Physics MCQs for NEET — Practice Questions with Answers

$\lambda =\frac{C}{v}=\frac{3\times {10}^8}{40\times {10}^6}=7.5 m$

 The waves that are used in communication are radio waves, microwaves, infrared radiation, and light. But in telecommunication microwaves are used.

Light is an electromagnetic wave and does not require any medium to travel while the sound is a mechanical wave and requires a medium to travel.

$J=\frac{P}{A}=\frac{3.9\times {10}^{26}}{4\mathrm{\pi }\left(6.96\times {10}^8\right)}=6.4\times {10}^7 \frac{W}{m^2}$

$Wave impedance; z=\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}=\sqrt{\frac{4\mathrm{\pi }\times {10}^{-7}}{8.85\times {10}^{-12}}}=376.6 \Omega$

$\lambda =\frac{c}{\nu }=\frac{3\times {10}^8}{1057\times {10}^6}=0.28 m=28 cm\Rightarrow radiowaves$

$$\begin{gathered} {\lambda }_{m}T=cons\tan t=0.2896 cmK \\ {\lambda }_m=\frac{0.2896}{2.7}=0.10 cm=1 mm\Rightarrow micro waves \end{gathered}$$

 The penetrating power of X-rays increases with the increase in frequency because energy is directly proportional to frequency.

The kinetic energy of electrons bombarding on the target depends on the accelerating voltage. The shortest wavelength of x rays emitted depends on the kinetic energy of the electrons. Therefore, the shortest wavelength depends on the voltage applied.

 X rays, having an extremely high frequency and low wavelength, would not be suitable to use as RADAR because they would penetrate the obstacle instead of reflecting and coming back.