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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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Verified Answer
The correct answer is:
$F A T^2$
Let energy $E=F^x A^y T^3$
Writing the dimensions on both sides
$\begin{aligned} & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]^x\left[\mathrm{LT}^{-2}\right]^y[\mathrm{~T}]^2} \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^x \mathrm{~L}^{x+y} \mathrm{~T}^{-2 x-2 y+z}\right]}\end{aligned}$
On comparing both sides
$\begin{aligned} & x=1 \\ & x+y=2 \\ & \Rightarrow \quad y=2-x=2-1=1 \\ & \text { and } \quad-2 x-2 y+z=-2 \\ & \Rightarrow-2 \times 1-2 \times 1+z=-2 \\ & \Rightarrow \quad z=-2+2+2 \\ & =2 \\ & \end{aligned}$
$\therefore$ Dimensional formula of energy
$=F A T^2$
Writing the dimensions on both sides
$\begin{aligned} & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]^x\left[\mathrm{LT}^{-2}\right]^y[\mathrm{~T}]^2} \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^x \mathrm{~L}^{x+y} \mathrm{~T}^{-2 x-2 y+z}\right]}\end{aligned}$
On comparing both sides
$\begin{aligned} & x=1 \\ & x+y=2 \\ & \Rightarrow \quad y=2-x=2-1=1 \\ & \text { and } \quad-2 x-2 y+z=-2 \\ & \Rightarrow-2 \times 1-2 \times 1+z=-2 \\ & \Rightarrow \quad z=-2+2+2 \\ & =2 \\ & \end{aligned}$
$\therefore$ Dimensional formula of energy
$=F A T^2$
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