Question

If velocity is given as $v=\alpha -\beta x$, then dimensional formula of $\beta {\alpha }^2$ is

• A. $\left[M^{0}L{T}^{-2}\right]$
• B. $\left[M^1{L}^2{T}^{-2}\right]$
• C. $\left[M^0{L}^2{T}^{-3}\right]$
• D. $\left[M^0{L}^3{T}^{-3}\right]$

Answer 1

The correct option is C $\left[M^0{L}^2{T}^{-3}\right]$

Dimention of $\left[\beta x\right]=\left[L{T}^{-1}\right]\Rightarrow \left[\beta \right]=\left[T^{-1}\right]$
Dimention of $\left[\alpha \right]=\left[L{T}^{-1}\right]\Rightarrow \left[{\alpha }^2\right]=\left[L^2{T}^{-2}\right]$
$\therefore \left[{\alpha }^2\beta \right]=\left[L^2{T}^{-3}\right]$