NSEA 2024 previous year paper · official paper and PYQ-pattern practice

NSEA 2024 PYQ-pattern questions

Showing 10 of the 15 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 · Kinematics

   A celestial body is observed to move with position given by x = 4t² + 3t + 2 meters, where t is in seconds. What is the instantaneous velocity at t = 3 seconds?

   • A. 21 m/s
   • B. 24 m/s
   • C. 27 m/s
   • D. 30 m/s
   Answer and solution

   Answer: C

   Velocity v = dx/dt = 8t + 3. At t = 3 s, v = 8(3) + 3 = 24 + 3 = 27 m/s.

2. Q2 · Mechanical Properties of Solids and Fluids

   The atmospheric pressure on the surface of Mars is approximately 600 Pa. What force does this pressure exert on a rover's solar panel of area 2 m²?

   • A. 300 N
   • B. 600 N
   • C. 1200 N
   • D. 2400 N
   Answer and solution

   Answer: C

   Force F = Pressure × Area = 600 Pa × 2 m² = 1200 N.

3. Q3 · Trigonometry

   If sin θ + cos θ = √2, then the value of sin 2θ is:

   • A. 0
   • B. 1
   • C. √2
   • D. 1/2
   Answer and solution

   Answer: B

   Squaring both sides: sin²θ + cos²θ + 2sinθcosθ = 2. Since sin²θ + cos²θ = 1, we get 2sinθcosθ = 1, hence sin 2θ = 1.

4. Q4 · Binomial Theorem

   In the expansion of (2x + 1/x)⁶, the coefficient of the constant term (x⁰) is:

   • A. 20
   • B. 60
   • C. 160
   • D. 240
   Answer and solution

   Answer: C

   The general term is ⁶Cᵣ(2x)⁶⁻ʳ(1/x)ʳ = ⁶Cᵣ 2⁶⁻ʳ x⁶⁻²ʳ. For x⁰, 6-2r=0 so r=3. Coefficient = ⁶C₃ × 2³ = 20 × 8 = 160.

5. Q5 · Mathematical Reasoning

   If the parallax angle of a star doubles, by what factor does its measured distance change?

   • A. Doubles
   • B. Halves
   • C. Quadruples
   • D. Remains the same
   Answer and solution

   Answer: B

   Distance d is inversely proportional to parallax p (d ∝ 1/p). If p doubles, d becomes half its original value.

6. Q6 · Solar System

   The synodic period of Mars is approximately 780 days. Given Earth's orbital period is 365 days, what is the sidereal period of Mars in years?

   • A. 1.88 years
   • B. 2.14 years
   • C. 1.52 years
   • D. 3.14 years
   Answer and solution

   Answer: A

   Using 1/Psyn = |1/PE - 1/PM|, we have 1/780 = 1/365 - 1/PM. Solving: 1/PM = 1/365 - 1/780 ≈ 0.00274 - 0.00128 = 0.00146, so PM ≈ 685 days ≈ 1.88 years.

7. Q7 · Stellar Astronomy Basics

   A main sequence star has a surface temperature of 12,000 K and a luminosity 400 times that of the Sun. If the Sun's surface temperature is 5,800 K, what is the approximate radius of this star in solar radii?

   • A. 3.2 R☉
   • B. 4.5 R☉
   • C. 6.8 R☉
   • D. 9.1 R☉
   Answer and solution

   Answer: B

   Using Stefan-Boltzmann law L ∝ R²T⁴, we have 400 = (R/R☉)² × (12000/5800)⁴. Solving: (R/R☉)² = 400/(12000/5800)⁴ = 400/20.25 ≈ 19.75, giving R ≈ 4.4 solar radii.

8. Q8 · History of Astronomy

   Which astronomer first provided observational evidence that the Milky Way is not the only galaxy in the universe by resolving Cepheid variables in the Andromeda Nebula?

   • A. William Herschel
   • B. Edwin Hubble
   • C. Henrietta Leavitt
   • D. Harlow Shapley
   Answer and solution

   Answer: B

   Edwin Hubble used the 100-inch Hooker telescope in 1924 to resolve Cepheid variables in M31 (Andromeda), proving it was a separate galaxy beyond the Milky Way.

9. Q9 · Gravitation

   A satellite orbits a planet of mass M at a distance R from its center with orbital period T. If the planet's mass suddenly doubles while the satellite's orbit remains unchanged, what would be the new orbital period?

   • A. T/√2
   • B. T/2
   • C. T
   • D. √2 T
   Answer and solution

   Answer: A

   From Kepler's third law, T² ∝ R³/M. If M doubles at constant R, then T_new = T/√2.

10. Q10 · Thermal Physics and Thermodynamics

    A star radiates energy at a rate of 4.0 × 10²⁶ W and has a surface temperature of 6000 K. If its temperature increases to 12000 K while its radius remains constant, what will be the new power radiated?

    • A. 6.4 × 10²⁷ W
    • B. 3.2 × 10²⁷ W
    • C. 1.6 × 10²⁷ W
    • D. 8.0 × 10²⁶ W
    Answer and solution

    Answer: A

    From Stefan-Boltzmann law, P ∝ T⁴. When T doubles, P increases by factor of 2⁴ = 16. New power = 16 × 4.0 × 10²⁶ = 6.4 × 10²⁷ W.