QJEEM25Ph24e4: JEE Main 2025 Physics - Class 12 JEEM25Ph24e4 Solution: The electric flux is where and are linear and surface charge…

Let's analyze the given expression step by step.

We are given the electric flux:

$$\varphi =\alpha \sigma +\beta \lambda$$

where:

$$\sigma$$ is a surface charge density with units of charge per unit area (C/m²).

$$\lambda$$ is a linear charge density with units of charge per unit length (C/m).

For the equation to be dimensionally consistent, the two terms on the right must have the same units as the flux $$\varphi$$.

For the term $$\alpha \sigma$$:

Since $$\sigma$$ has units $${\text{C/m}}^2$$, the constant $$\alpha$$ must have units that convert $$\sigma$$ into the same units as the flux.

For the term $$\beta \lambda$$:

Since $$\lambda$$ has units $$\text{C/m}$$, the constant $$\beta$$ carries its own units to ensure the term matches the flux’s dimensions.

Since both terms add to give the flux, they must share the same overall dimension. Thus, the dimensions of $$\alpha \sigma$$ and $$\beta \lambda$$ must be equal:

$$[\alpha ]\cdot \frac{Q}{L^2}=[\beta ]\cdot \frac{Q}{L}$$

Here, $$Q$$ represents the unit of charge and $$L$$ represents a unit of length. Canceling the common factors, we find:

$$\frac{[\alpha ]}{[\beta ]}=L$$

That is, the ratio $$\frac{\alpha }{\beta }$$ has the dimension of length.

Now, let’s match this with the given options:

Option A: Displacement (commonly measured as a length)

Option B: Charge (has the unit Coulomb)

Option C: Electric Field (has units of V/m or N/C)

Option D: Area (has units of length squared, $$L^2$$)

Since $$\frac{\alpha }{\beta }$$ has the dimensions of a length, it logically corresponds to a displacement.

Thus, the correct answer is:

Option A – displacement.