Units & Measurement and Dimensions

Given, time-period of oscillation $T$ is

$ T \propto p^\alpha S^\beta E^\gamma \text { or } T=k p^\alpha S^\beta E^\gamma $

Now, subtituting dimensions of $T, p, S$ and $E$, we have

$ \left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=k\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^\alpha\left[\mathrm{ML}^{-3}\right]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]^\gamma $

Equating powers of similar terms, we have

$ \begin{array}{r} \alpha+\beta+\gamma=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -\alpha-3 \beta+2 \gamma=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ -2 \alpha-2 \gamma=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Solving, we get $\alpha=\frac{-5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$

Area of rectangle, A = lb = 15.2 $\times$ 2.9 = 44.08 cm²

Minimum possible measurement of scale = 0.1 cm

As we know, the product should have significant figures that are present in the measurement with least number of significant figures which is 2 in this case. So, area measured by scale = 44 cm².

As, RC is the time constant, so it has the dimension of time and V has the dimensions of $L{{dl} \over {dt}}$.

$\therefore$ Dimensions of ${L \over {RCV}},\left[ {{L \over {RCV}}} \right] = {{[L]} \over {[RC][V]}}$

${{[L]} \over {[T][L]\left[ {{A \over T}} \right]}} = {{[L][T]} \over {[T][L][A]}} = {1 \over {[A]}}$

So, $\left( {{L \over {RCV}}} \right)$ has the dimensions $[{A^{ - 1}}]$, i.e. same as the dimensions of ${1 \over {current}}$.