Units & Dimensions

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22 of 149

Question 22 of 149

LL, CC and RR represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula ML2T−4A−2ML^2 T^{-4} A^{-2} corresponds to __________.

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Answer: (A) RLC\dfrac{R}{\sqrt{LC}}

Dimensional formula of inductance LL is obtained from E=12LI2E = \dfrac{1}{2}LI^2:

[L]=[E][I]2=ML2T−2A2=ML2T−2A−2[L] = \dfrac{[E]}{[I]^2} = \dfrac{ML^2T^{-2}}{A^2} = ML^2T^{-2}A^{-2}

Dimensional formula of capacitance CC is obtained from E=12Q2CE = \dfrac{1}{2}\dfrac{Q^2}{C}:

[C]=[Q]2[E]=A2T2ML2T−2=M−1L−2T4A2[C] = \dfrac{[Q]^2}{[E]} = \dfrac{A^2T^2}{ML^2T^{-2}} = M^{-1}L^{-2}T^4A^2

Dimensional formula of resistance RR is obtained from P=I2RP = I^2R:

[R]=[P][I]2=ML2T−3A2=ML2T−3A−2[R] = \dfrac{[P]}{[I]^2} = \dfrac{ML^2T^{-3}}{A^2} = ML^2T^{-3}A^{-2}

The dimensional formula of LC\sqrt{LC} is:

[LC]=(ML2T−2A−2)(M−1L−2T4A2)=T2=T[\sqrt{LC}] = \sqrt{(ML^2T^{-2}A^{-2})(M^{-1}L^{-2}T^4A^2)} = \sqrt{T^2} = T

Evaluating the dimensional formula of RLC\dfrac{R}{\sqrt{LC}}:

[RLC]=ML2T−3A−2T=ML2T−4A−2\left[\dfrac{R}{\sqrt{LC}}\right] = \dfrac{ML^2T^{-3}A^{-2}}{T} = ML^2T^{-4}A^{-2}

This matches the given dimensional formula.

Answer: RLC\dfrac{R}{\sqrt{LC}}

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