AP EAMCET 2022 previous year paper practice · PYQ-pattern set

AP EAMCET 2022 PYQ-pattern questions

Showing 10 of the 13 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 · Trigonometric Ratios up to Transformations

   If tan A + cot A = 5, then the value of tan²A + cot²A is

   • A. 21
   • B. 23
   • C. 25
   • D. 27
   Answer and solution

   Answer: B

   Squaring: (tan A + cot A)² = 25 ⇒ tan²A + cot²A + 2 = 25. Hence tan²A + cot²A = 23.

2. Q2 · Random Variables and Probability Distributions

   A random variable X has the probability distribution P(X = k) = λk²/3 for k = 1, 2, 3. Then the value of λ is

   • A. 3/7
   • B. 3/14
   • C. 7/14
   • D. 14/3
   Answer and solution

   Answer: B

   For probability distribution, ΣP(X = k) = 1. Hence λ(1² + 2² + 3²)/3 = 1 → λ(1 + 4 + 9)/3 = 1 → λ(14/3) = 1 → λ = 3/14.

3. Q3 · Hyperbola

   The equation of the hyperbola whose foci are (±5, 0) and the length of the transverse axis is 8, is

   • A. x²/16 - y²/9 = 1
   • B. x²/9 - y²/16 = 1
   • C. x²/25 - y²/16 = 1
   • D. x²/4 - y²/21 = 1
   Answer and solution

   Answer: A

   Length of transverse axis = 2a = 8, so a = 4. Foci are at (±ae, 0) = (±5, 0), so ae = 5 giving e = 5/4. Using b² = a²(e²-1) = 16(25/16 - 1) = 9, the equation is x²/16 - y²/9 = 1.

4. Q4 · Integration

   ∫(4x³ + 6x² - 2x + 5) dx equals:

   • A. x⁴ + 2x³ - x² + 5x + C
   • B. x⁴ + 3x³ - x² + 5x + C
   • C. 4x⁴ + 6x³ - 2x² + 5x + C
   • D. x⁴ + 2x³ + x² + 5x + C
   Answer and solution

   Answer: A

   Integrating term-by-term: x⁴ + 2x³ - x² + 5x + C.

5. Q5 · Limits and Continuity

   If f(x) = {(x² - 16)/(x - 4), x ≠ 4; k, x = 4} is continuous at x = 4, then k equals:

   • A. 4
   • B. 8
   • C. 16
   • D. 0
   Answer and solution

   Answer: B

   For continuity at x = 4: lim(x→4) (x² - 16)/(x - 4) = lim(x→4) (x + 4) = 8. Thus k = 8.

6. Q6 · Work, Energy and Power

   A body of mass 4 kg is moving with a velocity of 10 m/s. What is its kinetic energy?

   • A. 100 J
   • B. 200 J
   • C. 400 J
   • D. 800 J
   Answer and solution

   Answer: B

   Kinetic energy KE = ½mv² = ½(4)(10)² = 200 J.

7. Q7 · Motion in a Straight Line

   A particle moves along a straight line such that its velocity v at time t is given by v = 4t - t². At what time will the particle attain maximum velocity?

   • A. 1 s
   • B. 2 s
   • C. 3 s
   • D. 4 s
   Answer and solution

   Answer: B

   For maximum velocity, dv/dt = 0. Here dv/dt = 4 - 2t = 0, giving t = 2 s.

8. Q8 · Mechanical Properties of Fluids

   Two soap bubbles of radii 2 cm and 3 cm coalesce to form a single bubble under isothermal conditions. The radius of the new bubble is

   • A. √13 cm
   • B. 5 cm
   • C. √26 cm
   • D. 6 cm
   Answer and solution

   Answer: A

   Under isothermal conditions, the total volume of air inside the bubbles remains constant (assuming atmospheric pressure dominates). V₁ + V₂ = V. (4/3)πr₁³ + (4/3)πr₂³ = (4/3)πR³, so R³ = r₁³ + r₂³ = 8 + 27 = 35, giving R = ∛35 ≈ 3.27 cm. But this doesn't match options. Alternatively, for soap bubbles with surface tension, total surface energy is conserved: 2×4πr₁²×T + 2×4πr₂²×T = 2×4πR²×T (factor 2 because bubble has two surfaces), so r₁² + r₂² = R², giving R = √(4+9) = √13 cm.

9. Q9 · Magnetism and Matter

   A bar magnet of magnetic moment 5 A·m² is placed perpendicular to a magnetic field of 0.4 T. What is the torque acting on it?

   • A. 1.0 N·m
   • B. 2.0 N·m
   • C. 2.5 N·m
   • D. 3.0 N·m
   Answer and solution

   Answer: B

   Torque on a bar magnet in a magnetic field is τ = MB sin θ, where θ is the angle between the magnetic moment and field. For perpendicular orientation, sin 90° = 1, so τ = 5 × 0.4 × 1 = 2.0 N·m.

10. Q10 · Electrostatic Potential and Capacitance

    Two capacitors of capacitance 4 μF and 6 μF are connected in series across a 100 V supply. What is the potential difference across the 4 μF capacitor?

    • A. 40 V
    • B. 60 V
    • C. 50 V
    • D. 70 V
    Answer and solution

    Answer: B

    In series, charge is same on both capacitors. The voltage divides inversely as capacitance: V₁/V₂ = C₂/C₁. So V₁ = V × C₂/(C₁+C₂) = 100 × 6/(4+6) = 60 V.