Question

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

• A. $ [ML^{2} T^{-2}] $
• B. $ [L^{-1/2} T^{2}] $
• C. $ [L^{-1/2} T^{-1}] $
• D. $ [LT^{-2}] $

Answer 1

Answer: (B)

Given, $F=A\sqrt{x}+Bt^{2}$
Substituting dimension of force, distance and time in given expression
$[MLT^{-2}]=A[L]^{1/2}+B[T]^{2}$
Now, according to principle of homogeneity dimension of $F$ will be same as dimension of $A\sqrt{x}$ and that of $Bt^{2}$
$\therefore A\left[L\right]^{1 /2}=\left[MLT^{-2}\right]$
$\Rightarrow A=\frac{\left[MLT^{-2}\right]}{\left[L^{1/ 2}\right]}$
$=\left[ML^{+1/ 2} T^{-2}\right]$
Similarly, $B\left[T\right]^{2}=\left[MLT^{-2}\right]$
$B=\left[MLT^{-4}\right]$
$\therefore $ Dimension of $\frac{A}{B}=\frac{\left[ML^{1/ 2}T^{-2}\right]}{\left[ML T^{-4}\right]}$
$=\left[L^{-1/ 2} T^{2}\right]$