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NDA Physics
Complete NDA Physics Study Guide & Revision Notes
By SenaPrep Defence Team Updated July 2026 12 min read
Cracking the UPSC National Defence Academy (NDA) written examination requires a highly focused preparation timeline. In this comprehensive study guide, our defence education experts analyze the core concepts of NDA physics guide, score 100 in physics GAT, physics syllabus notes, outline high-yield preparation tips, and provide step-by-step examples.
Part 1: Physics core concepts & weightage
Physics contributes 23-25 questions in the General Science segment. Key areas include Mechanics, Properties of Matter, Oscillations/Waves, Optics, Heat/Thermodynamics, Electricity/Magnetism, and Modern Physics applications.
Syllabus Breakdown & Marks Allocation
Subject Segment
Total Questions
Weightage (Marks)
Preparation Priority
Paper 1: Mathematics
120 Questions
300 Marks
Critical (Sectional Cutoff)
Paper 2: GAT (English)
50 Questions
200 Marks
High Yield (4 Marks/Q)
Paper 2: GAT (General Knowledge)
100 Questions
400 Marks
High Scoring (Science & GS)
Part 2: Key Revision Strategies for NDA physics guide, score 100 in physics GAT, physics syllabus notes
Success in this topic requires consistent daily practice. Here are the core routines recommended by top defence educators:
Build Foundational Concepts: Rely on Class 11 and 12 NCERT books for math/science topics, and read standard grammar texts for English.
Solve Previous Year Papers (PYQs): Go through at least the last 5-10 years of UPSC question sets to analyze repeat question structures.
Speed & Time Management: Practice mock tests with a timer to improve your answering speed. You have only 75 seconds per math question in the real exam.
Track Weak Areas: Bookmark difficult problems and review your mistake history to continuously raise your score.
Part 3: Practical Questions & Solved Examples
Below is a preview of the type of questions asked by the UPSC. Test your current understanding of NDA physics guide, score 100 in physics GAT, physics syllabus notes and review the step-by-step methods:
Use the interactive test panel below to solve live questions pulled directly from the SenaPrep exam bank.
Interactive Practice Quiz
Test your preparation live! Choose an option for each question to see instant feedback and deep step-by-step explanations.
Question 1 of 5 Live App Practice
The presence of magnetic field can be determined using which one of the following instruments?
Detailed Solution:
Correct Answer
✓ Option (c) is correct. The magnetic needle or compass is used to determine the presence and direction of a magnetic field.
📖 Governing Law / Core Concept
The behavior of a magnetic needle is governed by the principles of magnetism. A magnetic needle is essentially a small magnet that aligns itself with the magnetic field lines of the Earth due to the torque acting on it.
🧪 Step-by-Step Breakdown
When a magnetic field is applied, the magnetic needle experiences a torque given by:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Where:
\(\tau\): Torque (N·m)
m: Magnetic moment of the needle (A·m²)
B: Magnetic field strength (T)
\(\theta\): Angle between \(\vec{m}\) and \(\vec{B}\)
As the needle turns, it reaches an equilibrium pointing in the direction of the magnetic field.
Motor: Converts electrical energy into mechanical energy; does not specifically measure magnetic presence.
⚡ Mnemonic / Speed-Run
To remember how a magnetic needle responds, think of the acronym "MAGNET":
M: Magnetic field interaction
A: Aligns with field
G: Gauge of direction
N: Needle reacts to field
E: Equilibrium direction
T: Torque influences movement
6. Visual Suggestion
6. Visual Suggestion
A diagram showing a magnetic needle aligning itself with the Earth's magnetic field lines. Indicate the north and south poles of the needle and illustrate how the needle deflects when placed near a source of external magnetic field.
📖 Factual Verification & Reference
📖 Factual Verification & Reference:
Verified against NCERT Class 11 Physics, Chapter 4 (Magnetism and Matter).
Question 2 of 5 Live App Practice
The focal length of the objective lens of a telescope is 50 cm. If the magnification of the telescope is 25, then the focal length of the eye- piece is
Detailed Solution:
Solution
Option (c) is correct.
Explanation:
\[ M = \frac{f_o}{f_e} \]
Where \( M \) is the magnification, \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece. Given, \( f_o = 50 \, \text{cm} \) and \( M = 25 \).
Thus:
\[
25 = \frac{50}{f_e}
\]
Rearranging gives:
\[
f_e = \frac{50}{25} = 2 \, \text{cm}
\]
Thus the focal length of the eyepiece \( f_e \) is 2 cm.
📌 Hints / Properties Used:
Magnification formula of a telescope: \( M = \frac{f_o}{f_e} \)
📊 Visual Diagram Suggestion:
A diagram illustrating a simple telescope, showing the objective and eyepiece lenses, and the path of light through them, labeling the focal lengths.
📖 Factual Verification & Reference:
Verified against NCERT Class XI Physics, Chapter on Optics.
Question 3 of 5 Live App Practice
The light energy escaping from the Sun can be spread by
Detailed Solution:
Correct Answer
✓ Option (a) is correct. A shower of rain drops can disperse light.
📖 Governing Law / Core Concept
Light scattering occurs when light interacts with particles. The phenomenon is governed by Rayleigh scattering and diffraction principles.
🧪 Step-by-Step Breakdown
When sunlight passes through raindrops, refraction occurs at the surface, causing dispersion into a spectrum of colors.
Light enters the raindrop and refracts.
Internally reflects off the opposite side of the droplet.
Exits the droplet, refracting again and spreading out into a rainbow.
🔍 Option Analysis
Option B: A plane mirror reflects light but does not spread it.
Option C: A convex lens focuses light rather than dispersing it.
Option D: A combination of lenses is complex and doesn't disperse like raindrops.
⚡ Mnemonic / Speed-Run
Remember: "Raindrops Rainbows" - Drops scatter and create rainbows!
6. Visual Suggestion
6. Visual Suggestion
Illustrate the refraction and dispersion of light as it passes through a raindrop, showing incident light, refracted paths, and resulting spectrum (rainbow).
📖 Factual Verification & Reference
📖 Factual Verification & Reference:
Verified against NCERT Class XI Physics, Chapter on Light and Optics.
Question 4 of 5 Live App Practice
The image of an object formed by a plane mirror is
Detailed Solution:
Correct Answer
✓ Option (b) is correct. The image in a plane mirror is erect, virtual, and of the same size as the object.
📖 Governing Law / Core Concept
The behavior of images formed by mirrors is governed by the laws of reflection. A plane mirror reflects light according to the principle that the angle of incidence is equal to the angle of reflection.
🧪 Step-by-Step Breakdown
When an object is placed in front of a plane mirror:
\[ \text{Image Characteristics:} \]
1. \text{Erect}
2. \text{Virtual}
3. \text{Same Size}
The distance of the image from the mirror equals the distance of the object from the mirror, confirming the virtual nature and size equivalence of the image.
🔍 Option Analysis
Options A, C, and D describe characteristics inconsistent with the properties of images formed by plane mirrors.
⚡ Mnemonic / Speed-Run
ERC: Erect, Real, and Virtual - remember the characteristics.
Visualize the virtual setup: a line from the object to the mirror and back.
6. Visual Suggestion
6. Visual Suggestion
A ray diagram showing an object with rays reflecting off a plane mirror, indicating the path of incident and reflected rays, and highlighting the position of the virtual image behind the mirror.
📖 Factual Verification & Reference
📖 Factual Verification & Reference:
Verified against NCERT Class XII Physics, Chapter 10 (Wave Optics).
Question 5 of 5 Live App Practice
When a light beam falls on a triangular glass prism, a band of colours is obtained. Which one of the following statements is correct in this regard?
Detailed Solution:
Correct Answer
✓ Option (c) is correct. Violet light bends the most due to its highest refractive index in glass.
📖 Governing Law / Core Concept
Snell's Law describes the refraction of light:
\[\sin(\theta_1) n_1 = \sin(\theta_2) n_2\]
where \(n\) is the refractive index and \(\theta\) is the angle of incidence/refraction.
🧪 Step-by-Step Breakdown
In a prism, dispersion occurs because of different refractive indices for various wavelengths of light:
\[ n = \frac{c}{v} \]
Where:
\(n\): Refractive index of the medium
\(c\): Speed of light in vacuum (approximately \(3 \times 10^8 \, m/s\))
\(v\): Speed of light in the medium
Since violet light has a shorter wavelength, it experiences a higher refractive index than red light:
\[\text{Higher } n \Rightarrow \text{Greater bend}\]
Thus, violet light bends more than red light when passing through the prism.
🔍 Option Analysis
Option (a) is incorrect as red light does not have the highest refractive index. Option (b) contradicts the relationship, stating that red light bends the most which is not true. Option (d) is also incorrect since violet light has the highest refractive index, meaning it bends the most.
⚡ Mnemonic / Speed-Run
To remember the order of bending: Violet < Red (Bending: Highest to Lowest)!
6. Visual Suggestion
6. Visual Suggestion
A ray diagram should illustrate a triangular prism, with incident light splitting into its component colors: violet at the bottom bending the most and red at the top bending the least, clearly showing the spectrum of visible light.
📖 Factual Verification & Reference
📖 Factual Verification & Reference:
Verified against NCERT Class XII Physics, Chapter on Optics.
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