Match List I with List - II.
| List - I | List - II |
| (A) Electric Field | (I) \( [LTA] \) |
| (B) Electric Flux | (II) \( [L^{2}] \) |
| (C) Electric Dipole Moment | (III) \( [ML^{3}T^{-3}A^{-1}] \) |
| (D) Area Vector Element | (IV) \( [MLT^{-3}A^{-1}] \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
The dimensional formula of a quantity follows from its defining relation, written in terms of mass [M], length [L], time [T] and current [A]. Charge is current times time, so \( [q] = [AT] \).
(A) Electric Field — this matches (IV) \( [MLT^{-3}A^{-1}] \). The electric field is force per unit charge.
\( \displaystyle [E] = \frac{[F]}{[q]} = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}] \)
(B) Electric Flux — this matches (III) \( [ML^{3}T^{-3}A^{-1}] \). Flux is field times area, \( \Phi = \vec{E}\cdot\vec{S} \).
\( [\Phi] = [MLT^{-3}A^{-1}][L^{2}] = [ML^{3}T^{-3}A^{-1}] \)
(C) Electric Dipole Moment — this matches (I) \( [LTA] \). The dipole moment is charge times separation, \( p = q \times 2a \).
\( [p] = [AT][L] = [LTA] \)
(D) Area Vector Element — this matches (II) \( [L^{2}] \). An area element \( d\vec{S} \) is an area, so its dimension is \( [L^{2}] \).
Note: A similar question was also asked in CUET (UG) 2026 on 31-05-2026 (Slot 2).
NCERT (2026-27):
Summary – …
