Units & Dimensions

JEE Main level, single correct, medium

116 of 149

Question 116 of 149

If momentum(P)(P), area (A)(A) and time (T)(T) are taken to be the fundamental quantities then the dimensional formula for energy is :

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Answer: (C) [PA1/2T−1]\left[{PA}^{1/2}{T}^{-1}\right]

Let dimension formula of energy will be
E=AaTbPcE={A}^{a}{T}^{b}{P}^{c}
P=M1L1T−1P={M}^{1}{L}^{1}{T}^{-1}
A=L2A={L}^{2}
T=T1T={T}^{1}
M1L2T−2=MCL2a+cTb−c{M}^{1}{L}^{2}{T}^{-2}={M}^{C}{L}^{2a+c}{T}^{b-c}
by comparison C=1C=1.........(1).........\left(1\right)
2a+c=22a+c=2......(2)......\left(2\right)
b−c=−2b-c=-2........(3)........\left(3\right)
c=1,a=1/2,b=−1c=1,a=1/2,b=-1
E=A1/2T−1P1E={A}^{1/2}{T}^{-1}{P}^{1}

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