Physics Basics – Kerala PSC (General Awareness & Current Affairs)
These notes are designed for quick revision and to give you a solid foundation of core physics concepts that frequently appear in the Kerala PSC exam.
1. Simple Explanation
Physics is the study of the fundamental nature of matter, energy, and the forces that govern their interactions. For the PSC, focus on the classical domain (Newtonian mechanics, thermodynamics, waves, electromagnetism, optics) and a few modern concepts (atomic structure, radioactivity).
2. Key Points – Core Topics
| Topic | What You Need to Know | Essential Formula(s) |
|---|---|---|
| Kinematics | Motion of objects (displacement, velocity, acceleration) | (v = u + at) (s = ut + \frac12 at^2) (v^2 = u^2 + 2as) |
| Dynamics | Forces & Newton’s Laws | (F = ma) Net force = sum of all forces |
| Work & Energy | Work done by a force, kinetic & potential energy, conservation | (W = Fd\cos\theta) (K = \frac12 mv^2) (U = mgh) (K_i + U_i = K_f + U_f) |
| Power | Rate of doing work | (P = \frac{W}{t} = Fv\cos\theta) |
| Simple Harmonic Motion (SHM) | Oscillatory motion (e.g., spring, pendulum) | (x = A\cos(\omega t + \phi)) (\omega = \sqrt{\frac{k}{m}}) |
| Thermodynamics | Temperature, heat, laws of thermodynamics | (Q = mc\Delta T) (PV = nRT) |
| Waves | Wave properties, sound waves | (v = f\lambda) |
| Electromagnetism | Static electricity, magnetic field, Faraday’s law | (E = \frac{V}{d}) (B = \frac{\mu_0 I}{2\pi r}) (\mathcal{E} = -\frac{d\Phi}{dt}) |
| Optics | Reflection, refraction, lenses | (n_1\sin\theta_1 = n_2\sin\theta_2) Lens formula: (\frac1f = \frac1d_o + \frac1d_i) |
| Atomic & Nuclear | Structure of atom, types of radioactivity | (E = mc^2) Half‑life concept |
Tip: For PSC, focus on conceptual clarity and common formulas. Memorize the key equations and understand their applications.
3. Worked Example
Problem
A car starts from rest and accelerates uniformly to a speed of 25 m/s in 10 s.
Find:
- Its acceleration.
- The distance covered in that time.
- The work done by the engine if the net force is constant.
Solution
Step 1 – Acceleration
Using (v = u + at) (where (u = 0) because it starts from rest):
[ 25 = 0 + a(10) \quad\Rightarrow\quad a = \frac{25}{10} = 2.5\ \text{m/s}^2 ]
Step 2 – Distance
Use (s = ut + \frac12 at^2):
[ s = 0 + \frac12(2.5)(10)^2 = 0.5 \times 2.5 \times 100 = 125\ \text{m} ]
Step 3 – Work Done
Work (W = Fd).
Net force (F = ma = 2.5m).
Distance (d = 125\ \text{m}).
[ W = (2.5m)(125) = 312.5,m\ \text{J} ]
If the mass of the car is, say, 1000 kg:
[ W = 312.5 \times 1000 = 312,500\ \text{J} ]
Answers
- Acceleration = 2.5 m/s²
- Distance = 125 m
- Work done = 312.5 m J (or 312.5 kJ for a 1 kg mass)
4. Quick Revision Summary
| Concept | Formula | Quick Check |
|---|---|---|
| Velocity | (v = \frac{s}{t}) | Units m/s |
| Acceleration | (a = \frac{v-u}{t}) | Positive → speeding up |
| Newton’s 2nd Law | (F = ma) | Direction of motion |
| Work | (W = Fd\cos\theta) | Positive if force & displacement align |
| Kinetic Energy | (K = \frac12 mv^2) | Same units as work |
| Potential Energy (gravity) | (U = mgh) | Height in meters |
| Power | (P = \frac{W}{t}) | Watts (W) |
| SHM Angular Frequency | (\omega = \sqrt{k/m}) | Radians per second |
| Ideal Gas | (PV = nRT) | (R = 8.314\ J/(mol·K)) |
| ** |