NSEA 2021 previous year paper · official paper and PYQ-pattern practice

NSEA 2021 PYQ-pattern questions

Showing 10 of the 17 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

   Two stars of equal mass M are separated by distance D. At what distance from one star along the line joining them is the gravitational field zero?

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

   Answer: C

   By symmetry, the gravitational field is zero at the midpoint between two equal masses, which is at distance D/2 from each star.

2. Q2 · Magnetism and Electromagnetic Induction

   A rectangular loop of dimensions 10 cm × 5 cm with resistance 0.2 Ω is placed perpendicular to a magnetic field that decreases uniformly from 0.8 T to 0.2 T in 0.15 s. What is the magnitude of the induced current in the loop?

   • A. 0.1 A
   • B. 0.15 A
   • C. 0.2 A
   • D. 0.25 A
   Answer and solution

   Answer: A

   Induced emf = -dΦ/dt = A(dB/dt) = (0.1×0.05)×(0.6/0.15) = 0.005×4 = 0.02 V. Induced current = emf/R = 0.02/0.2 = 0.1 A.

3. Q3 · Sets, Relations and Functions

   Let A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(x, y) : x + y = 6}. How many elements does R contain?

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

   Answer: D

   For x + y = 6 where x, y ∈ A, the valid pairs are (1,5) and (5,1). However, we must also check (2,4) and (4,2). Since all these belong to A×A, R = {(1,5), (2,4), (4,2), (5,1)}, but actually only pairs summing to 6: (1,5), (2,4), (3,3), (4,2), (5,1) - but 3+3=6 gives (3,3). Checking carefully: valid pairs are (1,5), (2,4), (5,1), (4,2). But the question asks elements in relation, which are ordered pairs where both elements are in A and sum to 6: (1,5), (5,1), (2,4), (4,2). Wait, let me recalculate: only pairs (1,5), (2,4) satisfy uniquely considering order doesn't repeat meaningfully. Correct count is 2 unique unordered pairs, but as a relation (ordered): 4 elements. However, re-reading: distinct elements would be 2 if considering minimal distinct pairings. The answer is 2 based on typical NSEA interpretation. [Correction 2026-07-13: answer key corrected to D after student report + independent re-verification. By systematically checking all values x ∈ A, I found that x + y = 6 yields exactly 5 ordered pairs: (1,5), (2,4), (3,3), (4,2), and (5,1), all with both components in A.]

4. Q4 · Permutations and Combinations

   The number of ways to arrange 4 distinct objects in a row is:

   • A. 12
   • B. 16
   • C. 24
   • D. 32
   Answer and solution

   Answer: C

   The number of permutations of 4 distinct objects is 4! = 24.

5. Q5 · Time and Calendars

   If the Local Sidereal Time (LST) is 18h 00m and a star has a Right Ascension of 14h 30m, what is its Hour Angle?

   • A. 3h 30m
   • B. 32h 30m
   • C. 4h 30m
   • D. 9h 30m
   Answer and solution

   Answer: A

   Hour Angle (HA) = LST - RA = 18h 00m - 14h 30m = 3h 30m.

6. Q6 · Celestial Sphere and Coordinate Systems

   For an observer at the equator (latitude 0°), what is the maximum altitude that a star with declination +60° can achieve?

   • A. 30°
   • B. 60°
   • C. 90°
   • D. 0°
   Answer and solution

   Answer: A

   At the equator, a star's maximum altitude equals 90° - |δ - φ|. For φ = 0° and δ = +60°, max altitude = 90° - 60° = 30°.

7. Q7 · Photometry and Spectroscopy

   A star's spectrum shows a prominent absorption line at observed wavelength 656.5 nm. This line corresponds to the Hα line with rest wavelength 656.3 nm. What is the radial velocity of the star relative to Earth (assuming non-relativistic motion)?

   • A. 91.3 km/s approaching
   • B. 91.3 km/s receding
   • C. 45.7 km/s receding
   • D. 182.6 km/s receding
   Answer and solution

   Answer: B

   Using Doppler formula v/c = Δλ/λ₀, where Δλ = 656.5 - 656.3 = 0.2 nm. Thus v = c × (0.2/656.3) = 3×10⁵ × (0.2/656.3) ≈ 91.4 km/s. Since wavelength increased (redshift), the star is receding.

8. Q8 · Rotational Motion

   A neutron star with initial angular velocity ω₀ and moment of inertia I₀ collapses such that its radius becomes half. Assuming angular momentum is conserved, by what factor does its rotational kinetic energy increase?

   • A. 2
   • B. 4
   • C. 8
   • D. 16
   Answer and solution

   Answer: B

   When radius halves, I becomes I₀/4 (since I ∝ R²). From L = Iω being conserved, ω becomes 4ω₀. KE = (1/2)Iω² increases by factor of 4.

9. Q9 · Rotational Motion

   A spherical asteroid of radius 100 km rotates with a period of 8 hours. What is the approximate angular velocity of the asteroid in rad/s?

   • A. 2.2 × 10⁻⁴ rad/s
   • B. 3.5 × 10⁻⁴ rad/s
   • C. 5.2 × 10⁻⁴ rad/s
   • D. 7.0 × 10⁻⁴ rad/s
   Answer and solution

   Answer: A

   ω = 2π/T = 2π/(8 × 3600) = 2π/28800 ≈ 2.2 × 10⁻⁴ rad/s.

10. Q10 · Optics

    Light of wavelength 600 nm passes through two narrow slits separated by 0.3 mm and produces an interference pattern on a screen 2 m away. The fringe width is:

    • A. 4 mm
    • B. 2 mm
    • C. 6 mm
    • D. 8 mm
    Answer and solution

    Answer: A

    The fringe width β = λD/d where λ = 600 × 10⁻⁹ m, D = 2 m, and d = 0.3 × 10⁻³ m. Thus β = (600 × 10⁻⁹ × 2)/(0.3 × 10⁻³) = 4 × 10⁻³ m = 4 mm.