If the energy (E), velocity (v) and forc...

If the energy (E), velocity (v) and force (F) are taken as fundamental quantities, then what is the dimensional formula for mass?

a = 1, b =-2

a = 1, b=2

b = -2, c=-1

c = 0, a=1

Text Solution

Verified by Experts

The correct Answer is:

A, C

Let us assume that
`mpropE^(a)v^(b)F^(c)`
`m=kE^(a)v^(b)F^(c)" "......(i)`
where k is the constant of proportionality and is a dimensionless constant.
The dimensions of
Mass m= [M]
energy (E) `=[ML^(2)T^(-2)]`
Velocity (v) `=[LT^(-1)]`
Force (F) `=[MLT^(-2)]`
Putting these dimensions is eqn. (i), we get
`[M]=[ML^(2)T^(-2)]^(a)[LT^(-1)]^(b)[MLT^(-2)]^(c)`
Equating the dimensions on both sieds, we get
`a+c=1" ".....(ii)`
`2a+b+c=0" "......(iii)`
`-2a-b-2c=0" "........(iv)`
Solving (ii), (iii) and (iv), we get
a=1, b=2, c=0 `:.[m]=k[E][v]^(-2)`
The constant of proportionally k is found to be, 1. Therefore the formula becomes
`:.m=Ev^(-2)`

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If energy (E). velocity (V) and force (F) be taken as fundamental quantity, then what are the dimensions of mass:

`EV^2`

`EV^(-2)`

`FV^(-1)`

`FV^(-2)`

If the energy ( E) ,velocity (v) and force (F) be taken as fundamental quantities,then the dimension of mass will be

`[Fv^(-2)]`

`[Fv^(-1)]`

`[Ev^(-2)]`

`[Ev^(2)]`

If energy E , velocity v and time T are taken as fundamental quanties, the dimensional formula for surface tension is

`[Ev^(-2) T^(-2)]`

`[E^(2) vT^(-2)]`

`[Ev^(-2) T^(-1)]`

`[E^(-2) v^(-2) T^(-1)]`