Question

Dimensional methods provide three major advantages in verification, derivation and changing the system of units. Any empirical formula that is derived based on this method has to be verified and proportionality constants found by experimental means. The presence or absence of certain factors – non dimensional constants or variables – cannot be identified by this method. So, every dimensionally correct relation cannot be taken as perfectly correct.
If \[\alpha \]kilogram, \[\beta \] meter and \[\gamma \] second ae the fundamental units, 1 cal can be expressed in new units as – \[[1cal=4.2J]\]
\[\begin{align}
& \text{A) }{{\alpha }^{-1}}{{\beta }^{2}}\gamma \\
& \text{B) }{{\alpha }^{-1}}{{\beta }^{-2}}\gamma \\
& \text{C) 4}\text{.2}{{\alpha }^{-1}}\beta \\
& \text{D) 4}\text{.2}{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}} \\
\end{align}\]

Answer 1

Hint: We need to understand the fundamental relation between the basic quantities such as mass, length and time with the energy. This relation can be extended to any unit of energy joules or calories and we can solve the problem easily.

Complete answer:
We know that the dimensional analysis is a technique to understand the relation of a given physical quantity with the basic physical quantities. We are given a set of new physical relations in terms of \[\alpha ,\beta \text{ and }\gamma \] which are to be related with the actual dimensions of the energy.
We are given that the unit calories is related to the joules in SI system as –
\[1cal=4.2J\]
Now, let us find out the basic physical quantities to which the joule is related to. We know that energy or work done is the product of the force and displacement, therefore we can get the units of energy in terms of these as –
\[\begin{align}
& E=F.s \\
& \therefore J=Nm \\
\end{align}\]
Now, we know that ‘m’ is the length which is already a basic physical quantity. The unit ‘N’ can be derived from Newton's laws of motion as the product of mass and acceleration. We can write the unit Joule further in terms of more basic quantities as –
\[\begin{align}
& E=F.s \\
& \Rightarrow E=mas \\
& \Rightarrow J=kgm{{s}^{-2}}m \\
& \therefore J=kg{{m}^{2}}{{s}^{-2}} \\
\end{align}\]
From the above equation we get the unit joule in terms of three basic units of mass, length and time. Now, let us convert this into new units. It is given that 4.2 J contributes to 1 cal. To balance the both sides of the dimensional formula, let us consider ‘x’ as the connection between the two as –
\[\begin{align}
& 4.2J=1cal \\
& \Rightarrow 4.2(kg{{m}^{2}}{{s}^{-2}})=x(\alpha kg{{(\beta m)}^{2}}{{(\gamma s)}^{-2}}) \\
& \Rightarrow x=4.2\dfrac{(kg{{m}^{2}}{{s}^{-2}})}{(\alpha kg{{(\beta m)}^{2}}{{(\gamma s)}^{-2}})} \\
& \therefore x=4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}} \\
\end{align}\]
This is the required solution for the new dimension for energy in calories.

The correct answer is option D.

Note:
The dimensional formula is the easiest approach to any physical quantity. It gives the basic physical quantities related to the given quantity so that we can understand the nature of the parameter and use them appropriately in situations.