Question

What is the dimensions of $\frac{A}{B}$ in the relation $F=A\sqrt{x}+B{t}^2$, where $F$ is the force, $x$ is the distance and $t$ is the time?

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

Answer 1

The correct option is B $\left[L^{-1/2}{T}^2\right]$

Given relation is $F=A\sqrt{x}+B{t}^2$
$\Rightarrow \left[F\right]=\left[A\sqrt{x}\right]=\left[B{t}^2\right]$
$\Rightarrow \frac{\left[A\right]}{\left[B\right]}=\frac{\left[t^2\right]}{[x{]}^{\frac{1}{2}}}=\left[L^{-1/2}{T}^2\right]$