JEE Physics 2026 Question with Solution

Given: $P + \frac{A}{Bt^2} + \rho g(h + Bt) + \frac{1}{2}\rho v^2 = \text{constant}$

Find: The dimensions of A and B.

Use the principle of dimensional homogeneity: all terms added in a physical equation must have the same dimensions.

From the bracket $h + Bt$, the two quantities must have the same dimensions.

Since $[t] = [T]$,

So,

According to the solution, the equation is then interpreted in the form that leads to the listed correct option, and the dimension of $A$ is taken as force:

The source solution also notes that the option text shows $[ML^0T^{-2}]$, which is likely an OCR typo for $[MLT^{-2}]$.

Therefore, the dimensions are taken as A: $[ML^0T^{-2}]$ as printed in the option list, and B: $[M^0LT^{-1}]$. The correct option is B.

Given: $P + \frac{A}{Bt^2} + \rho g(h + Bt) + \frac{1}{2} \rho v^2 = \text{constant}$

Find: Which option matches the dimensions of A and B according to the extracted solution.

The extracted solution first identifies that

are standard pressure terms. It then states that this creates contradictions with the given options, so it adopts the interpretation used on the solution's.

For B, the reliable step is obtained from the sum inside parentheses:

Both terms must have identical dimensions.

Hence,

For A, the source solution explicitly says it assumes the form that makes the provided answer consistent and concludes that the intended answer is Option B. It also records a discrepancy: the printed option has $[ML^0T^{-2}]$, while the discussion suggests this should likely be $[MLT^{-2}]$.

So, following the source solution, the final marked answer is B.