Question

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

Answer 1

We know the dimensions of
Moment of inertia, \([I]=[ML^2]\)
Force, \([F] =[MLT^{-2]}\)
Velocity, \([v]=[LT^{-1}]\)
Work, \([W]=[ML^2T^{-2}]\)
Length, \([L]=[L]\)
Given that
\(\displaystyle [Q]=\left[\dfrac{IFv^2}{WL^3}\right]\)
Substituting values in \([Q]\)
\(\Rightarrow [Q]=\left[\dfrac{[ML^2]\times [MLT^{-2}]\times [L^2T^{-2}]}{[ML^2T^{-2}]\times [L^3]}]\right]\)
\(\Rightarrow [Q]=[MT^{-2}]\) ......(i)
Further,
The dimension of surface tension \((T')\) is,
\([T']=\left[\dfrac{F}{L}\right]\)
\(\Rightarrow [T']=\left [\dfrac{MLT^{-2}}{L}\right]=\left[MT^{-2}\right]\)
......(ii)
The dimension of surface energy \((U)\) is
\([U]=\left[\dfrac{\text{Work done}}{\text{ Area}}\right]\)
\(\Rightarrow [U]=\left[\dfrac{ML^2T^{-2}}{L^2}\right]\)
\(\Rightarrow [U]=[MT^{-2}]\) ....(iii)

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

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