Question

A force is represented by $F = ax^2 + b t^{1/2}$ Where $x =$ distance and $t =$ time. The dimensions of $\frac{b^2}{a}$ are:

• A. $[ML^3T^{-3}]$
• B. $[MLT^{-2}]$
• C. $[ML^{-1}T^{-1}]$
• D. $[ML^2T^3]$

Answer 1

Answer: (A)

$[ML^3T^{-3}]$

Answer 2

Given:

$F = ax^2 + bt^{1/2}$

The dimensions of $a$ are given by:

$[a] = \left[ \frac{F}{x^2} \right] = [MLT^{-2}][L]^{-2} = [ML^{-1}T^{-2}]$

The dimensions of $b$ are given by:

$[b] = \left[ \frac{F}{t^{1/2}} \right] = [MLT^{-2}][T^{-1/2}] = [MLT^{-5/2}]$

Now, the dimensions of $\frac{b^2}{a}$ are:

$\left[ \frac{b^2}{a} \right] = \frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]} = [ML^3T^{-3}]$