Question

The dimensions of $\frac{a}{b}$ in the equation $P=\frac{a-t^2}{bx},$ where $P$ is pressure, $x$ is distance and $t$ is time, are

• A. $\left[M^{2}L{T}^{-3}\right]$
• B. $\left[M{T}^{-2}\right]$
• C. $\left[L{T}^{-3}\right]$
• D. $\left[M{L}^3{T}^{-1}\right]$

Answer 1

Answer: (B)

$\left[M{T}^{-2}\right]$

Since terms separated by + or - have same dimensions, thus $\left[P\right]=\left[\frac{a}{bx}\right]=\left[\frac{t^2}{bx}\right]$
The first equality gives, $\left[P\right]=M{L}^{-1}{T}^{-2}=\left[a\right]\left[b^{-1}\right]L^{-1}$
or, $\left[\frac{a}{b}\right]=M{T}^{-2}$