Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | \( \left[M^{0}L^{-2}T^{0}A^{1}\right] \) | (I) | Conductance |
| (B) | \( \left[M^{-1}L^{-2}T^{3}A^{2}\right] \) | (II) | Electrical conductivity |
| (C) | \( \left[ML^{3}T^{-3}A^{-2}\right] \) | (III) | Current density |
| (D) | \( \left[M^{-1}L^{-3}T^{+3}A^{+2}\right] \) | (IV) | Specific resistance |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Each quantity is reduced to the base dimensions M, L, T and A, starting from the resistance \( R = V/I \). Since \( V = W/q \), the dimension of the potential difference is \( [ML^{2}T^{-3}A^{-1}] \), so
\( [R] = \frac{[ML^{2}T^{-3}A^{-1}]}{[A]} = [ML^{2}T^{-3}A^{-2}] \)
(A) \( [M^{0}L^{-2}T^{0}A^{1}] \) — this matches (III) Current density. Current density is current per unit area, \( J = I/A \), which has the dimension \( [A][L^{-2}] \).
(B) \( [M^{-1}L^{-2}T^{3}A^{2}] \) — this matches (I) Conductance. Conductance is the reciprocal of resistance, \( G = 1/R \), so its dimension is the inverse of \( [ML^{2}T^{-3}A^{-2}] \).
(C) \( [ML^{3}T^{-3}A^{-2}] \) — this matches (IV) Specific resistance. Specific resistance (resistivity) is \( \rho = RA/l \), so its dimension is that of resistance multiplied by length, \( [ML^{2}T^{-3}A^{-2}][L] \).
(D) \( [M^{-1}L^{-3}T^{+3}A^{+2}] \) — this matches (II) Electrical conductivity. Conductivity is the reciprocal of resistivity, \( \sigma = 1/\rho \), so its dimension is the inverse of that in (C).
Note: A similar question was also asked in CUET (UG) 2023 on 15-06-2023 (Slot 2).
NCERT (2026-27):
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