Tamil Nadu HSE Second Year Physics common mistakes
Tamil Nadu HSE Second Year Physics · Understand the traps and how to avoid them.
Substitutes charge values in microcoulomb or nanocoulomb straight into F = (1/4πε₀)(q₁q₂/r²) - e.g. writes q₁ = 2 and q₂ = 3 for 2 µC and 3 µC - and reports the force in newtons with no 10⁻⁶ factor; also drops the factor 1/K when the question says the charges are in a medium of dielectric constant K.
Textbook numericals often print symbols without units, and the prefixes µ (10⁻⁶) and n (10⁻⁹) get treated as labels rather than powers of ten. When r is also given in cm, two missing conversions can be in play at once and the error is invisible because the arithmetic 'works'.
What to do instead
Write a conversion line beneath the data, every time: 2 µC = 2 × 10⁻⁶ C, 3 µC = 3 × 10⁻⁶ C, 5 cm = 5 × 10⁻² m. Then substitute. If the medium has relative permittivity K, insert 1/(4πε₀K) rather than 1/(4πε₀). Sanity-check the magnitude: two µC charges a few centimetres apart give a force of order 1–100 N, never 10⁹ N or 10⁻⁹ N.
When a dielectric slab is inserted into a capacitor, uses the same Q and the same V at the same time, or applies C′ = Kε₀A/d without first deciding whether the battery is still connected.
Only one formula for capacitance is well rehearsed, but the discharge/charge situation divides into two genuinely different cases: with the battery connected, V is held constant and Q rises; with the battery removed, Q is held constant and V falls. Candidates memorise the formula and skip the physics of the case.
What to do instead
Before any substitution, write one line: 'battery connected → V fixed' or 'isolated → Q fixed'. Then compute C′ = Kε₀A/d and follow through with Q′ = C′V or V′ = Q/C′ and U = ½CV². A quick consistency check: energy stored always increases when a dielectric is inserted with the battery connected, and decreases when the capacitor is isolated.
Writes Kirchhoff loop equations with inconsistent signs - taking the emf of a cell and the IR drop across it with the same sign, or assigning all emfs positive regardless of which terminal the loop arrow enters - so the second equation contradicts the first and the solved current comes out negative without comment.
The sign rules are usually stated in words but never practised with a marked traversal direction. Candidates read the printed number beside a cell as 'positive', ignoring whether the loop passes from the − terminal to the + terminal or the reverse.
What to do instead
Draw the loop as a closed arrow and mark an assumed current direction on each branch before writing anything. Apply one fixed rule throughout: emf is positive when the cell is traversed from − to +, and the IR term is negative when the resistor is traversed along the assumed current. Finish by interpreting the sign - a negative current simply means the true direction is opposite to the assumption - and check each junction with ΣI = 0.
Uses the wrong angle in F = qvB sin θ - for instance taking θ = 0° when a charge enters perpendicular to a field that is drawn horizontally - or reports the force direction from Fleming's left-hand rule without reversing it for a negative charge.
The formula is remembered as a product with a sine factor but θ is understood as the angle in the diagram rather than the angle between the velocity vector and the magnetic field vector. Two-dimensional textbook figures make the perpendicular case look 'horizontal', and the charge-sign reversal is an extra step that is easily forgotten.
What to do instead
State in words what θ is before substituting: 'θ is the angle between v and B; here v ⊥ B, so θ = 90° and F = qvB'. Get the direction from the vector product v × B (or Fleming's left hand with the forefinger along B and the middle finger along the conventional current), then apply the rule that a negative charge feels the force in the opposite direction. Remember F is always perpendicular to both v and B, so it can never change the speed, only the direction.
Adds the voltages in a series LCR circuit arithmetically, writing V = V_R + V_L + V_C, and mixes peak and rms values - for example dividing one voltage by √2 but not the others when finding impedance or power.
Every previous circuit problem involved scalar addition of DC drops, and the phasor relation is a genuinely different rule that appears only in this chapter. Peak and rms values also look interchangeable until a numerical answer turns out wrong by √2.
What to do instead
Start from the phasor relations and write them before substituting: V = √(V_R² + (V_L − V_C)²) and Z = √(R² + (X_L − X_C)²), with tan φ = (X_L − X_C)/R. Convert every given quantity to the same convention - all rms or all peak - in a single line at the top of the answer, and use average power P = V_rms I_rms cos φ = I_rms² R. If V_L > V_C the circuit is inductive; if V_C > V_L it is capacitive.
Substitutes every distance as positive in the lens and mirror formulae - writing 1/f = 1/v − 1/u for a concave mirror, or accepting a positive v for a real image - and mixes centimetres, millimetres and nanometres when computing fringe width from β = λD/d.
The New Cartesian sign convention requires a deliberate decision about direction before substitution, and candidates instead plug in magnitudes and adjust the answer afterwards. Wavelength is almost always quoted in nm while D and d are in cm, so three different prefixes appear in one formula.
What to do instead
Write the convention at the top of the answer: origin at the pole or optical centre, distances measured along the direction of incident light are positive. Use the correct formula for the system - 1/f = 1/v + 1/u for mirrors, 1/f = 1/v − 1/u for lenses - and let the signs produce the result, so a negative v automatically means a virtual image. For interference, convert first: λ = 500 nm = 5 × 10⁻⁷ m, D and d in metres, then β = λD/d.
Uses the work function in electron-volts while h is in J·s in hν = φ₀ + K_max, or forgets that the maximum kinetic energy equals eV₀ where V₀ is the magnitude of the stopping potential, producing an answer wrong by a factor of 1.6 × 10⁻¹⁹.
Work functions are tabulated in eV because the numbers are convenient, and the stopping potential is quoted in volts, so the joule-based equation is fed mixed units without any visible cue that something is inconsistent.
What to do instead
Choose one unit system before starting and say so. Either convert φ₀ to joules by multiplying by 1.6 × 10⁻¹⁹ C, or work entirely in eV, using hc ≈ 1240 eV·nm so that φ₀ = hc/λ₀ comes out directly in eV. Always write K_max = eV₀ with V₀ positive, and check the answer against the threshold: if the incident wavelength is longer than λ₀, no emission is possible and K_max must be zero, not negative.
Treats half-life and mean life as the same constant - writing T½ = 1/λ - or handles a decay time that is not a whole number of half-lives by repeatedly halving, and balances nuclear equations by matching only the mass numbers while ignoring the atomic numbers.
λ, T½ and τ are introduced together and all describe the same decay, so the three relations blur. The exponential form N = N₀e^(−λt) looks harder than repeated halving, so candidates avoid it even when t/T½ is not an integer.
What to do instead
Write the three relations side by side and keep them together: T½ = 0.693/λ, τ = 1/λ, so τ ≈ 1.44 T½ and mean life is always longer than half-life. For any time use N = N₀e^(−λt), or N = N₀(1/2)^(t/T½) which is exact even for fractional t/T½. In nuclear equations, verify both conservation laws on each side: the sum of mass numbers A and the sum of atomic numbers Z must be equal, and check the emitted particle (α: A−4, Z−2; β⁻: Z+1; γ: no change in A or Z).
In a transistor circuit, substitutes into β = I_C/I_E, or states that both junctions are forward-biased in the active region, and in a diode circuit assumes a fixed 0.7 V drop regardless of whether the diode is silicon, germanium or ideal.
Three currents, three symbols and two configurations are learned in one block, so α and β get attached to the wrong current ratio. Diode drop is memorised as a single number without the material being checked.
What to do instead
Derive the ratios from the currents rather than memorising labels: α = I_C/I_E (always less than 1), β = I_C/I_B (typically 20–200), and the conversions β = α/(1−α), α = β/(1+β), with I_E = I_B + I_C as the check. For the active region used in amplification, the emitter-base junction is forward-biased and the collector-base junction reverse-biased. For diodes, read the given material or the word 'ideal': about 0.7 V for silicon, about 0.3 V for germanium, 0 V for ideal.
Computes the modulation index as m = A_m/A_c but substitutes peak-to-peak values for one amplitude and peak values for the other, and states that the bandwidth of an AM signal equals the modulating frequency; also assigns FM broadcast to sky-wave propagation.
The definitions of A_m and A_c are not fixed in the candidate's mind, so any 'amplitude' from the graph is used. Bandwidth and propagation mode are memorised as isolated facts without the frequency ranges that decide which mode applies.
What to do instead
Fix the definition first: m = A_m/A_c where both are peak (or both peak-to-peak) values of the modulating and carrier waves, and m must be ≤ 1 for undistorted modulation. Bandwidth of AM equals 2f_m (upper and lower sidebands); FM bandwidth is approximately 2(Δf + f_m). Attach frequencies to propagation modes: ground wave below about 2 MHz, sky wave roughly 3–30 MHz by ionospheric reflection, space wave (line of sight) for VHF and UHF such as TV and FM.