JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Units & Measurement

The quantity being tested is inductance divided by the product of resistance, capacitance and voltage. Writing each dimension in terms of mass, length, time and current: inductance $[L]=[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}{\mathrm{A}}^{-2}]$, resistance $[R]=[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}]$, capacitance $[C]=[{\mathrm{M}}^{-1}{\mathrm{L}}^{-2}{\mathrm{T}}^4{\mathrm{A}}^2]$ and voltage $[V]=[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}]$. Multiplying the last three gives $[RCV]=[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}{\mathrm{A}}^{-1}]$. Dividing, every mass, length and time factor cancels and the current exponent is $-2-(-1)=-1$, so $\frac{[L]}{[RCV]}=[{\mathrm{A}}^{-1}]$. The combination therefore carries the dimension of inverse current.