Physics: Mechanics, Electrodynamics, Optics and Modern Physics
The Physics section of Assam CEE tests NCERT Class 11–12 level mechanics, electrodynamics, optics and modern physics. Numerical single-correct MCQs dominate, so dimensional checks and limiting cases often eliminate options faster than full calculation.
Core rules
- Constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as. Resolve vectors into components before applying.
- Newton's laws: draw a free-body diagram. Static friction satisfies f ≤ μsN; kinetic friction is f = μkN.
- Work–energy theorem: W_net = ΔK. Conservative forces have potential energy; friction does not.
- Electrostatics: Coulomb's law F = kq₁q₂/r². Field lines run from positive to negative; E inside a conductor in electrostatic equilibrium is zero.
- Optics: mirror 1/f = 1/v + 1/u, lens 1/f = 1/v − 1/u, with the standard Cartesian sign convention. Power P = 1/f in metres.
- Modern physics: E = hc/λ; photoelectric K_max = hν − φ.
Worked example
A ball is thrown at 20 m/s at 30° above horizontal from level ground. Find its range using g = 10 m/s².
Horizontal velocity u_x = 20 cos30° = 17.32 m/s; vertical u_y = 20 sin30° = 10 m/s. Time of flight T = 2u_y/g = 2 s. Range = u_xT = 17.32 × 2 = 34.6 m. The reasoning is that vertical and horizontal motions are independent: the vertical component alone fixes flight time, and the horizontal component then fixes range. Maximum height would be H = u_y²/(2g) = 5 m.
Traps that cost marks
- Substituting g = 9.8 when options are built on g = 10.
- Saying speed is zero at maximum height: only the vertical velocity component is zero.
- Reversing signs in the mirror or lens formula.
- Assuming static friction always equals μsN rather than f ≤ μsN.
- In photoelectric effect, intensity changes photocurrent, not stopping potential.
- Applying R = u²sin2θ/g on unequal launch and landing heights.