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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

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a

$F^{2} A T$

b

$F A T$

c

$F A^{2} T$

d

$F A T^{2}$

answer is B.

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Detailed Solution

$E = K F^{a} A^{b} T^{c}$

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

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

$∴ a = 1, a + b = 2 \Rightarrow b = 1$

and $- 2 a - 2 b + c = - 2 \Rightarrow c = 2$

$∴ E = K F A T^{2}$

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