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Question

Physics Question on Units and measurement

Dimension of electrical resistance is

A

[ML2T3A1][ML^2 T^{-3} A^{-1} ]

B

[ML2T3A2][ML^2 T^{-3} A^{-2} ]

C

[ML3T3A2][ML^3 T^{-3} A^{-2}]

D

[ML1L3T3A2][ML^{-1} L^3 T^3 A^2 ]

Answer

[ML2T3A2][ML^2 T^{-3} A^{-2} ]

Explanation

Solution

From ohm's law V=IRV=I R where VV is voltage, RR is resistance and II is current. R=VI\therefore R=\frac{V}{I} Dimensions of R= Dimensions of V Dimensions of IR=\frac{\text { Dimensions of } V}{\text { Dimensions of } I} =[ML2T3A1][A]=\frac{\left[ ML ^{2} T ^{-3} A ^{-1}\right]}{[ A ]} =[ML2T3A2]=\left[ ML ^{2} T ^{-3} A ^{-2}\right] Alternative : From definition of time constant t=RCt=R C, where RR is resistance CC is capacitance. R=tC=[T][M1L2T4A2]\therefore R=\frac{t}{C}=\frac{[ T ]}{\left[ M ^{-1} L ^{-2} T ^{4} A ^{2}\right]} R=[ML2T3A2]R=\left[ ML ^{2} T ^{-3} A ^{-2}\right]