Match List I with List II. The force F acting on a particle in a...

According to the principle of homogeneity of dimensions:

1. The argument of an exponential function must be dimensionless.
$[Bt] = [\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^0] \Rightarrow [B] = \dfrac{1}{[\mathrm{T}]} = [\mathrm{T}^{-1}]$
Thus, (A) matches with (IV).

2. Quantities added together must have the same dimensions. In the denominator $(D + x^2)$:
$[D] = [x^2] = [\mathrm{L}^2]$
Thus, (B) matches with (III).

3. Every term in an equation must have the same dimensions as the whole. For the first term:
$[A e^{-Bt}] = [F] \Rightarrow [A] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]$
Then, \left[\dfrac{A}{B}\right] = \dfrac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]}{[\mathrm{T}^{-1}]} = [\mathrm{M} \mathrm{L} \mathrm{T}^{-1}]$<br>Thus, (C) matches with (II).<br><br>4. For the second term:<br>\left[\dfrac{C}{D + x^2}\right] = [F] \Rightarrow \dfrac{[C]}{[\mathrm{L}^2]} = [\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]$
$[C] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-2}] \times [\mathrm{L}^2] = [\mathrm{M} \mathrm{L}^3 \mathrm{~T}^{-2}]$
Thus, (D) matches with (I).

The correct matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).

Answer: (A)-(IV), (B)-(III), (C)-(II), (D)-(I)