JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
Match List I with List II.
Choose the correct answer from the options given below
| List - I | List - II | ||
|---|---|---|---|
| (a) | Capacitance, C | (i) | M¹L¹T⁻³A⁻¹ |
| (b) | Permittivity of free space, ε ₀ | (ii) | M⁻¹L⁻³T⁴A² |
| (c) | Permeability of free space, μ ₀ | (iii) | M⁻¹L⁻²T⁴A² |
| (d) | Electric field, E | (iv) | M¹L¹T⁻²A⁻² |
- (a) (a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)
- (b) (a) → (iii), (b) → (iv), (c) → (ii), (d) → (i)
- (c) (a) → (iv), (b) → (ii), (c) → (iii), (d) → (i)
- (d) (a) → (iv), (b) → (iii), (c) → (ii), (d) → (i)
Answer and solution
Answer (a) (a) → (iii), (b) → (ii), (c) → (iv), (d) → (i)
Solution
Physics sir · 4 steps
Build all four from force, charge and voltage, and let c² = 1/(μ₀ε₀) hand you μ₀.
- Start with basics. Charge is q = It, so [q] = AT. Force is MLT⁻², and voltage is work per charge: [V] = ML²T⁻²/AT = ML²T⁻³A⁻¹.
- Use each definition. Capacitance is C = q/V. From Coulomb's law, ε₀ = q²/(4π F r²). For μ₀, use c² = 1/μ₀ε₀, so μ₀ = 1/(ε₀ c²). Electric field is E = F/q.
Quantity Working Dimensions Match C AT/ML²T⁻³A⁻¹ M⁻¹L⁻²T⁴A² (iii) ε₀ A²T²/(MLT⁻² × L²) M⁻¹L⁻³T⁴A² (ii) μ₀ 1/(M⁻¹L⁻³T⁴A² × L²T⁻²) MLT⁻²A⁻² (iv) E MLT⁻²/AT MLT⁻³A⁻¹ (i) - Quick cross-check. A parallel-plate capacitor has C = ε₀ A/d, so C is ε₀ times a length. That is exactly why (iii) has L⁻² while (ii) has L⁻³.
- So. The matching is (a) → (iii), (b) → (ii), (c) → (iv), (d) → (i), which is option (a).
The idea to keep. Write charge as AT, then get every electrical quantity from a formula you trust. For μ₀, do not hunt for a formula: use μ₀ = 1/(ε₀ c²).
Answer from the NTA final answer key. Solution by PrepWiser.