Question

The dimensions of ‘$k$’ in the relation of $V = kavt$ is:
(Where $'V'$volume of a liquid passing through any point in time is $t$, $'a'$ is area of cross section, is the velocity of the liquid)
A. ${M^1}{L^2}{T^{ - 1}}$.
B. ${M^1}{L^1}{T^{ - 1}}$.
C. ${M^0}{L^0}{T^{ - 1}}$
D. ${M^0}{L^0}{T^0}$.

Answer 1

Hint: In the solution use the measurable unit of the volume, cross sectional area, time, and velocity for the determination of their dimensional formula and equate with the equation that is given in the question so that we can obtain correct answer.

Complete step by step solution:
Given:
The volume of the liquid passing through time $t$ is $V$.
The area of cross section is $a$.
The velocity of the liquid is $v$.

The dimensional formula of volume is,
$V = \left[ {{M^3}{L^0}{T^0}} \right]$ (1)

Here, $M$ is the mass, $L$ is the length and $T$ is the time.

The dimensional formula of the time is,
$T = \left[ {{M^0}{L^0}{T^1}} \right]$ (2)

The dimensional formula of the cross sectional area is,
$a = \left[ {{M^0}{L^2}{T^0}} \right]$ (3)

The dimensional formula of the velocity is,
$V = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$ (4)

By substituting the dimensional formulas of $V$, $a$, $v$ and $t$ from (1), (2), (3) and (4) in the given relation of$V$, the dimensions of $k$ becomes,

$\begin{array}{l}
V = kavt\\
\left[ {{M^0}{L^3}{T^0}} \right] = k\left[ {{M^0}{L^2}{T^0}} \right]\left[ {{M^0}{L^1}{T^1}} \right]\left[ {{M^0}{L^0}{T^1}} \right]\\
k = \left[ {{M^0}{L^{\left( {3 - 2 - 1 - 0} \right)}}{T^{\left( {0 - 0 + 1 - 1} \right)}}} \right]\\
k = \left[ {{M^0}{L^0}{T^0}} \right]
\end{array}$

Therefore, the option (d) is the correct answer that is ${M^0}{L^0}{T^0}$.

Note: Remember the measurable unit of volume, cross section area, time and velocity that are ${m^3}$, ${m^2}$, $s$ and $m/s$, and obtain the dimensional formula according to these units. The incorrect measurable unit can give the different answers which may not present in the given options.