Maharashtra HSC Physics common mistakes
Maharashtra HSC Physics · Understand the traps and how to avoid them.
When computing moment of inertia about an axis parallel to one through the centre of mass, the candidate adds md directly to I_cm instead of md², then substitutes the wrong value into I = I_cm + md².
The parallel-axis theorem is memorised as a formula string without tracking the dimensional meaning of each term; because I has units of mass×length², the d² is dropped under time pressure when the candidate is juggling numbers rather than dimensions.
What to do instead
Before every parallel-axis calculation, write the units of each term beside it (kg·m² for both I_cm and md²) and reject any term whose units do not match. Drill three cases (rod about end, disc about rim, ring about tangent) until the d² appears automatically, and always state the axis explicitly at the start of the solution.
In rolling-without-slipping problems the candidate uses v = rω at the contact point or forgets to combine translational and rotational kinetic energy, giving total KE as ½mv² only.
The candidate treats the rolling body as a point particle because translational formulas are more familiar, and does not visualise that the contact point is instantaneously at rest while the centre moves at v = rω relative to it.
What to do instead
Draw the velocity arrows at top, centre and bottom of the rolling body each time. Then write KE_total = ½mv² + ½Iω², substitute I and ω = v/r, and check the result against the known fractions (½mv² for a ring, ⅗mv² for a solid sphere, ¾mv² for a hollow sphere).
Applying Bernoulli's equation across a pipe of varying cross-section while ignoring the pressure drop due to viscous flow, or applying it to a horizontal pipe and then claiming pressure is equal at points of different speed.
Bernoulli's theorem is learned as a single plug-in equation and the conditions of validity (steady, non-viscous, incompressible, streamline flow) are not checked before use, so the candidate applies it where energy is being lost to viscosity.
What to do instead
At the top of every fluid problem, list the four validity conditions and tick or cross them. If viscosity matters, use Poiseuille's relation or the continuity equation alone; use Bernoulli only when the conditions hold, and always pair it with the continuity equation A₁v₁ = A₂v₂ to eliminate one unknown.
Using the rms speed formula v_rms = √(3RT/M) with M in grams per mole or T in degrees Celsius, producing answers off by a factor of about 32 or a wrong temperature dependence.
Textbook statements mix molar mass in kg/mol with molecular mass in u, and the candidate copies whichever number appears without converting to SI, since the formula itself looks unit-free.
What to do instead
Fix a habit: convert M to kg/mol by dividing the gram value by 1000, and convert T to kelvin by adding 273.15, writing both conversions on the line above the substitution. Sanity-check with the ratio v_rms(H₂)/v_rms(O₂) = 4 at the same temperature.
Reporting the efficiency of a Carnot engine as a percentage without converting temperatures to kelvin, or using η = 1 − T₂/T₁ with T₂ and T₁ in Celsius so that a negative or absurd efficiency appears.
The thermodynamic temperature scale is treated as a formality, and because the formula is a pure ratio the candidate assumes any consistent unit will cancel, which is false for the Celsius offset.
What to do instead
Adopt a rule that no temperature enters a thermodynamic formula until it has been rewritten in kelvin, and always compute the efficiency both as a fraction and as a percentage. Test the formula on a case where T₂ = 0 K to confirm η = 1, which only works in kelvin.
In SHM problems, the candidate uses ω = 2πf but plugs the time period in seconds as f, or computes the phase at t = 0 without the initial phase constant, so the displacement equation misses φ.
The general solution x = A sin(ωt + φ) is reduced to x = A sin ωt because most textbook examples start from the mean position, and the candidate does not check the initial condition before dropping φ.
What to do instead
At the start of every SHM problem, write the initial conditions (x₀ and v₀) and choose the form of the equation that fits them: sine if starting at the mean, cosine if starting at an extreme, general form otherwise. Then verify by differentiating once and checking v₀.
In Young's double-slit problems, the candidate measures the fringe width as the distance from the central maximum to the first bright fringe instead of between two consecutive bright fringes, or substitutes the slit separation d and screen distance D in the wrong places.
The formula β = λD/d is memorised without the meaning of each symbol, and diagrams in the candidate's mind collapse the geometry so that D and d are swapped.
What to do instead
Label the physical setup on paper before substituting: d is the small slit gap in millimetres, D is the large slit-to-screen distance in metres. State that β is the spacing between successive bright or successive dark fringes, and check that β comes out in the range of a fraction of a millimetre to a few millimetres for visible light.
Adding capacitances in series as if they were resistors, or computing the equivalent capacitance of a mixed network without first identifying which branches are in series and which are in parallel.
The resistor rules are practised far more often, and the reciprocal nature of series capacitance feels counterintuitive, so the candidate defaults to the resistor pattern under exam pressure.
What to do instead
Redraw the circuit with nodes labelled A, B, C and colour branches that share the same two nodes as parallel. Then apply 1/C_series = Σ1/C and C_parallel = ΣC, and verify with the physical rule that series capacitors always give a smaller equivalent than the smallest individual capacitor.
In potentiometer and Wheatstone bridge numericals, the candidate balances the bridge using the wrong ratio (P/Q = S/R instead of P/Q = R/S) or ignores the end corrections and internal resistance of the cell.
The balance condition is remembered as a shape rather than derived, so the position of the unknown resistance in the ratio is guessed, and end corrections are skipped because they seem like small details.
What to do instead
Derive the balance condition once from zero galvanometer current (V_A = V_B) so the ratio follows from the circuit, not memory. Then always write the balanced form with the unknown isolated on one side, and include end correction e in ℓ = ℓ₁ + e whenever the problem supplies it.
For a step-down transformer, the candidate writes V_s/V_p = N_p/N_s and then reports a secondary current larger than the primary current while also claiming power is conserved without checking the turns ratio direction.
The turns-ratio equation is memorised without attaching it to the physical direction of energy transfer, and the candidate mixes the voltage ratio with the current ratio, which is its reciprocal.
What to do instead
Write the two relations side by side every time: V_s/V_p = N_s/N_p and I_s/I_p = N_p/N_s, then confirm that V_s I_s ≈ V_p I_p. For a step-down transformer, state explicitly that N_s < N_p, so voltage falls and current rises, and check the numerical answer against that statement.