BSEB Inter Physics common mistakes
BSEB Inter Physics · Understand the traps and how to avoid them.
In numerical problems on Coulomb's law and electric field, students substitute distances in centimetres directly into the formula (e.g., F = 9×10^9 q1q2/r² with r = 10) and report forces that are 10^4 times too large, instead of first converting every length to metres.
Bihar Board numericals often state separations as '10 cm' or '20 cm', and candidates trained on formula-memorisation rush to substitution without tracking units; the SI constant 9×10^9 makes the error invisible because the answer still 'looks like a number'.
What to do instead
Adopt a fixed three-step habit for every electrostatics numerical: write the given quantities with units in a column, convert each to SI (cm→10^-2 m, μC→10^-6 C, nC→10^-9 C) in the same line, then substitute. After computing, do a rough order-of-magnitude check: two 1 μC charges 10 cm apart must give about 0.9 N, not 9×10^3 N.
When drawing or using circuit diagrams, students treat an ammeter as if it has high resistance and a voltmeter as if it has low resistance, or connect the voltmeter in series with the component, then cannot explain why the readings contradict the expected values.
The ideal-instrument rules are memorised verbally ('ammeter in series, voltmeter in parallel') but the resistance reason behind them is not, so under exam pressure the two get swapped; practical-record questions also test this and candidates reproduce a circuit they once copied without checking.
What to do instead
Anchor the rule to its purpose: an ammeter must not change the current, so it needs very low resistance (ideally zero); a voltmeter must not draw current, so it needs very high resistance (ideally infinite). Redraw every circuit labelling each meter as 'low R, in series' or 'high R, in parallel' before solving, and state the ideal assumption explicitly when calculating.
Applying Fleming's left-hand rule or the right-hand grip rule, students point the fingers/thumb in the wrong sequence and so report the force on a current-carrying conductor in the wrong direction, or get the magnetic field direction around a straight wire reversed.
The two rules are learned as gestures without a written mapping of which finger means field, current and force, and the left/right distinction (motor vs generator) is confused; in a written exam there is no diagram to fall back on.
What to do instead
Write the mapping once and reuse it: left hand - First finger = Field, seCond finger = Current, thuMb = Motion/force; right-hand grip - thumb = current, curled fingers = field. For every such question, sketch the wire and field arrows on the answer sheet and mark the direction with an arrow before writing the final word, so the reasoning is visible and self-checking.
In AC circuit problems, students add the voltages across a resistor and an inductor (or capacitor) arithmetically, e.g., writing V = V_R + V_L = 100 V, instead of combining them as phasors to get V = √(V_R² + V_L²).
Series DC practice makes 'voltages add' feel automatic, and the phase difference of 90° between V_R and V_L is known as a fact but not applied as a vector sum; the impedance triangle is memorised separately from the voltage triangle.
What to do instead
Always draw the phasor diagram first: I along the reference axis, V_R along I, V_L leading by 90°, V_C lagging by 90°. Then read off V = √(V_R² + (V_L − V_C)²) and Z = √(R² + (X_L − X_C)²) from the same triangle. Check with a limiting case: with only R and L, the supply voltage must always exceed each individual voltage.
Using the lens/mirror formula, students insert object distance as positive (u = +20 cm) or forget the sign convention for image distance and focal length, producing a virtual image where a real one is expected or a wrong magnification sign.
The Cartesian sign convention (distances measured from the pole/optical centre, against the incident light negative) is stated in the text but rarely practised with actual numbers, and different textbooks' mirror/lens sign tables blur together in memory.
What to do instead
Before substituting, draw the principal axis with the pole at the origin and mark the incident-light direction with an arrow. Then write u, v, f with signs on the diagram itself, using the rule 'measured opposite to incident light = negative'. For mirrors use 1/v + 1/u = 1/f; for lenses 1/v − 1/u = 1/f. Interpret the sign of v at the end: negative v for a lens means a virtual image on the same side as the object.
In photoelectric-effect and de Broglie questions, students use the intensity of incident light to predict whether emission occurs or to compute the maximum kinetic energy, instead of using frequency and the work function.
Wave intuition (brighter light → more energetic electrons) conflicts with the photon model, and the threshold-frequency condition is remembered as a phrase rather than as K_max = hν − φ; the exam often phrases the question around 'increased intensity' precisely to test this.
What to do instead
Fix the two-line rule in memory: intensity controls the number of photoelectrons (saturation current); frequency controls their maximum kinetic energy. Solve with K_max = hc/λ − φ and check emission only via ν > ν₀ (λ < λ₀). For de Broglie use λ = h/p = h/√(2mK), and always convert eV to joules (1 eV = 1.6×10^-19 J) before dividing by h.
Writing nuclear equations, students balance mass numbers but not atomic numbers (or vice versa), so a beta-minus decay is written with the atomic number unchanged, or an alpha decay is shown with the daughter's Z too high by one.
The conservation laws are known in words but not applied as a bookkeeping check, and the nature of the beta particle (an electron emitted from the nucleus when a neutron converts to a proton) is not linked to the Z change.
What to do instead
For every nuclear equation, draw two columns - top row mass numbers, bottom row atomic numbers - and confirm both sides balance before moving on. Remember the signatures: alpha decay lowers A by 4 and Z by 2; beta-minus raises Z by 1 with A unchanged; beta-plus or electron capture lowers Z by 1. Write the antineutrino in beta-minus decay if the question asks for the complete equation.
In semiconductor questions, students state that doping increases conductivity because it adds free electrons to an n-type and free electrons to a p-type as well, and they describe a p-n junction as 'two separate materials joined' rather than as a single crystal with a depletion region.
The terms 'n-type' and 'p-type' are learned as labels without the underlying majority/minority carrier picture, and the depletion region, barrier potential and forward/reverse bias behaviour are memorised as isolated sentences that are easy to mix up.
What to do instead
Build one mental picture and reuse it: pentavalent doping gives extra electrons (n-type, electrons are majority carriers, holes minority); trivalent doping creates holes (p-type, holes majority). At the junction, diffusion leaves fixed ions forming a depletion layer with a barrier potential (~0.7 V for Si, ~0.3 V for Ge); forward bias reduces the barrier, reverse bias widens it. Answer any question by naming majority carriers first, then the bias effect.
In practical-physics questions, students write the least count of a vernier callipers or screw gauge from memory as '0.1 cm' or '0.01 cm' without computing it from the scale data given, and they omit the zero error correction when the question supplies a zero-error value.
Least count is memorised as a single number from the standard instrument rather than as a formula (smallest main-scale division ÷ number of vernier divisions, or pitch ÷ number of circular-scale divisions), and zero error is treated as an optional extra rather than a required correction.
What to do instead
Always recompute: vernier least count = 1 main scale division / number of vernier divisions; screw gauge least count = pitch / number of circular scale divisions. Then apply the reading as main scale + (vernier/circular division × LC) and subtract the zero error with its sign (positive zero error is subtracted, negative is added). Show the corrected value with the same number of decimals as the least count.
In reasoning and 'assertion–reason' or short-answer items, students give a correct final statement but no physical justification, or justify with a formula they cannot derive, so they lose the reasoning marks even when the conclusion is right.
Preparation focuses on numerical answers and definitions, while Bihar Board papers award marks for stating the law, the condition of applicability and the reasoning chain; candidates also confuse similar-sounding laws (e.g., Biot–Savart vs Ampere's circuital law) when asked to justify.
What to do instead
Practise a fixed answer skeleton for reasoning items: (1) state the relevant law or principle by name, (2) write its mathematical form, (3) state the condition or approximation under which it applies, (4) apply it to the given situation in one sentence. For assertion–reason items, decide separately whether each statement is true, then whether the reason actually explains the assertion, and write both judgements explicitly.