Question

A force is given by $F=at+b{t}^2$, where $t$ is time, the dimensions of $a$ and $b$ are respectively :

• A. $\left[ML{T}^{-4}\right]$ and $\left[ML{T}^{-1}\right]$
• B. $\left[ML{T}^{-1}\right]$ and $\left[ML{T}^0\right]$
• C. $\left[ML{T}^{-3}\right]$ and $\left[ML{T}^{-4}\right]$
• D. $\left[ML{T}^{-3}\right]$ and $\left[ML{T}^0\right]$

Answer 1

Answer: (C)

$\left[ML{T}^{-3}\right]$ and $\left[ML{T}^{-4}\right]$

$\Rightarrow$ Since $F=at+b{t}^2---\left(1\right)$

$\therefore$ Dimension of $at$ and $b{t}^2$ must be equal to force only.

Hence $\left[F\right]=\left[M^1{L}^1{T}^{-2}\right]---\left(2\right)$

$F$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \left[at\right]=a\left[T\right]=\left[F\right]$

$\therefore a\left[T\right]=\left[M^1{L}^1{T}^{-2}\right]$

$\therefore \left[a\right]=\left[M^1{L}^1{T}^{-3}\right]$

and per $b{t}^2=b\left[T^2\right]=\left[M^1{L}^1{T}^{-2}\right]$

$\therefore \left[b\right]=\left[M^1{L}^1{T}^{-4}\right]$