Question

If density $D$ , frequency $F$ , and velocity $V$ are taken as fundamental quantities then the dimensional formula for kinetic energy should be
(A) $\left[ {D{F^{ - 3}}{V^5}} \right]$
(B) $\left[ {{D^{ - 2}}{F^2}{V^{ - 3}}} \right]$
(C) $\left[ {{D^{ - 3}}{F^5}V} \right]$
(D) $\left[ {DF{V^{ - 3}}} \right]$

Answer 1

Hint : To solve this question, we have to assume a dimensional formula of the kinetic energy in the form of three unknown variables. Then we need to write the dimensional formula for the three quantities, the density, the frequency, and the velocity. Then we have to put them into the assumed dimensional formula to get the final answer.

Formula used: The formulae used to solve this question are given by
$\Rightarrow D = \dfrac{M}{V}$ , here $D$ is the density, $M$ is the mass, and $V$ is the volume.
$\Rightarrow F = \dfrac{1}{T}$ , here $F$ is the frequency, and $T$ is the time period.
$\Rightarrow V = \dfrac{d}{t}$ , here $V$ is the velocity, $d$ is the displacement covered in time $t$ .
$\Rightarrow K = \dfrac{1}{2}m{V^2}$ , here $K$ is the kinetic energy of a body of mass $m$ moving with velocity $V$ .

Complete step by step answer
Let us assume the dimensional formula of the kinetic energy be represented as follows
$\Rightarrow \left[ K \right] = \left[ {{D^x}{F^y}{V^z}} \right]$ ………………………...(1)
Now, we know that the density is given by the formula
$\Rightarrow D = \dfrac{M}{V}$
Taking the dimensions of both the sides, we get
$\Rightarrow \left[ D \right] = \dfrac{{\left[ {{M^1}{L^0}{T^0}} \right]}}{{\left[ {{M^0}{L^3}{T^0}} \right]}}$
$\Rightarrow \left[ D \right] = \left[ {{M^1}{L^{ - 3}}{T^0}} \right]$ ………………………...(2)
Now the frequency is given by
$\Rightarrow F = \dfrac{1}{T}$
Taking the dimensions of both the sides, we get
$\Rightarrow \left[ F \right] = \dfrac{1}{{\left[ {{M^0}{L^0}{T^1}} \right]}}$
$\Rightarrow \left[ F \right] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]$ ………………………...(3)
The velocity is given by
$\Rightarrow V = \dfrac{d}{t}$
Taking the dimensions of both the sides, we get
$\Rightarrow \left[ V \right] = \dfrac{{\left[ {{M^0}{L^1}{T^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}} \right]}}$
$\Rightarrow \left[ V \right] = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$ ………………………...(4)
Finally, the kinetic energy is given by
$\Rightarrow K = \dfrac{1}{2}m{V^2}$
Taking the dimensions of both the sides, we get
$\Rightarrow \left[ K \right] = \left[ {{M^1}{L^0}{T^0}} \right]{\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^2}$
$\Rightarrow \left[ K \right] = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$ ………………………...(5)
Substituting (2) (3) (4) and (5) in (1) we get
$\Rightarrow \left[ {{M^1}{L^2}{T^{ - 2}}} \right] = {\left[ {{M^1}{L^{ - 3}}{T^0}} \right]^x}{\left[ {{M^0}{L^0}{T^{ - 1}}} \right]^y}{\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^z}$
$\Rightarrow \left[ {{M^1}{L^2}{T^{ - 2}}} \right] = \left[ {{M^x}{L^{ - 3x + z}}{T^{ - y - z}}} \right]$
Equating the exponents of mass, length and time of both sides, we get
$\Rightarrow x = 1$ ………………………...(6)
$\Rightarrow - 3x + z = 2$ ………………………...(7)
$\Rightarrow - y - z = - 2$ ………………………...(8)
On solving (6) (7) and (8) we get
$\Rightarrow x = 1,y = - 3,z = 5$
So from (1) the dimensional formula for the kinetic energy becomes
$\Rightarrow \left[ K \right] = \left[ {D{F^{ - 3}}{V^5}} \right]$
Thus, the dimensional formula of the kinetic energy is $\left[ {D{F^{ - 3}}{V^5}} \right]$ .
Hence, the correct answer is option A.

Note
While deriving a formula using dimensional analysis, be careful while writing the dimensions of each quantity. We can use any physical formula of each quantity to find its dimensions. Always prefer to use the formula which relates the quantity with more fundamental quantities.