In a system of units if force (F), acceleration (A) and time (T) are taken as fundamental units then the dimensional formula of energy is

Q: In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is

b (b) $E = K{F^a}{A^b}{T^c}$ $\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {ML{T^{ - 2}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$ $\left[ {M{L^2}{T^{ - 2}}} \right] = \left[ {{M^a}{L^{a + b}}{T^{ - 2a - 2b + c}}} \right]$ $\therefore $ $a = 1$, $a + b = 2$ $⇒$ $b = 1$ and $ - 2a - 2b + c = - 2\;\; \Rightarrow \;c = 2$ $\therefore $ $E = KFA{T^2}$.

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is

A$F{A^2}T$
B$FA{T^2}$
C${F^2}AT$
D$FAT$

Medium

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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