JEE Main 20137 April
- Exam
- JEE Main
- Year
- 2013
- Date
- 7 April
- Subject
- Physics
- Chapter
- Units and Dimensions
JEE Main 2013 · 7 April · Physics · Units and Dimensions
Let [ ∈ _ 0 ] denote the dimensional formula of the permittivity of...
Let denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :
Options
- A\left[{∈}_{0}\right] = \left[{\text{M}}^{-1} {\text{L}}^{2} {\text{T}}^{-1} {\text{A}}^{- 2}\right]
- B\left[{∈}_{0}\right] = \left[{\text{M}}^{-1} {\text{L}}^{2} {\text{T}}^{-1} \text{A}\right]
- C\left[{∈}_{0}\right] = \left[{\text{M}}^{-1} {\text{L}}^{- 3} {\text{T}}^{2} \text{A}\right]
- D\left[{∈}_{0}\right] = \left[{\text{M}}^{-1} {\text{L}}^{- 3} {\text{T}}^{4} {\text{A}}^{2}\right]
Correct answer
D. \left[{∈}_{0}\right] = \left[{\text{M}}^{-1} {\text{L}}^{- 3} {\text{T}}^{4} {\text{A}}^{2}\right]
Step-by-step solution
\text{F} = \dfrac{1}{4 π {ε}_{0}} \dfrac{{\text{q}}_{1} {\text{q}}_{2}}{{\text{R}}^{2}}{ε}_{0} = \dfrac{{\text{q}}_{1} {\text{q}}_{2}}{4 π {\text{FR}}^{2}}{ε}_{0} = \dfrac{{\text{C}}^{2}}{{\text{N . m}}^{2}} = \dfrac{{\left[\text{AT}\right]}^{2}}{\text{ML}{\text{T}}^{- 2} {\text{. L}}^{2}} = \left[{\text{M}}^{-1} {\text{L}}^{- 3} {\text{T}}^{4} {\text{A}}^{2}\right]