Q.If momentum ($P$), area ($A$) and time ($T$) are taken to be fundamental quantities, then energy has the dimensional formula

The key idea here is Dimensional Analysis, where we express the dimensions of a derived quantity in terms of new fundamental quantities.

1. First, write the dimensional formulas of the given fundamental quantities ($P, A, T$) and the quantity whose dimensions are sought (Energy, $E$) in terms of the standard fundamental dimensions ($M, L, T$):

   • Momentum ($P$) = $M L T^{-1}$
   • Area ($A$) = $L^2$
   • Time ($T$) = $T$
   • Energy ($E$) = $M L^2 T^{-2}$

2. Assume the dimensional formula for energy in terms of $P, A, T$ is $E = P^a A^b T^c$. Substitute the standard dimensions:

   $M L^2 T^{-2} = (M L T^{-1})^a (L^2)^b (T)^c$

   $M L^2 T^{-2} = M^a L^{a+2b} T^{-a+c}$

3. Equate the powers of $M, L, T$ on both sides:

   • For $M$: $a = 1$
   • For $L$: $a + 2b = 2$
   • For $T$: $-a + c = -2$ …

When we talk about "fundamental quantities," we're referring to a set of independent physical quantities from which all other physical quantities can be derived. Traditionally, we use Mass (M), Length (L), and Time (T) as our fundamental quantities. However, the choice of fundamental quantities is arbitrary; any set of independent quantities can serve this purpose, as long as they can be used to express all other quantities.

This problem asks us to treat momentum ($P$), area ($A$), and time ($T$) as the new fundamental quantities. Our goal is to express the dimensions of energy ($E$) in terms of these new fundamental quantities. The core idea behind dimensional analysis is that any physical equation must be dimensionally consistent. This means the dimensions on both sides of an equation must be identical.

We will assume that energy can be expressed as a product of powers of these new fundamental quantities, like $E \propto P^a A^b T^c$. By comparing the standard dimensions of energy with the combined dimensions of $P^a A^b T^c$, we can solve for the exponents $a$, $b$, and $c$.

Here's how we approach it step-by-step:

1. Identify the standard dimensions of the derived quantity (Energy) and the new fundamental quantities.

   • Energy ($E$): Energy is work done, which is force times distance. Force is mass times acceleration. So, $[E] = [M L T^{-2}] \cdot [L] = [M L^2 T^{-2}]$.
   • Momentum ($P$): Momentum is mass times velocity. So, $[P] = [M L T^{-1}]$.
   • Area ($A$): Area is length squared. So, $[A] = [L^2]$.
   • Time ($T$): Time is a fundamental quantity in both systems. So, $[T] = [T]$.

2. Express Energy as a product of powers of the new fundamental quantities.

   We assume that the dimensional formula for energy in terms of $P$, $A$, and $T$ can be written as:

where $a$, $b$, and $c$ are the exponents we need to determine.

3. Substitute the standard dimensions into the equation.

Now, replace each quantity with its standard dimensional formula in terms of M, L, T:

4. Simplify the right-hand side by combining the powers of M, L, and T.

5. Equate the powers of M, L, and T on both sides of the equation. …

Method: Dimensional Analysis Using Given Fundamental Quantities

Concept:

When standard fundamental quantities (mass, length, time) are replaced by new ones (here momentum $P$, area $A$, time $T$), we express the derived quantity's dimension in terms of these new fundamentals by relating their standard dimensional formulas.

Steps:

1. Write standard dimensional formulas

   • Momentum: $[P] = M^1 L^1 T^{-1}$
   • Area: $[A] = L^2$
   • Time: $[T] = T^1$
   • Energy: $[E] = M^1 L^2 T^{-2}$

2. Assume energy dimension in new system

   Let $[E] = P^a \, A^b \, T^c$

3. Substitute standard dimensions

4. Equate powers of M, L, T
   • For $M$: $1 = a$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that the given set is non-standard

Students often jump straight to writing the standard dimensional formula for energy:

$[E] = [M L^2 T^{-2}]$

and then try to express $M$ and $L$ in terms of $P$, $A$, and $T$ — but they do it incorrectly or incompletely.

How to avoid:

• First, write the given fundamental quantities and their own dimensions in terms of $M$, $L$, $T$:
  • Momentum: $[P] = [M L T^{-1}]$
  • Area: $[A] = [L^2]$
  • Time: $[T] = [T]$
• Then, express $M$ and $L$ in terms of $P$, $A$, $T$ by solving the system of equations.

Mistake 2: Incorrectly solving the dimensional equations

A common error is to assume $M = P \cdot T$ or $L = \sqrt{A}$ without checking consistency with all three equations.

How to avoid:

Set up the system properly:

From $[P] = [M L T^{-1}]$ and $[A] = [L^2]$, we get:

• $L = A^{1/2}$
• Substitute into $P$: $[P] = [M \cdot A^{1/2} \cdot T^{-1}]$ $\Rightarrow M = P \cdot A^{-1/2} \cdot T$

Now substitute $M$ and $L$ into $[E] = [M L^2 T^{-2}]$:

So the correct answer is (D).

Mistake 3: Mixing up powers of $T$ when substituting

Students often lose track of the exponent on $T$ when moving from $P$ to $M$.

How to avoid:

Write every step explicitly with exponents:

• $P^1 = M^1 L^1 T^{-1}$
• $A^1 = L^2$
• $T^1 = T^1$

Then solve for $M$ and $L$ by equating exponents — treat it like solving three linear equations in three unknowns.

Mistake 4: Not checking the final answer against all options …