The dimensional formula of $\frac{1}{\mu_{0} \varepsilon_{0}}$ is . . . . . . .

In the equation $y = x^2 \cos^2 \left( 2 \pi \frac{\beta \gamma}{\alpha} \right)$,the units of $x, \alpha, \beta$ are $m, s^{-1}$,and $(ms^{-1})^{-1}$ respectively. The units of $y$ and $\gamma$ are:

An expression for a dimensionless quantity $P$ is given by $P = \frac{\alpha}{\beta} \log_{e} \left( \frac{kt}{\beta x} \right)$,where $\alpha$ and $\beta$ are constants,$x$ is distance,$k$ is the Boltzmann constant,and $t$ is the temperature. Then the dimensions of $\alpha$ will be:

$A$ physical quantity $z$ depends on four observables $a, b, c$ and $d$ as $z = \frac{a^2 b^{2/3}}{\sqrt{c} d^3}$. The percentage errors in the measurement of $a, b, c$ and $d$ are $2\%, 1.5\%, 4\%$ and $2.5\%$ respectively. The percentage error in $z$ is $......\%$. (in $.5$)

The dimension of stopping potential $V_{0}$ in the photoelectric effect in terms of Planck's constant $h$,speed of light $c$,gravitational constant $G$,and ampere $A$ is:

The velocity $v$ of a particle at time $t$ is given by $v = at + \frac{b}{t + c}$,where $a, b,$ and $c$ are constants. What are the dimensions of $a, b,$ and $c$ respectively?