1.

If momentum /(p), area a and time (t) are taken to be the fundamental quantities then the energy has the dimensional formula:

Answer»

<P>`[p^(1)a^(-1)t^(1)]`
`[p^(0)a^(1)t^(1)]`
`[p^(1)a^(-1/2)t^(1)]`
`[p^(1)a^(1/2)t^(-1)]`

Solution :Here `p^(1)=MLT^(-1)`
`:.M=(p^(1))/(L^(1)T^(-1))`
ALSO `a^(1)=L^(2)` or `L=a^(1//2)`
`:.` Energy `=ML^(2)T^(-2)`
`=(P^(1))/(L^(1)T^(-1))xxaxxT^(-2)`
`(P^(1)xxaxxT^(-2))/(a^(1//2)T^(-1))`
`=p^(1)a^(1/2)T^(-1)`
`:.(d)` is CORRECT.


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