Physics And Measurement Question 161
Question: In a system of units if force (F), acceleration and time (T) are taken as fundamental units then the dimensional formula of energy is?
[BHU 2005]
Options:
A) $ FA^{2}T $
B) $ FAT^{2} $
C) $ F^{2}AT $
D) $ FAT $
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=KF^{a}A^{b}T^{c} $
$ [ ML^{2}{{T}^{-2}} ]={{[ ML{{T}^{-2}} ]}^{a}}{{[ L{{T}^{-2}} ]}^{b}}{{[ T ]}^{c}} $
$ [ ML^{2}{{T}^{-2}} ]=[ M^{a}{{L}^{a+b}}{{T}^{-2a-2b+c}} ] $
$ \therefore $ $ a=1 $ , $ a+b=2 $
therefore $ b=1 $ and $ -2a-2b+c=-2\ \ \Rightarrow \ c=2 $
$ \therefore $ $ E=KFAT^{2} $ .