Question

If a composite physical quantity in terms of moment of inertia $\left(I\right)$, force $\left(F\right)$, velocity $\left(v\right)$, work $\left(W\right)$ and length $\left(L\right)$ is defined as $Q=\frac{IF{v}^2}{W{L}^3}$, then $Q$ may be

• A. Surface tension
• B. Surface energy
• C. Both (a) and (b)
• D. None of these

Answer 1

The correct option is C Both (a) and (b)

We know the dimensions of
Moment of inertia, $\left[I\right]=\left[M{L}^2\right]$
Force, $\left[F\right]=[ML{T}^{-2]}$
Velocity, $\left[v\right]=\left[L{T}^{-1}\right]$
Work, $\left[W\right]=\left[M{L}^2{T}^{-2}\right]$
Length, $\left[L\right]=\left[L\right]$
Given that
$\left[Q\right]=\left[\frac{IF{v}^2}{W{L}^3}\right]$
Substituting values in $\left[Q\right]$
$\Rightarrow \left[Q\right]=\left[\frac{\left[M{L}^2\right]\times \left[ML{T}^{-2}\right]\times \left[L^2{T}^{-2}\right]}{\left[M{L}^2{T}^{-2}\right]\times \left[L^3\right]}\right]]$
$\Rightarrow \left[Q\right]=\left[M{T}^{-2}\right]$ ......(i)
Further,
The dimension of surface tension $\left(T^{\prime }\right)$ is,
$\left[T^{\prime }\right]=\left[\frac{F}{L}\right]$
$\Rightarrow \left[T^{\prime }\right]=\left[\frac{ML{T}^{-2}}{L}\right]=\left[M{T}^{-2}\right]$
......(ii)
The dimension of surface energy $\left(U\right)$ is
$\left[U\right]=\left[\frac{\text{Work done}}{\text{Area}}\right]$
$\Rightarrow \left[U\right]=\left[\frac{M{L}^2{T}^{-2}}{L^2}\right]$
$\Rightarrow \left[U\right]=\left[M{T}^{-2}\right]$ ....(iii)

From (i), (ii), and (iii), we can see that $Q$ may be surface tension or surface energy.
Hence, option (c) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - From this problem, it is}\\ \text{evident that if dimensions are given,}\\ \text{the physical quantity may or may}\\ \text{not be unique.}\\ \text{.}\end{array}$