Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question The dimension of μ 0 ε 0​ is equal to that of: Options: Velocity Charge Magnetic field Current (Given: μ 0 \mu_0  = vacuum permeability, ε 0 \varepsilon_0 = vacuum permittivity.) 🖼️ Question Image ✍️ Short Solution Let’s analyze the dimensions of μ 0 ε 0 . Step 1 — Recall known relations in electromagnetism From Maxwell’s equations, the speed of light c c c in vacuum is related to μ 0 \mu_0 μ 0 ​ and ε 0 \varepsilon_0 ε 0 ​ as: c = 1 μ 0 ε 0 . c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}. ​ 1 ​ . Now take the reciprocal and modify it: μ 0 ε 0 = 1 ε 0 c . \sqrt{\frac{\mu_0}{\varepsilon_0}} = \frac{1}{\varepsilon_0 c}. But let’s instead focus on the dimensional relation . We know that: μ 0 ε 0 = 1 c . So, μ 0 ε 0 = μ 0 μ 0 ε 0 = μ 0 c . Thus, the dimension of μ 0 ε 0 = (dimension of μ₀) × (dimension of velocity) . Step 2 — Dimensional formula of μ₀ From the definition of magnetic field B B : B = μ 0 I 2 π r ...

  ❓ Question Given below are two statements: Assertion (A): Refractive index of glass is higher than that of air. Reason (R): Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index. 🖼️ Question Image ✍️ Short Solution Let’s carefully analyze both the Assertion (A) and Reason (R) to see which statements are true and whether R correctly explains A. 🔹 Assertion (A): Refractive index of glass is higher than that of air The refractive index (μ) tells us how much light slows down in a medium compared to vacuum. It is defined as: μ = c v​ where c c  = speed of light in vacuum, and v v v = speed of light in the medium. In air , v ≈ c v \approx c , so μ ≈ 1. In glass , v < c v < c , so μ > 1. Hence, refractive index of glass is indeed higher than that of air. ✅ Assertion (A) is TRUE. 🔹 Reason (R): Optical density ∝ mass density ⇒ higher μ Optical density refer...

  ❓ Question The unit of 2 I ε 0 c​ is to be found, where I I  = intensity of an electromagnetic wave, ε 0 \varepsilon_0 ​ = permittivity of free space, and c c  = speed of light. 🖼️ Question Image ✍️ Short Solution Step 1: Expressing E 0 E_0 ​ I = 1 2 ε 0 c E 0 2 ⇒ E 0 = 2 I ε 0 c So, 2 I ε 0 c = E 0 ε 0 c . Step 2: Dimensional Analysis We know the SI units : E 0 E_0 ​ : electric field intensity → V/m = N/C = kg m s − 3 A − 1 ε 0 \varepsilon_0 ​ : permittivity → C 2 N − 1 m − 2 = A 2 s 4 kg − 1 m − 3 c c : speed of light → m/s \text{m/s} Now substitute: Unit of  E 0 ε 0 c = ( kg m s − 3 A − 1 ) ( A 2 s 4 kg − 1 m − 3 ) ( m s − 1 ) \text{Unit of } \frac{E_0}{\varepsilon_0 c} = \frac{(\text{kg·m·s}^{-3}\text{·A}^{-1})} {(\text{A}^2\text{·s}^4\text{·kg}^{-1}\text{·m}^{-3})(\text{m·s}^{-1})} Simplify powers of base units step-by-step: kg: 1 − ( − 1 ) = 2 1 - (-1) = 2 m: 1 − ( − 3 + 1 ) = 1 + 2 = 3 1 - (-3 + 1) = 1 + 2 =...