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

Answer 1

Answer: (C)

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