Question

If momentum (p), area (A) and time (T) are taken to be
fundamental quantities, then energy has the dimensional
formula
(a) [pA⁻¹T¹] (b) [p²AT]
(c) [pA⁻¹/²T] (d) [ pA¹/²T]

Answer 1

Answer:

$\boxed{\sf [Energy] = [p A^{\frac{1}{2} } T^{-1}] }$

To find:

To express Energy in terms of momentum (p), area (A) and time (T).

Explanation:

Dimension Formula of:

$\sf Momentum \: (p) = [MLT^{-1} ] \\ \\ \sf Area\ (A)=[L^{2} ] \\\\ \sf Time \ (T) = [T] \\ \\ \sf Energy = [ML^{2} T^{-2} ]$

$\sf [Energy] = [Momentum]^{a} [Area]^{b} [Time]^{c} \\ \\ \sf \implies [Energy] = [p]^{a} [A]^{b} [T]^{c} \ \ \ \ \ \ \ \ \ \ .....Equation \ 1\\ \\ \sf \implies [ML^{2} T^{-2} ]= [MLT^{-1}]^{a} [L^{2} ]^{b}[T]^{c} \\ \\ \sf \implies [ML^{2} T^{-2} ]=[M^{a}L^{a+2b}T^{-a+c}] \ \ \ \ \ \ \ \ \ \ .....Equation \ 3$

From the equation 3 we can solve for a, b and c:

$\sf \implies a=1$

$\sf \implies a+2b=2 \\ \sf \implies 1 +2b = 2 \\ \sf \implies2b = 2-1 \\ \sf \implies2b =1 \\ \sf \implies b= \frac{1}{2}$

$\sf \implies -a+c=-2 \\ \sf \implies -1 + c=-2 \\ \sf \implies c=-2 +1 \\\sf \implies c=-1$

Now to get expression of energy in terms of momentum (p), area (A) and time (T) we will put the value of a, b, c in Equation 1

$\sf [Energy] = [p]^{a} [A]^{b} [T]^{c} \\ \\ \sf \implies [Energy] = [p]^{1} [A]^{\frac{1}{2} } [T]^{-1} \\ \\ \sf \implies [Energy] = [p A^{\frac{1}{2} } T^{-1}]$