Angular Impulse Latent Heat Resistivity EMF Dimensions

 

❓ Question

Match the following physical quantities in List-I with their correct dimensional formulae in List-II.

List-I

(A) Angular Impulse
(B) Latent Heat
(C) Electrical Resistivity
(D) Electromotive Force

List-II

$$(1)\ [M^0L^2T^{-2}]$$
$$(2)\ [ML^2T^{-3}A^{-1}]$$
$$(3)\ [ML^2T^{-1}]$$
$$(4)\ [ML^3T^{-3}A^{-2}]$$

---

✍️ Solution

(A) Angular Impulse

Angular impulse is equal to the change in angular momentum:

$$J_\theta=\Delta L$$

Angular momentum is

$$L=mvr$$

Therefore,

$$[L]=[M][LT^{-1}][L]$$
$$\boxed{[L]=[ML^2T^{-1}]}$$

Hence,

$$\boxed{A\rightarrow 3}$$

---

(B) Latent Heat

Specific latent heat is energy required per unit mass:

$$L=\frac{Q}{m}$$

Therefore,

$$[L] = \frac{[ML^2T^{-2}]}{[M]}$$
$$\boxed{[L]=[M^0L^2T^{-2}]}$$

Hence,

$$\boxed{B\rightarrow 1}$$

---

(C) Electrical Resistivity

We know

$$R=\rho\frac{l}{A}$$

Therefore,

$$\rho=\frac{RA}{l}$$

Now,

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

and

$$[V]=[ML^2T^{-3}A^{-1}]$$

Thus,

$$[R]=[ML^2T^{-3}A^{-2}]$$

Hence,

$$[\rho] = [R]\frac{[L^2]}{[L]}$$
$$\boxed{[\rho]=[ML^3T^{-3}A^{-2}]}$$

Therefore,

$$\boxed{C\rightarrow 4}$$

---

(D) Electromotive Force

EMF is work done per unit charge:

$$\mathcal{E}=\frac{W}{q}$$

Therefore,

$$[\mathcal{E}] = \frac{[ML^2T^{-2}]}{[AT]}$$
$$\boxed{[\mathcal{E}]=[ML^2T^{-3}A^{-1}]}$$

Hence,

$$\boxed{D\rightarrow 2}$$

---

✅ Final Matching

$$\boxed{A\rightarrow3,\quad B\rightarrow1,\quad C\rightarrow4,\quad D\rightarrow2}$$

Final Answer

$$\boxed{(A)-(3),\ (B)-(1),\ (C)-(4),\ (D)-(2)}$$