1.

In the relation y=rsin (omegat-kx) the dimensions of (omega)/(k) are :

Answer»

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

Solution :`y=rsin(omegat-kx)`
Here `omegat=angle:.omega=(1)/(T)=T^(-1)`
SIMILARLY `kx`=angle`impliesk=(1)/(L)=L^(-1)`
`:.(omega)/(k)=(T^(-1))/(L^(-1))=LT^(-1)`
or Simply `(omega)/(k)` REPRESENT WAVE velocity as
`(omega)/(k)=(2piv)/(2pi//lambda)=vlambda`
Hence correct choice is `(b)`.


Discussion

No Comment Found

Related InterviewSolutions