JEE PYQ: Units & Measurements - Question ID d38a2aa51d3f (JEE Main 2024)

Core Formula & Concept:

Latent heat ($L$) is defined as the amount of heat energy ($Q$) required to change the phase of a unit mass ($m$) of a substance without changing its temperature. Mathematically, this relationship is expressed as:

$Q = mL$

Here:

• $Q$ is the heat energy supplied or released,
• $m$ is the mass of the substance,
• $L$ is the latent heat.

To find the dimensional formula of latent heat, we need to express $L$ in terms of fundamental dimensions: mass ($M$), length ($L$), and time ($T$). We will use the principle of dimensional homogeneity, which states that the dimensions on both sides of a physical equation must be the same.

Step-by-Step Derivation:

Step 1: Write the defining equation for latent heat

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

Step 2: Determine the dimensional formula of heat energy ($Q$)

Heat energy is a form of energy. The dimensional formula of energy is derived from the work-energy principle: $\text{Energy} = \text{Force} \times \text{Displacement}$ The dimensional formula of force is $[\mathrm{MLT}^{-2}]$ and displacement is $[\mathrm{L}]$. Therefore: $[Q] = [\mathrm{MLT}^{-2}] \times [\mathrm{L}] = [\mathrm{ML}^2 \mathrm{T}^{-2}]$

Step 3: Determine the dimensional formula of mass ($m$)

Mass has the dimensional formula: $[m] = [\mathrm{M}]$

Step 4: Substitute the dimensions into the latent heat equation

$[L] = \frac{[Q]}{[m]} = \frac{[\mathrm{ML}^2 \mathrm{T}^{-2}]}{[\mathrm{M}]}$

Step 5: Simplify the expression

$[L] = [\mathrm{M}^{1-1} \mathrm{L}^2 \mathrm{T}^{-2}] = [\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$

Thus, the dimensional formula of latent heat is $[\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$.

Common Traps & Exam Tip:

Common Mistake 1: Students often confuse latent heat with specific heat capacity. Specific heat capacity has the dimensional formula $[\mathrm{L}^2 \mathrm{T}^{-2} \mathrm{K}^{-1}]$ because it involves temperature change. Latent heat, however, does not involve temperature change, so its dimensional formula does not include temperature ($K$ or $\Theta$).

Common Mistake 2: Some students incorrectly assume that latent heat has the same dimensions as energy ($[\mathrm{ML}^2 \mathrm{T}^{-2}]$). However, latent heat is energy per unit mass, so the mass dimension cancels out, leading to $[\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$.

Exam Tip: Always remember that when a physical quantity is defined as energy per unit mass, the mass dimension reduces to $[\mathrm{M}^0]$. This is a key insight for solving dimensional analysis problems involving specific quantities (e.g., specific heat, latent heat, gravitational potential).

The correct answer is Option C: $[\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$.