If mu0 and 0 are the permeability and permittivity of free space, respectively, then the dimension of ((1/mu00)) is:

Q16 jee_main_2025_28_jan_morning Energy Density of EM Waves

Due to presence of an em-wave whose electric component is given by

mathrmE = 100sin (omega t - kx)mathrmNC^-1

, a cylinder of length

200~mathrmcm

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

A. $25sin (omega t - kx)mathrmNC^-1$
B. $200sin (omega t - kx)mathrmNC^-1$
C. $400sin (omega t - kx)mathrmNC^-1$
D. $50sin (omega t - kx)mathrmNC^-1$

Solution

### Related Formula

textEnergy Density = frac12 epsilon_0 mathrmE^2

textTotal Energy = textEnergy Density times textVolume

### Core Logic Since both cylinders hold equal amounts of electromagnetic energy:

left(textEnergyright)_1 = left(textEnergyright)_2

frac12 epsilon_0 mathrmE_1^2 cdot c pi mathrmR_1^2 times mathrmL_1 = frac12 epsilon_0 mathrmE_2^2 cdot c pi mathrmR_2^2 times mathrmL_2

Since the lengths are identical (

mathrmL_1 = mathrmL_2

), this simplifies to:

mathrmE_1^2 mathrmR_1^2 = mathrmE_2^2 mathrmR_2^2 implies mathrmE_1 mathrmR_1 = mathrmE_2 mathrmR_2

Given the second cylinder has half the diameter (and radius) of the first (

mathrmR_2 = fracmathrmR_12

):

100 times mathrmR_1 = mathrmE_2 times fracmathrmR_12

mathrmE_2 = 200 mathrmN/C

### Step 1: Final Equation Match The wave equation adjusts its amplitude factor to

200sin (omega t - kx)mathrmNC^-1

, which matches option (2). ### Pattern Recognition When energy is constant and volume scales down inversely by a factor of 4 (due to

mathrmR^2

), the electric field strength must increase by a factor of

sqrt4 = 2

to maintain balance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q4 jee_main_2025_03_april_morning Radiation Pressure

The radiation pressure exerted by a

450mathrm~W

light source on a perfectly reflecting surface placed at

2mathrm~m

away from it, is :

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

Solution

### Related Formula For a perfectly reflecting surface, the radiation pressure

P_textrad

is given by:

P_textrad = frac2Ic

where,

I

= intensity of the light source,

c

= speed of light

approx 3 times 10^8mathrm~m/s

. ### Core Logic Let's first calculate the intensity

I

of the point source at a distance

r = 2mathrm~m

:

I = fractextPowertextArea = fracP4pi r^2

Substitute the given values (

P = 450mathrm~W

and

r = 2mathrm~m

):

I = frac4504pi times 2^2 = frac45016pimathrm~W/m^2

### Step 1: Calculating Radiation Pressure Now, substitute

I

into the radiation pressure formula:

P_textrad = frac2 times left(frac45016piright)3 times 10^8 = frac90016pi times 3 times 10^8

P_textrad = frac30016pi times 10^8 = frac754pi times 10^8mathrm~N/m^2

Using

pi approx 3.1416

:

P_textrad = frac754 times 3.1416 times 10^8 = frac7512.566 times 10^-8

P_textrad approx 5.968 times 10^-8mathrm~Pascals approx 6 times 10^-8mathrm~Pascals

### Pattern Recognition Remember: Perfectly absorbing surface

implies P = I/c

. Perfectly reflecting surface

implies P = 2I/c

. Always pay close attention to the surface's properties mentioned in the prompt! ### Evaluation Rubric / Model Answer null ### 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

frac10pi

and relative permittivity of

frac10.0885

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

(mu_0 = 4pi times 10^-7text H / m, epsilon_0 = 8.85 times 10^-12text F / m, c = 3 times 10^8text m / s)

Numerical Answer. Answer: 6 to 6

Solution

### Related Formula

v = frac1sqrtmu epsilon = frac1sqrtmu_0mu_r cdot epsilon_0epsilon_r = fraccsqrtmu_r epsilon_r

### Core Logic Given parameters:

mu_r = frac10pi

epsilon_r = frac10.0885

Let's substitute these into the refractive index radical term

sqrtmu_r epsilon_r

:

mu_r epsilon_r = frac10pi times frac10.0885

### Step 1: Simplify Numerical Ratio Using standard approximations

pi approx 3.14

:

mu_r epsilon_r = frac103.1415 times 0.0885 approx frac100.278 approx 36

Taking the square root:

sqrtmu_r epsilon_r = sqrt36 = 6

Therefore,

v = fracc6 implies c = 6v

. Velocity in a vacuum is exactly **6** times faster. ### Pattern Recognition The expression

sqrtmu_r epsilon_r

is identical to the definition of refractive index

n

. Simplifying indices down to a perfect square (

36 implies 6

) clarifies the ratio immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q3 jee_main_2025_07_april_evening Intensity of EM Waves

The unit of

sqrtfrac2Iepsilon_0c

is: (

I

= intensity of an electromagnetic wave,

c

: speed of light) [cite: 10]

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

Solution

### Related Formula

I = frac12epsilon_0 E_0^2 c

[cite: 670] ### Core Logic Rearranging the equation for intensity

I

to express the peak electric field amplitude

E_0

: [cite: 670]

E_0^2 = frac2Iepsilon_0 c

E_0 = sqrtfrac2Iepsilon_0 c

[cite: 670] Thus, the given quantity is simply the magnitude of the peak electric field

E_0

[cite: 672]. The standard SI unit of an electric field is Newtons per Coulomb (

mathrmNcdot C^-1

) or Volts per meter (

mathrmVcdot m^-1

)[cite: 673]. Matching with the given structural choices,

mathrmNC^-1

is the correct unit[cite: 31, 669]. ### Pattern Recognition Recognize the standard configuration for energy flux density (intensity)

I = u_textavgc

[cite: 670]. Identifying that

sqrtfrac2Iepsilon_0 c

resolves to the electric field dimension directly yields the solution unit

mathrmNC^-1

or

mathrmV/m

[cite: 670, 672, 673]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q4 jee_main_2025_07_april_evening Impedance of Free Space

The dimension of

fracmu_0epsilon_0

is equal to that of: (

mu_0=

Vacuum permeability and

epsilon_0=

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 = fracmu_0 N^2 Al implies mu_0 propto L

[cite: 675]

C = fracepsilon_0 Ad implies epsilon_0 propto C

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

fracmu_0epsilon_0 propto fracLC

[cite: 678] We know that the time constant for an

LR

circuit is

tau = fracLR

and for a

RC

circuit is

tau = RC

[cite: 679]. Equating these time dimensions: [cite: 679]

fracLR = RC implies fracLC = R^2

[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

sqrtfracmu_0epsilon_0

represents the intrinsic impedance of free space, which has the value

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]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

If $mu_0$ $\mu_0$ and $varepsilon_0$ $\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $left(frac1mu_0varepsilon_0right)$ is:

Solution & Explanation

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Electromagnetic Waves Questions — jee_main_2025_02_april_evening

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If mu_0$\mu_0$ and varepsilon_0$\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of left(frac1mu_0varepsilon_0right)$\left(\frac{1}{\mu_0\varepsilon_0}\right)$ is:

Solution & Explanation

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

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If $mu_0$ $\mu_0$ and $varepsilon_0$ $\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $left(frac1mu_0varepsilon_0right)$ is: