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100+ Free NSEA Practice Questions

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2026 Statistics

Key Facts: NSEA Exam

60

Total Questions

IAPT

120 mins

Exam Time

IAPT

₹300

Exam Fee

IAPT 2026

Nov 21

2026 Exam Date

IAPT

NSEA is a 2-hour offline exam with 60 MCQs (48 single-correct and 12 multi-correct) organized by IAPT and HBCSE. It tests Class 11-12 Physics and Mathematics, along with astronomy and astrophysics fundamentals. The fee is ₹300, and qualifying leads to the INAO.

Sample NSEA Practice Questions

Try these sample questions to test your NSEA exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Kepler's second law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This law is a direct consequence of which conservation principle?
A.Conservation of linear momentum
B.Conservation of angular momentum
C.Conservation of mechanical energy
D.Conservation of mass
Explanation: Kepler's second law (law of equal areas) is a direct consequence of the conservation of angular momentum. Since the gravitational force is radial (directed toward the Sun), the torque acting on the planet is zero. Therefore, the angular momentum of the planet about the Sun is constant, which mathematically translates to a constant areal velocity.
2A hypothetical planet has twice the mass of Earth and twice the radius of Earth. If the escape velocity from Earth's surface is $11.2\text{ km/s}$, what is the escape velocity from the surface of this hypothetical planet?
A.5.6 km/s
B.11.2 km/s
C.15.8 km/s
D.22.4 km/s
Explanation: The escape velocity is given by $v_{esc} = \sqrt{2GM/R}$. For the hypothetical planet, $M' = 2M_E$ and $R' = 2R_E$. Substituting these values gives $v'_{esc} = \sqrt{2G(2M_E)/(2R_E)} = \sqrt{2GM_E/R_E} = v_{esc}$. Therefore, the escape velocity remains $11.2\text{ km/s}$.
3A satellite is in a circular orbit around Earth at an altitude equal to Earth's radius $R_E$. If the orbital speed of a satellite near the Earth's surface is $v_0 \approx 7.9\text{ km/s}$, what is the orbital speed of this satellite?
A.3.95 km/s
B.5.59 km/s
C.7.90 km/s
D.11.17 km/s
Explanation: The orbital speed in a circular orbit is $v = \sqrt{GM/r}$. Near the surface, $r \approx R_E$ and $v_0 = \sqrt{GM/R_E}$. At altitude $h = R_E$, the orbital radius is $r = R_E + h = 2R_E$. The orbital speed is $v = \sqrt{GM/(2R_E)} = v_0 / \sqrt{2}$. Calculating this gives $7.9 / 1.414 \approx 5.59\text{ km/s}$.
4What is the time period of a simple pendulum of length $L$ if it is placed inside a spacecraft orbiting the Earth in a stable circular path of radius $2R_E$?
A.$2\pi\sqrt{L/g}$
B.$2\pi\sqrt{L/(2g)}$
C.$2\pi\sqrt{L/(g/4)}$
D.Infinite
Explanation: A spacecraft in a stable orbit is in a state of free fall, meaning the effective gravitational acceleration ($g_{eff}$) inside the spacecraft is zero. Since the time period of a simple pendulum is $T = 2\pi\sqrt{L/g_{eff}}$, as $g_{eff} \to 0$, the time period $T \to \infty$. Practically, the pendulum will not oscillate.
5A cosmic particle of mass $m$ undergoes a head-on elastic collision with a stationary helium nucleus of mass $4m$. What fraction of the particle's initial kinetic energy is transferred to the helium nucleus?
A.16%
B.36%
C.64%
D.80%
Explanation: For a head-on elastic collision where mass $m_1 = m$ collides with a stationary target of mass $m_2 = 4m$, the final velocity of the target is $v_2 = \frac{2m_1}{m_1 + m_2} v_1 = \frac{2m}{5m} v_1 = 0.4 v_1$. The kinetic energy transferred to the target is $K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (4m) (0.4 v_1)^2 = 0.64 (\frac{1}{2} m v_1^2) = 0.64 K_1$. Alternatively, the fraction of kinetic energy retained by the incoming particle is $((m - 4m)/(m + 4m))^2 = (-3/5)^2 = 9/25 = 0.36$. Since total energy is conserved, the fraction transferred is $1 - 0.36 = 0.64$, which is 64%.
6Star A has radius $R$ and surface temperature $2T$. Star B has radius $2R$ and surface temperature $T$. Assuming both behave as perfect blackbodies, what is the ratio of the luminosity of Star A to that of Star B ($L_A / L_B$)?
A.1
B.2
C.4
D.8
Explanation: According to the Stefan-Boltzmann law, the luminosity of a spherical star is $L = 4\pi R^2 \sigma T^4$. Thus, the luminosity ratio is $L_A / L_B = (R_A / R_B)^2 \times (T_A / T_B)^4 = (R / 2R)^2 \times (2T / T)^4 = (1/4) \times 16 = 4$.
7A star has a surface temperature of $6000\text{ K}$. Assuming it radiates as a blackbody, at what wavelength does its spectral radiance peak? (Wien's displacement constant $b \approx 2.898 \times 10^{-3}\text{ m K}$)
A.350 nm
B.483 nm
C.580 nm
D.720 nm
Explanation: By Wien's displacement law, $\lambda_{max} T = b$. Therefore, $\lambda_{max} = b / T = 2.898 \times 10^{-3}\text{ m K} / 6000\text{ K} \approx 4.83 \times 10^{-7}\text{ m} = 483\text{ nm}$. This wavelength lies in the blue-green part of the visible spectrum.
8A disk-shaped galaxy of mass $M$ and radius $R$ is rotating with constant angular velocity $\omega$. If it collapses uniformly under internal gravitational forces to half its original radius while preserving its disk shape and mass, what is its new angular velocity?
A.$\omega / 2$
B.$2\omega$
C.$4\omega$
D.$8\omega$
Explanation: By conservation of angular momentum, $I_1 \omega_1 = I_2 \omega_2$. For a uniform disk, the moment of inertia is $I = \frac{1}{2} M R^2$. When the radius becomes $R_2 = R/2$, the new moment of inertia is $I_2 = \frac{1}{2} M (R/2)^2 = I_1 / 4$. Therefore, $\omega_2 = 4 \omega_1 = 4\omega$.
9An astronomical telescope has an objective lens of focal length $100\text{ cm}$ and an eyepiece of focal length $2.0\text{ cm}$. Under normal adjustment (focused at infinity), what is the magnification and the total length of the telescope tube?
A.Magnification = 50, Tube Length = 98 cm
B.Magnification = 50, Tube Length = 102 cm
C.Magnification = 200, Tube Length = 102 cm
D.Magnification = 200, Tube Length = 98 cm
Explanation: For an astronomical telescope in normal adjustment, the angular magnification is given by $M = f_o / f_e = 100 / 2.0 = 50$. The length of the telescope tube is the distance between the two lenses, which is $L = f_o + f_e = 100 + 2.0 = 102\text{ cm}$.
10What is the theoretical angular resolution (in radians) of the Hubble Space Telescope for visible light of wavelength $\lambda = 500\text{ nm}$, given its primary mirror diameter is $D = 2.4\text{ m}$?
A.$1.25 \times 10^{-7}\text{ rad}$
B.$2.54 \times 10^{-7}\text{ rad}$
C.$5.08 \times 10^{-7}\text{ rad}$
D.$1.22 \times 10^{-6}\text{ rad}$
Explanation: According to Rayleigh's criterion, the angular resolution (minimum resolvable angle) is $\theta = \frac{1.22 \lambda}{D}$. Substituting the values: $\theta = \frac{1.22 \times 500 \times 10^{-9}\text{ m}}{2.4\text{ m}} = 2.54 \times 10^{-7}\text{ rad}$.

About the NSEA Exam

The National Standard Examination in Astronomy (NSEA) is the first stage of the Indian Astronomy Olympiad program, organized by the Indian Association of Physics Teachers (IAPT) in collaboration with the Homi Bhabha Centre for Science Education (HBCSE). It is a national-level offline examination for students in classes 10, 11, or 12. The exam syllabus is broadly equivalent to the CBSE senior secondary level (Class 11 and 12) in Physics and Mathematics, along with basic concepts of astronomy and astrophysics. Qualifying students advance to the Indian National Astronomy Olympiad (INAO) and subsequent stages leading to the International Olympiad on Astronomy and Astrophysics (IOAA).

Questions

60 scored questions

Time Limit

2 hours

Passing Score

Minimum Admissible Score (MAS), typically 50% of the average of the top ten scores

Exam Fee

₹300 (IAPT (Indian Association of Physics Teachers) & HBCSE)

NSEA Exam Content Outline

40%

Physics

Mechanics (laws of motion, work-energy, circular and rotational motion, gravitation), Heat and Thermodynamics (gas laws, blackbody radiation, Wien's law), Optics (ray optics, telescopes, wave optics, diffraction), Electromagnetism, and Modern/Nuclear Physics.

30%

Mathematics

Calculus (differentiation, integration, simple differential equations), Coordinate Geometry (conic sections like ellipses and hyperbolas for orbits), Trigonometry (properties of triangles, trigonometric equations), Vectors, and Complex Numbers.

30%

Astronomy and Astrophysics

Celestial Sphere (coordinate systems, altitude, azimuth, RA, declination), Kepler's Laws and orbital dynamics, Stellar Astrophysics (star properties, magnitudes, H-R diagram, stellar life cycles), Galaxies and Cosmology (Hubble's law, expanding universe), and Observational Telescopes (resolving power, magnification, light gathering power).

How to Pass the NSEA Exam

What You Need to Know

  • Passing score: Minimum Admissible Score (MAS), typically 50% of the average of the top ten scores
  • Exam length: 60 questions
  • Time limit: 2 hours
  • Exam fee: ₹300

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

NSEA Study Tips from Top Performers

1Master Class 11 and 12 NCERT Physics, especially Mechanics, Gravitation, Optics (Telescopes), and Thermodynamics (Blackbody Radiation).
2Build a strong math base in Calculus and Coordinate Geometry (ellipse equations are highly relevant for planetary orbits).
3Study celestial coordinates (Right Ascension, Declination, Altitude, Azimuth) and spherical trigonometry concepts.
4Understand the H-R diagram, stellar classification (OBAFGKM), stellar evolution stages, and absolute/apparent magnitude math (Pogson's ratio).
5Practice previous years' NSEA question papers to adapt to the +3/-1 and multi-correct marking schemes.

Frequently Asked Questions

What is the eligibility criteria for NSEA 2026-27?

Candidates must be Indian citizens, born between July 1, 2007, and June 30, 2012 (both dates inclusive). They must reside and study in India, and should not have completed their Class 12 board exams before November 30, 2026. Students in Class 10, 11, or 12 are eligible.

What is the exam pattern for NSEA?

NSEA is a paper-based exam lasting 2 hours. It typically contains 60 multiple-choice questions. Part 1 has 48 single-correct questions (+3 marks for correct, -1 mark for incorrect). Part 2 has 12 multi-correct questions (+6 marks if all correct options are marked, with no negative marking).

What is the registration fee and how do I register?

The enrollment fee is ₹300 per subject. Students can enroll through their schools if the school is a registered NSE centre, or register individually at a nearby approved IAPT centre. Online registration generally runs from August 21 to September 14.

How do I qualify for the next stage (INAO)?

To qualify for the Indian National Astronomy Olympiad (INAO), a candidate must score above the Minimum Admissible Score (MAS), which is 50% of the average of the top ten scores nationwide. Top scorers from each state/union territory are selected based on a state-wise quota, along with a national merit list.

Is astronomy required in school to write NSEA?

No. Astronomy is not a separate school subject in CBSE/ICSE curriculums. The exam assumes a strong command of Class 11 and 12 Physics and Mathematics, and tests basic astronomical concepts that can be self-studied, such as Kepler's laws, coordinate systems, and stellar magnitudes.