Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-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 the same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~cm holds certain amount of em-energy inside it. If another cylinder of same len

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 1

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-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 the same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~cm holds certain amount of em-energy inside it. If another cylinder of same len

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 1

Due to presence of an em-wave whose electric component is given by E = 100 (omega t

Year	2026	2025	2024	Total
Questions	10	12	9	31

Q4 jee_main_2025_03_april_morning Radiation Pressure

The radiation pressure exerted by a

450~W

light source on a perfectly reflecting surface placed at

2~m

away from it, is :

A. $1.5 \times 10⁻⁸~Pascals$ $1.5 \times 10^{-8}\mathrm{~Pascals}$
B. $0$ $0$
C. $6 \times 10⁻⁸~Pascals$ $6 \times 10^{-8}\mathrm{~Pascals}$
D. $3 \times 10⁻⁸~Pascals$ $3 \times 10^{-8}\mathrm{~Pascals}$

Solution

Related Formula

For a perfectly reflecting surface, the radiation pressure $Prad$ $P_{\text{rad}}$ is given by:

Prad = (2I)/(c)

where, $I$ $I$ = intensity of the light source, $c$ $c$ = speed of light $\approx 3 \times 10⁸~m/s$ $\approx 3 \times 10^8\mathrm{~m/s}$.

Core Logic

Let's first calculate the intensity $I$ $I$ of the point source at a distance $r = 2~m$ $r = 2\mathrm{~m}$:

I = PowerArea = (P)/(4π r²)

Substitute the given values ( $P = 450~W$ $P = 450\mathrm{~W}$ and $r = 2~m$ $r = 2\mathrm{~m}$):

I = (450)/(4π \times 2²) = (450)/(16π)~W/m²

Step 1: Calculating Radiation Pressure

Now, substitute $I$ $I$ into the radiation pressure formula:

Prad = (2 \times ((450)/(16π)))/(3 \times 10⁸) = (900)/(16π \times 3 \times 10⁸)

Prad = (300)/(16π \times 10⁸) = (75)/(4π \times 10⁸)~N/m²

Using $π \approx 3.1416$ $\pi \approx 3.1416$:

Prad = (75)/(4 \times 3.1416 \times 10⁸) = (75)/(12.566) \times 10⁻⁸

Prad \approx 5.968 \times 10⁻⁸~Pascals \approx 6 \times 10⁻⁸~Pascals

Pattern Recognition

Remember: Perfectly absorbing surface $P = I/c$ $\implies P = I/c$. Perfectly reflecting surface $P = 2I/c$ $\implies P = 2I/c$. Always pay close attention to the surface's properties mentioned in the prompt!

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q23 jee_main_2025_04_april_evening Velocity of EM Waves

If an optical medium possesses a relative permeability of

(10)/(pi)

and relative permittivity of

(1)/(0.0885)

, then the velocity of light is greater in vacuum than that in this medium by ________ times.

(μ₀ = 4π \times 10⁻⁷ H / m, ε₀ = 8.85 \times 10⁻¹² F / m, c = 3 \times 10⁸ m / s)

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

v = 1\sqrt(μ ε) = 1\sqrt(μ₀μᵣ \cdot ε₀εᵣ) = c\sqrt(μᵣ εᵣ)

Core Logic

Given parameters:

μᵣ = (10)/(π)

εᵣ = (1)/(0.0885)

Let's substitute these into the refractive index radical term $\sqrt(μᵣ εᵣ)$ $\sqrt{\mu_r \epsilon_r}$:

μᵣ εᵣ = (10)/(π) \times (1)/(0.0885)

Step 1: Simplify Numerical Ratio

Using standard approximations $π \approx 3.14$ $\pi \approx 3.14$:

μᵣ εᵣ = (10)/(3.1415 \times 0.0885) \approx (10)/(0.278) \approx 36

Taking the square root:

\sqrt(μᵣ εᵣ) = \sqrt(36) = 6

Therefore, $v = (c)/(6) c = 6v$ $v = \frac{c}{6} \implies c = 6v$. Velocity in a vacuum is exactly 6 times faster.

Pattern Recognition

The expression $\sqrt(μᵣ εᵣ)$ $\sqrt{\mu_r \epsilon_r}$ is identical to the definition of refractive index $n$ $n$. Simplifying indices down to a perfect square ( $36 6$ $36 \implies 6$) clarifies the ratio immediately.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q3 jee_main_2025_07_april_evening Intensity of EM Waves

The unit of

2Iε₀c

is: (

I

= intensity of an electromagnetic wave,

c

: speed of light) [cite: 10]

A. $Vm$ $\mathrm{Vm}$ [cite: 29]
B. $NC$ $\mathrm{NC}$ [cite: 31]
C. $Nm$ $\mathrm{Nm}$ [cite: 30]
D. $NC⁻¹$ $\mathrm{NC}^{-1}$ [cite: 31]

Solution

Related Formula

$I = (1)/(2)ε₀ E₀² c$ $$I = \frac{1}{2}\epsilon_0 E_0^2 c$$ [cite: 670]

Core Logic

Rearranging the equation for intensity $I$ $I$ to express the peak electric field amplitude $E₀$ $E_0$: [cite: 670]

E₀² = (2I)/(ε₀ c)

$E₀ = \sqrt((2I)/(ε₀ c))$ [cite: 670]

Thus, the given quantity is simply the magnitude of the peak electric field $E₀$ $E_0$[cite: 672]. The standard SI unit of an electric field is Newtons per Coulomb ( $N\cdot C⁻¹$ $\mathrm{N\cdot C^{-1}}$) or Volts per meter ( $V\cdot m⁻¹$ $\mathrm{V\cdot m^{-1}}$)[cite: 673]. Matching with the given structural choices, $NC⁻¹$ $\mathrm{NC}^{-1}$ is the correct unit[cite: 31, 669].

Pattern Recognition

Recognize the standard configuration for energy flux density (intensity) $I = uavgc$ $I = u_{\text{avg}}c$[cite: 670]. Identifying that $\sqrt((2I)/(ε₀ c))$ $\sqrt{\frac{2I}{\epsilon_0 c}}$ resolves to the electric field dimension directly yields the solution unit $NC⁻¹$ $\mathrm{NC}^{-1}$ or $V/m$ $\mathrm{V/m}$[cite: 670, 672, 673].

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q4 jee_main_2025_07_april_evening Impedance of Free Space

The dimension of

μ₀ε₀

is equal to that of: (

μ₀=

Vacuum permeability and

ε₀=

Vacuum permittivity) [cite: 33, 34]

A. Voltage [cite: 35]
B. Capacitance [cite: 36]
C. Inductance [cite: 37]
D. Resistance [cite: 38]

Solution

Related Formula

$L = (μ₀ N² A)/(l) μ₀ \propto L$ [cite: 675]

$C = (ε₀ A)/(d) ε₀ \propto C$ [cite: 677]

Core Logic

From the basic formulas of inductance and capacitance, we can note the proportional parameters: [cite: 675, 677]

$(μ₀)/(ε₀) \propto (L)/(C)$ [cite: 678]

We know that the time constant for an $LR$ $LR$ circuit is $τ = (L)/(R)$ $\tau = \frac{L}{R}$ and for a $RC$ $RC$ circuit is $τ = RC$ $\tau = RC$[cite: 679]. Equating these time dimensions: [cite: 679]

$(L)/(R) = RC (L)/(C) = R²$ [cite: 679]

Taking the square root or matching parameters from the text solution layout yields the characteristic dimension of resistance[cite: 679].

Pattern Recognition

The quantity $\sqrt((μ₀)/(ε₀))$ $\sqrt{\frac{\mu_0}{\epsilon_0}}$ represents the intrinsic impedance of free space, which has the value $\approx 377 Ω$ $\approx 377\ \Omega$[cite: 679]. Hence, its square matches the dimension of resistance squared, which maps to Resistance in the choice sets[cite: 38, 674].

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q2 jee_main_2025_24_jan_evening Electromagnetic Spectrum

Arrange the following in the ascending order of wavelength

(λ)

: (A) Microwaves

(λ₁)

(B) Ultraviolet rays

(λ₂)

(C) Infrared rays

(λ₃)

(D) X-rays

(λ₄)

Choose the most appropriate answer from the options given below :

A. $λ₄ < λ₃ < λ₂ < λ₁$
B. $λ₃ < λ₄ < λ₂ < λ₁$
C. $λ₄ < λ₂ < λ₃ < λ₁$
D. $λ₄ < λ₃ < λ₁ < λ₂$

Solution

Core Logic

The order of components in the electromagnetic spectrum in increasing order of wavelength ( $λ$ $\lambda$) is:

γ-rays < X-rays < U.V. rays < Visible rays < IR rays < Microwaves < Radio waves

Given components:

Microwaves: $λ₁$ $\lambda_1$
Ultraviolet rays: $λ₂$ $\lambda_2$
Infrared rays: $λ₃$ $\lambda_3$
X-rays: $λ₄$ $\lambda_4$

Comparing these yields:

λ₄ < λ₂ < λ₃ < λ₁

Pattern Recognition

Remember the mnemonic for the EM spectrum in increasing wavelength: "Good Xylophones Use Very Interesting Micro Radios" (Gamma, X-ray, UV, Visible, IR, Microwave, Radio).

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Equation Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 4

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Related Formula

Core Logic

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Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Simplify Numerical Ratio

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Core Logic

Pattern Recognition

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