Question

The dimensions of $\frac{\alpha }{\beta }$ in the equation $F=\frac{\alpha -t^2}{\beta v^2}$, where $F$ is the force, $v$ is velocity and $t$ is time, is

• A. $\left[ML{T}^{-1}\right]$
• B. $\left[M{L}^{-1}{T}^{-2}\right]$
• C. $\left[M{L}^3{T}^{-4}\right]$
• D. $\left[M{L}^2{T}^{-4}\right]$

Answer 1

Answer: (C)

$F=\frac{\alpha -t^2}{\beta v^2}$
Dimensionally, $\alpha =\left[T^2\right]$
$\left[ML{T}^{-2}\right]=\frac{\left[T^2\right]}{\beta \left[L^2{T}^{-2}\right]}$
$\beta =\frac{T^2}{\left[ML{T}^{-2}\cdot L^2{T}^{-2}\right]}$
$\Rightarrow \beta =\left[M^{-1}{L}^{-3}{T}^6\right]$
Dimensions of $\frac{\alpha }{\beta }=\frac{T^2}{M^{-1}{L}^{-3}{T}^6}$
$=\left[M{L}^3{T}^{-4}\right]$