Question

If momentum (p), area (1) and time (t) are taken to be fundamental quantities, then energy has the dimensional formula

• A. p¹ A^-1 t^-1
• B. p² A¹ t¹
• C. p¹ A^-1/2 t¹
• D. p¹ A^1/2 t^-1

Answer 1

Answer: (D)

p¹ A^1/2 t^-1

Answer 2

Let, energy, $E = kp^{a}A^{b}t^{c}$

Where k is a dimensionless constant of proportionality Equating dimensions on both sides of (i), we get $\lbrack ML^{2}T^{- 2}\rbrack = \lbrack MLT^{- 1}\rbrack^{a}\lbrack M^{0}L^{2}T^{0}\rbrack^{b}\lbrack M^{0}L^{0}T\rbrack^{c} = \lbrack M^{a}L^{a + 2b}T^{- a + c}\rbrack$

Applying the principle of homogeneity of dimension

We get

a = 1 (i)

a + 2b = 2 (ii)

-a + c = -2 (iii)

On solving eqs. (ii), (iii) and (iv) we get

$a = 1,b = \frac{1}{2},c = 1$

$\therefore \lbrack E\rbrack = \lbrack P^{1}A^{1/2}t^{- 1}\rbrack$