Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| (Physical Quantity) | (Dimensions) | ||
| A. | Polarisation | I. | \( [\mathrm{L^{-1}A}] \) |
| B. | Electric flux | II. | \( [\mathrm{ML^{3}T^{-3}A^{-1}}] \) |
| C. | Magnetic intensity | III. | \( [\mathrm{MLT^{-2}A^{-2}}] \) |
| D. | Magnetic permeability | IV. | \( [\mathrm{L^{-2}TA}] \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-IV, B-II, C-I, D-III
Write each quantity through its defining relation and SI unit, then reduce the unit to M, L, T and A (use \( 1\,\mathrm{C} = 1\,\mathrm{A\,s} \)).
(A) Polarisation — this matches (IV) \( [\mathrm{L^{-2}TA}] \). Polarisation is the dipole moment per unit volume, so its unit is C m per m3 = C m−2.
\( \displaystyle P = \frac{p}{V} \Rightarrow \frac{\mathrm{C\,m}}{\mathrm{m^{3}}} = \mathrm{A\,s\,m^{-2}} = [\mathrm{L^{-2}TA}] \)
(B) Electric flux — this matches (II) \( [\mathrm{ML^{3}T^{-3}A^{-1}}] \). Flux is field times area, and the field is force per unit charge.
\( \displaystyle \phi_E = EA = \frac{F}{q}\,A = \frac{[\mathrm{MLT^{-2}}][\mathrm{L^{2}}]}{[\mathrm{AT}]} = [\mathrm{ML^{3}T^{-3}A^{-1}}] \)
(C) Magnetic intensity — this matches (I) \( [\mathrm{L^{-1}A}] \). For a solenoid \( H = nI \) (turns per metre times current), so H is measured in A m−1.
(D) Magnetic permeability — this matches (III) \( [\mathrm{MLT^{-2}A^{-2}}] \). From the force per unit length between two parallel currents, \( \mu_0 \) has the unit N A−2.
\( \displaystyle \frac{F}{l} = \frac{\mu_0I_1I_2}{2\pi d} \Rightarrow \mu_0 = \frac{2\pi d\,F}{l\,I_1I_2} \Rightarrow \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{A^{2}}]} = [\mathrm{MLT^{-2}A^{-2}}] \)
NCERT (2026-27):
Summary – …
