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 $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
$A,B$
$A,C$
$A,D$
$B,C$
Frequency is the function of density $(\rho )$, length $(a)$ and surface tension $(T)$. Then its value is
Turpentine oil is flowing through a tube of length $l$ and radius $r$. The pressure difference between the two ends of the tube is $P .$ The viscosity of oil is given by $\eta=\frac{P\left(r^{2}-x^{2}\right)}{4 v l}$ where $v$ is the velocity of oil at a distance $x$ from the axis of the tube. The dimensions of $\eta$ are
$Assertion$ : Specific gravity of a fluid is a dimensionless quantity.
$Reason$ : It is the ratio of density of fluid to the density of water
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in $SI$ units.
($1$) The relation between $[E]$ and $[B]$ is
$(A)$ $[ E ]=[ B ][ L ][ T ]$ $(B)$ $[ E ]=[ B ][ L ]^{-1}[ T ]$ $(C)$ $[ E ]=[ B ][ L ][ T ]^{-1}$ $(D)$ $[ E ]=[ B ][ L ]^{-1}[ T ]^{-1}$
($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[ T ]^{-2}$ $(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[ T ]^2$ $(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^2[ T ]^{-2}$ $(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^{-2}[ T ]^2$
Give the answer or quetion ($1$) and ($2$)
The frequency $(v)$ of an oscillating liquid drop may depend upon radius $(r)$ of the drop, density $(\rho)$ of liquid and the surface tension $(s)$ of the liquid as : $v=r^{ a } \rho^{ b } s ^{ c }$. The values of $a , b$ and $c$ respectively are