The value of gravitational acceleration in the $C.G.S.$ system is $980 \; cm/s^2$. Find the value of $g$ in the $M.K.S.$ system.

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 -
$(A)$ $\alpha + p = 2\beta$
$(B)$ $p + q - r = \beta$
$(C)$ $p - q + r = \alpha$
$(D)$ $p + q + r = \beta$

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 velocity of light $(c)$,gravitational constant $(G)$,and Planck's constant $(h)$ are chosen as fundamental units,then the dimensions of mass in this new system are:

$A$ calorie is a unit of heat or energy and it equals about $4.2 \; J$ where $1 \; J = 1 \; kg \; m^2 \; s^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha \; kg$, the unit of length equals $\beta \; m$, and the unit of time equals $\gamma \; s$. Show that a calorie has a magnitude $4.2 \; \alpha^{-1} \beta^{-2} \gamma^2$ in terms of the new units.

The frequency $(v)$ of an oscillating liquid drop may depend upon the radius $(r)$ of the drop,the density $(\rho)$ of the 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: