AP EAMCET 2023 previous year paper practice · PYQ-pattern set

AP EAMCET 2023 PYQ-pattern questions

Showing 10 of the 14 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · Partial Fractions

   The partial fraction decomposition of (5x - 2)/(x² - x - 2) is of the form A/(x - 2) + B/(x + 1). Then A + B equals

   • A. 3
   • B. 4
   • C. 5
   • D. 6
   Answer and solution

   Answer: C

   Factoring: x² - x - 2 = (x - 2)(x + 1). So 5x - 2 = A(x + 1) + B(x - 2). Setting x = 2 gives 8 = 3A, so A = 8/3. Setting x = -1 gives -7 = -3B, so B = 7/3. Hence A + B = 15/3 = 5.

2. Q2 · Properties of Triangles

   In a triangle ABC, if a = 7, b = 8, c = 9, then cos A equals

   • A. 1/2
   • B. 2/3
   • C. 3/4
   • D. 5/6
   Answer and solution

   Answer: B

   Using cosine rule: cos A = (b² + c² - a²)/(2bc) = (64 + 81 - 49)/(2·8·9) = 96/144 = 2/3.

3. Q3 · Pair of Straight Lines

   The equation ax² + 2hxy + by² = 0 represents a pair of perpendicular lines if

   • A. a + b = 0
   • B. a - b = 0
   • C. h² = ab
   • D. h = 0
   Answer and solution

   Answer: A

   For a pair of perpendicular lines in the form ax² + 2hxy + by² = 0, the condition is coefficient of x² + coefficient of y² = 0, i.e., a + b = 0.

4. Q4 · Ellipse

   If the eccentricity of an ellipse is 3/5 and the distance between its foci is 6, then the length of the semi-major axis is

   • A. 4
   • B. 5
   • C. 6
   • D. 10
   Answer and solution

   Answer: B

   Distance between foci is 2ae = 6, and e = 3/5. Therefore, 2a(3/5) = 6, which gives a = 5.

5. Q5 · Applications of Derivatives

   The function f(x) = 2x³ - 9x² + 12x + 5 is strictly increasing in the interval:

   • A. (-∞, 1) ∪ (2, ∞)
   • B. (1, 2)
   • C. (-∞, ∞)
   • D. (0, 3)
   Answer and solution

   Answer: A

   f'(x) = 6x² - 18x + 12 = 6(x² - 3x + 2) = 6(x - 1)(x - 2). f'(x) > 0 when x < 1 or x > 2, so f is increasing on (-∞, 1) ∪ (2, ∞).

6. Q6 · Units and Measurements

   A physical quantity P is given by P = (A³B²)/(C√D). If the percentage errors in A, B, C, and D are 2%, 1%, 3%, and 4% respectively, then the maximum percentage error in P is:

   • A. 10%
   • B. 11%
   • C. 13%
   • D. 15%
   Answer and solution

   Answer: C

   Maximum error in P = 3(2%) + 2(1%) + 3% + (1/2)(4%) = 6% + 2% + 3% + 2% = 13%.

7. Q7 · Laws of Motion

   A block of mass 8 kg is placed on a horizontal surface. A horizontal force of 32 N is applied on it. If the coefficient of kinetic friction is 0.25, what is the acceleration of the block? (Take g = 10 m/s²)

   • A. 1.5 m/s²
   • B. 2.0 m/s²
   • C. 2.5 m/s²
   • D. 4.0 m/s²
   Answer and solution

   Answer: A

   Friction force f = μN = 0.25×8×10 = 20 N. Net force = 32 - 20 = 12 N. Acceleration a = F/m = 12/8 = 1.5 m/s².

8. Q8 · Thermodynamics

   An ideal gas undergoes a cyclic process ABCA where AB is isothermal expansion, BC is adiabatic expansion, and CA is constant volume process. If the work done by the gas in the isothermal process is 600 J and work done on the gas in the adiabatic process is 200 J, then the heat rejected during the constant volume process is

   • A. 200 J
   • B. 400 J
   • C. 600 J
   • D. 800 J
   Answer and solution

   Answer: A

   AB isothermal: Q_AB=W_AB=600 J, ΔU=0. BC adiabatic: Q=0, work by gas=-200 J so ΔU_BC=+200 J. Over the cycle ΔU=0, so ΔU_CA=-200 J. At constant volume W=0, thus Q_CA=ΔU_CA=-200 J, meaning 200 J is rejected. Verified via net work = Q_net (400 J).

9. Q9 · Moving Charges and Magnetism

   A proton enters a uniform magnetic field of 0.5 T with a velocity of 2 × 10⁶ m/s perpendicular to the field. What is the radius of the circular path? (Mass of proton = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C)

   • A. 4.2 cm
   • B. 8.4 cm
   • C. 2.1 cm
   • D. 1.05 cm
   Answer and solution

   Answer: A

   The radius of circular path is r = mv/(qB). Substituting values: r = (1.67×10⁻²⁷ × 2×10⁶)/(1.6×10⁻¹⁹ × 0.5) = 4.2×10⁻² m = 4.2 cm.

10. Q10 · Dual Nature of Radiation and Matter

    The work function of a metal is 2.5 eV. What is the maximum wavelength of light that can cause photoelectric emission from this metal? (Take hc = 1240 eV·nm)

    • A. 496 nm
    • B. 620 nm
    • C. 372 nm
    • D. 744 nm
    Answer and solution

    Answer: A

    For photoelectric emission to occur, the photon energy must at least equal the work function: hf = φ₀. Thus, λ_max = hc/φ₀ = 1240/2.5 = 496 nm.