Karnataka II PUC Physics (KSEAB) common mistakes
Karnataka II PUC Physics (KSEAB) · Understand the traps and how to avoid them.
When a capacitor is charged and then the battery is disconnected before the plates are separated, the student recomputes the new field or potential by holding V constant (using E = V/d or C = ε₀A/d with V fixed), instead of recognising that Q is now fixed and E = Q/(ε₀A) is unchanged while V = Q/C rises.
Most textbook drills are on isolated capacitors or on capacitors kept connected to a battery, and the two regimes (constant V vs constant Q) are memorised as formulas rather than as a physical constraint. The phrase 'battery disconnected' is read as a minor detail rather than the key condition that freezes charge.
What to do instead
Before any capacitor manipulation, write one line: 'Q constant (isolated)' or 'V constant (battery connected)'. Then derive only from Q = CV plus the geometry formula C = ε₀A/d (or C = ε₀A/(d − t) with a slab). Practise the four standard cases explicitly: battery connected and d increased (V fixed, C falls, Q falls, E falls, U falls); battery removed and d increased (Q fixed, C falls, V rises, E unchanged, U rises). Check the sign of the energy change against the mechanical work done.
While solving Kirchhoff or Wheatstone-bridge numericals, the candidate assigns the same current to two series-looking branches that actually meet a junction, or applies the loop rule but drops the internal-resistance term (writing ΣE = ΣIR and ignoring r), so the final current is off by a factor of E/(R + r).
Circuit diagrams in the exam are drawn compactly and students trace loops visually rather than marking nodes. Internal resistance is often written in small print next to the cell symbol and is mentally treated as part of the connecting wire.
What to do instead
Redraw every circuit with labelled nodes A, B, C and mark an arrow plus a symbol (I₁, I₂) on each branch before writing any equation. At each junction write the current-conservation equation first. In every loop, write the cell as (E − Ir) or include +Ir on the resistance side. For Wheatstone networks, first test the balance condition P/Q = R/S; if balanced, delete the galvanometer branch entirely, which usually collapses the problem to simple series–parallel reduction.
In problems on a charged particle entering a uniform magnetic field, the candidate equates magnetic force to centripetal force but substitutes the speed component parallel to B, or uses the full velocity when the velocity makes an angle θ with B, producing a wrong radius and pitch for the helical path.
The formula r = mv/(qB) is memorised without the condition v ⊥ B, and the helical-motion case (velocity at angle θ, giving r = mv sinθ/(qB) and pitch = v cosθ · T) is treated as a separate fact rather than as a resolution of velocity into components.
What to do instead
Always resolve v into v∥ = v cosθ (along B, unchanged, gives pitch) and v⊥ = v sinθ (perpendicular, gives circular radius) before applying r = mv⊥/(qB). Use T = 2πm/(qB) for the period, noting it is independent of speed. For direction, apply the right-hand rule for a positive charge and reverse for negative charges (electron, proton sign check). Sanity-check units: r in metres requires m in kg, v in m/s, q in coulomb, B in tesla.
When finding the direction of induced current, the candidate applies Lenz's law by looking at the change in flux but then states the induced current direction that would increase the flux, or confuses the direction of induced current with the direction of the induced magnetic field inside the loop.
Lenz's law is stated verbally as 'opposes the change', and under time pressure students skip the intermediate step of identifying whether flux is increasing or decreasing, then jump straight to a direction. The induced field inside the loop and the current direction (given by the right-hand grip rule) are conflated.
What to do instead
Use a fixed four-step routine: (1) find the original B direction through the loop; (2) decide whether flux is increasing or decreasing; (3) the induced B must oppose that change (into the page if flux out of page is increasing); (4) convert induced B to current direction with the right-hand grip rule. Do this for the standard cases: magnet approaching/leaving a coil, coil rotating in a field, and a rod sliding on rails. Also state the magnitude separately from Faraday's law, ε = −dΦ/dt, keeping the minus sign as the Lenz indicator.
In ray-optics numericals with a lens or mirror, the candidate substitutes object distance as positive (u = +20 cm) or forgets the sign convention for focal length of a concave lens, then reports an inverted or virtual image that contradicts the physical setup.
The Cartesian sign convention (distances measured from the pole/optical centre, positive in the direction of incident light) is stated once in the textbook but is not reinforced by most practice problems, so students default to 'all distances positive' arithmetic.
What to do instead
Write the sign convention at the top of every optics answer: light travels left to right; distances to the right of the pole/optical centre are positive, to the left negative; heights above the principal axis positive. Then for a real object u is always negative. Use 1/v − 1/u = 1/f for lenses and 1/v + 1/u = 1/f for mirrors, and check the sign of f: convex lens/mirror f positive, concave lens f negative, concave mirror f negative. Verify the answer by asking whether the image should be real or virtual from the diagram described.
In Young's double-slit problems, the candidate uses the single-slit formula for fringe width, or forgets that fringe width β = λD/d changes when the whole apparatus is immersed in a liquid of refractive index μ (λ becomes λ/μ), and reports the same fringe width.
The symbols d (slit separation), D (screen distance) and a (slit width) look interchangeable, and the single-slit central-maximum width (2λD/a) is memorised alongside the double-slit fringe width without distinguishing the physical setup. The medium change is treated as irrelevant because 'the geometry has not changed'.
What to do instead
Label the setup before substituting: two slits → β = λD/d; one slit → width of central maximum = 2λD/a. For a medium, replace λ by λ/μ in every interference/diffraction formula and note that the pattern shrinks. Also remember that the fringe width is measured between consecutive bright (or consecutive dark) fringes, and that the path difference at the nth bright fringe is nλ and at the nth dark fringe is (2n − 1)λ/2.
In photoelectric-effect numericals, the candidate uses the intensity of incident light to compute the maximum kinetic energy of photoelectrons, or applies hν = φ₀ + K_max with ν replaced by the threshold frequency.
The roles of intensity and frequency are learned as a list ('intensity affects current, frequency affects energy') but the Einstein equation is then applied mechanically without checking which quantity is given. Threshold frequency and incident frequency are both denoted ν in different lines of the solution.
What to do instead
Before calculating, write two separate statements: saturation photocurrent ∝ intensity (number of photons per second), and K_max = hν − hν₀ = hc/λ − hc/λ₀ (energy per photon). Then substitute only the incident frequency/wavelength into the energy equation. Convert wavelength to frequency with c = νλ and keep h = 6.63 × 10⁻³⁴ J s, or work in eV using hc = 1240 eV nm. Check that K_max ≥ 0; a negative result means no emission, not a negative energy.
In nuclear physics numericals, the candidate computes the mass defect in atomic mass units but then uses E = mc² with m in u and c = 3 × 10⁸ m/s without converting u to kg, obtaining an energy wrong by many orders of magnitude; or confuses the binding energy of a nucleus with the binding energy per nucleon.
The conversion 1 u = 931.5 MeV/c² is a single memorised line that is easily dropped when the problem is phrased in terms of 'mass defect in u'. Binding energy and binding energy per nucleon are both called 'binding energy' in casual reading.
What to do instead
Adopt a fixed template: Δm in u → multiply by 931.5 MeV to get energy in MeV (or convert u to kg with 1 u = 1.66 × 10⁻²⁷ kg if SI is demanded). Always state whether the answer is total binding energy (Δm × 931.5 MeV) or binding energy per nucleon (divide by mass number A). For fission/fusion energy release, compute (total mass of reactants − total mass of products) × 931.5 MeV and confirm the sign is positive for a spontaneous reaction.
While analysing a diode or transistor circuit, the candidate treats an ideal diode as always conducting, or assumes the base-emitter junction of an npn transistor in the active region is reverse biased, and then reports a collector current that cannot flow.
Diode and transistor behaviour is learned as a set of statements rather than as bias conditions that depend on the surrounding circuit. The forward/reverse state of each junction is not checked against the applied potentials before applying the equations.
What to do instead
For every diode, first determine the potential difference across it from the rest of the circuit: if the anode is at higher potential than the cathode by more than the cut-in voltage (≈0.7 V for Si), it conducts and can be replaced by a 0.7 V drop; otherwise it is an open circuit. For a transistor, state the bias of each junction explicitly: active region requires emitter–base forward and collector–base reverse; saturation requires both forward; cut-off requires both reverse. Then apply I_C = βI_B and I_E = I_B + I_C only in the active region.
In the practical/experimental section, the candidate describes the metre-bridge or potentiometer procedure without stating the balancing condition and the precautions (clean ends of the wire, tight connections, jockey tapped gently, readings taken away from the ends), and reports a null point obtained by sliding the jockey with the circuit closed continuously.
Experimental skills are studied as a list of apparatus names rather than as a sequence of operations with reasons, and the balancing condition (P/Q = l/(100 − l) for metre bridge) is memorised without the requirement that the galvanometer deflection be exactly zero.
What to do instead
Write each experiment as: aim, circuit description, balancing condition with the formula, observations table, and precautions with a reason for each. For the metre bridge, state P/Q = l/(100 − l) with l in cm from the left gap, and note that the null point should lie between 40 cm and 60 cm for best accuracy (interchange resistances if not). For the potentiometer, state that the potential gradient must be set so the null point falls on the wire, and that the driver cell emf must exceed the emf being measured. Mention that the jockey is tapped, not dragged, to avoid changing the wire's resistance by heating.