In the expression $A=B+\frac{C}{D+E}$,the dimensions of physical quantities $B$ and $C$ are $[L^{1} M^{0} T^{-1}]$ and $[L^{1} M^{0} T^{0}]$ respectively. The dimensions of quantities $A, D$ and $E$ are

Which of the following relations cannot be deduced using dimensional analysis? [the symbols have their usual meanings]

$A$ quantity $x$ is given by $(IF v^{2} / WL^{4})$ in terms of moment of inertia $I$,force $F$,velocity $v$,work $W$,and length $L$. The dimensional formula for $x$ is the same as that of:

In terms of potential difference $V$,electric current $I$,permittivity $\varepsilon_0$,permeability $\mu_0$,and speed of light $c$,the dimensionally correct equation$(s)$ is(are):
$(A)$ $\mu_0 I^2 = \varepsilon_0 V^2$
$(B)$ $\varepsilon_0 I = \mu_0 V$
$(C)$ $I = \varepsilon_0 cV$
$(D)$ $\mu_0 cI = V$

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: $[position] = [X^\alpha]$; $[speed] = [X^\beta]$; $[acceleration] = [X^p]$; $[linear momentum] = [X^q]$; $[force] = [X^r]$. Then -
$(A)$ $\alpha + p = 2\beta$
$(B)$ $p + q - r = \beta$
$(C)$ $p - q + r = \alpha$
$(D)$ $p + q + r = \beta$

If energy $(E)$,velocity $(V)$,and time $(T)$ are chosen as the fundamental quantities,the dimensional formula of surface tension will be