Consider a modified Bernoulli equation. ( P +frac A Bt 2 )+ρ g ( h + Bt )+1/2 ρ V 2 = constant If t has the dimension of time then the dimensions of A and B are \ \ \ \ , \ \ \ \ respectively.

[ ML 0 T -1 ] and [ M 0 LT -1 ]

[ ML 0 T -2 ] and [ M 0 LT -2 ]

[ ML 0 T -2 ] and [ M 0 LT -1 ]

Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 21 January, Morning Shift — Question 31

Consider a modified Bernoulli equation.

\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^{2}}\right)+\rho \mathrm{g}(\mathrm{~h}+\mathrm{Bt})+\frac{1}{2} \rho \mathrm{~V}^{2}=\text { constant }

If t has the dimension of time then the dimensions of A and B are $\_\_\_\_$ , $\_\_\_\_$ respectively.

Option A:
$\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$ and $\left[\mathrm{M}^{0} \mathrm{LT}\right]$
Option B:
$\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$ and $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$
Correct
Option C:
$\left[\mathrm{ML}^{0} \mathrm{~T}^{-2}\right]$ and $\left[\mathrm{M}^{0} \mathrm{LT}^{-2}\right]$
Option D:
$\left[\mathrm{ML}^{0} \mathrm{~T}^{-2}\right]$ and $\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]$

$\Rightarrow[P]=\left[\frac{A}{B t^{2}}\right]$

\Rightarrow[\mathrm{h}]=[\mathrm{Bt}] \end{gathered}$$

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

Putting B is equation (1) $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\frac{\mathrm{A}}{\mathrm{LT}^{-1} \times \mathrm{T}^{2}}\right]$ $[\mathrm{A}]=\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]$

Answer key and solution verified before publishing.

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