Question

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

$\left[M{L}^3{T}^{-4}\right]$

The given equation is $F=\frac{\alpha -t^2}{\beta v^2}$

Dimension of $\alpha$ is same as that of square of time i.e. $\alpha =\left[T^2\right]$

Dimensoin of $\beta$ is $\beta =\frac{\alpha -t^2}{F{v}^2}=\frac{\left[T^2\right]}{\left[ML{T}^{-2}\right]\left[L^2{T}^{-2}\right]}=\left[M^{-1}{L}^{-3}{T}^6\right]$

Thus $\frac{\alpha }{\beta }=\frac{\left[T^2\right]}{\left[M^{-1}{L}^{-3}{T}^6\right]}=\left[M{L}^3{T}^{-4}\right]$