If momentum $ (P)$, area $(A)$ and time $(T)$ are taken to be the fundamental quantities then the dimensional formula for energy is :

Question

The area enclosed between the parabola $ y^2 = 4x $ and the line $ y = 2x - 4 $ is:

Answer 1

We are given that momentum $ (P) $, area $ (A) $, and time $ (T) $ are taken as the fundamental quantities. We need to find the dimensional formula for energy in terms of these quantities.

The dimensional formula for energy is generally given as $ [M^1L^2T^{-2}] $ in terms of mass $(M)$, length $(L)$, and time $(T)$.

Let's express each of these in terms of the given fundamental quantities:

Momentum $ (P) $ is defined as $ P = M \cdot V $, where $ M $ is mass and $ V $ is velocity. The dimensional formula for velocity $ V $ is $ [LT^{-1}] $. Therefore, the dimensional formula for momentum $ P $ is:
[P] = [M^1L^1T^{-1}]
Area $ (A) $ is defined as the square of length, so its dimensional formula is:
[A] = [L^2]
Time $ (T) $ is already given as a fundamental quantity, so:
[T] = [T]

Now, to find the dimensions of energy $ E $ in terms of $ P, A, $ and $ T $, assume:

[E] = [P^x A^y T^z]

Substituting the dimensions of $ P, A, $ and $ T $ into the expression, we have:

[M^1L^2T^{-2}] = [M^1L^1T^{-1}]^x [L^2]^y [T]^z

Expanding the right side, we get:

= [M^xL^xT^{-x}] [L^{2y}] [T^z]

Combining powers of the same base, we have:

= [M^xL^{x+2y}T^{-x+z}]

Equating the powers on both sides, we get:

For $ M $: $ x = 1 $
For $ L $: $ x + 2y = 2 $
For $ T $: $-x + z = -2 $

Substituting $ x = 1 $ into the equations:

$ x + 2y = 2 $ becomes $ 1 + 2y = 2 $ implying $ y = \frac{1}{2} $
$-x + z = -2 $ becomes $-1 + z = -2 $ implying $ z = -1$

Therefore, the dimensional formula for energy in terms of $ P, A, $ and $ T $ is:

[E] = [P^1 A^{1/2} T^{-1}]

Thus, the correct answer is $[PA^{1/2}T^{-1}]$.

Answer 2