NEET Practice Questions (MCQs) with Answers & Solutions

When a plane electromagnetic wave is incident on a material surface, it transfers both energy and momentum. Therefore, both p (momentum) and E (energy) are not zero.

Maxwell's modified form of Ampere's circuital law includes the displacement current term and is given by: $$ \oint \\vec B . d \\vec l = \\\mu_0 i + \\\mu_0 \\\varepsilon_0 { \\frac{d \\\\varphi_E}{dt} } $$ This accounts for the changing electric field in addition to the current.

X-rays have wavelengths in the range of $10^{-8}$ to $10^{-12}$ meters. Therefore, the order of magnitude for the wavelength of X-rays is typically around $10^{-10}$ meters.

The wave number $k$ and angular frequency $ ext{ω}$ relationship is given by $k = rac{2 ext{π} n}{ ext{λ}}$ and $ ext{ω} = rac{2 ext{π} c}{ ext{λ}}$, where $n$ is the refractive index, $ ext{λ}$ is the wavelength, and $c$ is the speed of light. Given the equation $E_z = 100 ext{cos} (6 imes 10^8 t + 4x)$, we can identify $ ext{ω} = 6 imes 10^8$ rad/s and $k = 4 ext{m}^{-1}$. Using the relation $c = rac{ ext{ω}}{k}$, we find $c = rac{6 imes 10^8 ext{rad/s}}{4 ext{m}^{-1}} = 1.5 imes 10^8 ext{m/s}$. The refractive index $n$ can be calculated using $n = rac{c_{ ext{vacuum}}}{c} = rac{3 imes 10^8 ext{m/s}}{1.5 imes 10^8 ext{m/s}} = 2$.

The radiation pressure exerted by an electromagnetic wave on a perfectly reflecting surface is given by $ P = rac{2I}{c} $, where $ I $ is the intensity of the wave and $ c $ is the speed of light. Given that $ I = 10 ext{ W/m}^2 $ and the area $ A = 20 ext{ cm}^2 = 20 imes 10^{-4} ext{ m}^2 $, the force $ F $ can be calculated as $ F = PA = rac{2IA}{c} $. Substituting the given values, we get $ F = rac{2 imes 10 imes 20 imes 10^{-4}}{3 imes 10^8} ext{ N} ightarrow F = 1.33 imes 10^{-10} ext{ N} $. Thus, the correct option is $ 1.33 imes 10^{-10} ext{ N} $.

The intensity $ I $ of light at a distance $ r $ from an isotropic point source is given by $ I = rac{P}{4 ext{π}r^2} $, where $ P $ is the power of the source. Given $ P = 40 ext{ W} $ and $ r = 2 ext{ m} $, we have $ I = rac{40}{4 ext{π}(2)^2} = rac{40}{16 ext{π}} = rac{10}{4 ext{π}} ext{ W/m}^2 $. The rms value of the electric field $ E_{ ext{rms}} $ is related to the intensity by $ I = rac{1}{2} c ext{ε}_0 E_{ ext{rms}}^2 $, where $ c $ is the speed of light and $ ext{ε}_0 $ is the permittivity of free space. Rearranging and solving for $ E_{ ext{rms}} $, we get $ E_{ ext{rms}} = igg(rac{2I}{c ext{ε}_0}igg)^{1/2} $. Substituting the values, $ E_{ ext{rms}} = igg(rac{2 imes rac{10}{4 ext{π}}}{3 imes 10^8 imes 8.854 imes 10^{-12}}igg)^{1/2} ightarrow E_{ ext{rms}} ightarrow 17.3 ext{ V/m} $. Thus, the correct option is $ 17.3 ext{ V/m} $.

Gamma rays are a form of electromagnetic radiation with high energy and short wavelength. They are used in medicine to destroy cancer cells because they can penetrate deep into tissues and cause ionization, which damages the DNA of cancer cells, leading to their destruction. Thus, the correct option is gamma rays.