If the time period $t$ of the oscillation of a drop of liquid of density $d$,radius $r$,vibrating under surface tension $s$ is given by the formula $t = \sqrt{r^{2b} s^c d^{a/2}}$. It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$. The value of $b$ should therefore be

If the orbital velocity of a planet is given by $v = G^a M^b R^c$,then:

An artificial satellite is revolving around a planet of mass $M$ and radius $R$ in a circular orbit of radius $r$. From Kepler's third law about the period of a satellite around a common central body,the square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis that $T = \frac{k}{R}\sqrt{\frac{r^3}{g}}$,where $k$ is a dimensionless constant and $g$ is the acceleration due to gravity.

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 for $T$ is:

If momentum $(P)$,area $(A)$,and time $(T)$ are taken to be fundamental quantities,then the dimensional formula for energy is:

Let force $F = A \sin(Ct) + B \cos(Dx)$,where $x$ and $t$ are displacement and time respectively. The dimensions of $\frac{C}{D}$ are the same as the dimensions of