From the following,the quantity (constructed from the basic constants of nature) that has the dimensions of length,as well as the correct order of magnitude for a typical atomic size,is:

The equation of state of a real gas is given by $(P+\frac{a}{V^2})(V-b)=RT$,where $P, V$ and $T$ are pressure,volume and temperature respectively and $R$ is the universal gas constant. The dimensions of $\frac{a}{b^2}$ are similar to that of:

Velocities $(V)$ and accelerations $(a)$ in two systems of units $1$ and $2$ are related as $V_2 = \frac{n}{m^2} V_1$ and $a_2 = \frac{a_1}{mn}$ respectively. Here $m$ and $n$ are constants. Dimensionally,the relations between distances ($S_1$ and $S_2$) and times ($t_1$ and $t_2$) in the two systems are respectively:

$A$ book with many printing errors contains four different formulas for the displacement $y$ of a particle undergoing a certain periodic motion:
$(a) \; y = a \sin \left(\frac{2 \pi t}{T}\right)$
$(b) \; y = a \sin v t$
$(c) \; y = \left(\frac{a}{T}\right) \sin \frac{t}{a}$
$(d) \; y = (a \sqrt{2}) \left(\sin \frac{2 \pi t}{T} + \cos \frac{2 \pi t}{T}\right)$
($a =$ maximum displacement of the particle,$v =$ speed of the particle,$T =$ time-period of motion). Rule out the wrong formulas on dimensional grounds.

$A$ liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let $R$ be the radius of its largest horizontal section. $A$ small disturbance causes the drop to vibrate with frequency $v$ about its equilibrium shape. By dimensional analysis,the ratio $\frac{v}{\sqrt{\sigma / \rho R^3}}$ can be (Here,$\sigma$ is surface tension,$\rho$ is density,$g$ is acceleration due to gravity and $k$ is an arbitrary dimensionless constant)

$A$ famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{0}$ of a particle in terms of its speed $v$ and the speed of light $c$. (This relation first arose as a consequence of special relativity due to Albert Einstein). $A$ boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes:
$m = \frac{m_{0}}{(1 - v^{2})^{1/2}}$
Guess where to put the missing $c$.