Consider a modified Bernoulli equation. left(mathrmP+fracmathrmAmathrmBt^2right)+rhomathrmg(mathrmh+mathrmBt)+frac12rhomathrmV^2=textconstant$\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^{2}}\right)+\rho\mathrm{g}(\mathrm{h}+\mathrm{Bt})+\frac{1}{2}\rho\mathrm{V}^{2}=\text{constant}$ If t has the dimension of time then the dimensions of A and B are ____, ____ respectively.

### Related Formula textPrinciple of Homogeneity: Only quantities with same dimensions can be added/subtracted.$$\text{Principle of Homogeneity: Only quantities with same dimensions can be added/subtracted.}$$ ### Core Logic From the term (h + Bt)$(h + Bt)$, h$h$ (height/length) and Bt$Bt$ must have the same dimension. [h] = [Bt]$[h] = [Bt]$ [B] = frac[h][t] = frac[L][T] = [L T^-1]$$[B] = \frac{[h]}{[t]} = \frac{[L]}{[T]} = [L T^{-1}]$$ So, B = [M^0 L T^-1]$B = [M^0 L T^{-1}]$. ### Step 1: Finding Dimension of A From the term left(P + fracABt^2right)$\left(P + \frac{A}{Bt^2}\right)$, pressure P$P$ and fracABt^2$\frac{A}{Bt^2}$ must have the same dimension. Dimension of Pressure P = fractextForcetextArea = [M L^-1 T^-2]$P = \frac{\text{Force}}{\text{Area}} = [M L^{-1} T^{-2}]$. [P] = left[fracABt^2right]$$[P] = \left[\frac{A}{Bt^2}\right]$$ [M L^-1 T^-2] = frac[A][L T^-1] times [T^2]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T^{-1}] \times [T^2]}$$ [M L^-1 T^-2] = frac[A][L T]$$[M L^{-1} T^{-2}] = \frac{[A]}{[L T]}$$ [A] = [M L^-1 T^-2] times [L T] = [M L^0 T^-1]$$[A] = [M L^{-1} T^{-2}] \times [L T] = [M L^0 T^{-1}]$$ So, A = [M L^0 T^-1]$A = [M L^0 T^{-1}]$. ### Pattern Recognition Apply the Principle of Homogeneity across any (X + Y)$(X + Y)$ terms inside brackets. Extract one unknown, plug it into the next additive bracket group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids