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

Answer 1

Answer: (B)

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