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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? |
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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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