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In relation y=asin(omegat-kx) dimension ...

In relation `y=asin(omegat-kx)` dimension of k will be……

A

`M^(0)L^(1)T^(1)`

B

`M^(0)L^(-1)T^(0)`

C

`ML^(0)T^(-1)`

D

`M^(0)L^(-1)T^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
In `y=asin(omegat-kx)`
kx is dimensionless
`:.kx=M^(0)L^(0)T^(0)`
`:.` Dimension of `k=(M^(0)L^(0)T^(0))/(x)`
`=(M^(0)L^(0)T^(0))/(M^(0)L^(1)T^(0))=M^(0)L^(-1)T^(0)`
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