Question

If density d, frequency f and velocity v are taken as fundamental quantities then the dimensional formula for kinetic energy should be

Answer 1

Here's your answer :
We know kinetic energy = 1/2 ×m× v ²
SI unit of mass = kilograms
SI unit of velocity = m/s
so dimensional formula of K.E= [ L²M¹T^-2]
Hope it helps you.

Answer 2

Answer:

The dimensional formula for kinetic energy should be $k.E = \rho^1 f^{-3} v^5$.

Explanation:

Given that,

If density ρ, frequency f and velocity v are taken as fundamental quantities then the dimensional formula for kinetic energy

The dimension formula of density frequency and velocity is

Density $\rho = [M^1L^{-3}]$

Frequency$f =[T^{-1}]$

Velocity $v = [L^1T^{-1}]$

Let us take the kinetic energy

$K.E=[M^1L^2T^{-2}]$

Now, Kinetic energy can be written as

$k.E = d^a f^b v^c$

$k.E=[M^1L^{-3}]^a [T^{-1}]^b [L^1T^{-1}]^c$

$[M^1L^2T^{-2}] = [M^1L^{-3}]^a [T^{-1}]^b [L^1T^{-1}]^c$

Equating the value of a,b and c

a = 1...(I)

-3 a+c=2....(II)

-b-c=-2.....(III)

Put the value of a in equation (II)

-3+c=2

c = 5

Now put the value of c in equation (III)

-b-5=-2

b = -3

Now, the kinetic energy is

$k.E = \rho^1 f^{-3} v^5$

Hence, The dimensional formula for kinetic energy should be $k.E = \rho^1 f^{-3} v^5$.