Question

If momentum P, area A nd time T are taken to be fundamental quantities. What is the dimensional formula of energy?

Answer 1

ANSWER:

• Dimensional formula of energy = [ P √A T⁻¹ ]

GIVEN:

• Momentum P, area A and time T are taken to be fundamental quantities.

TO FIND:

• Dimensional formula of energy.

EXPLANATION:

$\sf E \propto P^a \ A^b\ T^c$

Here E is the Energy.

$\sf E = k \ P^a \ A^b\ T^c$

Here k is the constant of proportionality and has no unit and dimensions.

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

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

$\sf [\ P\ ]^a = [\ M \ L\ T^{-1}\ ]^a$

$\sf [\ P\ ]^a = [\ M^a \ L^a \ T^{-a}\ ]$

$\sf [\ A\ ]^b = [\ L^2\ ]^b$

$\sf [\ A\ ]^b = [\ L^{2b}\ ]$

$\sf [\ T\ ]^c = [\ T\ ]^c$

$\sf [\ T\ ]^c = [\ T^c\ ]$

$\sf [\ M \ L^2\ T^{-2}\ ] = [\ M^a \ L^a \ T^{-a}\ ] [\ L^{2b}\ ] [\ T^c\ ]$

Add the powers of equal bases.

$\sf [\ M \ L^2\ T^{-2}\ ] = [\ M^a \ L^{a + 2b} \ T^{-a + c}\ ]$

Equate the powers of equal bases.

$\sf [\ M \ ] = [\ M^a \ ]$

a = 1

$\sf [\ L^2\ ] = [\ L^{a + 2b} \ ]$

a + 2b = 2

Substitute a = 1

1 + 2b = 2

2b = 1

b = 1/2

$\sf [\ T^{-2}\ ] = [ \ T^{-a + c}\ ]$

- a + c = - 2

Substitute a = 1

- 1 + c = - 2

c = - 1

$\sf [ \ E \ ]= [\ P \ {A}^{^1\!\!/\!_2} \ T^{ - 1} ]$

Dimensional formula of energy [ E ] = [ P √A T⁻¹ ]

Answer 2

$\mathcal{\gray{\underbrace{\blue{GIVEN:-}}}}$

⚡ If momentum P, area A and time T are taken to be fundamental quantities .

• momentum = P

• Area = A

• Time = T

$\mathcal{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The dimensional formula of energy .

$\mathcal{\gray{\underbrace{\blue{SOLUTION:-}}}}$

We have know that,

$\pink\bigstar\:\mathcal{\green{\boxed{\orange{Energy(E)\:=\:[M^1\:L^{2}\:T^{-2}]\:}}}}$

$\pink\bigstar\:\mathcal{\green{\boxed{\orange{Momentum(P)\:=\:[M^1\:L^{1}\:T^{-1}]\:}}}}$

$\pink\bigstar\:\mathcal{\green{\boxed{\orange{Area(A)\:=\:[M^0\:L^{2}\:T^{0}]\:}}}}$

$\pink\bigstar\:\mathcal{\green{\boxed{\orange{Time(T)\:=\:[M^0\:L^{0}\:T^{1}]\:}}}}$

⚡ According to the question,

$\rm{\green{Energy(E)\:\propto\:[P]^a\:[A]^b\:[T]^c\:}}$

$\rm{\green{Energy(E)\:=\:k\:[P]^a\:[A]^b\:[T]^c\:}}$----(1)

• k = Constant .

• [k] = [M⁰ L⁰ T⁰]

$\rm{\implies\:E\:=\:[M^0\:L^0\:T^0]\:[M^1\:L^1\:T^{-1}]^a\:[M^0\:L^2\:T^0]^b\:[M^0\:L^0\:T^1]^c\:}$

$\rm{\implies\:E\:=\:[M^{a}L^{a}T^{-a}]\:[L^{2b}]\:[T^{c}]\:}$

$\rm{\implies\:E\:=\:[M^{a}\:L^{a}\:L^{2b}\:T^{-a}\:T^{c}]\:}$

$\rm{\implies\:[M^1\:L^2\:T^{-2}]\:=\:[M^{a}\:L^{(a\:+\:2b)}\:T^{(-a\:+\:c)}]\:}$

⚡ Now, comparing L.H.S and R.H.S of the above equation we get

✍️ a = 1

✍️ a + 2b = 2

=> 1 + 2b = 2

=> 2b = (2 - 1)

=> b = 1/2

✍️ -a + c = -2

=> -1 + c = -2

=> c = -2 + 1

=> c = -1

⚡ Putting the value of a, b and c in Equation (1), we get

$\red\bigstar\:\rm{\pink{\boxed{\purple{\therefore\:[E]\:=\:[P]^1\:[A]^{1/2}\:[T]^{-1}\:}}}}$