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  • If momentum (P), area (A), and time (T) are taken to be fundamental quantities, then energy has the dimensional formula a)(P^{1}A^{-1}T^{1}) b) (P^{2}A^{1}T^{1}) c) (P^{1}A^{-1/2}T^{1}) d) (P^{1}A^{1/2}T^{-1})

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

a)(P^{1}A^{-1}T^{1})

b) (P^{2}A^{1}T^{1})

c) (P^{1}A^{-1/2}T^{1})

d) (P^{1}A^{1/2}T^{-1})

Answers (1)

The answer is the option d) (P^{1}A^{1/2}T^{-1})

Explanation: Let us consider [P^{a}A^{b}T^{c}] as the formula for energy for fundamental quantities P, A & T.

Thus, dimensional formula of-

Energy(E) = [P^{a}A^{b}T^{c}]

Momentum (P) = [MLT^{-1}]

Area (A) =[L^{2}]

Time(T) = [T^{1}]

Now, E = f.s

           = [MLT^{-2}L].[ ML^{2}T^{-2}]

            = [MLT^{-1}]^{a}[L^{2}]^{b}[T]^{c} . [M^{a}L^{2+2b}T^{-a+c}]

Now, let us compare the powers,

a=1\; \; \; \; \; \; \; \; a+2b=2\; \; \; \; \; \; \; \; \; -a+c=-2

                        \; \; \; \; \; \; \; \; 1+2b=2\; \; \; \; \; \; \; \; \; -1+c=-2

                        \; \; \; \; \; \; \; \; 2b=2-1\; \; \; \; \; \; \; \; \; c=-2+1

                        \; \; \; \; \; \; \; \; b=\frac{1}{2}\; \; \; \; \; \; \; \; \; c=-1

Thus, [P^{1}A^{1/2}T^{-1}] is the dimensional formula of energy.

Posted by

infoexpert22

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