Examiners love asking: *"A charge enters a magnetic field. Find the work done by the magnetic force after time $t$."*
The Trap: Trying to calculate displacement and force. Remember that $\vec{F} = q(\vec{v} \times \vec{B})$ is always strictly perpendicular to $\vec{v}$. Thus, $\vec{F} \cdot \vec{v} = 0$, and magnetic work is identically ZERO at all times. Kinetic energy and speed never change in a pure magnetic field.
The standard formula $\mathcal{E} = \frac{1}{2} B \omega L^2$ applies when the rod rotates about one of its ends. If the rod rotates about its midpoint, the EMF induced across the two ends is ZERO because the induced EMFs in both halves cancel each other out!
If the incident frequency $\nu < \nu_0$ (threshold frequency), photoelectric current is ZERO, regardless of whether intensity is doubled, quadrupled, or magnified a thousand-fold!
Students assume ionization energy monotonically increases across Period 2 ($Li < Be < B$). Incorrect! $Be (1s^2 2s^2)$ has a higher $IE_1$ than $B (1s^2 2s^2 2p^1)$ because removing an electron from a fully-filled stable $2s$ orbital requires more energy than removing a lone $2p$ electron.
Aldol condensation strictly requires at least one $\alpha$-hydrogen adjacent to the carbonyl group. Benzaldehyde ($C_6H_5CHO$) and Formaldehyde ($HCHO$) have NO $\alpha$-hydrogens; they undergo the Cannizzaro reaction with 50% concentrated alkali instead of Aldol condensation!
For dissociation ($AB_2 \rightarrow A^{2+} + 2B^-$), $i = 1 + (n - 1)\alpha$ where $n$ is the number of ions formed. For dimerization ($2A \rightarrow A_2$), $i = 1 + (\frac{1}{n} - 1)\alpha = 1 - \frac{\alpha}{2}$. Using the dissociation formula for dimerization causes negative marking.
Writing $\cos^{-1}(\cos(7\pi/6)) = 7\pi/6$ is wrong! The principal range of $\cos^{-1}(x)$ is strictly $[0, \pi]$. Since $7\pi/6 > \pi$, we must write $\cos(7\pi/6) = \cos(2\pi - 5\pi/6) = \cos(5\pi/6)$, which yields the correct principal value of $5\pi/6$.
For an $n \times n$ matrix, $\det(kA) = k^n \det(A)$, NOT $k \det(A)$. For a $3 \times 3$ matrix, multiplying by 2 increases the determinant by a factor of $2^3 = 8$!
For equation $x^2 - 6x - 2 = 0$, the signs in Newton's relation follow the equation directly: $a_{10} - 6a_9 - 2a_8 = 0 \implies a_{10} - 2a_8 = 6a_9$. Students frequently slip on the sign of the constant term.