In two different systems of units acceleration is represented by the same number, while velocity is represented by numbers in the ratio $1:3$. The ratio of unit of length and time are
A) $\dfrac{1}{3},\dfrac{1}{9}$
B) $\dfrac{1}{9},\dfrac{1}{3}$
C) $1,1$
D) $\dfrac{1}{3},\dfrac{1}{3}$

Hint: Replacement of fundamental units by the dimensional formula we are able to get the physical quantity. Proportionality constants cannot be determined by dimensional analysis.

Complete step by step answer:
Let us consider two systems of units to be 1 and 2 respectively.
Given, acceleration is represented in two different systems of units by the same number. That is, $\left[ {{a_1}} \right] = \left[ {{a_2}} \right]$ ………………(1)
Acceleration is a ratio of change in velocity to the time taken.
$a = \dfrac{{dv}}{{dt}}$
Representing acceleration, velocity and time in two different systems of units. We get
${a_1} = \dfrac{{{V_1}}}{{{T_1}}}$ and
${a_2} = \dfrac{{{V_2}}}{{{T_2}}}$
From equation (1), we get
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
Rearranging the above equation,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$…………………(2)
Given, in the question, velocity is represented in two different system of unit by numbers in the ratio 1:3
That is, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{1}{3}$
From this equation (2) becomes, $\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{3}$………………(3)
We have, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{1}{3}$…...........(4)
Dimensional formula for velocity is given by,$\left[ {L{T^{ - 1}}} \right]$
Now apply dimensional formula, in two different system of units to the equation (4) we get
$\dfrac{{\left[ {{L_1}{T_1}^{ - 1}} \right]}}{{\left[ {{L_2}{T_2}^{ - 1}} \right]}} = \dfrac{1}{3}$
$\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^{ - 1}} = \dfrac{1}{3}$
Above equation can also be written as,
$\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]\dfrac{1}{{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]}} = \dfrac{1}{3}$
$\rightarrow\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right] = \dfrac{1}{3} \times \left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]$
From equation (3) we have, $\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{3}$
Then, $\rightarrow \left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right] = \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$
$\therefore $ The ratio of unit of time and length are $\dfrac{1}{9}:\dfrac{1}{3}$

Thus, the correct option is (b).

Additional information:
The expression showing the powers to which the fundamental quantities must be raised to represent a physical quantity is called dimensional formula.
Constants having dimensional formulas are called Dimensional constants.
Eg : Planck’s constant, speed of light, Universal gravitational constant..
Physical quantities having no dimensional formula are called Dimensionless quantities.
Eg : Angle, Strain, Relative Density.
Limitations of Dimensional analysis.
Proportionality constants cannot be determined by dimensional analysis.
Formulae containing non- algebraic functions like sin, cos, log, exponential etc., cannot be derived.

Note: Acceleration is the ratio of change in velocity to the total time taken
Velocity is the ratio of change in displacement to the time taken.
Displacement is the shortest distance directed from initial position to final position.
Distance is the actual path length covered between initial and final positions.