Exams › JEE Main › Physics

Let [epsilon0] be the dimensional formula of the permittivity of free space. With M = mass, L = length, T = time and A = electric current, which expression is correct?

1. [epsilon0] = [M⁻¹ L⁻³ T⁴ A²]
2. [epsilon0] = [M L⁻³ T² A]
3. [epsilon0] = [M⁻¹ L² T⁻¹ A⁻²]
4. [epsilon0] = [M⁻¹ L² T⁻¹ A]

Correct answer: [epsilon0] = [M⁻¹ L⁻³ T⁴ A²]

Solution

Solving Coulomb's law for epsilon0 gives epsilon0 = q² / (F * r²). Substituting dimensions of charge, force and length yields the dimensional formula.

Related JEE Main Physics questions

• A body has a measured mass of 5.00 ± 0.05 kg and a measured volume of 1.00 ± 0.05 m³. What is the maximum possible percentage error in its density?
• A substance has a density of 4 g/cm³ in the CGS system. If a new unit system is used where the unit of length is 10 cm and the unit of mass is 100 g, what is the numerical value of the density in that system?
• Which of the following physical quantities has dimensions that do not match the other three?
• If the percentage uncertainties in measuring mass M, length L, and time T are 1%, 1.5%, and 3% respectively, what is the percentage error in a quantity having dimensions M L⁻¹ T⁻¹?
• If E, m, J and G denote energy, mass, angular momentum and the gravitational constant, respectively, then the dimensional formula of EJ²/(m⁵G²) is identical to that of which quantity?
• A rod was found to have a measured length of 3.50 cm. Which measuring device was used for this reading?

⚔️ Practice JEE Main Physics free + battle 1v1 →