If the time period $(T)$ of vibration of a liquid drop depends on surface tension $(S)$, radius $(r)$ of the drop and density $(\rho )$ of the liquid, then the expression of $T$ is
$T = k\sqrt {\rho {r^3}/S} $
$T = k\sqrt {{\rho ^{1/2}}{r^3}/S} $
$T = k\sqrt {\rho {r^3}/{S^{1/2}}} $
None of these
In a new system of units energy $(E)$, density $(d)$ and power $(P)$ are taken as fundamental units, then the dimensional formula of universal gravitational constant $G$ will be .......
The position of a particle at time $t$ is given by the relation $x(t) = \left( {\frac{{{v_0}}}{\alpha }} \right)\,\,(1 - {e^{ - \alpha t}})$, where ${v_0}$ is a constant and $\alpha > 0$. The dimensions of ${v_0}$ and $\alpha $ are respectively
If $P$ represents radiation pressure, $c$ represents speed of light and $Q$ represents radiation energy striking a unit area per second, then non-zero integers $x,\,y$ and $z$ such that ${P^x}{Q^y}{c^z}$ is dimensionless, are
If momentum $(P)$, area $(A)$ and time $(T)$ are taken to be fundamental quantities then energy has dimensional formula