Question 41:
easyFrequency of an E.M. waves is $10 MHz$ then its wavelength is:
(1999)
Using the wave equation $c = f\lambda$ , we can solve for wavelength $\lambda = \frac{c}{f}$ . Given $f = 10 MHz = 10^7 Hz$ and $c = 3 \times 10^8 m/s$ , we have $\lambda = \frac{3 \times 10^8}{10^7} = 30 m$ .
Question 42:
easyIf $\epsilon_0$ and $\mu_0$ are the electric permittivity and magnetic permeability in a free space, $\epsilon$ and $\mu$ are the corresponding quantities in medium, the refractive index of the medium is:
(1994)
The refractive index is defined as $n = \frac{c}{v}$ . We know $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ and $v = \frac{1}{\sqrt{\mu\epsilon}}$ . Substituting these expressions, we get $n = \frac{1 / \sqrt{\mu_0\epsilon_0}}{1 / \sqrt{\mu\epsilon}} = \sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}}$ .
Question 43:
easyThe frequency of electromagnetic wave, which best suited to observe a particle of radius $3 \times 10^{-4} cm$ is of the order of
(1991)
To effectively observe a particle, the resolving electromagnetic wave must have a wavelength of the order of the particle's size: $\lambda \approx 3 \times 10^{-4} cm = 3 \times 10^{-6} m$ . The corresponding frequency is $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{3 \times 10^{-6}} = 10^{14} Hz$ .
Question 44:
easyMatch List-I with List-II
\[
\begin{array}{ll@{\hspace{2cm}}ll}
\textbf{List-I (Electromagnetic waves)} & \textbf{List-II (Wavelength)} \\[0.5em]
\text{a. AM radio waves} & \text{(i) } 10^{-10}\text{ m} \\
\text{b. Microwaves} & \text{(ii) } 10^2\text{ m} \\
\text{c. Infra-red radiations} & \text{(iii) } 10^{-2}\text{ m} \\
\text{d. X-rays} & \text{(iv) } 10^{-4}\text{ m}
\end{array}
\]
Choose the correct answer from the options given below:
(2022)
Based on the standard electromagnetic spectrum: AM radio waves have the longest wavelength ( $\sim 10^2 m$ ). Microwaves are shorter ( $\sim 10^{-2} m$ ). Infrared is shorter still ( $\sim 10^{-4} m$ ). X-rays have very short wavelengths ( $\sim 10^{-10} m$ ). This matches option a.
Question 45:
easyThe E.M. wave with shortest wavelength among the following is:
(2020-Covid)
In the electromagnetic spectrum, gamma rays have the highest frequency and the most energy, which corresponds to the shortest wavelength among all the listed options.
Question 46:
easyWhich colour of the light has the longest wavelength?
(2019)
In the visible light spectrum (VIBGYOR), the wavelength progressively increases from the violet end to the red end. Thus, red light has the longest wavelength (approximately $700 nm$ ).
Question 47:
easyThe energy of the E.M. waves is of the order of $15 keV$ . To which part of the spectrum does it belong?
(2015 Pre)
The wavelength associated with $15 keV$ is $\lambda = \frac{hc}{E} = \frac{1240 eV\cdot nm}{15 \times 10^3 eV} \approx 0.08 nm$ . A wavelength of $0.08 nm$ (or $0.8 \mathring{A}$ ) falls squarely in the X-ray region of the electromagnetic spectrum.
Question 48:
easyThe condition under which a microwave oven heats up a food item containing water molecules most efficiently is:
(2013)
Microwave ovens heat food efficiently when the frequency of the microwaves matches the resonant frequency of water molecules, transferring energy via resonance.
Question 49:
easyIf $\lambda_v$ , $\lambda_x$ and $\lambda_m$ represent the wavelengths of visible light, X-rays and microwaves respectively, then:
(2005)
The electromagnetic spectrum in order of decreasing wavelength is radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. Therefore, the relation among the given wavelengths is $\lambda_m > \lambda_v > \lambda_x$ .
Question 50:
easyWhich of the following ray are not electromagnetic waves:
(2003)
X-rays, $\gamma$ -rays, and heat rays (infrared) are part of the electromagnetic spectrum. $\beta$ -rays are streams of high-energy electrons or positrons, which are particles, not electromagnetic waves.