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JEE Main 2025 (28 Jan, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2025 (28 Jan, Shift 1)

Q4JEE Main 2025 · 28 JAN · SHIFT 1Medium

Energy Density & Intensity

Due to presence of an em-wave whose electric component is given by E=100sin⁡(ωt−kx) N C−1E = 100 \sin(\omega t - kx)\text{ N C}^{-1} a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

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Answer: (c)

  1. Energy U=12ε0E02×πr2LU = \tfrac{1}{2}\varepsilon_0 E_0^2 \times \pi r^2 L
  2. Half diameter ⇒\Rightarrow area becomes 1/41/4, same LL.
  3. Same U⇒E02U \Rightarrow E_0^2 must become 4 times ⇒E0′=2×100=200\Rightarrow E_0' = 2 \times 100 = 200 N/C
  4. So the answer is (c) 200sin⁡(ωt−kx)200 \sin(\omega t - kx) N C−1^{-1}.
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