Find the dimension formula of $\frac{a}{b}$ in the equation $F=a \sqrt{x}+b t^2$, where $F$ is a force, $x$…

Find the dimension formula of $\frac{a}{b}$ in the equation $F=a \sqrt{x}+b t^2$, where $F$ is a force, $x$ is distance and $t$ is time.
  1. $\left[M^0 L^{-1 / 2} T^2\right]$
  2. $\left[M^0 L^0 T^{3 / 2}\right]$
  3. $\left[M^0 L^1 T^{-4}\right]$
  4. $\left[M^0 L^{-3 / 2} T^4\right]$

Solution

$F=a \sqrt{x}+b t^2$
By principle of homogeneity $\begin{aligned} & {[\mathrm{F}]=[\mathrm{a} \sqrt{\mathrm{x}}] \Rightarrow[\mathrm{a}]=\frac{[\mathrm{F}]}{[\sqrt{\mathrm{x}}]}} \\ & {[\mathrm{F}]=\left[\mathrm{bt}^2\right] \Rightarrow[\mathrm{b}]=\frac{[\mathrm{F}]}{\left[\mathrm{t}^2\right]}} \end{aligned}$ $\begin{aligned} \therefore\left[\frac{\mathrm{a}}{\mathrm{b}}\right]=\frac{[\mathrm{F}]}{[\sqrt{\mathrm{x}}]} \times \frac{\left[\mathrm{t}^2\right]}{[\mathrm{F}]} & =\frac{\left[\mathrm{t}^2\right]}{\left[\frac{1}{\frac{1}{2}}\right]}=\frac{\left[\mathrm{T}^2\right]}{\left[\frac{1}{\mathrm{~L}^2}\right]} \\ & =\left[\mathrm{M}^0 \mathrm{~L}^{-1 / 2} \mathrm{~T}^2\right]\end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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