The electric field in an electromagnetic wave is given by ${E}=\left(50\, {NC}^{-1}\right) \sin \omega({t}-{x} / {c})$

The energy contained in a cylinder of volume ${V}$ is $5.5 \times 10^{-12} \, {J}$. The value of ${V}$ is $......{cm}^{3}$ $\left(\right.$ given $\left.\in_{0}=8.8 \times 10^{-12} \,{C}^{2} {N}^{-1} {m}^{-2}\right)$

In a certain region uniform electric field $E$ and magnetic field $B$ are present in the opposite direction. At the instant $t = 0,$ a particle of mass $m$ carrying a charge $q$ is given velocity $v_o$ at an angle $\theta ,$ with the $y$ axis, in the $yz$ plane. The time after which the speed of the particle would be minimum is equal to

For the plane electromagnetic wave given by $\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$ and $\mathrm{B}=\mathrm{B}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, the ratio of average electric energy density to average magnetic energy density is

The mean intensity of radiation on the surface of the Sun is about $10^{8}\,W/m^2.$ The $rms$ value of the corresponding magnetic field is closet to

There exists a uniform magnetic and electric field of magnitude $1\, T$ and $1\, V/m$ respectively along positive $y-$ axis. A charged particle of mass $1\,kg$ and of charge $1\, C$ is having velocity $1\, m/sec$ along $x-$ axis and is at origin at $t = 0.$ Then the co-ordinates of particle at time $\pi$ seconds will be :-

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