Question 58: Units , Dimensions and Error Analysis

Question

Consider a modified Bernoulli equation.
$ \left(\mathrm{P}+\frac{A}{B t^2}ight)+ho g(h+B t)+\frac{1}{2} ho V^2=	ext { constant } $
If $t$ has the dimension of time then the dimensions of $A$ and $B$ are $\_\_\_\_$ , $\_\_\_\_$ respectively.

Accepted Answer

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}ight] $
Therefore, the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}ight]$.
In the term $\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^2}ight), \frac{\mathrm{A}}{\mathrm{Bt}^2}$ is added to P (Pressure).
The dimension of pressure is $\mathrm{P}=\frac{	ext { Force }}{	ext { Area }}$ :
$ [\mathrm{P}]=\frac{\left[\mathrm{MLT}^{-2}ight]}{\left[\mathrm{L}^2ight]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight] $
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}ight]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight] $
$\Rightarrow $$\frac{[\mathrm{A}]}{\left[\mathrm{LT}^{-1}ight]\left[\mathrm{T}^2ight]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight]$
$\Rightarrow $ $[\mathrm{A}]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight] 	imes[\mathrm{LT}]$
$\Rightarrow $ $[\mathrm{A}]=[\mathrm{M}]\left[\mathrm{L}^{-1+1}ight]\left[\mathrm{T}^{-2+1}ight]$
$\Rightarrow $ $ [\mathrm{A}]=\left[\mathrm{ML}^0 \mathrm{~T}^{-1}ight] $
Therefore, dimensions of A is $\left[\mathrm{ML}^0 \mathrm{~T}^{-1}ight]$ and the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}ight]$. Hence, the correct option is (D).

Answer

$\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}\right]$

Answer

$\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$

Answer

$\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}^{-2}\right]$

Answer

$\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$

Units , Dimensions and Error Analysis

Consider a modified Bernoulli equation. \left(\mathrm{P}+\frac{A}{B t^2}\right)+\rho g(h+B t)+\frac{1}{2} \rho V^2=\text { constant } If t has the dimension of time then the dimensions of A and B are \_\_\_\_ , \_\_\_\_ respectively.

Choose the correct answer: