InterviewSolution
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InterviewSolution
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Column-1 gives three physical quantities. Select the appropriate units for the choices given in column-II. Some of the physical quantities may have more than one choice. `{:("Column-I",,"Column-II"),("Capacitance",,"ohm-second"),("Inductance",,"Coulomb"^(2)//"Joule"),("Magnetic induction",,"coulomb"//"volt"),(,,"newton"//"ampere metre"),(,,"volt-second"//"ampere"):}` |
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Answer» (I) `q=CV i.e., [C]=[q//V]` so `[C]=[M^(-1)L^(-2)T^(4)A^(2)]` `U=1/2Li^(2), i.e., [L]=[U//i^(2)]` so `[L]=[ML^(2)T^(-2)A^(-2)]` `F=Bil sin theta, i.e., [B]=[F//il]` so `[B]=[MT^(-2)A^(-1)]` (II) Now the dimensions from given units are `ohmxx sec equiv [R][T] rarr [ML^(2)T^(-3)A^(-2)][T]` `=[ML^(2)T^(-2)A^(-2)]` `"coul"^(2)-"joule"^(-1) equiv [q^(2)/W] rarr ([A^(2)T^(2)])/([ML^(2)T^(-2)])` `=[M^(-1)L^(-2)T^(4)A^(2)]` `"coul (volt)"^(-1) equiv[q^(2)/W] rarr ([A^(2)T^(2)])/([ML^(2)T^(-2)])=[M^(-1)L^(-1)T^(4)A^(2)]` `"newton (amp-m)"^(-1) equiv [F/(il)] rarr ([MLT^(-2)])/([AL])=[MT^(-2)A^(-1)]` `"volt sec (amp)"^(-1) equiv [(WxxT)/(qxxA)] rarr [(ML^(2)T^(-2)T)/(ATA)]` `=[ML^(2)T^(-2)A^(-2)]` Comparing dimensions of II with I, we find that : Capacitance has unit `"coulomb"^(2)"-joule"^(-1)`, and coulomb `("volt")^(-1)`, inductance has units ohm-sec and volt-sec `("ampere")^(-1)`, and magnetic induction has unit newton `("ampere m")^(-1)`. |
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