Question

If the force is given by $ F = at + bt^2$ with $t$ as time. The dimensions of $a$ and $b$ are

• A. $\left[MLT^{-4}\right]$ and $\left[MLT^{-2}\right]$
• B. $\left[MLT^{-3}\right]$ and $\left[MLT^{-4}\right]$
• C. $\left[ML^2T^{-3}\right]$ and $\left[ML^2T^{-2}\right]$
• D. $\left[ML^2T^{-3}\right]$ and $\left[ML^3T^{-4}\right]$

Answer 1

Answer: (B)

Dimension of $at =$ Dimension of $F$
$\left[at\right] = \left[F\right]$
$\Rightarrow \left[a\right] = \left[\frac{F}{t}\right] $
$\left[b\right] = \left[\frac{MLT^{-2}}{T}\right]$
$ \Rightarrow \left[a\right] = \left[MLT^{-3}\right] $
Dimension of $bt^2$ = Dimension of $F$
$ \left[bt^{2}\right] =\left[F\right] \left[b\right] =\left[\frac{F}{t^{2}}\right] $
$ \left[b\right] = \left[\frac{MLT^{-4}}{T^{2}}\right] $
$\Rightarrow \left[b\right] = \left[MLT^{-4}\right]$