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NEET PHYSICSMedium

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

A

[F][A][T]

B

[F][A][T^2]

C

[F][A][T^{-1}]

D

[F][A^{-1}][T]

Step-by-Step Solution

Energy, E∝FaAbTcE \propto F^a A^b T^c. [E]=[Fa][Ab][Tc][E] = [F^a][A^b][T^c]. [ML2T−2]=[MLT−2]a[LT−2]b[T]c=[MaLa+bT−2a−2b+c][ML^2T^{-2}] = [MLT^{-2}]^a [LT^{-2}]^b [T]^c = [M^a L^{a+b} T^{-2a-2b+c}]. Comparing dimensions on both sides: a=1a = 1; a+b=2⇒b=1a+b = 2 \Rightarrow b = 1; −2=−2a−2b+c⇒−2=−2−2+c⇒c=2-2 = -2a - 2b + c \Rightarrow -2 = -2 - 2 + c \Rightarrow c = 2. Thus, [E]=[FAT2][E] = [FAT^2].

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