1.

In the formula `X=3YZ^(2)`, X and Z have dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y in MKSQ system?

Answer» As `r=CV, C=q/V=q^(2)/W [as V=W/q]`
So, `[X] rarr [C]=[Q^(2)/(ML^(2)T^(-2))]`
`=[M^(-1)L^(-2)T^(2)Q^(2)]`
and `F= Bil sin theta," "[B]=[F/(il)]`
So, `[Z] rarr [B]=[(MLT^(-2))/(QT^(-1)L)]=[MT^(-1)Q^(-1)]`
But as it is given
`X= 3YZ^(2), " "i.e., Y=X//(3Z^(2))`
So, `[Y]=([M^(-1)L^(-2)T^(2)Q^(2)])/([MT^(-1)Q^(-1)]^(2)) rarr[M^(-3)L^(-2)T^(4)Q^(4)]`


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