Common physics mistakes in JEE Main, and how to stop making them

Ask a candidate why they lost marks in a physics mock and the first answer is usually "silly mistakes". That label hides the useful part. Almost every "silly" mistake falls into one of five families, and each family has a habit that prevents it. Once you know which family your errors belong to, you can fix them in weeks rather than hoping they stop on their own.

Why small mistakes cost so much

JEE Main awards +4 for a correct answer and −1 for a wrong one. A question you understood but got wrong is therefore a swing of 5 marks against getting it right, and in recent sessions the numerical-value questions have carried negative marking as well. Check the current information bulletin for the exact scheme.

Physics makes this worse because a single slip early in a problem, such as a missing force or a wrong sign, carries through every later line. The algebra can be perfect and the answer still wrong.

Family 1: setup mistakes

• A missing force. Friction, normal reaction, tension in a second string, buoyancy and the pseudo force in an accelerating frame are the usual absentees. List every contact (anything touching the body) and every field force (gravity, electric, magnetic) before you write an equation.
• The wrong system. In pulley and block problems, decide whether you are writing F = ma for one block or for the whole system. Internal forces (the tension between two blocks) cancel only for the whole system.
• Mixing frames. Either work in the ground frame, or work in the accelerating frame and add a pseudo force −ma. Doing half of each is how a lift problem gives the wrong scale reading.
• Assuming the body moves. On a rough incline, check whether mg sin θ exceeds the maximum static friction μmg cos θ before you use kinetic friction. If it does not, the block stays put and friction equals mg sin θ.

Family 2: direction and sign mistakes

• Vectors added as numbers. Two 10 N forces at 120° give a resultant of 10 N, not 20 N.
• g with the wrong sign. Choose a positive direction, write it on the rough sheet and keep it. If up is positive, a = −g throughout the flight, including at the top.
• Optics signs. Using the Cartesian convention, distances measured against the incident light are negative. The focal length of a concave mirror and a concave lens is negative, and that of a convex lens is positive. Put the numbers in with their signs; do not fix signs at the end.
• Work and energy signs. Work by kinetic friction on a sliding block is negative. Work done by a gas is positive on expansion. In chemistry the convention for work is reversed, so keep the two apart.

Family 3: a formula used outside its conditions

Every formula carries an assumption. Write it next to the formula on your sheet, because the exam often builds the question on exactly the case where the assumption breaks.

Family 4: units and conversions

Convert to SI before substituting. These conversions cause most unit errors:

Family 5: reading and answer-entry mistakes

• Missing the key word. "Maximum", "after 2 s", "relative to the lift", "in mJ": underline these on the rough sheet as you read.
• Rounding too early. Keep two or three extra figures until the final step. In a numerical-value question, round only the final number, as the question instructs.
• Answering in the wrong unit. If the question asks for the answer in cm or mJ, convert your SI result at the end.

Our general guide to JEE Main mistakes covers exam-day errors such as time management, and the guide to numerical value questions covers answer entry in more detail.

Worked problems: spot the mistake

Problem 1: the forgotten friction. A 2 kg block is released from rest on an incline of 37° with μ = 0.5 (g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8). A candidate writes a = g sin θ = 6 m/s². Find the correct acceleration.

First check that it slides: tan 37° = 0.75, which is greater than μ = 0.5, so it does.
Net force along the incline: mg sin θ − μmg cos θ.
a = g(sin θ − μ cos θ) = 10(0.6 − 0.4) = 2 m/s².

Problem 2: constant-acceleration formula with variable acceleration. A particle starts from rest with a = 2t m/s². A candidate writes v = at = 6 × 3 = 18 m/s at t = 3 s. Find the correct speed.

Acceleration changes with time, so integrate: v = ∫2t dt = t² (v = 0 at t = 0).
At t = 3 s, v = 9 m/s.

Problem 3: the sign in the mirror formula. An object is 30 cm in front of a concave mirror of focal length 20 cm. A candidate takes f = +20 and gets a virtual image. Locate the image correctly.

With the Cartesian convention, u = −30 cm and f = −20 cm.
1/v = 1/f − 1/u = −1/20 + 1/30 = (−3 + 2)/60 = −1/60, so v = −60 cm.
The image is real, 60 cm in front of the mirror. Magnification m = −v/u = −(−60)/(−30) = −2: inverted and twice the size.

Problem 4 (numerical answer): the square-millimetre trap. A wire of length 2 m and cross-section 0.4 mm² has resistivity 1.7 × 10⁻⁸ Ω m. Find its resistance in mΩ.

A = 0.4 mm² = 0.4 × 10⁻⁶ m² = 4 × 10⁻⁷ m². (Taking 1 mm² as 10⁻³ m² is the usual error.)
R = ρL/A = (1.7 × 10⁻⁸ × 2)/(4 × 10⁻⁷) = (3.4 × 10⁻⁸)/(4 × 10⁻⁷) = 0.085 Ω = 85 mΩ.

Practice set: find the error

Each statement contains one mistake. Find it and give the correct result.

1. "At the highest point of a projectile's path, its acceleration is zero."
2. A block slides 2 m on a fixed rough floor against a friction force of 5 N. "Work done by friction = +10 J."
3. One mole of an ideal gas at 27 °C occupies 24.9 L. "P = nRT/V = 1 × 8.31 × 27/0.0249 ≈ 9,000 Pa."
4. Resistors of 3 Ω and 6 Ω are in parallel across a 9 V battery. "Current through the 3 Ω resistor = 9/9 = 1 A."
5. An object is 10 cm from a convex lens of focal length 20 cm. "Using 1/v + 1/u = 1/f, the image is real."
6. A body in SHM has amplitude 0.1 m and frequency 5 Hz. "Maximum speed = Aω = 0.1 × 5 = 0.5 m/s."
7. A 60 kg person stands on a scale in a lift accelerating downward at 2 m/s² (g = 10 m/s²). "Scale reading = m(g + a) = 720 N."

Answers

1. The acceleration is g downward throughout. Only the vertical component of velocity is zero at the top.
2. Friction opposes the motion, so the work is −10 J.
3. Temperature must be in kelvin: P = 8.31 × 300/0.0249 ≈ 1.0 × 10⁵ Pa.
4. Each parallel branch has the full 9 V across it: I = 9/3 = 3 A (and 1.5 A in the 6 Ω branch).
5. That is the mirror formula. With 1/v − 1/u = 1/f, u = −10 cm and f = +20 cm: 1/v = 1/20 − 1/10 = −1/20, so v = −20 cm. The image is virtual, erect and magnified 2 times (m = v/u = 2).
6. ω = 2πf = 10π rad/s, so v_max = 0.1 × 10π = π ≈ 3.14 m/s.
7. Accelerating downward reduces the normal force: N = m(g − a) = 60 × 8 = 480 N.

What to do next

• Go through your last three mock tests and put every physics error into one of the five families above.
• Add a "condition" column to your physics formula sheet and fill it for every formula.
• For two weeks, draw a free-body diagram for every mechanics problem, however easy it looks.
• Before submitting any numerical answer, re-read the unit the question asks for.
• Re-solve every wrong problem a week later without looking. If it is right, strike it off; if not, it stays on the list. The physics plan shows where this fits in a week.