NSEA 2023 previous year paper · official paper and PYQ-pattern practice

NSEA 2023 PYQ-pattern questions

Showing 10 of the 18 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 · Gravitation

   A satellite orbits Earth at a height equal to Earth's radius R above the surface. If the orbital period at Earth's surface (neglecting atmosphere) would be T₀, what is the orbital period at this height?

   • A. 2√2 T₀
   • B. 2T₀
   • C. 2√2 T₀/π
   • D. 4T₀/π
   Answer and solution

   Answer: A

   Using Kepler's third law T² ∝ r³, at height R, orbital radius is 2R. So T²/T₀² = (2R)³/R³ = 8, giving T = 2√2 T₀.

2. Q2 · Thermal Physics and Thermodynamics

   An ideal gas in a star's interior undergoes an adiabatic expansion where its volume increases by a factor of 8. If γ = 5/3, by what factor does its pressure decrease?

   • A. 16
   • B. 24
   • C. 32
   • D. 64
   Answer and solution

   Answer: C

   For adiabatic process PVᵞ = constant, so P₁V₁ᵞ = P₂V₂ᵞ. P₂/P₁ = (V₁/V₂)ᵞ = (1/8)^(5/3) = 1/(2³)^(5/3) = 1/2⁵ = 1/32.

3. Q3 · Sets, Relations and Functions

   If f(x) = 2x + 3 and g(x) = x² - 1, what is the value of (f ∘ g)(2)?

   • A. 7
   • B. 9
   • C. 11
   • D. 13
   Answer and solution

   Answer: B

   (f ∘ g)(2) = f(g(2)) = f(2² - 1) = f(3) = 2(3) + 3 = 9.

4. Q4 · Vectors

   Two vectors ā = 2î + 3ĵ - k̂ and b̄ = î - 2ĵ + 4k̂ are given. The dot product ā · b̄ equals:

   • A. -8
   • B. 0
   • C. 2
   • D. 8
   Answer and solution

   Answer: A

   ā · b̄ = (2)(1) + (3)(-2) + (-1)(4) = 2 - 6 - 4 = -8.

5. Q5 · Three Dimensional Geometry

   A line passes through the point (1, 2, 3) and is parallel to the vector 2î + 3ĵ + 6k̂. The symmetric form of the line is:

   • A. (x-1)/2 = (y-2)/3 = (z-3)/6
   • B. (x+1)/2 = (y+2)/3 = (z+3)/6
   • C. x/2 = y/3 = z/6
   • D. (x-2)/1 = (y-3)/2 = (z-6)/3
   Answer and solution

   Answer: A

   The symmetric form through (x₁,y₁,z₁) with direction ratios (a,b,c) is (x-x₁)/a = (y-y₁)/b = (z-z₁)/c. Hence (x-1)/2 = (y-2)/3 = (z-3)/6.

6. Q6 · Celestial Mechanics

   A satellite orbits Earth at an altitude equal to Earth's radius (RE = 6400 km). What is the orbital period of the satellite? (Take g = 10 m/s²)

   • A. 4.0 hours
   • B. 5.7 hours
   • C. 8.0 hours
   • D. 2.8 hours
   Answer and solution

   Answer: A

   Orbital radius r = 2RE = 12800 km. Using T = 2π√(r³/GM) and GM = gRE², we get T = 2π√(8RE³/gRE²) = 2π√(8RE/g) = 2π√(8×6.4×10⁶/10) ≈ 2π×2263 s ≈ 14200 s ≈ 4.0 hours.

7. Q7 · Photometry and Spectroscopy

   Two stars have apparent magnitudes m₁ = 3.0 and m₂ = 6.0. What is the ratio of the flux received from star 1 to that from star 2?

   • A. 2.5
   • B. 10.0
   • C. 15.85
   • D. 100.0
   Answer and solution

   Answer: C

   The magnitude difference is Δm = 6.0 - 3.0 = 3.0. Using the Pogson relation, flux ratio = 100^(Δm/5) = 100^(3/5) = 100^0.6 ≈ 15.85.

8. Q8 · Lunar Phases, Eclipses and Tides

   The Moon completes one orbit around Earth in approximately 27.3 days (sidereal month). Why is the time between successive Full Moons approximately 29.5 days?

   • A. The Moon's orbital velocity decreases near Full Moon
   • B. Earth's motion around the Sun requires the Moon to travel additional angular distance
   • C. The Moon's orbit precesses due to solar tidal forces
   • D. Atmospheric refraction delays the apparent Full Moon
   Answer and solution

   Answer: B

   During 27.3 days, Earth moves approximately 27° in its orbit around the Sun, so the Moon must travel an additional ~2.2 days to realign with the Sun-Earth line, making the synodic month 29.5 days.

9. Q9 · Laws of Motion

   A rocket in deep space ejects 150 kg of gas per second at a relative velocity of 2000 m/s. What is the thrust force experienced by the rocket?

   • A. 1.5 × 10⁵ N
   • B. 3.0 × 10⁵ N
   • C. 4.5 × 10⁵ N
   • D. 6.0 × 10⁵ N
   Answer and solution

   Answer: B

   Thrust force F = (dm/dt) × v_exhaust = 150 kg/s × 2000 m/s = 3.0 × 10⁵ N.

10. Q10 · Kinetic Theory of Gases

    In the solar corona, the temperature is approximately 2 × 10⁶ K. What is the root-mean-square speed of hydrogen atoms (mass = 1.67 × 10⁻²⁷ kg) at this temperature? (Boltzmann constant k = 1.38 × 10⁻²³ J/K)

    • A. 1.0 × 10⁵ m/s
    • B. 1.4 × 10⁵ m/s
    • C. 2.0 × 10⁵ m/s
    • D. 2.8 × 10⁵ m/s
    Answer and solution

    Answer: C

    v_rms = sqrt(3kT/m). 3kT = 3×1.38e-23×2e6 = 8.28e-17 J; divided by m=1.67e-27 gives 4.96e10; sqrt ≈ 2.2×10⁵ m/s, closest to 2.0×10⁵ m/s.