NSEA 2025 previous year paper · official paper and PYQ-pattern practice

NSEA 2025 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 · Units and Measurements

   The dimensional formula of universal gravitational constant G is [M^a L^b T^c]. What are the values of a, b, and c respectively?

   • A. -1, 3, -2
   • B. 1, -3, 2
   • C. -1, 3, 2
   • D. 1, 3, -2
   Answer and solution

   Answer: A

   From F = Gm₁m₂/r², we get G = Fr²/(m₁m₂). So [G] = [MLT⁻²][L²]/[M²] = [M⁻¹L³T⁻²], giving a = -1, b = 3, c = -2.

2. Q2 · Rotational Motion

   A pulsar rotates with angular velocity 100 rad/s. If it contracts uniformly to half its radius while conserving angular momentum, what is its new angular velocity?

   • A. 200 rad/s
   • B. 400 rad/s
   • C. 800 rad/s
   • D. 1600 rad/s
   Answer and solution

   Answer: B

   For a sphere I ∝ MR², so I ∝ R². By conservation of angular momentum Iω = constant, I₁ω₁ = I₂ω₂. If R becomes R/2, then I becomes I/4, so ω becomes 4ω = 400 rad/s.

3. Q3 · Kinetic Theory of Gases

   A gas mixture contains helium and nitrogen in equal molar proportions at temperature T. If the root mean square speed of helium atoms is 1350 m/s, what is the root mean square speed of nitrogen molecules? (Atomic mass of He = 4 u, molecular mass of N₂ = 28 u)

   • A. 450 m/s
   • B. 510 m/s
   • C. 675 m/s
   • D. 810 m/s
   Answer and solution

   Answer: B

   RMS speed is proportional to √(T/M). Since both gases are at the same temperature, v_rms(N₂)/v_rms(He) = √(M_He/M_N₂) = √(4/28) = √(1/7). Therefore, v_rms(N₂) = 1350/√7 ≈ 510 m/s.

4. Q4 · Optics

   In a Young's double slit experiment, the distance between slits is 0.5 mm and the screen is placed 2 m away. If the wavelength of light used is 600 nm, what is the fringe width?

   • A. 1.2 mm
   • B. 2.4 mm
   • C. 3.0 mm
   • D. 4.8 mm
   Answer and solution

   Answer: B

   Fringe width β = λD/d = (600×10⁻⁹ × 2)/(0.5×10⁻³) = 1200×10⁻⁶/0.5×10⁻³ = 2.4×10⁻³ m = 2.4 mm.

5. Q5 · Algebra

   If log₅(3x - 7) = 2, then the value of x is:

   • A. 32/3
   • B. 9
   • C. 11
   • D. 32
   Answer and solution

   Answer: A

   From log₅(3x - 7) = 2, we get 3x - 7 = 5² = 25, so 3x = 32, giving x = 32/3.

6. Q6 · Coordinate Geometry

   The equation of the line passing through the points (2, 3) and (5, 9) is:

   • A. y = 2x - 1
   • B. y = 2x + 1
   • C. y = 3x - 3
   • D. y = x + 1
   Answer and solution

   Answer: A

   Slope m = (9-3)/(5-2) = 2. Using y - 3 = 2(x - 2), we get y = 2x - 1.

7. Q7 · Probability and Statistics

   A meteor shower occurs such that on average 18 meteors are observed per hour. Assuming the meteors follow a Poisson distribution, what is the probability of observing exactly 2 meteors in a 10-minute interval?

   • A. 0.224
   • B. 0.149
   • C. 0.045
   • D. 0.090
   Answer and solution

   Answer: A

   In 10 minutes (1/6 hour), the expected number λ = 18/6 = 3. For Poisson distribution, P(k=2) = (3²×e⁻³)/2! = 9e⁻³/2 ≈ 0.224.

8. Q8 · Diurnal and Annual Motion

   At what latitude will a star with declination -40° be circumpolar?

   • A. 50°S or higher southern latitude
   • B. 40°S or higher southern latitude
   • C. 50°N or higher northern latitude
   • D. Never circumpolar
   Answer and solution

   Answer: A

   A star is circumpolar if its declination δ ≥ 90° - |latitude|. For δ = -40°, we need southern latitude φ such that -40° ≤ -(90° - φ), giving φ ≥ 50°S.

9. Q9 · Lunar Phases, Eclipses and Tides

   During a total lunar eclipse, an observer on the Moon would see the Earth blocking the Sun. What percentage of the Earth's disc would appear illuminated by refracted sunlight through Earth's atmosphere during totality?

   • A. Approximately 100% as a bright red ring around the edge
   • B. Approximately 50% on the side facing away from the Sun
   • C. Approximately 0% as the Earth appears completely dark
   • D. Approximately 25% on the polar regions only
   Answer and solution

   Answer: A

   During a total lunar eclipse viewed from the Moon, Earth blocks direct sunlight but its atmosphere refracts and scatters sunlight, creating a bright reddish ring around Earth's entire limb, illuminating approximately 100% of its edge.

10. Q10 · Telescopes and Instruments

    A Newtonian reflector telescope has a primary mirror of diameter 20 cm and focal length 100 cm. What is its f-number (focal ratio)?

    • A. f/2
    • B. f/5
    • C. f/10
    • D. f/20
    Answer and solution

    Answer: B

    The f-number is the ratio of focal length to aperture diameter: f-number = f/D = 100 cm / 20 cm = 5, expressed as f/5.