Physical World Units and Measurements - Result Question 9
9. If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:
(a) $[EV^{-1} T^{-2}]$
(b) $[EV^{-2} T^{-2}]$
(c) $[E^{-2} V^{-1} T^{-3}]$
(d) $[EV^{-2} T^{-1}]$
[2015]
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Answer:
Correct Answer: 9. (b)
Solution:
- (b) As we know that, surface tension(s) =
$\frac{\text{ Force }[F]}{\text{ Length }[L]}$
So, $[S]=\frac{[M L T^{-2}]}{[L]}=[MT^{-2}]$
Energy, $(E)=$ Force $\times$ Displacement,
$[E]=[ML^{2} T^{-2}]$
Velocity $(V)=\frac{\text{ displacement }}{\text{ time }}$
$ [V]=[LT^{-1}] $
Let surface tension expressed as,
$s=E^{a} V^{b} T^{c}$ where $a, b, c$ are constant.
Put the value
$ \frac{[MLT^{-2}]}{[L]}=[ML^{2} T^{-2}]^{a}[\frac{L}{T}]^{b}[T]^{c} $
From the principle of homogeneity, Equating the dimension of LHS and RHS
$[ML^{0} T^{-2}]=[M^{a} L^{2 a+b} T^{-2 a-b+c}]$
$\Rightarrow a=1,2 a+b=0,-2 a-b+c=-2$
$\Rightarrow a=1, b=-2, c=-2$
Hence, the dimensions of surface tension are $[EV^{-2} T^{-2}]$
Length, mass and time are arbitrarily chosen as fundamental quantities in mechanics. In fact any three quantities in mechanics can be termed as fundamental as all other quantities can be expressed in terms of these. If force (F) and acceleration (a) are taken as fundamental quantities, then mass will be defined as force $(F)$ and acceleration (a) will be termed as derived quantity.
