JEE PYQ: Units & Measurements - Question ID a1fbfe6197af (JEE Main 2026)

In the term $(h+B t), B t$ is added to $h$, where $h$ is the height or length which has the dimension: $[h]=[L]$.

Therefore, using principle of homogeneity the product Bt must also have the dimension of length: $[\mathrm{L}]$.

Given that t is time whose dimension is $[\mathrm{T}]$ :

$[\mathrm{B}][\mathrm{t}]=[\mathrm{L}]$

$\Rightarrow$ $[\mathrm{B}][\mathrm{T}]=[\mathrm{L}]$

$\Rightarrow$ $[\mathrm{B}]=\frac{[\mathrm{L}]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-1}\right]$

Therefore, the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$.

In the term $\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^2}\right), \frac{\mathrm{A}}{\mathrm{Bt}^2}$ is added to P (Pressure).

The dimension of pressure is $\mathrm{P}=\frac{\text { Force }}{\text { Area }}$ :

$[\mathrm{P}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

According to the principle of homogeneity, the dimension of $\frac{\mathrm{A}}{\mathrm{Bt}^2}$ must equal the dimension of P :

$\left[\frac{\mathrm{A}}{\mathrm{Bt}^2}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

$\Rightarrow$$\frac{[\mathrm{A}]}{\left[\mathrm{LT}^{-1}\right]\left[\mathrm{T}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

$\Rightarrow$ $[\mathrm{A}]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times[\mathrm{LT}]$

$\Rightarrow$ $[\mathrm{A}]=[\mathrm{M}]\left[\mathrm{L}^{-1+1}\right]\left[\mathrm{T}^{-2+1}\right]$

$\Rightarrow$ $[\mathrm{A}]=\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$

Therefore, dimensions of A is $\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$ and the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$. Hence, the correct option is (D).