Ncert Exercises & Solutions

If momentum (${p}$), area (${A}$), and time (${T}$) are taken to be fundamental quantities, then energy has the dimensional formula:

1.	$ {\left[{pA}^{-1} {~T}^1\right]} $	2.	$ {\left[{p}^2 {AT}\right]} $
3.	$ {\left[{pA}^{-1 / 2} {~T}\right]} $	4.	$ {\left[{pA}^{1 / 2} {~T}^{-1}\right]}$

(4) Hint: Use the dimensional analysis to find the correct relation.

Step 1: Define a relation for energy.

Given, fundamental quantities are momentum (p), area (A), and time (T).

We can write energy E as

$\begin{array}{l} E \propto p^{a} A^{b} T^{c} \\ E = {kp}^{a} A^{A} T^{c} \end{array}$

where k is dimensionless constant ot proportionality.

Step 2: Put all the dimensions.

$Dimensions of E = [E] = [{ML}^{2} T^{- 2}] and [p] = [{MLT}^{- 1}]$
$[A] = [L^{2}]$
$[T] = [T]$
$[E] = [K] [p]^{a} [A]^{b} [T]^{c}$

Putting all the dimensions, we get

$\begin{array}{l} {ML}^{2} T^{- 2} = [{MLT}^{- 1}]^{a} [L^{2}]^{b} [T]^{c} \\ = M^{a} L^{2 b + a} T^{- a} + c \end{array}$

Step 3: Compare the dimensions.

By the principle of homogeneity of dimensions,

$a = 1, 2 b + a = 2$
$\Rightarrow 2 b + 1 = 2$
$\Rightarrow b = 1 / 2 - a + c = - 2$
$\Rightarrow c = - 2 + a = - 2 + 1 = - 1$
$Hence, E = {pA}^{1 / 2} T^{- 1}$

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1.	\( {\left[{pA}^{-1} {~T}^1\right]} \)	2.	\( {\left[{p}^2 {AT}\right]} \)
3.	\( {\left[{pA}^{-1 / 2} {~T}\right]} \)	4.	\( {\left[{pA}^{1 / 2} {~T}^{-1}\right]}\)