Question

Match the LIST-I with LIST-II:
\begin{center} \begin{tabular}{|c|l|c|l|} \hline List-I & & List-II &
\hline A. & Magnetic induction & I. & $MLT^{-2}A^{-2}$
B. & Magnetic flux & II. & $ML^{2}T^{-2}A^{-2}$
C. & Magnetic permeability & III. & $ML^{0}T^{-2}A^{-1}$
D. & Self inductance & IV. & $ML^{2}T^{-2}A^{-1}$
\hline \end{tabular} \end{center} Choose the correct answer from the options given below:

• A. A-III, B-IV, C-II, D-I
• B. A-I, B-III, C-IV, D-II
• C. A-IV, B-III, C-I, D-II
• D. A-III, B-IV, C-I, D-II
  \bigskip

Hint:

Always derive dimensions using basic definitions like force, flux, and current to avoid memorization errors.

Answer 1

The Correct Option is D

Step 1: Magnetic induction ($B$).
Magnetic induction is force per unit current per unit length. Its dimensional formula is:
\[ [ B ] = ML^{0}T^{-2}A^{-1} \] Hence, A $\rightarrow$ III.
Step 2: Magnetic flux ($\Phi$).
Magnetic flux is given by $B \times \text{area}$.
\[ [ \Phi ] = (ML^{0}T^{-2}A^{-1}) \times L^2 = ML^{2}T^{-2}A^{-1} \] Hence, B $\rightarrow$ IV.
Step 3: Magnetic permeability ($\mu$).
From $B = \mu H$,
\[ [\mu] = \dfrac{[B]}{[H]} = \dfrac{ML^{0}T^{-2}A^{-1}}{A\,L^{-1}} = MLT^{-2}A^{-2} \] Hence, C $\rightarrow$ I.
Step 4: Self inductance ($L$).
Self inductance is flux per unit current.
\[ [L] = \dfrac{ML^{2}T^{-2}A^{-1}}{A} = ML^{2}T^{-2}A^{-2} \] Hence, D $\rightarrow$ II.
Step 5: Final conclusion.
The correct matching is A-III, B-IV, C-I, D-II, corresponding to option (4).