)) is zero, but at least one of the determinants \( \Delta_x, \Delta_y, \) or \( \Delta_z \) is non-zero. For this problem, we first find the values of \( \lambda \) that make \( \Delta = 0 \).

Step 1: Find the values of \( \lambda \) for which the determinant \( \Delta = 0 \).

The coefficient matrix is: \[ \Delta = \begin{vmatrix} 1 & \lambda & -2
1 & -1 & \lambda
1 & -2 & 3 \end{vmatrix} \]
Expanding along the first row: \[ \Delta = 1(-3 + 2\lambda) - \lambda(3 - \lambda) - 2(-2 + 1) = 0 \] \[ -3 + 2\lambda - 3\lambda + \lambda^2 + 2 = 0 \] \[ \lambda^2 - \lambda - 1 = 0 \]

Step 2: Apply the properties of quadratic equations.

The values \( \lambda_1 \) and \( \lambda_2 \) are the roots of the quadratic equation \( \lambda^2 - \lambda - 1 = 0 \).
For any quadratic equation \( a\lambda^2 + b\lambda + c = 0 \), the sum of the roots is given by \( -\frac{b}{a} \).
Here, \( a = 1 \) and \( b = -1 \). \[ \lambda_1 + \lambda_2 = -\frac{-1}{1} = 1 \]

Step 3: Verify inconsistency.

For inconsistency, we must ensure \( \Delta_x, \Delta_y, or \Delta_z \neq 0 \) for these roots. \[ \Delta_x = \begin{vmatrix} 1 & \lambda & -2
2 & -1 & \lambda
3 & -2 & 3 \end{vmatrix} = 1(-3 + 2\lambda) - \lambda(6 - 3\lambda) - 2(-4 + 3) \] \[ \Delta_x = -3 + 2\lambda - 6\lambda + 3\lambda^2 + 2 = 3\lambda^2 - 4\lambda - 1 \]
Substituting \( \lambda^2 = \lambda + 1 \) from our previous equation: \[ \Delta_x = 3(\lambda + 1) - 4\lambda - 1 = 3\lambda + 3 - 4\lambda - 1 = 2 - \lambda \]
Since the roots of \( \lambda^2 - \lambda - 1 = 0 \) are \( \frac{1 \pm \sqrt{5}}{2} \), \( \Delta_x \neq 0 \). Thus, the system is inconsistent for these values. Quick Tip: When a question asks for the sum or product of parameters that make a system inconsistent, don't waste time solving for the exact roots. Form the quadratic equation for \( \Delta = 0 \) and use Vieta's formulas (\( Sum = -b/a \)) to get the answer instantly.

Concept:
The Greatest Integer Function \( [f(x)] \) changes its value at points where \( f(x) \) becomes an integer. To integrate such a function, we must split the interval of integration at these critical points and evaluate the constant value of the function within each sub-interval.

Step 1: Identify the critical points for \( [x^2] \) in the interval \( [1, 2] \).

As \( x \) varies from \( 1 \) to \( 2 \), \( x^2 \) varies from \( 1^2 = 1 \) to \( 2^2 = 4 \).
The integers between \( 1 \) and \( 4 \) are \( 1, 2, 3, \) and \( 4 \).
The function \( [x^2] \) will change at:

\( x^2 = 1 \Rightarrow x = 1 \)
\( x^2 = 2 \Rightarrow x = \sqrt{2} \)
\( x^2 = 3 \Rightarrow x = \sqrt{3} \)
\( x^2 = 4 \Rightarrow x = 2 \)


Step 2: Split the integral based on these intervals.

We can rewrite the integral as: \[ I = \int_{1}^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^{2} [x^2] dx \]

Step 3: Evaluate the constant values of \( [x^2] \) in each sub-interval.


For \( 1 \leq x < \sqrt{2} \), \( 1 \leq x^2 < 2 \), so \( [x^2] = 1 \).
For \( \sqrt{2} \leq x < \sqrt{3} \), \( 2 \leq x^2 < 3 \), so \( [x^2] = 2 \).
For \( \sqrt{3} \leq x < 2 \), \( 3 \leq x^2 < 4 \), so \( [x^2] = 3 \).


Step 4: Perform the integration.
\[ I = \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx \] \[ I = [x]_{1}^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^{2} \] \[ I = (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) \] \[ I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} \] \[ I = 5 - \sqrt{2} - \sqrt{3} \] Quick Tip: For definite integrals involving \( [x^2] \), the critical points are always the square roots of integers. The result of such an integral can be visualized as the sum of areas of rectangles with heights equal to the integer values of the function.

Concept:
To evaluate an integral involving absolute values, we must resolve the modulus signs by identifying where the expression inside the modulus changes sign. This requires splitting the interval of integration at the "critical points" where the internal expressions equal zero.

Step 1: Resolve the inner modulus \( |x - 2| \) based on the interval \( [0, 4] \).

The function is \( f(x) = ||x - 2| - x| \). The inner modulus \( |x - 2| \) behaves differently at \( x = 2 \):

For \( 0 \leq x < 2 \), \( |x - 2| = -(x - 2) = 2 - x \).
For \( 2 \leq x \leq 4 \), \( |x - 2| = x - 2 \).


Step 2: Simplify the function in sub-intervals.

Case 1: \( 0 \leq x < 2 \) \[ f(x) = |(2 - x) - x| = |2 - 2x| = 2|1 - x| \]
This further splits at \( x = 1 \):

For \( 0 \leq x < 1 \), \( f(x) = 2(1 - x) = 2 - 2x \).
For \( 1 \leq x < 2 \), \( f(x) = 2(x - 1) = 2x - 2 \).


Case 2: \( 2 \leq x \leq 4 \) \[ f(x) = |(x - 2) - x| = |-2| = 2 \]

Step 3: Evaluate the definite integral by splitting it.
\[ I = \int_{0}^{1} (2 - 2x) dx + \int_{1}^{2} (2x - 2) dx + \int_{2}^{4} 2 dx \] \[ I = \left[ 2x - x^2 \right]_{0}^{1} + \left[ x^2 - 2x \right]_{1}^{2} + \left[ 2x \right]_{2}^{4} \] \[ I = (2 - 1) + (4 - 4 - (1 - 2)) + (8 - 4) \] \[ I = 1 + (0 - (-1)) + 4 \] \[ I = 1 + 1 + 4 = 6 \] Quick Tip: For nested modulus functions, always work from the "inside out." Once the inner modulus is simplified for a specific range, the resulting expression is much easier to manage.

Concept:

Based on the problem, we utilize the specific property of definite integrals:

\[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx \]

Concept:
The problem relates to the elastic properties of matter. The Bulk modulus (\(B\)) is defined as the ratio of volumetric stress (pressure) to volumetric strain. It is a measure of how resistant a substance is to compression. \[ B = \frac{\Delta P}{-\frac{\Delta V}{V}} \]
Where:

\(\Delta P\) is the change in pressure.
\(\frac{\Delta V}{V}\) is the volumetric strain (fractional change in volume).


Step 1: Identify the given values.

From the question:

Pressure (\(\Delta P\)) = \(2 \times 10^6\) Pa
Percentage decrease in volume = \(0.1%\)

Calculating volumetric strain (\(\frac{\Delta V}{V}\)): \[ \frac{\Delta V}{V} = \frac{0.1}{100} = 10^{-3} \]

Step 2: Calculate the Bulk Modulus.

Using the formula: \[ B = \frac{\Delta P}{\frac{\Delta V}{V}} \] \[ B = \frac{2 \times 10^6}{10^{-3}} \] \[ B = 2 \times 10^6 \times 10^3 \] \[ B = 2 \times 10^9 Pa \] Quick Tip: When the volume decrease is given as a percentage, the strain is simply \(\frac{% change}{100}\). If \(\Delta V/V\) is \(10^{-n}\), the exponent of \(10\) in the result will increase by \(n\).

Concept:
The problem is based on Hooke's Law, which states that within the elastic limit, the extension (\(\Delta L\)) of a material is directly proportional to the force (\(F\)) applied to it. \[ F \propto \Delta L \quad \Rightarrow \quad F = k\Delta L \]
where \(k\) is the force constant of the wire.

Step 1: Establish the relationship between Force and Extension.

Since the same wire is used, the proportionality remains constant: \[ \frac{F_1}{\Delta L_1} = \frac{F_2}{\Delta L_2} \]

Step 2: Substitute the given values.

From the question:

Initial force (\(F_1\)) = \(100\) N
Initial extension (\(\Delta L_1\)) = \(1\) mm
Final force (\(F_2\)) = \(300\) N

Plugging these into the ratio: \[ \frac{100}{1} = \frac{300}{\Delta L_2} \]

Step 3: Solve for the new extension.
\[ \Delta L_2 = \frac{300 \times 1}{100} \] \[ \Delta L_2 = 3 mm \] Quick Tip: For linear relationships like Hooke's Law, if the force becomes 'n' times, the extension also becomes 'n' times. Here, since \(300\) N is \(3\) times \(100\) N, the extension is simply \(3 \times 1\) mm = \(3\) mm.

Concept:
For a fluid flowing through a horizontal pipe, Bernoulli's principle states that the sum of pressure energy and kinetic energy per unit volume remains constant. \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
Where:

\(P\) is the pressure.
\(\rho\) is the density of the fluid.
\(v\) is the velocity of the fluid.


Step 1: Identify the given values.

From the question:

Initial pressure (\(P_1\)) = \(2000\) Pa
Initial speed (\(v_1\)) = \(2\) m/s
Final speed (\(v_2\)) = \(4\) m/s
Density of water (\(\rho\)) = \(1000\) kg/m\(^3\)


Step 2: Apply Bernoulli's equation.

Rearrange the formula to solve for \(P_2\): \[ P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) \]
Substitute the values: \[ P_2 = 2000 + \frac{1}{2}(1000) (2^2 - 4^2) \] \[ P_2 = 2000 + 500 (4 - 16) \]

Step 3: Calculate the final pressure.
\[ P_2 = 2000 + 500 (-12) \] \[ P_2 = 2000 - 6000 \] \[ P_2 = -4000 Pa \] Quick Tip: In a horizontal pipe, as the speed of the fluid increases, the pressure must decrease to satisfy Bernoulli's principle. Since the speed doubled, the kinetic energy term increased fourfold, leading to a significant drop in pressure.

Concept:
For an incompressible fluid flowing through a pipe of varying cross-section, the principle of continuity applies. It states that the mass flow rate remains constant throughout the pipe, which leads to the equation: \[ A_1 v_1 = A_2 v_2 \]
Where:

\(A\) is the cross-sectional area.
\(v\) is the velocity of the fluid at that section.


Step 1: Identify the given values.

From the provided problem data:

Area of wider section (\(A_1\)) = \(4\) cm\(^2\)
Velocity in wider section (\(v_1\)) = \(2\) m/s
Area of narrower section (\(A_2\)) = \(1\) cm\(^2\)


Step 2: Apply the Equation of Continuity.

Substitute the known values into the formula: \[ (4 cm^2) \times (2 m/s) = (1 cm^2) \times v_2 \] \[ 8 cm^2 \cdot m/s = 1 cm^2 \cdot v_2 \]

Step 3: Solve for the unknown velocity.
\[ v_2 = \frac{8}{1} m/s \] \[ v_2 = 8 m/s \] Quick Tip: The velocity of a fluid is inversely proportional to the cross-sectional area (\(v \propto 1/A\)). If the area decreases by a factor of 4 (from 4 to 1), the velocity must increase by a factor of 4 (\(2 \times 4 = 8\)).

Concept:
The excess pressure inside a soap bubble is due to surface tension. Unlike a liquid drop which has only one free surface, a soap bubble has two free surfaces (inner and outer). Therefore, the formula for excess pressure (\(\Delta P\)) is: \[ \Delta P = \frac{4T}{R} \]
Where:

\(T\) is the surface tension of the soap solution.
\(R\) is the radius of the bubble.


Step 1: Identify and convert the given values.

From the question:

Surface tension (\(T\)) = \(0.03\) N/m
Radius (\(R\)) = \(1\) cm = \(0.01\) m (converting to SI units)


Step 2: Apply the excess pressure formula.

Substitute the values into the equation: \[ \Delta P = \frac{4 \times 0.03}{0.01} \] \[ \Delta P = \frac{0.12}{0.01} \]

Step 3: Calculate the final result.
\[ \Delta P = 12 Pa \] Quick Tip: Always remember the difference between a liquid drop (\(\frac{2T}{R}\)) and a soap bubble (\(\frac{4T}{R}\)). The "double surface" of the soap bubble doubles the excess pressure compared to a drop of the same size and surface tension.

Concept:
The relationship between molecular formula and empirical formula is given by: \[ Molecular Formula = n \times (Empirical Formula) \]
To find the multiplier \(n\), we use the formula: \[ n = \frac{Molecular Mass}{Empirical Formula Mass} \]
Additionally, the molecular mass of a compound is twice its vapour density: \[ Molecular Mass = 2 \times Vapour Density \]

Step 1: Calculate the Empirical Formula Mass.

The empirical formula is CH\(_2\)O.

Mass of C = \(1 \times 12 = 12\)
Mass of H = \(2 \times 1 = 2\)
Mass of O = \(1 \times 16 = 16\)
Empirical Formula Mass = \(12 + 2 + 16 = 30\)


Step 2: Calculate the Molecular Mass.

Given vapour density is \(45\). \[ Molecular Mass = 2 \times 45 = 90 \]

Step 3: Determine the value of \(n\) and the Molecular Formula.
\[ n = \frac{90}{30} = 3 \]
Now, multiply the empirical formula by \(n\): \[ Molecular Formula = 3 \times (CH_2O) = C_3H_6O_3 \] Quick Tip: In competitive exams, if you've already calculated the molecular mass (\(90\)), you can quickly check the options to see which one sums to that mass. For (D): \((3 \times 12) + (6 \times 1) + (3 \times 16) = 36 + 6 + 48 = 90\).

Concept:
For an isothermal, reversible expansion of an ideal gas, the work done (\(W\)) is calculated using the formula: \[ W = -2.303 \, nRT \log \left( \frac{P_1}{P_2} \right) \]
Where:

\(n\) is the number of moles of the gas.
\(R\) is the gas constant (\(8.314 \, J/mol\[\cdot K\]\)).
\(T\) is the absolute temperature.
\(P_1\) and \(P_2\) are the initial and final pressures.


Step 1: Calculate the number of moles of Hydrogen (\(H_2\)).

Given mass = \(10 \, g\).
Molar mass of \(H_2 = 2 \, g/mol\). \[ n = \frac{10}{2} = 5 \, moles \]

Step 2: Substitute values into the work formula.

Given \(T = 273 \, K\), \(P_1 = 10 \, atm\), and \(P_2 = 1 \, atm\). \[ W = -2.303 \times 5 \times 8.314 \times 273 \times \log \left( \frac{10}{1} \right) \] \[ W = -2.303 \times 5 \times 8.314 \times 273 \times 1 \]

Step 3: Calculate the final value in kilojoules.
\[ W \approx -26145 \, J \]
Converting to kilojoules: \[ W \approx -26.145 \, kJ \]
Based on the options, the closest value is \(-26.09 \, kj\). Quick Tip: Work done by the system (expansion) is always negative according to IUPAC convention. Since pressure is reduced (expansion), you can immediately eliminate options (b) and (d).

Concept:
The relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta E\) or \(\Delta U\)) for a gaseous reaction is given by the formula: \[ \Delta H = \Delta E + \Delta n_g RT \]
Where:

\(\Delta n_g\) is the change in the number of moles of gaseous products and reactants.
\(R\) is the universal gas constant.
\(T\) is the absolute temperature.


Step 1: Calculate \(\Delta n_g\) for the given reaction.

The chemical equation is: \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

Moles of gaseous products (\(n_p\)) = \(2\)
Moles of gaseous reactants (\(n_r\)) = \(1 (for N_2) + 3 (for H_2) = 4\)
\[ \Delta n_g = n_p - n_r = 2 - 4 = -2 \]

Step 2: Substitute \(\Delta n_g\) into the relationship formula.

Using the formula \(\Delta H = \Delta E + \Delta n_g RT\): \[ \Delta H = \Delta E + (-2)RT \] \[ \Delta H = \Delta E - 2RT \] Quick Tip: To find \(\Delta n_g\) quickly, only sum the coefficients of substances in the gaseous state. If \(\Delta n_g\) is negative, \(\Delta H < \Delta E\); if positive, \(\Delta H > \Delta E\); and if zero, \(\Delta H = \Delta E\).

Concept:
Entropy (\(S\)) is a measure of the degree of randomness or disorder in a system. A negative entropy change (\(\Delta S < 0\)) indicates that the system is becoming more ordered.

Gas > Liquid > Solid (Order increases \(\rightarrow \Delta S\) is negative)
Solid > Liquid > Gas (Disorder increases \(\rightarrow \Delta S\) is positive)


Step 1: Analyze each process.


I) Sublimation of dry ice: Solid CO\(_2\) turns into gaseous CO\(_2\). Disorder increases, so \(\Delta S\) is positive.
II) Freezing of water: Liquid water turns into solid ice. The molecules become fixed in a lattice, increasing order. Thus, \(\Delta S\) is negative.
III) Crystallisation of the dissolved substance: Solute particles move from a random state in a solution to a highly ordered crystalline solid. Thus, \(\Delta S\) is negative.
IV) Burning of rocket fuel: Chemical combustion typically releases a large volume of gases from solid or liquid propellants. Disorder increases significantly, so \(\Delta S\) is positive.


Step 2: Conclusion.

Based on the analysis , processes II and III result in a decrease in randomness. Quick Tip: To quickly determine the sign of \(\Delta S\), look at the change in physical state. Phase changes towards "Solid" (Freezing, Condensation, Deposition) usually have \(-\Delta S\), while changes towards "Gas" (Melting, Evaporation, Sublimation) have \(+\Delta S\).

Concept:
The spontaneity of a chemical reaction is determined by the Gibbs free energy change (\(\Delta G\)). A reaction is spontaneous only when \(\Delta G < 0\). The relationship is given by the Gibbs-Helmholtz equation: \[ \Delta G = \Delta H - T\Delta S \]
Where:

\(\Delta H\) is the change in enthalpy.
\(T\) is the absolute temperature (always positive in Kelvin).
\(\Delta S\) is the change in entropy.


Step 1: Determine the conditions for spontaneity at all temperatures.

To ensure \(\Delta G\) is always negative regardless of the value of \(T\):

The enthalpy term (\(\Delta H\)) should be negative (exothermic).
The term \(-T\Delta S\) should also be negative. Since \(T\) is always positive, \(\Delta S\) must be positive.


Step 2: Evaluate the mathematical result.

If \(\Delta H\) is negative (\(-\)) and \(\Delta S\) is positive (\(+\)), then the equation becomes: \[ \Delta G = (-) - T(+) \] \[ \Delta G = (-) - (+) = Always Negative \]
Thus, based on the problem, the reaction will be spontaneous at any temperature. Quick Tip: Think of it this way: Nature favors lower energy (\(-\Delta H\)) and higher disorder (\(+\Delta S\)). When both conditions are met, the reaction has no choice but to be spontaneous!