Find dimensions of each quantity in List-I

We match each physical quantity with its dimensional formula.

(A) Angular Impulse

Angular impulse = torque $\times$ time.

Torque has dimensions: $[\tau] = [\text{force}]\,[\text{distance}] = (MLT^{-2})(L) = ML^2T^{-2}$

Therefore, $[\text{angular impulse}] = ML^2T^{-2}\times T = ML^2T^{-1}$

So, $\boxed{(A) \to (III)}$

(B) Latent Heat

Latent heat here means heat energy required per unit mass.

Heat (energy) has dimensions: $[Q] = ML^2T^{-2}$

Hence latent heat (specific latent heat): $[L] = \frac{ML^2T^{-2}}{M} = L^2T^{-2} = M^0L^2T^{-2}$

So, $\boxed{(B) \to (I)}$

(C) Electrical resistivity

Resistivity $\rho$ is given by: $\rho = R\frac{A}{l}$

Resistance: $R = \frac{V}{I}$

Potential difference has dimensions: $[V] = \frac{\text{work}}{\text{charge}} = \frac{ML^2T^{-2}}{AT} = ML^2T^{-3}A^{-1}$

Therefore, $[R] = \frac{ML^2T^{-3}A^{-1}}{A} = ML^2T^{-3}A^{-2}$

Now, $[\rho] = [R]\frac{L^2}{L} = (ML^2T^{-3}A^{-2})(L) = ML^3T^{-3}A^{-2}$

So, $\boxed{(C) \to (IV)}$

(D) Electromotive force

Electromotive force has same dimensions as potential difference: $[\text{emf}] = ML^2T^{-3}A^{-1}$

So, $\boxed{(D) \to (II)}$

Final matching

$(A)\to(III),\quad (B)\to(I),\quad (C)\to(IV),\quad (D)\to(II)$

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C

Our derived answer = C

Hence, they agree.