Nearly $10\%$ of the power of a $110 \mathrm{W}$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1 \mathrm{m}$ from the bulb to a distance of $5 \mathrm{m}$ is $a\times {10}^{-2} \mathrm{W} {\mathrm{m}}^{-2}$. The value of '$a$' will be

If $\overrightarrow{E}$ and $\overrightarrow{K}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by:

($\omega -$ angular frequency)

The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

 $E_x=E_0\sin \left(kz-\omega t\right)$

$B_y=B_0\sin \left(kz-\omega t\right)$

Then the correct relation between $E_0$ and $B_0$ is given by 

Match List I and List II

Choose the correct answer from the options given below :

Given below are two statements :

Statement I : Electromagnetic waves are not deflected by electric and magnetic field.

Statement II : The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as $E_0=\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}{B}_0$.

In the light of the above statements, choose the correct answer from the options given below:

Match List I and List II

Choose the correct answer from the option given below: